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Chapter 10. Parametric Equations and Polar Coordinates

Calculus

http://matrix.skku.ac.kr/Cal-Book1/Ch1/

http://matrix.skku.ac.kr/Cal-Book1/Ch2/

http://matrix.skku.ac.kr/Cal-Book1/Ch5/

미적분학 1 (Math & Coding) 강의 (학생들이 제출한 과제/평가) 기록 :

미적분학 2 (Math & Coding) 강의 (학생들이 제출한 과제/평가) 기록 :

Chapter 10. Parametric Equations and Polar Coordinates

10.1 Parametric Equations    http://youtu.be/hQGCZk1tpuA

문제풀이 by 문지호  http://youtu.be/uz1DkKVeD2k

문제풀이 by 임효정  http://youtu.be/Ybs68e0iMZI

10.2 Calculus with Parametric Curves   http://youtu.be/QFMSbGKhoX4

문제풀이 by 장찬영  http://youtu.be/yF5oZOQVnCE

10.3 Polar Coordinates    http://youtu.be/lKPJeAGw0ZA

문제풀이 by 계성곤  http://youtu.be/smAmDRK-tWY

문제풀이 by 황인철  http://youtu.be/4hoVKvk8dq0

10.4 Areas and Lengths in Polar Coordinates

문제풀이 by 곽주현  http://youtu.be/LRmasW9uqYY

10.5 Conic Section

문제풀이 by 변희성  http://youtu.be/ONItxvlsnb8

문제풀이 by 이한울  http://youtu.be/CZ9SHMtqVy4

10.1 Parametric Equations

The motion of a particle along a curve is shown in Figure 1. Since fails the Vertical Line Test, the curve cannot be described by . If there was such a relation it would not tell the whole story. Although such a function does tell us where the object has been, it does not tell us when the object was at a given point . Since the and -coordinates of the particle depend on time , we can write and . Such a pair of equations is often a convenient way of describing a curve.

Figure 1

If and are continuous functions of on an interval , then the set of ordered pairs is called a plane curve . The equations and are called parametric equations for , and is called a parameter. (See Figure 1.)

Although in most applications the parameter denotes time, it can also be represented by other letters and represent other quantities.

Example 1

Sketch the plane curve defined by the parametric equations

, .

Solution. Each value of gives a point on the curve, as shown in the table. For instance, if , then, , and so the corresponding point is . In Figure 2 we plot the points determined by several values of the parameter and we join them to produce a curve.

 －2 －3 8 －1 －2 3 0 －1 0 1 0 －1 2 1 0 3 2 3 4 3 8

Figure 2

If the position of a particle is represented by parametric equations, then as increases the particle moves along the curve in the direction of the arrows.

By eliminating the parameter from both equations and , we have

.

Thus, the curve represented by the given parametric equations is the parabola .    ■

It often happens that two different sets of parametric equations represent the same curve. For example, the parametric equations

and

have the same curve as Example 1. However, comparing the value of , we see that the above is traced out more rapidly (considering as time) than Example 1. Thus, in applications, different parametric representations can be used to represent various speeds at which objects travel along a given path.

When , the curve with parametric equations , has an initial point and a terminal point .

Figure 3

Example 2

Find the curve represented by the parametric equations

,   .

Solution. First of all, we solve for and from the given equations.

and

Now we utilize the identity to form an equation involving only and .

or .

From this rectangular equation we see that the graph is an ellipse centered at (0, 0), with vertex (0, 3) and (0, ). Note that the ellipse is traced out counterclockwise as varies from 0 to .

Figure 4

By eliminating from the above identity we have , which is a unit circle with center at the origin and radius 1. Here the parameter is the angle (in radians) shown in Figure 5. As increases from 0 to starting from the point , the point moves once around the circle in the counterclockwise direction.

Figure 5

Note that the parametric equations , , also represent the above unit circle. Again different sets of parametric equations can represent the same curve.

In general, the graph of an equation of the form can be represented by the parametric equations , .

Notice that curves with equations (the ones we are most familiar with-graphs of functions) can also be regarded as curves with parametric equations .

Note that we may not always be able to convert the parametric equations of a curve to a Cartesian equation.

Finding parametric equations for a graph

In the previous examples, we have been looking at techniques for sketching the graph represented by a set of parametric equations. Let us think about the reverse problem. How can we find a set of parametric equations for a given graph or a given physical description? We know that such a representation is not unique from the above discussion. In the following example, we will find two different parametric representations for a given graph.

Example 3

Find a set of parametric equations to represent the graph of , using the following parameters.

(a)

(b) The slope at the point .

Solution. (a) Let . Then we obtain

and  for all

(b) To express and in terms of the parameter, we proceed as follows.

⇒

Thus the parametric equation are

and  .        ■

Figure 6

In Figure 6, the resulting curve has a left-to-right orientation as determined by the direction of increasing values of slope .

The Cycloid

The Cycloid is the curve traced out by a point on the circumference of a circle as the circle rolls along a straight line. (See Figure 7.)

Figure 7

Example 4

Determine the curves traced by a point on the circumference of a circle of radius rolling along the -axis in a plane. One position of is the origin.

Solution. Let the parameter be the measure of circle’s rotation, and we let the point begin at the origin. When , is at the origin; when , is at a point ; and when , is back on -axis at . From Figure 9 we see that the distance it has rolled from the origin is

Therefore, the center of the circle is . Since is any point on the cycloid, from Figure 9, we have

,

.

Thus, the parametric equations of the cycloid are

,    ,    .

Figure 9

For one rotation of the circle described by , one arch of the cycloid is traced out.

Cycloid is related to one of the most famous (pairs) problems in the history of calculus. The first problem began with Galileo’s discovery that the time required to complete a full swing of a given pendulum is approximately the same, whether it makes a large movement at high speeds or a small movement at a lower speeds. Late in his life, Galileo (1564-1642) realized that he could use this principle to construct a clock. However, he was not able to conquer the mechanics of the actual construction. Christian Huygens (1629-1695) was the first to design and construct a working model of such a clock. In his work with pendulums, he realized that a pendulum does not take exactly the same time to complete swings of varying lengths. But in studying the problem, Huygens discovered that a ball rolling back and forth on an inverted cycloid does complete each cycle in exactly the same time.

The second problem, proposed by the Swiss mathematician John Bernoulli, in 1696, is called the brachistochrone problem (in Greek brachys means short and chronos means time). In the brachistochrone problem, a curve is to be determined along which a particle slides in the shortest time (under the influence of gravity) from a point to a lower point not directly beneath . He showed that among all possible curves that join to , the particle will take the least time sliding from to if the curve is part of an inverted arch of a cycloid.

The Dutch physicist Huygens had already shown that the cycloid is also the solution to the tautochrone problem. That is, no matter where a particle is placed on an inverted cycloid, it takes the same time to slide to the bottom. Huygens proposed that pendulum clocks (which he invented) swing in cycloidal arcs because then the pendulum takes the same time to make a complete oscillation whether it swings through a wide or a small arc.

Example 5

Sketch the curve

.

Solution.

var('t,x,y')

x=t^3-1

y=2*t

a=-2

b=2

p=parametric_plot((x,y),(t,a,b))

small=0.001

step=0.5

n=(b-a)/step

arr=sum([arrow((x(t=a+i*step),

y(t=a+i*step)),(x(t=a+i*step+small), y(t= a+i*step+small))) for i in range(1,n) ])

p+arr

Figure 10

Example 6

Sketch the curve

.

Solution.

t=var('t')

parametric_plot((2*cos(t)-cos(2*t), sin(2*t)-2*sin(t)), (t, 0, 2*pi))

Figure 11

■

Example 7

Sketch the curve for

, .

Solution.

var('t, x, y')

x=12*sin(t)^3

y=8*cos(t)-5*cos(2*t)-2*cos(3*t)-cos(4*t)

p=parametric_plot((x,y), (t, 1, 20))

p

Figure 12

10.1 EXERCISES (Parametric Equations)

http://matrix.skku.ac.kr/Cal-Book/part1/CS-Sec-10-1-Sol.html

http://youtu.be/Ybs68e0iMZI

http://youtu.be/uz1DkKVeD2k

1-8. (a) Find the Cartesian equation of the curve.

(b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.

1.  , ,

Solution.

(a) From the second equation, we have . Now from the first equation, we get

.

Notice that if , we have . Thus the Cartesian equation of the given curve is , .

(b)

var('t,x,y')

x=t^2-2*t

y=3-t

a=0

b=3

p=parametric_plot((x, y) , (t, a, b))

small=0.001

step=0.25

n=(b-a)/step

arr=sum([arrow((x(t=a+i*step), y(t=a+i*step)), (x(t=a+i*step+small), y(t=a+i*step+small))) for i in range(1,n) ])

p+arr

2. , ,

Solution. (a)

(b)

var('t,x,y')

x=sin(t)^2

y=cos(3*t)

a=0

b=pi

p=parametric_plot((x, y), (t, a, b))

small=0.001

step=pi/8

n=(b-a)/step

arr=sum([arrow((x(t=a+i*step), y(t=a+i*step)), (x( t= a+i*step + small), y(t=a+i*step+small))) for i in range(1, n) ])

p+arr

3. , ,

Solution.

