Chapter 12. Vector Valued Functions

12.1 Vector-Valued Functions and Space Curves

12.2 Calculus of Vector Functions

*12.4 Motion Along A Space Curve: Velocity and Acceleration

Calculus

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Chapter 12. Vector Valued Functions

12.1 Vector-Valued Functions and Space Curves http://youtu.be/0pvywjBjsQw

문제풀이 by 최양현 http://youtu.be/jvMI6OzdR_I

12.2 Calculus of Vector Functions

문제풀이 by 김동윤 http://youtu.be/VS5rPyOjP2I

12.3 Arc Length and Curvature

*12.4 Motion Along A Space Curve: Velocity and Acceleration

12.1 Vector-Valued Functions and Space Curves

A vector-valued function, or simply a vector function, assigns to each element in the set of real numbers an element in the set of vectors. We will consider vector functions in three-dimensional vector space. This means that a vector function on ℝ is a function . These specific types of vector functions are denoted by expressions such as

where , and are real-valued functions called the component functions of the vector . The domain of a vector function is the intersection of the domains of the component functions , and . Often we imagine to be the time and to be the position of a moving particle at time .

Example 1

If , , , then the component functions are , and , the domain of is the interval . ■

DEFINITION 1 Limit of a Vector Function

A vector function approaches to the limit as approaches to , written

, provided .

Another way of defining the limit is as follows:

if and only if for every there is a number such that whenever .

THEOREM 2

Let be a vector function. Then a vector is the limit of a vector function as if and only if

, and .

Proof. Suppose that

iff .

iff

iff

iff .

In particular, . The limit of a vector function is determined by computing the limits of its components. ■

exists iff limits of component functions exist.

Example 2

If , then find .

Solution. ■

Rules of limits of vector functions are the same as rules of limits of real-valued functions. We mentioned the following properties.

THEOREM 2 Rules of Limits

Let and be vector functions and be a constant. Let and be real-functions. Suppose that , and exist and . Then we have the following properties of limits:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

All these properties can be easily proof ed by taking the limit of the corresponding functions.

DEFINITION 4 Continuity

A vector function is said to be continuous at if

(i) is defined

(ii) exists

(iii) .

In view of Definition 1, one can easily show that is continuous at if and only if its component functions , , and are continuous at .

DEFINITION 5 Continuity

Let , and be continuous real-valued functions on an interval . A space curve is the set of all points in space, where

, ,

and varies throughout the interval . These equations are called parametric equations of and is called a parameter.

We think of as being traced out by a moving particle whose position at time is . The vector

from the origin to the particle’s position at time is the particle’s position vector. Thus, any continuous vector function defines a space curve that is traced out by the tip of the moving vector , as shown in Figure 1.

Figure 1

Example 3

Describe the curve defined by the vector function

.

Solution. The corresponding parametric equations are

, , .

These equations are parametric equations of a line passing through the point and parallel to the vector . Alternatively, we could observe that the function can be written as , where and . ■

Plane curves can also be described in vector notation. For instance, the curve given by the parametric equations and can also be described by the vector equation

,

where and . We recognize this as the equation of a parabola with vertex at .

Example 4

Sketch the curve whose vector equation is

.

Solution. The parametric equations for this curve are

, , .

Figure 2

The curve must lie on the circular cylinder , because the -coordinate and -coordinate of the tip of satisfy the cylinder’s equation

.

The point lies directly above the point , which moves counterclockwise around the circle in the -plane. The curve spirals upward around the cylinder as increases. That is, each time increases by , the curve completes one turn around the cylinder. The curve, shown in Figure 2, is called a helix. ■

http://matrix.skku.ac.kr/cal-lab/cal-12-1-Exmaple-4.html

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

r=vector([cos(t),sin(t),2*t])

s=implicit_plot3d(x^2+y^2==1,(x,-3,3),(y,-3,3),(z,0,4*pi),color='goldenrod',opacity=0.2)

c=parametric_plot3d([cos(t),sin(t),2*t],(t,0,2*pi),color='green')

t1=pi/4

t2=2*pi/3

t3=4*pi/3

t4=5*pi/2

Dt=0.1

Ar1=arrow3d(r(t=t1),r(t=t1+Dt),color='green')

Ar2=arrow3d(r(t=t2),r(t=t2+Dt),color='green')

Ar3=arrow3d(r(t=t3),r(t=t3+Dt),color='green')

Ar4=arrow3d(r(t=t4),r(t=t4+Dt),color='green')

s+c+Ar1+Ar2+Ar3+Ar4

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Example 5

Find a vector equation and parametric equations for the line segment that joins the point to the point .

Solution. We know that the vector equation for the line segment joining the tips of the vectors and and oriented from to is given by

, .

Here we take and to obtain a vector equation of the line segment from to . Thus

, .

The corresponding parametric equations are

, , , . ■

Example 6

Find a vector equation that represents the curve of intersection of the cylinder and the plane .

http://matrix.skku.ac.kr/cal-lab/cal-12-1-Exmaple-6.html

Solution. Figure 3 shows how the plane and the cylinder intersect, and Figure 4 shows the curve of intersection , which is an ellipse.

Figure 3 Figure 4

The projection of onto the -plane is the circle , . So we can write , , .

From the equation of the plane, we have

.

So we can write parametric equations for as

, , , .

The corresponding vector equation is

, .

This equation is called a parameterization of the curve . The arrows in Figure 3 indicate the direction in which is traced as the parameter increases. ■

Example 7

Find a vector function that represents the curve of intersection of the cylinder and the surface .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-Exm7.html

Solution. Let , . Then

Figure 5

The answer is .

Figure 5 illustrates the Vector Function.

x, y, z=var('x, y, z')

p1=implicit_plot3d(x^2+y^2-5==0, (x, -3, 3), (y, -3, 3), (z, -3, 3),

rgbcolor='orange', opacity=0.5)

p2=implicit_plot3d(x*y-z==0, (x, -3, 3), (y, -3, 3),

(z, -3, 3), rgbcolor='green', opacity=0.5)

show(p1+p2)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

A vector function that represents the curve of their intersection is .

var('t')

x1=sqrt(5) *cos(t)

y1=sqrt(5) *sin(t)

z1=5*sin(t) *cos(t)

p3=parametric_plot3d((x1, y1, z1), (t, 0, 2*pi), rgbcolor='blue')

show(p3)

■

http://matrix.skku.ac.kr/cal-lab/cal-12-1-Rotations.html

http://matrix.skku.ac.kr/cal-lab/cal-12-1-Rotations-A.html

http://matrix.skku.ac.kr/cal-lab/cal-12-1-Rotations-B.html

Many surfaces are obtained by revolving a plane curve about various axes. For example if the upper semi-circle with and is rotated about the -axis, we get the sphere. We can easily write a parametrization of such surfaces. If we take a curve on an interval . Then we think of this as a curve in -plane.

12.1 EXERCISES (Vector-Valued Functions and Space Curves)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-12-1-Sol.html

1-2. Find the domain of the vector functions.

1. .

Solution. .

2. .

Solution. and . Thus and .

3-5. Find the limit.

3. .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-3.html

Solution.

var('t')

(limit(sin(t)/t, t=0), limit(cos(t), t=0), limit(t*ln(t), t=0))

Answer : (1, 1, 0)

4. .

5. .

6-9. Sketch curves with the given vector equations. Indicate with an arrow the direction in which increases.

6. .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-Exs-6.html

Solution.

var('t')

r=vector([cos(2*t),sin(3*t),t])

a=0

b=2*pi

C=parametric_plot3d(r,(t,a,b),color='green',thickness=2)

t1=pi/10

Dt=0.1

t2=2*pi/3

t3=5*pi/3

Ar1=arrow3d(r(t=t1),r(t=t1+Dt),color='green')

Ar2=arrow3d(r(t=t2),r(t=t2+0.05),color='green')

Ar3=arrow3d(r(t=t3),r(t=t3+Dt),color='green')

C+Ar1+Ar2+Ar3

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

7. .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-Exs-7.html

Solution.

var('t')

r=vector([t^2,t,t^3])

a=0

b=2*pi

C=parametric_plot3d(r,(t,a,b),color='green',thickness=2)

t1=pi/10

Dt=0.1

t2=2*pi/3

t3=5*pi/3

Ar1=arrow3d(r(t=t1),r(t=t1+Dt),color='green',thickness=4)

Ar2=arrow3d(r(t=t2),r(t=t2+0.05),color='green',thickness=4)

Ar3=arrow3d(r(t=t3),r(t=t3+Dt),color='green',thickness=4)