(a)

(b)

var('t')

parametric_plot((sin(t)^2,cos(t)^2),(t,0,1/2)).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-1, ymax=2)

parametric_plot((sin(t)^2,cos(t)^2),(t,0,1)).show(aspect_ratio=1, xmin=-2, xmax=2,

ymin=-1, ymax=2)

parametric_plot((sin(t)^2,cos(t)^2),(t,0,2)).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-1, ymax=2)

4. , , .7

Solution.

(a)

(b)

var('t')

parametric_plot((ln(t),t-1),(t,1,3/2)).show(aspect_ratio=1, xmin=-1, xmax=1, ymin=-1,

ymax=1.5)

parametric_plot((ln(t),t-1),(t,1,2)).show(aspect_ratio=1, xmin=-1, xmax=1, ymin=-1, ymax=1.5)

parametric_plot((ln(t),t-1),(t,1,5/2)).show(aspect_ratio=1, xmin=-1, xmax=1, ymin=-1, ymax=1.5)

5. ,

Solution.

6. , ,

Solution.

7. , ,

Solution. (a) Since .

(b)

var('tau')

parametric_plot((cos(tau)^2, cos(tau)),tau,0,4*pi). show(aspect_ratio=1, xmin=0, xmax=0.7, ymin=0, ymax=1)

8. ,

9. Find a parametric equation for the path of a particle that moves along in the manner described below.

(a) Once around clockwise, starting at (3, 1).

(b) Twice around counterclockwise, starting at (3, 1).

(c) Halfway around counterclockwise, starting at (1, 3).

(d) Graph the semicircle traced by the particle.

Solution. The circle has center (1, 1)and radius 2, so it can be represented by .

This represen-ta-tion gives us the circle with a counterclockwise orientation starting (3, 1).

(a) To get a clockwise orientation, we can replace by in the equations to get, .

(b) To get twice around in the counterclockwise direction, we use the original equations with the domain expanded to .

(c) To start at (1, 3) using the original equations, we must have ; that is, and . We use the original equations with the domain to be a counterclockwiseorientation starting (1, 3).

(d) Graph it by using Sage.

var('t')

parametric_plot((1+2*cos(t),1 + 2*sin(t)), (t,0,pi))

10.(a)Find parametric equation for the ellipse .

(b) Sketch the ellipse when , and

(c) How does the shape of the ellipse change as and varies?

Solution. (a) Let , and to obtain with as possible parametric equations for the ellipse .

(b)

var('t')

parametric_plot((3*cos(t),sin(t)),(t,0,2*pi)).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-4, ymax=4)

parametric_plot((3*cos(t),2*sin(t)),(t,0,2*pi)).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-4, ymax=4)

parametric_plot((3*cos(t),4*sin(t)),(t,0,2*pi)).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-4, ymax=4)

(c) As b increases, the ellipse stretches vertically.

11. Find two different sets of parametric equations to represent the curve , .

Solution. ⅰ) ,    ⅱ) .

12. Suppose that the position of a particle at time is given by , ,    and the position of the second particle is given by , , .

(a)Graph the paths of both particles. How many points of intersection are there?

(b)Are any of these points of intersection collision points? In other words, are the particles ever at the same place at same time? If so, find the collision points.

(c) Describe what happens if the path of the second particle is given by , .

Solution. (a) There are 4-points of intersection :

var('t')

p1=parametric_plot((sin(t),3*cos(t)),(t,0,2*pi)).show(aspect_ratio=1, xmin=-5, xmax=3, ymin=-3, ymax=3)

p2=parametric_plot((-1+2*cos(t),1+sin(t)),(t,0,2*pi)).show(aspect_ratio=1, xmin=-5, xmax=3, ymin=-3, ymax=3)

(b) A collision point occurs when for the same . Thus we get the following equations :

From [2], . Substituting into [1],

we get

.

Note that satisfies [1], [2], but does not.

So the only collision point occurs when , and this gives the point .

[We could check our work by graphing together as functions of and on another plot, as functions of . If we do so, we see that the only value of for which both pairs of graphs intersect is .]

(c) The equation , is the circle centered at instead of . There are still 4 intersection points. However there are no collision points, since (*) in part

(b) becomes .

13. Investigate the family of curves defined by the parametric equations

.

How does the shape change as changes? In particular, you should identify values of for which the basic shape of the curve changes.

Solution.

var('t')

@interact

def _(c=(0,2)):

p=parametric_plot((sin(t)*(c-sin(t)), cos(t)*(c-sin(t))), (t,0,2*pi))

show(p,xmin=-3, xmax=2, ymin=-2, ymax=2)

10.2 Calculus with Parametric Curves

We know how to represent a curve by a set of parametric equations. It is natural to ask how to use calculus to study parametric curves. Therefore, we will study problems involving tangents, area, arc length, and the areas formed by revolutions of parametric curves.

Tangents

By eliminating the parameter , curves defined by parametric equations and can also be expressed in the form . By substituting and in the equation , we get . If , , and are differentiable, then by using the chain rule, can be differentiated with respect to . Then we have  .

Solving for , we get

provided  .

Thus, the tangent to a parametric curve (without having to eliminate the parameter ) can be calculated from . Since the slope of the tangent line to the curve at is . Equation can be written as

if

If the motion of a particle is described by a parametric curve, then and are the vertical and horizontal velocities of the particle respectively. The is the slope of the tangent, which is the ratio of the vertical and horizontal velocities.

From , note that the curve has a horizontal tangent when (provided that ) and it has a vertical tangent when (provided that ). This is useful in sketching parametric curves.

Differentiating again with respect to , we get the second derivative

.

Example 1

Find the slope and concavity at point for the curve

and   .

Solution. Since

and

and furthermore is corresponding to .

Figure 1

Therefore the slope is

.

Moreover, when , the second derivative is

Thus we conclude that the graph is concave upward at (2, 4), as shown in Figure 1.       ■

Example 2

(a) Find the tangent to the circle at the point where .

(b)At what points is the tangent horizontal? When is it vertical?

Solution. (a) The slope of the tangent line is

.

When , we have and .

Therefore, the slope of the tangent line at the point where to is and its equation is

The tangent line is sketched in Figure 2.

Figure 2

(b)The tangent line is horizontal when , which occurs when and . That is, or . The corresponding points on the circle are and .

The tangent line is vertical when (provide that ), that is or . The corresponding points on the circle are and .   ■

Area

If then the area under a curve from to is . When the curve is given by parametric equations , and is traversed once as increases from to , then

or if is the leftmost endpoint.

Example 3

Find the area of the region enclosed by an ellipse

(for constants ).

Figure 3

Solution. The parametric equation of an ellipse is . Due to symmetry, it is clear that area inside the ellipse is 4 times the area inside the first quadrant. Thus we have

.    ■

Arc Length

We know how parametric equations can be used to describe the path of a particle moving in a plane. Now we develop a formula for determining the distance traveled by the particle along its path. Recall that the formula for the arc length of a curve given by over the interval is

.

If is represented by the parametric equations and , , where with is traversed only once, as increases from to and , , we have

or

This formula for arc length can be shown to be valid for any smooth curve that does not intersect itself.

Example 4

Find the arc length of the astroid , .

Figure 4

Solution. Looking at Figure 4, one arc is described by the parameter in the interval . Differentiating and with respect to we get

and .

Due to symmetry, it is clear that area inside the ellipse is 4 times the area inside the first coordinates. Thus we have

.

var('x, t')

x= (cos(t)*sin(t))

12*integral(x, t, 0, pi/2)

Example 5

Find the length of the curve

,   .

Solution.

var('x, y, t')

x=3*t^2-2;

y=-t+4;

parametric_plot((x, y), (t, -1, 1))

Figure 5

dxdt=diff(x, t)

dydt=diff(y, t)

integral(sqrt(dxdt^2+dydt^2), t, -1, 1)

Surface Area of a Surface of Revolution

As we did for arc length, we can use the formula for the surface area of a surface of revolution in rectangular form to develop a formula for a surface area in parametric form.

Suppose a curve can be represented by the parametric equations , , . The general formula and can also be used for parametric curves with replaced by

.

Assuming and are continuous and , the area of the surface of revolution obtained by rotating the curve about the -axis, is given by

.

The surface area of the surface of revolution obtained by rotating the curve about the -axis, is given by

.

Example 6

Find the surface area of the surface of revolution obtained by rotating about the -axis.

Solution. Since , , and , we get

(where )

.

s=revolution_plot3d(sin(x), (x, 0, pi/2), parallel_axis='x', show_curve=True, opaxity=0.3, colour='green')

s.show(aspect_ratio=(1,1,1))

Figure 6

10.2 EXERCISES (Calculus with Parametric Curves)

http://matrix.skku.ac.kr/Cal-Book/part1/CS-Sec-10-2-Sol.html

1-2. Find .

1.

Solution.

2.

Solution.

3-6. Find an equation of the tangent to the curve at the point.

3.

http://matrix.skku.ac.kr/cal-lab/cal-9-2-3.html

Solution. .

At the point with parameter value , the slope is

.

Slope of the tangent at is .

Hence, the equation of the tangent line is .

var('t')

x=t^4+3*t^3

y=t-t^2

T=diff(y)/diff(x)

print T (t=-1)

4.

Solution. At the point with parameter value , the slope is .

Slope of the tangent at is .

Hence, the equation of the tangent line is .

5.

Solution. At the point with parameter value , the slope is .

The point corresponds to the parameter value , so the slope of the tangent at the point is .

Hence, the equation of the tangent line is .

6.

Solution. At the point with parameter , the slope is

.

If . The slope of the tangent at is . Hence, the equation of the tangent line is .

var('t')

x=sin(t)

y=sin(t+t*sin(t))

T=diff(y)/diff(x)

print T (t=0)

p1=parametric_plot((sin(t),t+t*sin(t)),(t,0,2*pi))

p2=plot(t,color="red")

show(p1+p2)

7-10. Find . For what values of is the curve concave upward or downward?

7.

Solution. , .

⇒  Since and for all .

Hence, the curve is downward everywhere.

var('t')

x=t+exp(t)

y=t-exp(t)

parametric_plot([x, y], (t, -5, 2))

df1x=diff(x)

df1y=diff(y)

show(df1y/df1x)

df1=df1y/df1x

df2=diff(df1)

show((df2/df1x).factor())

8.

Solution. , .

⇒ If , . Thus if .

Hence, the curve is concave upward on ,

and is concave downward on .

9. , ,

Solution.   ,

.

If , . Thus if , .

Hence, the curve is upward on , and the  curve is downward on . Note that this curve is an ellipse.

var('t')

x= 3*cos(t)

y= 2*sin(t)

df1x=diff(x)

df1y=diff(y)

show(df1y/df1x)

df1=df1y/df1x

df2=diff(df1)

show(df2)

show((df2/df1x). simplify_trig())

10.

11-12. Find the point on the curve where the tangent is horizontal or vertical.

11.

Solution. so

so

The curve has horizontal tangents at ,

and vertical tangents at .

var('t')

x=t^3+t^2-8*t

y=t^3-t

print solve(diff(x)==0,t)

print solve(diff(y)==0,t)

print x (t=-2)

print y (t=-2)

print x (t=4/3)

print y (t=4/3)

print x (t=-1/3*sqrt(3))

print y (t=-1/3*sqrt(3))

print x (t=1/3*sqrt(3))

print y (t=1/3*sqrt(3))

parametric_plot([x, y], (t, -3, 3))

[t == -2,t == (4/3)]

[t == -1/3*sqrt(3), t == 1/3*sqrt(3)]

12

-6

-176/27

28/27

23/9*sqrt(3) + 1/3

2/9*sqrt(3)

-23/9*sqrt(3) + 1/3

-2/9*sqrt(3)

12.

Solution. , so for . Points at which the tangent is horizontal and , so for .

The curve has horizontal tangents at , and vertical tangents at .

13.Show that the curve , has two tangents at and find their equations.

Solution. .

Now is 0 when , so there are two tangents at the points since both

correspond to the origin. The tangent corresponding to has slope , and its equation is . The

tangent corresponding to has slope , and its equation is .

14.At what point does the curve , cross itself? Find the equations of both tangents at that point.

Solution.

var('t')

parametric_plot(((1-2*cos(t)^2), tan(t)*(1-2*cos(t)^2)), (t, -2, 2)).

show(aspect_ratio=1, xmin=-5, xmax=5, ymin=-7, ymax=7)

From the figure it is clear that at the curve does cross itself. Moreover, is when , , , .

The tangent corresponding to has slope or . Then the equations of tangents at are .

15.At what points on the curve is the tangent parallel to the line with equations ?

http://matrix.skku.ac.kr/cal-lab/cal-9-2-15.html

Solution. Given latter function’s slope is ,

so we should find when the former function’s slope is .

, .

We can get is or .

When is or , the former function’s tangent parallel to the latter’s.

var('t')

x1=-7*t

y1=12*t-5

x2=t^3+4*t

y2=6*t^2

d1=diff(y1)/diff(x1)

d2=diff(y2)/diff(x2)

print solve(d1==d2,t)

Answer :  [t == (-4/3), t == -1]

16.Find the area bounded by the curve , and the line .

Solution. When or , .

When , . When , .

var('x,y,t')

x=t-1/t;

y=t+1/t;

parametric_plot((x,y),(t,0,5)).show(aspect_ratio=1, xmin=-5, xmax=5, ymin=0, ymax=5)

(∵)

integral((5/2-y)*diff(x,t),t,1/2,2)

17. Find the area bounded by the curve , , and the lines and .

Solution. The curve intersects the -axis when . Then, the corresponding values of are . and are 0 and respectively.

var('t')

parametric_plot((cos(t), e^t), (t, 0, pi)).show(aspect_ratio=1, xmin=-5, xmax=10, ymin=-5, ymax=15)

var('x, y, t')

x=cos(t)

y=e^t

integral((9-y)*diff(x, t), t, 0, pi)

18. Find the area of a region enclosed by the astroid .

Solution.

var('t')

parametric_plot((cos(t)^3, sin(t)^3), (t, 0, 2*pi)).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-2, ymax=2)

( )

x, y, t=var('x, y, t')

x=cos(t)^3

y=sin(t)^3

4*integral(y*diff(x, t), t, pi/2, 0)

19. Find the arc length of the curve defined by

.

Solution. .

Then, .

Hence, .

20. Find the arc length of the curve defined by

.

Solution. ,

var('x,y,t')

x=e^t*sin(t)

y=e^t*cos(t)

dxdt=diff(x,t)

dxdt

e^t*sin(t) + e^t*cos(t)

dydt=diff(y,t)

dydt

-e^t*sin(t) + e^t*cos(t)

L=integral(sqrt(dxdt^2+dydt^2),t,0,pi)

L

21. Find the arc length of the curve defined by

, ;

Solution.

var('x,y,t')

x=2*ln(t);

y=t+(1/t);

parametric_plot((x,y),(t,1,4)).show(aspect_ratio=1, xmin=-1, xmax=3, ymin=-2, ymax=5)

dxdt=diff(x,t)

dxdt

2/t

dydt=diff(y,t)

dydt

-1/t^2 + 1

L=integral(sqrt(dxdt^2+dydt^2),t,1,4)

L

22. Find the arc length of the curve defined by

, ; .

Solution.

var('x, y, t')

x=t^2*cos(t);

y=t^2*sin(t);

parametric_plot((x,y),(t,0,2*pi)).show(aspect_ratio=1, xmin=-20, xmax=40, ymin=-30, ymax=10)

dxdt=diff(x, t)

dxdt

-t^2*sin(t) + 2*t*cos(t)

dydt=diff(y, t)

dydt

t^2*cos(t) + 2*t*sin(t)

L=integral(sqrt(dxdt^2+dydt^2), t, 0, 2*pi)

L

Answer :   1/3*(4*pi^2 + 4)^(3/2) - 8/3

23. Find the length of one arch of the cycloid

, .

Solution. .

Then, .

Hence, .

24-26. Find the area of the surface obtained by rotating the given curve about the -axis.

24. , ;

Solution. .

Then, .

Hence, .

25. , ;

http://matrix.skku.ac.kr/cal-lab/cal-10-2-25.html

Solution. .

Then, .

Hence,

.

var('r, t, x, y, k')

x=r*(t-sin(t))

y=r*(1-cos(t))

dxdt=diff(x,t)

dxdt

-(cos(t) - 1)*r

dydt=diff(y,t)

dydt

r*sin(t)

L1=integral(16*pi*r^2*(1-k^2),k,-1,1)

L1

26. , ;

27-29. Find the area of the surface generated by rotating the given curve about -axis.

27. , ;

http://matrix.skku.ac.kr/cal-lab/cal-9-2-27.html

Solution. ,

var('t, x, y')

x=t^2

y=r^3

dxdt=diff(x,t)

dydt=diff(y,t)

S=intergral(2*pi*x*sqrt(dxdt^2+dydt^2),t,0,1)

S

Answer : 32/1215*(sqrt(36*pi^2 + 4)*(243*pi^4 + 9*pi^2 - 2) + 4)*pi

28. , ;

29. , ;

10.3 Polar Coordinates

A coordinate system is a system which uses one or more numbers, or coordinates, to uniquely determine the position of a point in the coordinate plane. So far we have represented plane curves as collections of points In the rectangular coordinate system (Cartesian coordinate system), where and represent the directed distances from the coordinate axes to the point . In some cases the polar coordinate system introduced by Newton is a more useful and convenient coordinate system.

Choose a point in the plane that is called the pole (or origin) and label it . The polar axis is a ray (half-line) drawn starting at . Usually the polar axis is drawn horizontally to the right and aligns with the positive -axis in Cartesian coordinates.

Figure 1

For any point in the plane, let be the directed distance from to and let be the directed angle (usually measured in radians) between the polar axis and the line as in Figure 1. Then each point in the plane can be assigned polar coordinates . We assume that an angle is positive if measured in the counterclockwise direction from the polar axis and negative in the clockwise direction. If , then . Therefore, the pole for any value of can be  represented by .

When is negative, the meaning of polar coordinates is extended as follows. Observe that the points and (See Figure 2.) lie on the same line through and at the same distance from , but on opposite sides of . If , the point lies in the same quadrant as ; if , it lies in the quadrant on the opposite side of the pole. Thus, represents the same point as . Since an angle represents a complete counterclockwise rotation, the point represented by polar coordinates is also represented by

and for any integer .

Figure 2

Example 1

Plot the points whose polar coordinates are given.

(a)       (b)       (c)

Solution. The points are plotted in Figure 3. In part (c), the point is located three units from the pole in the fourth quadrant because the angle is in the second quadrant and is negative.