C+Ar1+Ar2+Ar3

8. .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-Exs-8.html

Solution.

var('t')

r=vector([t,t,sin(t)])

a=0

b=2*pi

C=parametric_plot3d(r,(t,a,b),color='green',thickness=2)

t1=pi/10

Dt=0.1

t2=2*pi/3

t3=5*pi/3

Ar1=arrow3d(r(t=t1),r(t=t1+Dt),color='green',thickness=4)

Ar2=arrow3d(r(t=t2),r(t=t2+0.05),color='green',thickness=4)

Ar3=arrow3d(r(t=t3),r(t=t3+Dt),color='green',thickness=4)

C+Ar1+Ar2+Ar3

9. .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-Exs-9.html

Solution.

var('t')

x(t)=t^2*cos(t);y(t)=t^2*sin(t);z(t)=sin(t);r(t)=(x(t), y(t), z(t))

dr(t)=(diff(x(t), t), diff(y(t), t, ), diff(z(t), t));

t0=2

p1=r(t0)

p2=dr(t0)/abs(dr(t0))+r(t0)

p=parametric_plot3d((x(t), y(t), z(t)), (t, 1, 5))

utv=line3d([p1, p2], rgbcolor=(1, 0, 0), arrow_head=True, thickness=3)

show(p+utv)

10-11. Find vector equations and parametric equations for the line segment to .

10. , .

Solution. ,

.

11. , .

Solution. ,

.

12. Show that the curve with parametric equation , , lies on the cone .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-12.html

Solution. We have .

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

p1=parametric_plot3d((t^2*cos(t), t^2*sin(t), t^2), (t, -pi, pi), color='red')

p2=implicit_plot3d(z^2==x^2+y^2, (x, -10, 10), (y, -10, 10), (z, 0, 30), color='yellow', opacity=0.5)

show(p1+p2)

We can see that the curve (red) lies on the cone (yellow).

13. Show that the curve with parametric equations , , is the curve of intersection of the surfaces and .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-13.html

Solution.

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

p1=parametric_plot3d((2*sin(t), 2*cos(t), t^2), (t, -pi, pi), color='red');

p2=plot3d(x^2/4, (x, -2, 2), (y, -2, 2), color='green', opacity=0.5)

p3=implicit_plot3d(x^2+y^2==4, (x, -2, 2), (y, -2, 2), (z, -2, 2), color='yellow', opacity=0.5)

show(p1+p2+p3)

We can see that the curve (red) is intersection of two surfaces (yellow, green).

14. Graph the curve with parametric equations

, , .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-14.html

Solution.

var('t')

parametric_plot3d(((1+sin(4*t))*cos(t), (1+sin(4*t))*sin(t), (1+sin(4*t))), (t, -2*pi, 2*pi))

15. Show that the curve with parametric equations

, , passes through the points and but not through the point .

Solution. If , , and if , . So, it passes through the points and . But, y-axis of the curve cannot be negative, so it does not pass through .

16-18. Find vector functions that represent the curves of intersection of the two surfaces.

16. The cylinder and the surface .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-16.html

Solution. Let , . Then .

So,, is the vector function of intersection of two surfaces.

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

p1=parametric_plot3d((sqrt(7)*cos(t), sqrt(7)*sin(t), 7*sin(2*t)), (t, -pi, pi), color='red');

p2=plot3d(2*x*y, (x, -3, 3), (y, -3, 3), color='green', opacity=0.5);

p3=implicit_plot3d(x^2+y^2==7, (x, -3, 3), (y, -3, 3), (z, -3, 30), color='yellow', opacity=0.5);

show(p1+p2+p3);

See the graph.

17. The planes and .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-17.html

Solution. Let . Then , .

Subtracting two equations, , so we have .

Substituting it to first equation,

, so .

So, is the vector function of intersection of two planes.

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

p1=parametric_plot3d((t, (t-10)/7, 15*(t-3)/14), (t, -3, 3), color='red');

p2=implicit_plot3d(3*x-6*y-2*z==15, (x, -3, 3), (y, -3, 3), (z, -21, 6), color='green', opacity=0.5)

p3=implicit_plot3d(2*x+y-2*z==5, (x, -3, 3), (y, -3, 3), (z, -21, 6), color='yellow', opacity=0.5)

show(p1+p2+p3)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

See the graph.

18.The paraboloid and the parabolic cylinder .

http://matrix.skku.ac.kr/cal-lab/cal-13-1-18.html

Solution. Put . Then , and .

So, is the vector function of intersection of two surfaces.

We can see the answer is correct by drawing the graph of surfaces and vector function.

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

p1=parametric_plot3d((t^2, t, t^4+4*t^2), (t, -sqrt(3), sqrt(3)), color='red')

p2=plot3d(x^2+4*y^2, (x, -3, 3), (y, -3, 3), color='green', opacity=0.5)

p3=implicit_plot3d(x-y^2==0, (x, -3, 3), (y, -3, 3), (z, -3, 45), color='yellow', opacity=0.5)

show(p1+p2+p3)

19. If two objects travel through space along two different curves, it is important to know whether they will collide (Will a missile hit its moving target? Will two aircraft collide?) The curves might intersect, but we need to know whether the objects are in the same position at the same time. Suppose the trajectories of two particles are given by the vector functions.

, for .

Do the particles collide?

Solution. Two particles collide means has a solution. But for the second coordinate, . Hence has no real root, and hence there would be no collisions.

20.Two particles travel along the space curves

, .

Do the particles collide? Do their paths intersect?

http://matrix.skku.ac.kr/cal-lab/cal-13-1-20.html

Solution. The first coordinate of the two particles cannot be the same, and hence there would be no collisions.

To see if there is an intersection, we draw a graph of and .

var('t')

p1=parametric_plot3d((t, t^2, t^3), (t, -5, 5), color='red')

p2=parametric_plot3d((1+t,2+5*t,3+12*t),(t,-5, 5));

show(p1+p2)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

By rotating the resultant graph, we can see one intersection.

12.2 Calculus of Vector Functions

In this section, we introduce differentiation and integration of vector functions.

Derivatives

The derivative of a vector function at a point can be defined in exactly same way as for one variable functions.

DEFINITION 1 Derivative of a Vector Function

The derivative of a vector function is

for all for which the limit exists.

The next theorem gives us a way for computing the derivative of a vector function .

THEOREM 2 Differentiation

If , then is differentiable at , if , and are differentiable at . Moreover,

.

Proof.Note that

.

By Theorem 2 of Section 12.1, exists if and only if

,

and

exists.

This proves that is differentiable if and only if , and are differentiable. Also . ■

A vector function is differentiable if it is differentiable at every point of its domain. The second derivative of a vector function is the derivative of , that is, . Higher order derivatives of may be defined in a similar way.

Example 1

Find the second derivative of .

Solution. The first derivative of is

.

The second derivative of is . ■

The curve traced by is smooth if is continuous and never 0; that is, if , and have continuous first derivatives that are not simultaneously 0. A curve that is made up of a finite number of smooth curves is called piecewise smooth.

Example 2

Determine whether the semicubical parabola is smooth.

Solution. Since , we have and therefore, the curve is not smooth. The point that corresponds to is , and we see from the graph in Figure 1 that there is a sharp corner at (1, 0). Any curve with this type of behavior,an abrupt change in direction,is not smooth.

Figure 1 The curve is not smooth.

■

(a) The secant vector (b) The tangent vector

Figure 2

The geometric significance of the definition of derivative is shown in Figure 2. The points and have position vectors and , and the vector is represented by the vector . For , the scalar multiple has the same direction as the vector . As the vector approaches a vector that is tangent to the curve at . The vector is called the tangent vector to the curve defined by at the point , provided that exists and . The tangent line to at is defined to be the line through parallel to the tangent vector . Look at the sage worksheet which gives a geometric meaning of as the limit of the difference quotient

.

For a vector , we want to find a tangent vector at . (See Figure 3.) It is a behavior of tangent vectors as values near . ■

http://matrix.skku.ac.kr/cal-lab/cal-12-2-Exm-2.html

t=var('t')

r=vector((cos(t),sin(2*t),t))

rprime=diff(r,t)

a=0.6

p = parametric_plot3d(r,(t,0,2),color='red',thickness=3)+point3d(r(t=a),size=15)

p = p + sum([arrow3d(r(t=a),r(t=a)+(r(t=a+h)-r(t=a))/h,width=0.3) for h in [0.01..1,step=0.1]])

p = p + arrow3d(r(t=a),r(t=a)+rprime(t=a),color='green',width=0.5)

show(p)

Figure 3

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

DEFINITION 3 Unit Tangent Vector

Let be a smooth parameterized curve.