Figure 3

In the Cartesian coordinate system, every point has only one representation. This is not true with polar coordinates. For instance, the point in Example 1(a) could be written as or or (See Figure 4.)

Figure 4

To establish the relationship between polar and rectangular coordinates, we let the polar axis coincide with the positive axis and the pole with the origin, as shown Figure 5.

Figure 5

Let the point have as Cartesian coordinates and as polar coordinates. Since lies on a circle of radius , it follows that . Moreover, for , we have

, .

Hence

and  .

In case of , we can show that the same relationship holds. This relationship allows us to covert coordinates from one system to the other when we align polar axis with positive -axis.

Note that for given values of and , as increases through the interval , each value of occurs twice. So in converting from Cartesian coordinates to polar coordinates, determine such that the point lies in the correct quadrant.

Example 2

Find the Cartesian coordinates represent for polar coordinates .

Solution. Since and , equation  gives

, and  .

Therefore, the point is in the Cartesian coordinates.

Example 3

Find the polar coordinates represent for Cartesian coordinates .

Solution. If we choose to be positive, then where we have

,and .

Since the point lies in the second quadrant, we can choose . Thus, one possible answer is ; another is .

Polar Curves

A polar curve is represented by the polar equation , or more generally, . Its graph consists of all points that have at least one polar representation satisfying the polar equation. Curve sketching in polar coordinates is similar to that of rectangular coordinates. We rely heavily on point-by-point plotting, aided by intercepts and symmetry. We can also convert to rectangular coordinates and then make the sketch, however, sometimes this can be difficult.

Example 4

(a) Draw the curve with the polar equation .

(b) Find a Cartesian equation for this curve.

Solution. (a) In Figure 6, we find the values of for some convenient values of and plot the corresponding points . Then, we join these points to sketch the curve which appears to be a circle. We have used only values of between 0 and . If we allow to increase beyond , we obtain the same points again.

 ... ...

Figure 6 Table of values and graph of

Figure 7

(b) We convert the given equation into the Cartesian equation. From , we have . So the equation becomes , which gives

or .

Completing the square, we obtain

which is an equation of a circle with center and radius 1.   ■

Example 5

Sketch the curve , .

Figure 8  in Cartesian coordinates

Solution. As in Example 4, we first sketch , , in the Cartesian coordinates in Figure 8. As increases from 0 to , Figure 10 shows that decreases from 1 to 0 and so we draw the corresponding portion of the polar curve in Figure 9 (indicated by ①). As increases from to , goes from to . This means that the distance from increases from 0 to 1, but instead of being in the first quadrant this portion of the polar curve (indicated by ②), it lies on the opposite side of the pole in the third quadrant. The remainder of the curve is drawn in a similar fashion, with the arrows and numbers indicating the order in which the portions are traced out. The resulting curve has three loops and is called a three-leaved rose.

Figure 9  Three-leaved rose

Shapes of the function and

 a cycle ,  : odd ,  : even leaves leaves (symmetric about -axis) leaves leaves (symmetric about -axis)

Symmetry

When making a table of points, it is often helpful to use at least one value from each of four quadrants. In the above Table, we see that no points are plotted in Quadrants III and IV because of is negative for these values of . Thus, on the interval from to the curve is traced twice. Moreover we also see that the curve is symmetric with respect to the line . Therefore, if we had known about this symmetry, we could have plotted fewer points.

When we sketch polar curves, it is helpful to utilize the symmetry. The following three rules will be useful in sketching polar curves.

(a) If then the curve is symmetric about the polar axis(or -axis).

(b)If or then the curve is sy-mme-tric about the pole. (That is, the curve remains unchanged if we rotate it through 180º about the origin.)

(c) If , then the curve is symmetric about the vertical line (or -axis). (See Figure 11.)

Figure 11

Since , the curves sketched in Examples 4 and 5 are symmetric about the polar axis. Thus, in Example 4, we should plot only the points for and then reflect it about the polar axis to obtain the complete circle. The three-leaved rose is symmetric about the poler axis because . Since, the curve is symmetric about .

Tangents to Polar Curves

To find the slope of a tangent line to a polar curve, consider a differentiable function . Here we regard as a parameter, and write its parametric equations as

.

Then, using the parametric form of given in Section 10.2, we have

From this formula, we can make the following observations:

1.Solutions to yield horizontal tangents. (provided )

2. Solutions to yield vertical tangents. (provided )

If and are simultaneously zero, then no conclusions can be drawn about tangent lines. Notice that if we are looking for tangent lines at the pole, then , simplifies to

if  .  (See Figure 12.)

Figure 12

Example 6

(a) For the cardioid , find the slope of the tangent line when . (See Figure 13.)

Figure 13

(b) Find the points on the cardioid where the tangent line is horizontal or vertical.

Solution. Using with , we have

.

(a) The slope of the tangent at the point where is

.

(b) Observe that when .

x, t= var('x, t')

show(plot(cos(t)*(1 - 2*sin(t))))

p = cos(t)*(1 - 2*sin(t))

solve(p==Integer(0), t)

when .

Therefore, there are horizontal tangents at the points , and vertical tangents at and . When , both and are 0, we must be careful. Using L’Hospital’s Rule, we have

.

By symmetry, .

Thus, there is a vertical tangent line at the pole.

THEOREM 2

Let be an angle between radius at point on a curve . Then

if .

Figure 14

Proof.  From Figure 14, implies that

.

Since the slope of tangent line at the point is and by ,

and

Now substituting these values in we get the required result in this theorem. ■

Let two curves and meet at the point . Say and are angles between radius and curves and , respectively. Then the angle between two curves is and satisfies

where and comes from . (See Figure 15.)

Figure 15

Example 7

Show two curves and meet at the point , and find angle between the curves at the point.

Solution. By solving , we obtain , then and , hence the point satisfies two equations. Now we call and be angles of the tangent lines of and at the point , respectively. Then we have , , respectively. Putting it in , the angle between two curves satisfies

.

Hence the angle at the point that satisfies is .     ■

Example 8

Sketch the curve with the given polar equation.

Solution.

var('theta')

polar_plot(2-cos(2*theta), (0, 2*pi)).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

Figure 16

10.3 EXERCISES (Polar Coordinates)

http://matrix.skku.ac.kr/Cal-Book/part1/CS-Sec-10-3-Sol.html

1-4. Plot the point whose polar coordinates are given. Then, find the Cartesian coordinates of the point.

1.

Solution. Graph of polar functions

def Polar(r, theta):

#converts Polar to Cartesian Coordinates

CartC = ([r*cos(theta), r*sin(theta)]);

return CartC;

pt=Polar(-1,pi);

vector(pt)

list_plot([pt], aspect_ratio=1, xmin=-1, xmax=2, ymin=-1, ymax=1)

2.

Solution.

def Polar(r, theta):

#converts Polar to Cartesian Coordinates

CartC = ([r*cos(theta), r*sin(theta)]);

return CartC;

pt=Polar(4, pi/4);

vector(pt)

Answer :   (x, y)=(2 sqrt {2}, 2 sqrt {2})

3.

Solution.

4.

Solution.

5-8. The Cartesian coordinates are given. Find two other pairs of polar coordinates of the point, one with  and the other .

5.

Solution. , and .

Since is in the fourth quadrant, the polar coordinates are

: , ⑵ : .

6.

Solution. , and .

Since is in the second quadrant, the polar coordinates are

⑴ : , ⑵ :

7.

Solution. , and .

Since is in the fourth quadrant, the polar coordinates are

⑴ : , ⑵ : .

8.

Solution. , and .

Since is in the second quadrant, the polar coordinates are

⑴ : , ⑵ : .

9. Find a formula for the distance between the points with polar coordinates and .

Solution. Let

Let be the distance between and .

10-12. Find a polar equation for the curve represented by the given Cartesian equation.

10.

Solution.

11.

Solution.

12.

Solution.

13-15. Find a Cartesian equation for the curve represented by the given polar equation.

13.

Solution.

, a circle of radius centered at . The first two equations are actually equivalent since .

But gives the point when .

Thus, the single equation is equivalent to the compound condition ().

14.

Solution.

15.

Solution.

16-25. Sketch the curve with the given polar equation.

16.

17.

Solution.

var('theta')

polar_plot(sin(theta), (0, pi)).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-2, ymax=2)

18.

Solution.

r=var('r');

polar_plot(3,(0, 2*pi)).show(aspect_ratio=1, xmin=-5, xmax=5, ymin=-5, ymax=5)

19.

Solution.

theta=var('theta');

polar_plot(theta, (0, 2*pi)).show(aspect_ratio=1, xmin=-10, xmax=10, ymin=-10, ymax=10)

20.

Solution.

var('theta')

polar_plot(1-2*cos(theta), (0, 2*pi)).show(aspect_ratio=1, xmin=-4, xmax=4, ymin=-4, ymax=4)

21.

Solution.

var('theta')

polar_plot(sin(theta/2), (0, 2*pi)).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-2, ymax=2)

22. ,

Solution. Graph of polar functions

var('theta')

polar_plot(sin(4*theta), (0, 2*pi)).show(aspect_ratio=1, xmin=-1, xmax=1, ymin=-1, ymax=1)

23. ,

Solution. r(t) = 1+2*cos(2*t) in

Graph of polar functions

http://matrix.skku.ac.kr/cal-lab/sage-grapher-polar.html

var('theta')

polar_plot(1+2*cos(2*theta), (0, 6*pi)).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-1, ymax=1)

24.

25.

26. Show that the polar curve (called a conchoid) has the line as a vertical asymptote by showing that . Use this fact to sketch the conchoid.

Solution. .

Now, ,

so .

Also, ,

so .

Therefore, is a vertical asymptote.

var('theta')

polar_plot(4+2*sec(theta), (0, 2*pi)).show(aspect_ratio=1, xmin=-5, xmax=15, ymin=-10, ymax=10)

27. Show that the curve (also a conchoid) has the line as a horizontal asymptote by showing that Use this fact to help sketch conchoid.

Solution. .

Now,  ,

so .

Also, ,

so .

Therefore,

is a horizontal asymptote.

var('theta')

polar_plot(2-csc(theta), (0, 2*pi)).show(aspect_ratio=1, xmin=-15, xmax=15, ymin=-10, ymax=10)

28.Show that the curve (called a cissoid of Diocles) has the line as a vertical asymptote. Show also that the curve lies entirely within the vertical strip . Use these facts to sketch the cissoid.

29-32. Find the slope of the tangent line at the given point.

29.

Solution.

30.

Solution.

(vertical tangent line)

31.

Solution.

32.

33-36. Find the points on the given curve where the tangent line is horizontal or vertical.

33.

Solution.

.

.

So the tangent is vertical at and .

.

So the tangent is horizontal at and .

34.

35.

36.

37.Show that the polar equation represents a circle and find its center and radius.

Solution.

.

Then the polar equation represents acircle and its center : , radius : .

10.4 Areas and Lengths in Polar Coordinates

The formula for the area of a region whose boundary is given by a polar equation is considered in this section. The development of a formula for the area of a polar region parallels that for the area of regions in the rectangular coordinates system, but use sectors of a circle, instead of rectangles, as the basic element of area. We know that the area of a sector of a circle with radius and the radian measure of the central angle is

(See Figure 1.)

Figure 1

Since the area of a sector is proportional to its central angle: .

Consider the region (See Figure 2.) bounded by the polar curve and by the rays and . Assume that is a continuous and non-negative function and . We partition the interval into equal subintervals

with equal width . Then is divided into smaller regions by rays with central angle . Choose in the th subinterval , then the area of the th region is approximated by the area of the sector of a circle with central angle and radius . (See Figure 3.)

Figure 2                          Figure 3

Using

.

Thus an approximation to the total area of is

.

As the Riemann sums for the function , given by the sums in converge to

.

Thus formula for the area of the polar region is

.

Since , Equation may be written as

.

The area given by or may be thought of as being swept out by a rotating ray through that starts with angle and ends with angle . We can use the same formula to find the area of a region bounded by the graph of a non-positive continuous function. However the formula is not necessarily valid if takes on both positive and negative values in the interval .

Sometimes the most difficult part of finding the area of a polar region is determining the limits of integration. A good sketch of the region helps.

Figure 4

Consider a region bounded by two polar curves with polar equations, , , and . (See Figure 4.) We assume that and are continuous and on and . The area of such a region that is bounded by two curves is found by subtracting the area inside from the area inside . Using , we have

.

To find all points of intersection of two polar curves, it is recommended to draw the graphs of both curves, because a single point has many representations in polar coordinates.

Example 1

Find the area of the region bounded by the limaçon with equation .

Solution.  The curve is sketched in Figure 5.

Figure 5

From we have

.

var('x, t')

x= (2+ 2*cos(t))^2

integral(x, (t, 0, pi))

Example 2

Find the entire area of the lemniscate with equation .

Figure 6

Solution.       The curve is sketched in Figure 6. Figure 6 is symmetric with respect to both and -axis, the whole area = 4the area in the first quadrant. Since when . If varies from to , the radius vector sweeps over the area . From we have

.

var('x, a')

f(x)= 2*a^2* (cos(2*x))

print integral(f(x), x, 0, pi/4)        # a^2

Figure 7

Example 3

Find the area of the region inside the circle and outside the cardioid .

Solution. The cardioid and the circle are shown in Figure 7 with the desired region shaded. The points of intersection of the two curves are given by which gives , so . The required area is obtained by subtracting the area inside the cardioid between from the area inside the circle from to Thus

.

Using symmetry of the region about the vertical axis , we can write

.

var('x, a')

f(x)= 12 *((sin(x))^2) + 4*sin(x) -1

print integral(f(x), x, 7*pi/6, 3*pi/2)     #

Because a point may be represented in different ways in polar coordinates, care must be taken in determining the points of intersection of two polar graphs. For example, consider the points of intersection of the graphs of and . As with cartesian equations, let us solve two equation simultaneously to find the points of intersection. Then we would obtain the following.

If you sketch the graphs, you will see that there is a third point of intersection that did not show up when we solved two polar equations simultaneously. This is one reason we stress sketching a graph when finding the area of a polar region. Because it does not occur with the same coordinates in the two graphs, we did not find the third point.  The point which occurs with the coordinates on the graph of is the same point which occurs with the coordinates on the graph of .

We can compare the problem of finding points of intersection of two polar graphs with that of finding collision points of two satellites in intersecting orbits about the earth. The satellites will not collide because they reach the points of intersection at different times (values). A collision will occur only at the points of intersection are a “simultaneous point”-those reached at the same time (value).

Example 4

Find the points of intersection of the cardioid and the cardioid .

Figure 8  ,

Solution. By solving the equations and , we get (See Figure 8.) and, therefore, . This gives us or .

var('x')

f(x)= sin (x + pi/4)

print solve (f == 0, x)

plot(f(x), x, 0, 2*pi)           #

So the values of in satisfying both equations are . Thus the two points of intersection are , . From Figure 8, we can see that the pole also is a point of intersection.    ■

Example 5

Find the area of the region that is bounded by the given curve and lies in the specified sector.

Solution.

var('theta')

polar_plot(theta^2, (0, pi), fill=True).show(aspect_ratio=1, xmin=-10, xmax=3, ymin=-2, ymax=5)

Figure 9

r=theta^2

integral(1/2*r^2, theta, 0, pi)

Arc Length

The formula for the length of a polar arc can be obtained from arc length formula for a curve described by parametric equations. Consider as a parameter of a polar curve , . Assume that is continuous. Then the parametric equations of the curve are

and  .

Differentiating with respect to , we have

and  .

Then

.

Thus the arc length of the polar curve is

.

Example 6

Find the length of the cardioid .

Solution. The full length of the cardioid (See Figure 10.) is given by the parameter interval . Using we have

var('x, t')

x= 2*(cos(t/2))

2*integral(x, t, 0, pi)

Figure 10

Example 7

Find the length of the curve.

Solution.

var('theta')

r=cos(theta)

polar_plot(r, (theta, 0, pi/3))

Figure 11

dr=diff(r, theta)

integral(sqrt(r^2+dr^2), theta, 0, pi/3)

Example 8

Find the approximated length of the curve , with numerical approximation.

Solution.

var('theta')

r= 1 - cos(theta)

dr=diff(r, theta)

h = integral(sqrt(r^2+dr^2), theta, 0, pi/3); h

h.n()

#print numerical_integral(sqrt(r^2+dr^2), theta, 0, pi/3)[0]

Answer :   0.535898384862246( -2*sqrt(3) + 4 )

10.4 EXERCISES (Areas and Lengths in Polar Coordinates)

http://matrix.skku.ac.kr/Cal-Book/part1/CS-Sec-10-4-Sol.html

1-4. Find the area of the region that is bounded by the given curve and lies in the specified sector.

1. ,

Solution.

var('theta')

polar_plot(theta, (0, pi/2), fill=True).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

r=theta;

A=integral(1/2*r^2,theta,0,pi/2);

2. ,

Solution.

var('theta')

polar_plot(e^(theta/3), (pi, 2*pi), fill=True).show(aspect_ratio=1, xmin=-5, xmax=10, ymin=-5, ymax=5)

r=e^(theta/3);

A=integral(1/2*r^2,theta,pi,2*pi);

3. ,

Solution.

var('theta')

polar_plot(sqrt(theta), (0, pi/3), fill=True).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-2, ymax=2)

r=sqrt(theta);

A=integral(1/2*r^2,theta,0,pi/3);

4. ,

5-9. Find the area bounded by one loop of the given curve.

5.

Solution.

var('theta')

polar_plot(2*cos(2*theta), (0, 2*pi), fill=True).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

r=2*cos(2*theta);

A=integral(1/2*r^2,theta,pi/4,3*pi/4);

6.

Solution.

var('theta')

polar_plot(3*sin(3*theta), (0, pi), fill=True).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

(3 leaves)

r=3*sin(3*theta);

A=integral(1/2*r^2,theta,0,pi/3);

7.

Solution.

var('theta')

polar_plot(2*cos(4*theta), (0, 2*pi), fill=True).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

r=2*cos(4*theta);

A=integral(1/2*r^2,theta,pi/8,3*pi/8);

8.

Solution.

var('theta')

polar_plot(2+sin(2*theta), (0, 2*pi), fill=True).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

r=2+sin(2*theta);

A=integral(1/2*r^2,theta,7*pi/6,11*pi/6);

9. Find the area between a large loop and the enclosed small loop of the curve .

Solution.

s

var('theta')

polar_plot(1+2*cos(theta), (0, 2*pi), fill=True).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

r=1+2*cos(theta);

B=integral(1/2*r^2,theta,0,2*pi);

B

From figure, the area between a large loop and the enclosed small loop of the curve is  .