If , then the unit tangent vector of is

.

Example 3

Find the unit tangent vector of at the point .

Solution. According to Theorem 2, we differentiate each component of :

.

Since and , the unit tangent vector at the point is

. ■

Example 4

For the curve , find and sketch the position vector and the tangent vector .

Solution.We have and . The curve is a plane curve and elimination of the parameter from the equations , gives . In Figure 4, we draw the position vector starting at the origin and the tangent vector starting at the corresponding point . ■

Figure 4

Example 5

Find parametric equations for the tangent line to the helix with parametric equations , , at .

Solution. The vector equation of the helix is , , .

Hence . At , the corresponding point is , so the tangent vector, at is . The tangent line is the line through parallel to the vector . Hence its parametric equations are

, , . ■

Figure 5

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

The following are the standard properties of derivatives of vector function.

THEOREM 4 Rules of Differentiation

Suppose that and are differentiable vector functions, is a scalar, and is a differentiable real-valued function. Then

(ⅰ)

(ⅱ)

(ⅲ)

(ⅳ)

(ⅴ)

(ⅵ) (Chain Rule).

We will prove (ⅳ) and leave the remaining proofs as exercises.

Proof. Let and .

Then, by the product rule

. ■

Example 6

Find and , where , .

Solution. Since , .

And since ,

. ■

THEOREM 5

If is a vector function and , where is a constant for all , then

,

that is, and are orthogonal vectors for all .

Proof. We recall that . It follows from Formula (iv) of Theorem 4 that

.

Then , which says that is orthogonal to . ■

Figure 6

Geometrically, Theorem 5 says that if a curve lies on a sphere with center at origin, then the tangent vector is always perpendicular to the radial vector . (See Figure 6.)

Example 7

Find the derivative of the vector function.

.

http://matrix.skku.ac.kr/cal-lab/cal-13-2-Exm6.html

Solution.

t=var('t')

x(t)=arccos(t)

y(t)=t^2

z(t)=1/sqrt(1+t^2)

(diff(x(t), t), diff(y(t), t), diff(z(t), t))

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer : (-1/sqrt(-t^2 + 1), 2*t, -t/(t^2 + 1)^(3/2)) ■

Integrals of Vector Functions

A differentiable vector function is antiderivative of a vector function on an interval if . If then an antiderivative of is where and are antiderivatives of and , respectively. Every antiderivative of on has the form where is an arbitrary constant vector. The set of all antiderivative of on is the indefinite integral of on .

DEFINITION 6 Indefinite Integral of a Vector Function

Let be a vector function and , where , and are anti- derivatives of and , respectively. Then the indefinite integral of is

,

where is an arbitrary constant vector.

, and in Definition 6 can be obtained by finding indefinite integrals of , and , respectively.

Example 8

If , then

,

where is a vector constant of integration. ■

Definite integrals of a vector function are evaluated by applying the Fundamental Theorem of Calculus to each component of a vector function.

DEFINITION 7 Definite Integral of a Vector Function

If is integrable (that is, it has anti- derivative) on , then the definite integral of from to is given by

.

This means that an integral of a vector function is obtained by integrating each component function.

The Fundamental Theorem of Calculus can be extended to continuous vector functions as follows :

where is an anti-derivative of , that is, . We use the notation for indefinite integral of , that is, the set of all anti-derivatives of .

Example 9

If , then

. ■

Properties of the Integrals of Vector Functions

Let and be continuous vector functions on , a scalar. Then the following hold:

(a) ,

(b) ,

(c) .

12.2 EXERCISES (Calculus of Vector Functions)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-12-2-Sol.html

1-5. Find and draw the position vector and the tangent vector for the given value of .