10-12. Sketch the curve and find the area that it encloses.

10. ,

Solution.

var('theta')

polar_plot(2*sin(3*theta), (0, pi), fill=True).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

r=2*sin(3*theta);

A=3*integral(1/2*r^2 ,theta,0,pi/3);

A

11.

Solution.

var('theta')

polar_plot(3*(1-cos(2*theta)), (0, 2*pi), fill=True).show(aspect_ratio=1, xmin=-8, xmax=8, ymin=-8, ymax=8)

r=3*(1-cos(2*theta));

A=integral(1/2*r^2,theta,0,2*pi);

A

12.

13-15. Find the area of the region that lies inside the first curve and outside the second curve.

13. ,

Solution.

var('theta')

p1=polar_plot(2*sin(theta), (0, 2*pi), fill=True).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-2, ymax=2)

p2=polar_plot(1, (0, 2*pi), fill=True).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-2, ymax=2)

r=2*sin(theta)

A=2*integral(1/2*(r^2-1), theta, pi/6, pi/2);

A

14. ,

Solution.

var('theta')

p1=polar_plot(3*sin(theta),(theta,0,2*pi), color="red", fill=True, fillcolor="red")

p2=polar_plot(1+cos(theta),(theta,0,2*pi), fill=True, fillcolor="blue")

show(p1+p2, aspect_ratio=1, ymin=-5, ymax=5, xmin=-5, xmax=5)

A=2*integral(1/2*((3*cos(theta))^2-(1+cos(theta))^2),theta,0,pi/3);

A

15. ,

Solution.

var('theta')

p1=polar_plot(sqrt(2*cos(2*(theta))), (- pi/4, pi/4), fill=True)

p2=polar_plot(sqrt(-2*cos(2*(theta))), ( pi/4, (3*pi)/4), fill=True)

p3=polar_plot(sqrt(-2*cos(2*(theta))), ( -3*pi/4, -pi/4), fill=True)

p4=polar_plot(sqrt(2*cos(2*(theta))), (3*pi/4, 5*pi/4), fill=True)

p5=polar_plot(1,(0, 2*pi), fill=True)

show(p1+p2+p3+p4+p5,aspect_ratio=1, xmin=-2, xmax=2, ymin=-2, ymax=2)

A=4*integral(1/2*(2*cos(2*theta)-1), theta, 2*pi/6, 4*pi/6);

A

Answer :  -2/3*pi – 2*sqrt(3)  #

16-17. Find all points of intersection of the given curves and find the area of the region that lies inside both curves.

16. ,

Solution.

or

var('theta')

p1=polar_plot(sin(3*(theta)), (0, pi), fill=True, fillcolor='pink', color='red')

p2=polar_plot(sin(theta),(0, pi), fill=True, fillcolor='skyblue', color='blue', ymin=0)

p1+p2

The two curves intersect at , and .

r=sin(3*theta);

A=2*integral(1/2*sin(theta)^2,theta,0,pi/4)+2*integral(1/2*r^2,theta,pi/4,pi/2);

A

17. ,

Solution.

var('theta')

p1=polar_plot(cos(2*(theta)), (0, 2*pi), fill=True).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-2, ymax=2)

p2=polar_plot(4*sin(2*(theta)),(0, 2*pi), fill=True).show(aspect_ratio=1, xmin=-4, xmax=4, ymin=-4, ymax=4)

A=8*2*integral(1/2*sin(2*theta), theta, 0, pi/8);

A

18-21. Find the length of the polar curve.

18. ,

19. ,

20.Graph the curve and

find its length.

Solution.

var('theta')

p1=polar_plot((cos(theta/3))^2, (0, 6*pi), fill=True).show(aspect_ratio=1, xmin=-2, xmax=2, ymin=-2, ymax=2)

var('r')

r=(cos(theta/3))^2;

drd(theta)=diff(r,theta);

drd(theta)

-2/3*sin(1/3*theta)*cos(1/3*theta)

L=4*integral(cos(theta/3)*sqr t(cos(theta/3)^2+4/9*sin(theta/3)^2), theta, 0, 3*pi/2);

L

21. Show that the area of the surface generated by rotating the polar curve with about the polar axis is

.

Solution. From

http://mathworld.wolfram.com/SurfaceofRevolution.html

we have the formula

.

Since ,

.

10.5 Conic Section

A conic section (or a conic) is a curve obtained by from the intersection between a cone (more precisely, a right circular conical surface) with a plane. The three types of conic sections are the hyperbola, the parabola, and the ellipse. The circle is a special case of the ellipse, and is of sufficient interest in its own right that it is sometimes called the fourth type of conic section.

The conic sections were named and studied as long ago as 200 BC, when Apollonius of Perga undertook a systematic study of their properties.

In this section we give the geometric definitions of parabolas, ellipses, and hyperbolas and derive their standard equations. (See Figure 1.)

Figure 1  Conics

Parabolas

The parabola is the set of all points in a plane that are equidistant from a fixed point (called the focus) and a fixed line (called the directrix). The focus does not lie on the directrix. The midpoint between the focus and the directrix is called the vertex and the line passing through the focus and the vertex is called the axis of the parabola.

In the 16th century Galileo showed that the path of a projectile that is shot into the air at an angle to the ground is a parabola. The parabola has many important applications, from automobile headlight reflectors to the design of ballistic missiles. They are frequently used in physics, engineering, and many other areas.

Figure 2                      Figure 3

When vertex is at the origin and its directrix is parallel to the ‐axis (See Figure 3.), we obtain a particularly simple equation for a parabola. Suppose that the focus is the point . Then the equation of the directrix is . For any point on the parabola, the distance from to the focus is

and the distance from to the directrix is . ( in Figure 3.) According to the definition of a parabola, these distances are equal:

.

By squaring and simplifying the above equation we get

.

Thus the equation of the parabola with focus and directrix is

.

The standard equation of a parabola is , where . The parabola opens upward if but downward if .(See Figures 4(a) and 4(b).) Since is a even function of , the graph is symmetric with respect to the ‐axis, which is the axis of the parabola.

An equation of the parabola with focus and directrix is

which is obtained by interchanging and in .

The parabola opens to the right if and to the left if (See Figures 4(c) and 4(d).). In both cases the graph is symmetric with respect to the ‐axis.

Figure 4

Example 1

Draw the graph of the parabola and find the focus and directrix.

Solution.       If we write the equation as and compare it with , we see that , so . Thus, the focus is and the directrix is . The sketch is shown in Figure 5.

Figure 5

Ellipses

The ellipse was first studied by Menaechmus, investigated by Euclid, and named by Apollonius. The focus and conic section directrix of an ellipse were considered by Pappus. In 1602, Kepler believed that the orbit of Mars was oval; he later discovered that it was an ellipse with the Sun at one focus. In fact, Kepler introduced the word “focus” and published his discovery in 1609. In 1705, Halley showed that the comet now named after him moved in an elliptical orbit around the Sun. (MacTutor Archive.) An ellipse rotated about its minor axis gives an oblate spheroid, while an ellipse rotated about its major axis gives a prolate spheroid.

An is the set of points in a plane satisfying the sum of whose distances from two fixed points and is a constant. (See Figure 6.) These two fixed points are called the foci (plural of focus). One of Kepler’s laws states that the orbits of the planets in the solar system are ellipses with the Sun at one focus. An interesting reflection property of the ellipses, similar to parabolas, is of practical use. For example, if a source of light or sound is placed at one focus of a surface with elliptical cross‐sections, then all the light or sound is reflected off the surface to the other focus. In lithotripsy, a treatment for kidney stones, a reflector with elliptical cross‐section is placed in such a way that the kidney stone is at one focus. High‐intensity sound waves generated at the other focus are reflected to the stone to destroy it without damaging surrounding tissue.

and are focuses

Figure  6                                    Figure 7

Suppose the foci are the points and on the ‐axis so that the origin is halfway between the foci. (See Figure 7.) Let be the sum of the distances from a point on the ellipse to the foci. Let be any point on the ellipse. Then . That is

or  .

By squaring on both sides, we get

Simplifying

.

By squaring again, we have

Which is

.

Note that . Since we have , so and, therefore, . Let . Then the equation of the ellipse becomes

or

.

Since , it follows that . By putting in , the ‐intercepts are . The corresponding points and are called the vertices of the ellipse and the line segment joining the vertices is called the major axis. The ‐intercepts are obtained by putting in  . The graph of the ellipse is symmetric about both ‐axis and ‐axis since is even in both and . When then and the foci coincide. Consequently the ellipse becomes a circle with radius . Thus the ellipse

,

has foci at , and vertices . Here . (See Figure 8.)

By interchanging and in , the ellipse with foci on the ‐axis at , and vertices at is

,    .

Here . (See Figure 9.)

Figure 8                               Figure 9

Example 2

Find the foci and sketch the graph of .

Solution. The given equation becomes the standard form of an ellipse by dividing it by :

Thus and Therefore the ‐intercepts are and the ‐intercepts are . Also, , so and the foci are . The graph is sketched in Figure 10.

Figure 10

Example 3

Determine an equation of the ellipse with foci and vertices .