1. ;

http://matrix.skku.ac.kr/cal-lab/cal-13-2-1.html

Solution. (-sin(t), 1)

So, we can see that ;

2. , ;

http://matrix.skku.ac.kr/cal-lab/cal-13-2-2.html

Solution.

var('t')

x(t)=sin(t);y(t)=t

r(t)=(x(t), y(t))

dr(t)=(diff(x(t), t), diff(y(t), t));dr(t)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer : (cos(t), 1)

So, we can see that ;

t0=0

p1=r(t0)

p2=dr(t0)/abs(dr(t0))+r(t0)

p=parametric_plot((x(t), y(t)), (t, 0, 2*pi))

utv=line([p1, p2], rgbcolor=(1, 0, 0), thickness=2)

show(p+utv)

3. , ;

http://matrix.skku.ac.kr/cal-lab/cal-13-2-3.html

Solution.

var('t')

x(t)=t^3;y(t)=t^2

r(t)=(x(t), y(t))

dr(t)=(diff(x(t), t), diff(y(t), t));dr(t)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer : (3*t^2, 2*t)

So, we can see that .

t0=1

p1=r(t0)

p2=dr(t0)/abs(dr(t0))+r(t0)

p=parametric_plot((x(t), y(t)), (t, 0, 2))

utv=line([p1, p2], rgbcolor=(1, 0, 0), thickness=2)

show(p+utv)

4. , ;

http://matrix.skku.ac.kr/cal-lab/cal-13-2-4.html

Solution.

var('t')

x(t)=exp(2*t)

y(t)=exp(-3*t)

r(t)=(x(t), y(t))

dr(t)=(diff(x(t), t), diff(y(t), t))

dr(t)

Answer : (2*exp(2*t), -3*exp(-3*t))

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

So, we can see that ;

t0=0

p1=r(t0)

p2=dr(t0)/abs(dr(t0))+r(t0)

p=parametric_plot((x(t), y(t)), (t, -2, 2))

utv=line([p1, p2], rgbcolor=(1, 0, 0), thickness=2)

show(p+utv)

5. , ;

http://matrix.skku.ac.kr/cal-lab/cal-13-2-5.html

Solution.

var('t')

x(t)=3*sin(t)

y(t)=2*cos(t)

r(t)=(x(t), y(t))

dr(t)=(diff(x(t), t), diff(y(t), t))

dr(t)

Answer : (3*cos(t), -2*sin(t))

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

So, we can see that ;

t0=pi/3

p1=r(t0)

p2=dr(t0)/abs(dr(t0))+r(t0)

p=parametric_plot((x(t), y(t)), (t, -2*pi, 2*pi))

utv=line([p1, p2], rgbcolor=(1, 0, 0), thickness=2)

show(p+utv)

6-10. Find the derivatives of the vector functions.

6. .

Solution. .

7. .

Solution. .

8. .

Solution. .

9. .

Solution. .

10. .

Solution. .

11-13. Find the unit tangent vector at the point with the given value of the parameter .

11. , .

Solution. , ,

Hence .

12. , .

Solution. . .

Therefore, .

13. , .

Solution. , ,

.

14. If , find , , , and .

Solution. , .

Therefore, , .

.

15. If , find and .

Solution..

.

.

. . .

http://matrix.skku.ac.kr/cal-lab/cal-12-2-Exs-15.html

var('t')

r=vector([e^t*cos(t),e^t*sin(t),e^t])

rprime=diff(r,t)

show(rprime)

show(rprime.norm())

T=r/rprime.norm()

·show(T(t=0))

rrprime=diff(r,t,2)

show(rrprime(t=0))

S=rprime.dot_product(rrprime)

show(S.simplify_trig())

Answer :

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

16-19. Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

16. ; .

Solution. We can see that .

For , . So,.

Hence .

17. ; .

Solution. We can see that .

For ,

.

Therefore,

Hence .

18. , , ; .

Solution. We can see that by solving , , .

For , .

So, . Hence

.

19. , , ; .

Solution. We can see that .

For , .

So, . Hence

.

20-21. Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustrate by graphing both the curve

and the tangent line together.

20. , , ; .

http://matrix.skku.ac.kr/cal-lab/cal-13-2-20.html

Solution. We can see that .

var('t,s')

x(t)=t; y(t)=2*cos(t); z(t)=2*sin(t);

r(t)=(x(t), y(t), z(t));

dr(t)=(diff(x(t), t), diff(y(t), t), diff(z(t), t));

t0=pi/3;

pe=r(t0)+dr(t0)*s;

print pe

(1/3*pi + s, -sqrt(3)*s + 1, s + sqrt(3))

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

So, parametric equation is .

p1=parametric_plot3d(r(t), (t, -pi, pi));

p2=parametric_plot3d(pe, (s, -pi, pi), color='red');

show(p1+p2);