Solution. and . Then . Hence, the equation of the ellipse is  or . ■

Hyperbolas

The definition of a hyperbola is similar to that of an ellipse. A hyperbola is an open curve with two branches, the intersection of a plane with both halves of a double cone. A hyperbola is the set of all points in a plane the difference of whose distances from two fixed points and (the foci) is a constant.(See Figure 11.)

Figure  11  is on the hyperbola when

Hyperbola occur frequently as graphs of equations in chemistry, physics, biology, and economics(Boyle’s Law, Ohm’s law, supply and demand curves). A particularly significant application of hyperbolas was found in the navigation systems developed in World Wars I and II.

Notice that the derivation of the equation of a hyperbola is also similar to the one given earlier for an ellipse. You can show that, when the foci are on the ‐axis at and the difference of distances is , the equation of the hyperbola is

where . The ‐intercepts are again and the vertices are the points and . When we put in we get , which means there is no real solution. Therefore, we don’t have ‐intercept.

Since is even in both and , the graph of hyperbola is symmetric with respect to both - and -axes.

Rewriting we have

or , so . Thus the hyperbola exists (is real) only when or , and it consists of two parts, called its branches.

The dashed lines and in Figure 12 are the asymptotes to the hyperbola. Both branches of the hyperbola come arbitrarily close to the asymptotes.

Figure 12

If the foci of a hyperbola are on the ‐axis, then by interchanging the roles of and we obtain the following information, which is illustrated in Figure 13.

Similarly, the hyperbola

has foci , on the ‐axis, vertices at , and as asymptotes. Here , ( is obtained by interchanging and : see Figure 13.)

Figure  13                Figure  14

Example 4

Sketch the graph and find the foci and asymptotes of the hyperbola .

Solution. Rewriting in form , we have with and . Since , the foci are . The asymptotes are the lines and . The graph is shown in Figure 14.

Example 5

Determine the foci and equation of the hyperbola with vertices and asymptotes .

Solution. With reference to , we have and . Thus, and . Then the foci are and the equation of the hyperbola is .   ■

Shifted Conics

By replacing and by and , in the standard equations , , , and , we obtain the shift conics.

Example 6

Find an equation of the ellipse with foci and vertices .

Solution. The line segment joining the vertices is the major axis and has length 6, so . The distance between the foci is 2, so . Thus, . Since the center of the ellipse is , replace by and by in . Thus the equation of the ellipse is

.

(See Figure 15.)

Figure  15     ■

Example 7

Find the foci and draw the conic .

Solution. Rearranging the terms and completing the squares, we get

.

With reference to , , and . The foci are and and the vertices are and . The asymptotes are

.     ■

Conic Sections in Polar Coordinates

We defined the parabola in terms of a focus and directrix, but we defined  the ellipse and hyperbola in terms of two foci. Now we will give a unified treatment of all the three types of conic sections in terms of a focus and directrix. We know that the ellipse and hyperbolas take simple forms when the origin lies at their center. As it happens, there are many important applications of conics in which it is more convenient to use one of the foci as the reference point(the origin) for the coordinate system. Therefore, if we move one of the foci to the origin, a conic section has a simple polar equation which provides a convenient description of the motion of planets, satellites, and comets. In this section we will see that polar equations of conics have a simple form if one of the foci lies at the pole.

The following theorem use concepts of eccentricity to classify the three basic types of conics.

THEOREM 1

Let be a fixed point (focus) and be a fixed line (directrix) in a plane. And let be a fixed positive number (called the eccentricity). Then a conic section is the set of all points in the plane such that

.

Furthermore the conic is an ellipse if , a parabola if and a hyperbola if . (See Figure 16.)

Figure  16

Proof.  If , then . Therefore, the given condition simply becomes the definition of a parabola.

If , then we consider the focus to lie at the origin and the directrix to lie to the right of the origin, as shown in Figure 16. For the point , we have  ,    .

Then  , or ,  or

.

By solving for , the polar equation of the conic (shown in Figure 16) is obtained as .

Squaring on both sides and converting it to rectangular coordinates, we get

or

.

Rearranging the terms and completing the square, we get

.

or

.

where

.

This is the equation of an ellipse if .

The foci of an ellipse are at a distance from the center, where

.

This shows that

.

From and , the eccentricity is given by .

If , then then equation can be rewritten in the form

which represents a hyperbola with where .            ■

THEOREM 2

A polar equation of the form

or

represents a conic section with eccentricity , where is distance between the focus at the pole and its corresponding directrix.

In Figure 17(b) the directrix , is to the left of the focus. In Figure 17(c) and 17(d) the directrix is parallel to the polar axis as ,

(a)                 (b)

(c)                 (d)

Figure 17

Example 8

Find a polar equation for a parabola that has its focus at the origin and whose directrix is .

Solution. From Figure 17(c) and using Theorem 2 with and the equation of the parabola is   .     ■

Example 9

Find the eccentricity, locate the directrix, identify and sketch the conic given by the polar equation  .

Solution. Rewriting

from Theorem 2 this represents an ellipse with . Since , we have

so the directrix has Cartesian equation . When ; when . So the vertices have polar coordinates and . The ellipse is sketched in Figure 18.

Figure 18

Example 10

Sketch the conic .

Solution. The rewritten equation

represents a hyperbola with the eccentricity .

Since and the directrix has equation . The vertices occur when and , so they are and . When we get the ‐intercepts with in both cases. As then or and . Thus, the asymptotes are parallel to the rays and . The hyperbola is sketched. (See Figure 19.)

Figure 19

Example 11

Find the vertices and foci of the ellipse and sketch its graph.

Solution.

var('x,y,a,b,c')

ellipse=implicit_plot(x^2/7+y^2/3==1,(x,-3,3),(y,-3,3)).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

Figure 20

solve(a^2==7,a)

[a == -sqrt(7), a == sqrt(7)]

solve(b^2==3,b)

[b == -sqrt(3), b == sqrt(3)]

solve(c^2==7-3,c)

[c == -2, c == 2]

var('x, y')

ellipse=implicit_plot(x^2/7+y^2/3==1,(x,-3,3),(y,-3,3))

f1=point((-2, 0), pointsize=20, rgbcolor=(1,0,0));

f2=point((2, 0), pointsize=20, rgbcolor=(1,0,0));

v1=point((0,-sqrt(3)), pointsize=20, rgbcolor=(0,0,1));

v2=point((0,sqrt(3)), pointsize=20, rgbcolor=(0,0,1));

v3=point((-sqrt(7), 0), pointsize=20, rgbcolor=(0,0,1));

v4=point((sqrt(7), 0), pointsize=20, rgbcolor=(0,0,1));

show(ellipse+f1+f2+v1+v2+v3+v4,aspect_ratio=1,xmin=-3,xmax=3, ymin=-3, ymax=3)

Figure 21

10.5 EXERCISES (Conic Section)

1-4. Sketch the parabola with the given equation. Show and label its vertex, focus, axis, and directrix.

1.

Solution.

var('x,y')

implicit_plot((y-2)^2==3*(x-3), (x, -10, 10), (y, -10, 10)).show(aspect_ratio=1, xmin=0, xmax=12, ymin=-4, ymax=8)

⇒ vertex : / focus : / axis : / directrix :

2.

Solution.

var('x, y')

parabola=implicit_plot(6*y+x^2==0, (x, -3, 3), (y, -3, 3)).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

⇒ vertex : / focus : / axis : / directrix : .

3.

Solution.

var('x, y')

parabola=implicit_plot((y-3)^2+2*(x+1)==0, (x, -5, 1), (y, 0, 7)).show(aspect_ratio=1, xmin=-5, xmax=1,

ymin=0, ymax=7)

⇒ vertex : / focus : / axis : / directrix :

4.

5-8. Find the vertices and foci of the ellipse and sketch its graph.

5.

Solution.

var('x,y,a,b,c')

ellipse=implicit_plot(x^2/6+y^2/4==1,(x,-3,3),(y,-3,3)).show(aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

solve(a^2==6,a)

[a == -sqrt(6), a == sqrt(6)]

solve(b^2==4,b)

[b == -2, b == 2]

solve(c^2==6-4,c)

[c == -sqrt(2), c == sqrt(2)]

var('x,y,a,b,c')

ellipse=implicit_plot(x^2/6+y^2/4==1,(x,-3,3),(y,-3,3))

f1=point((-sqrt(2),0), pointsize=20, rgbcolor=(1,0,0));

f2=point((sqrt(2),0), pointsize=20, rgbcolor=(1,0,0));

v1=point((-sqrt(6),0), pointsize=20, rgbcolor=(0,0,1));

v2=point((sqrt(6),0), pointsize=20, rgbcolor=(0,0,1));

v3=point((0,-2), pointsize=20, rgbcolor=(0,0,1));

v4=point((0,2), pointsize=20, rgbcolor=(0,0,1));

show(ellipse+f1+f2+v1+v2+v3+v4,aspect_ratio=1, xmin=-3, xmax=3, ymin=-3, ymax=3)

6.

7.

8.

9-11. Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph.

9.

Solution.

var('x,y,a,b,c')

hyperbola=implicit_plot(x^2/36-y^2/16==1,(x,-15,15),(y,-15,15)).show(aspect_ratio=1, xmin=-15,

xmax=15, ymin=-15, ymax=15)

solve(a^2==36,a)

[a == -6, a == 6]

solve(b^2==16,b)

[b == -4, b == 4]

solve(c^2==36+16,c)

[c == -2*sqrt(13), c == 2*sqrt(13)]

var('x,y,a,b,c')

hyperbola=implicit_plot(x^2/36-y^2/16==1,(x,-15,15),(y,-15,15))

f1=point((-2*sqrt(13),0), pointsize=20, rgbcolor=(1,0,0));

f2=point((2*sqrt(13),0), pointsize=20, rgbcolor=(1,0,0));

v1=point((-6,0), pointsize=20, rgbcolor=(0,0,1));

v2=point((6,0), pointsize=20, rgbcolor=(0,0,1));

show(hyperbola+f1+f2+v1+v2,aspect_ratio=1, xmin=-15, xmax=15, ymin=-15, ymax=15)

10.

Solution.

var('x,y,a,b,c')

hyperbola=implicit_plot(y^2/81-x^2/25==1,(x,-15,15),(y,-20,20)).show(aspect_ratio=1, xmin=-15,

xmax=15, ymin=-20, ymax=20)

solve(a^2==81,a)

[a == -9, a == 9]

solve(b^2==25,b)

[b == -5, b == 5]

solve(c^2==81+25,c)

[c == -sqrt(106), c == sqrt(106)]

x,y,a,b,c=var('x,y,a,b,c')

hyperbola=implicit_plot(y^2/81-x^2/25==1,(x,-15,15),(y,-20,20))

f1=point((0,-sqrt(106)), pointsize=20, rgbcolor=(1,0,0));

f2=point((0,sqrt(106)), pointsize=20, rgbcolor=(1,0,0));

v1=point((0,-9), pointsize=20, rgbcolor=(0,0,1));

v2=point((0,9), pointsize=20, rgbcolor=(0,0,1));

show(hyperbola+f1+f2+v1+v2,aspect_ratio=1, xmin=-15, xmax=15, ymin=-20, ymax=20)

11.

12-14. Identify the type of conic section whose equation is given and find the vertices and foci.

12.

Solution.   ⇒

⇒ : Ellipse

var('x,y,a,b,c')

ellipse=implicit_plot(x^2/3+(y-1)^2/1==1,(x,-2,2),(y,-2,4)).show(aspect_ratio=1,

xmin=-2, xmax=2, ymin=-2, ymax=4)

solve(a^2==3,a)

[a == -sqrt(3), a == sqrt(3)]

solve(b^2==1,b)

[b == -1, b == 1]

solve(c^2==3-1,c)

[c == -sqrt(2), c == sqrt(2)] : vertices (,1) / focus (,1)

var('x,y')

ellipse=implicit_plot(x^2/3+(y-1)^2/1==1,(x,-2,2),(y,-2,4))

f1=point((-sqrt(2),1), pointsize=20, rgbcolor=(1,0,0));

f2=point((sqrt(2),1), pointsize=20, rgbcolor=(1,0,0));

v1=point((-sqrt(3),1), pointsize=20, rgbcolor=(0,0,1));

v2=point((sqrt(3),1), pointsize=20, rgbcolor=(0,0,1));

v3=point((0,0), pointsize=20, rgbcolor=(0,0,1));

v4=point((0,2), pointsize=20, rgbcolor=(0,0,1));

show(ellipse+f1+f2+v1+v2+v3+v4,aspect_ratio=1,xmin=-2, xmax=2, ymin=-1, ymax=3)

13.

Solution.

: Parabola with vertices and focus

var('x,y')

parabola=implicit_plot((y-2)^2==8*(x-1),(x,-1,8),(y,-4,8))

f1=point((1,2), pointsize=20, rgbcolor=(1,0,0));

v1=point((3,2), pointsize=20, rgbcolor=(0,0,1));

show(parabola+f1+v1,aspect_ratio=1, xmin=-1, xmax=8,  ymin=-4, ymax=8)

14.

Solution. : Hyperbola with vertices

and focus ,

var('x,y')

hyperbola=implicit_plot((y-1)^2/5-(x+2)^2/4==1,(x,-10,87),(y,-8,10))

f1=point((-2,4), pointsize=20, rgbcolor=(1,0,0));

f2=point((-2,-2), pointsize=20, rgbcolor=(1,0,0));

v1=point((-2,1+sqrt Theorem), pointsize=20, rgbcolor=(0,0,1));

v2=point((-2,1-sqrt Theorem), pointsize=20, rgbcolor=(0,0,1));

show(hyperbola+f1+f2+v1+v2,aspect_ratio=1, xmin=-10, xmax=7, ymin=-8,ymax=10)

15-22. Find an equation for the conic that satisfies the given conditions.

15. Parabola, vertex , focus

Solution.

var('x,y')

implicit_plot(x^2+12*y,(x,-10,10),(y,-10,10))

16. Parabola, vertex , focus

Solution.

17. Ellipse, foci , vertices

Solution.   ,   ,

var('x,y')

implicit_plot(x^2/25+y^2/16-1,(x,-10,10),(y,-10,10))

18. Ellipse, foci , vertices

Solution.

19. Ellipse, center , focus vertex

Solution.

20. Hyperbola, foci , vertices

Solution.

var('x,y')

implicit_plot(-1*x^2/5+y^2/4-1,(x,-10,10),(y,-10,10))

21. Hyperbola, foci , asymptotes

Solution.

22. Hyperbola, focus , asymptotes and

Solution.

23-28. Write a polar equation of a conic with the focus at the origin and the given data.

23. Hyperbola, eccentricity , directrix

Solution.

var('r,theta')

solve(3-8*cos(theta)==0,theta)

r=6/(3-8*cos(theta))

polar_plot

(r, theta,-1*arccos(3/8)+0.1,arccos(3/8)-0.1)

24. Hyperbola, eccentricity , directrix .

Solution.

var('r,theta')

solve(3+5*sin(theta)==0,theta)

r=20/(3+5*sin(theta))

polar_plot(r, theta,-0.5,2.5)

25. Ellipse, eccentricity , directrix

Solution.

var('r,theta')

solve(3-2*cos(theta)==0,theta)

r=6/(3-2*cos(theta))

polar_plot(r, theta,-0.5,0.5)

26. Ellipse, eccentricity , directrix

Solution.

27. Parabola, eccentricity , directrix

Solution.

28. Parabola, eccentricity , directrix

Solution.

29-32. Find the eccentricity, identify the conic, give an equation of the directrix, and sketch the conic.

29.

Solution. , , Parabola, directrix :

30.

Solution. , , Parabola, directrix :

31.

Solution. , , Ellipse, directrix :

32.

Solution. , , Ellipse, directrix :

After semester <PBL report>

http://matrix.skku.ac.kr/Cal-Book1/Calculus-1/index.htm

http://matrix.skku.ac.kr/Cal-Book1/Calculus-2/index.htm

Part I  Single Variable Calculus

http://matrix.skku.ac.kr/Cal-Book/part1/part1.html

Part II  Multivariate Calculus

***********************************

[Grapher]

http://matrix.skku.ac.kr/cal-lab/sage-grapher-integral2.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-inverse.html

http://matrix.skku.ac.kr/cal-lab/cal-Newton-method.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-derivatives.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-integral.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-integral2.html

http://matrix.skku.ac.kr/cal-lab/SKKU-Cell-Matrix-Calculator.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-butterfly.html
http://matrix.skku.ac.kr/cal-lab/sage-grapher-cochleoid.html
http://matrix.skku.ac.kr/cal-lab/sage-grapher-dewdrop.html
http://matrix.skku.ac.kr/cal-lab/sage-grapher-epicycloid.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-flower.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-Fermat-Spiral.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-Freeth-Nephroid.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-Durer-Shell-Curves.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-Newton-Diverging-Parabolas.html

http://matrix.skku.ac.kr/cal-lab/sage-grapher-Talbot-Curve.html

[Engineering Math 공학수학 강의록]

2019학년도 1학기 성대 반도체공학과 공학수학 1 강의록  (영문)

http://matrix.skku.ac.kr/2019-EM/EM-1-W1/

http://matrix.skku.ac.kr/2019-EM/EM-1-W2/

http://matrix.skku.ac.kr/2019-EM/EM-1-W3/

http://matrix.skku.ac.kr/2019-EM/EM-1-W4/

http://matrix.skku.ac.kr/2019-EM/EM-1-W5/

http://matrix.skku.ac.kr/2019-EM/EM-1-W6/

http://matrix.skku.ac.kr/2019-EM/EM-1-W7/

중간고사

http://matrix.skku.ac.kr/2019-EM/EM-1-W9/

http://matrix.skku.ac.kr/2019-EM/EM-1-W10/

http://matrix.skku.ac.kr/2019-EM/EM-1-W11/

http://matrix.skku.ac.kr/2019-EM/EM-1-W12/

http://matrix.skku.ac.kr/2019-EM/EM-1-W13/

http://matrix.skku.ac.kr/2019-EM/EM-1-W14/

http://matrix.skku.ac.kr/2019-EM/EM-1-W15/

기말고사

12. Math for Big Data (빅데이터를 위한수학)

[강의 동영상 & 실습실]

13. [Math, Art &3D printing] (Math & Coding)

http://matrix.skku.ac.kr/mathlib/

http://matrix.skku.ac.kr/artsurf/

http://matrix.skku.ac.kr/3d-print/

http://matrix.skku.ac.kr/3d-print-e/

Calculus