21. , , ; .

http://matrix.skku.ac.kr/cal-lab/cal-13-2-21.html

Solution. We can see that .

var('t,s')

x(t)=exp(t); y(t)=sin(t); z(t)=2*exp(-3*t)

r(t)=(x(t), y(t), z(t));

dr(t)=(diff(x(t), t), diff(y(t), t), diff(z(t), t))

t0=0

pe=r(t0)+dr(t0)*s;

print pe

(1 + s, s, 2 -6*s)

So, parametric equation is .

p1=parametric_plot3d(r(t), (t, -2, 2));

p2=parametric_plot3d(pe, (s, -2, 2), color='red);

show(p1+p2);

22-24. Determine whether the following curves are smooth.

22. .

Solution. . Since , this curve is not smooth.

23. .

Solution. .

First coordinate of cannot be zero for any real number , so this curve is smooth.

24. .

Solution. .

Second coordinate of cannot be zero for any real number , so this curve is smooth.

25-30. Evaluate the integrals.

25. .

http://matrix.skku.ac.kr/cal-lab/cal-12-2-Exs-25.html

Solution. .

26. .

Solution. .

27..

Solution. .

28. .

Solution. where is a constant vector.

29. .

Solution. where is a constant vector.

30. .

Solution. .

31. Find if and .

Solution. , where is a constant vector.

Since , we finally have .

32. Find if and .

http://matrix.skku.ac.kr/cal-lab/cal-12-2-Exs-33.html

Solution.

, where is a constant vector.

Since , .

So, we finally have .

33. If and , use Theorem 4(Rules of Differentiation) to find

Solution. , .

, .

So, .

var('t')

u=vector([t,t^2,t^3])

v=vector([sin(t),sin(2*t),t])

r=u.dot_product(v)

rprime=diff(r,t)

show(rprime);

Answer :

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

34. If and , use Theorem 2(Rules of differentiation) to find .

http://matrix.skku.ac.kr/cal-lab/cal-12-2-Exs-34.html

Solution. ,

,

,

.

var('t')

u=vector([t,1-t,3+t^2])

v=vector([3-t,t-2,t^2])

r=u.cross_product(v)

rprime=diff(r,t);

show((expand(rprime[0]),expand(rprime[1]),expand(rprime[2])))

Answer :

35. Show that if is a vector function such that

exists, then .

Solution. Using Theorem 4 (Rules of Differentiation),

.

36. If , show that

.

Solution. We already know that . Differentiating both sides,

.

So, we finally have .

37. Find the unit tangent vector to the curve , , at any time .

http://matrix.skku.ac.kr/cal-lab/cal-14-1-19.html

Solution.

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

x(t)=4*sin(-t)

y(t)=2*cos(2*t)

z(t)=5*t

r(t)=(x(t),y(t),z(t))

r(t)

Answer : (4*sin(-t), 2*cos(2*t), 5*t)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

dr(t)=(diff(x(t),t),diff(y(t),t),diff(z(t),t)); # tangent vector(dr/dt)

dr(t)

(-4*cos(-t), -4*sin(2*t), 5)

dr(t)/abs(dr(t))+r(t) # unit tangent vector((dr/dt)/abs(dr/dt))

Answer: (-4*cos(-t)/sqrt(16*sin(2*t)^2 + 16*cos(-t)^2 + 25) + 4*sin(-t), + 16*cos(-t)^2 + 25) + 2*cos(2*t), 5*t + 5/sqrt(16*sin(2*t)^2 + 16*cos(-t)^2 + 25))

And we can draw a curve with given unit tangent vector.

t0=2

p1=r(t0)

p2=dr(t0)/abs(dr(t0))+r(t0)

p=parametric_plot3d((x(t),y(t),z(t)),(t,0,2*pi))

utv=line3d([p1,p2],rgbcolor=(1,0,0),arrow_head=True, thickness=5)

show(p+utv)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

12.3 Arc Length and Curvature

The length of a space curve is defined in exactly the same way as a plane curve. Suppose that a curve has the vector equation , , or, equivalently, the parametric equations

, ,

where , and are continuous. If the curve is traced exactly once as increases from to , then its length of is defined as

.

Notice that for space curves ,

and hence the arc length formula can be also be written as

.

Example 1

Find the length of the arc of the circular helix with vector equation from the point to the point .

http://matrix.skku.ac.kr/cal-lab/cal-12-3-Exm-1.html

Solution. The arc from to is described by the parameter interval . Hence from Equation , we have

.

var('t')

r=vector([2*cos(t),2*sin(t),t])

a=0

b=2*pi

rprime=diff(r,t)

L=integral(rprime.norm(),t,a,b)

show(L)

Answer : ■

Figure 1

A single curve can be represented by more than one vector function. Fortunately, we have the same arc length no matter which parameterization of the curve is used in . We can easily show this as follows. Suppose that

, and also ,

where is such that and , and is continuous. Then we have .

Thus from Theorem 4 (vi) in Section 12.2,

.

The technique of change of variables from calculus gives us

.

Now let be a curve given by a vector function

, ,

where is continuous and is traced exactly once as increases from to . The arc length function is the real valued function defined by

.

Thus is the length of the part of between and . (See Figure 2.)

Figure 2

Differentiating both sides of using the Fundamental Theorem of Calculus, we obtain

.

Since arc length is independent of a particular coordinate system parameterization and arises from the shape of the curve, parameterizion of a curve with respect to arc length is often useful. If a curve is already given in terms of a parameter and is the arc length function given by , then we may be able to solve for as a function of . Hence the curve can be reparametrized in terms of by substituting for . It is the arc length parameterization.

Example 2

Reparametrize the helix with respect to arc length measured from in the direction of increasing .

Solution. The initial point corresponds to the parameter value . From Example 1, we have

and so .

Solving this equation for gives . Therefore, the required reparameterization is obtained by substituting for :

. ■

Curvature of Curves

Let be a smooth curve defined by the vector function . Recall that the unit tangent vector is given by

and points in the direction of the curve . From Figure 3, we observe that changes direction gradually when bends slightly, but it changes direction rapidly when bends more sharply.

Figure 3

DEFINITION 1

The curvature of a curve is the magnitude of the rate of change of the unit tangent vector with respect to arc length. That is,

where is the unit tangent vector.

If a smooth curve is given in terms of parameter rather than the arc length parameter , we can compute the curvature as follows. We use Theorem 4 (vi) in Section 12.2 to write

and .

However from , thus we have the following formula for calculating curvature

.

Example 3

Find the curvature of a circle of radius .

Solution. We can take a circle of radius , centered at the origin, and then a parameterization is

.

Then

and .

From this, we find the unit tangent vector

and .

Hence by , the curvature is

since . ■

We see from Example 3 that the curvature of a circle is constant, which is the reciprocal of its radius. So small circles have large curvature and large circles have small curvature. We can see directly from the definition of curvature that the curvature of a straight line is 0 because the unit tangent vector points in the same direction, that is, is constant.

The following theorem provides another formula for curvature in terms of a general parameter .

THEOREM 2

If is smooth curve defined by the vector function , then the curvature can be expressed as

.

Proof.

It follows from definition of the unit tangent vector and that

.

Differentiating with respect to , we get

.

However the cross product of a vector with itself is a zero vector, so we have

.

.

Since for all so and are orthogonal by Theorem 5 in Section 12.2. Hence, by the definition of cross product and .

.

Then

.

Thus

. ■

Example 4

Find the curvature of the twisted cubic where and are positive.

Solution. We have

.

Hence

.

Therefore

and

.

By Theorem 2,

. ■

For a plane curve with equation , we take as the parameter, and then a parameterization is .

Then and

and hence , and . Thus by Theorem 2, we have

.

Figure 4

Let us look at this in another form.

If as , then .

The arc length function is

. Then,

(by 1st FTC)

.

Thus , which means the curvature at any point on a curve , is .

Curvature in Parametric Form

Using Theorem 2, we can show that the curvature of a plane parametric curve , is

.

Using Theorem 2, we can show that the curvature of a plane polar function is

.

A Circle of Curvature

In Figure 5 a osculating circle on at a point is called a circle of curvature. By Example 3, a radius of a circle of curvature is

where .

Figure 5

If and (concave downward), then the coordinates of center of a circle of curvature is (, ). Also since

, , .

Then the center of the circle of curvature can be obtained by

.

Example 5

Find the radius of curvature for the curve .

Solution. Since and ,

. ■

Example 6

Find the center of curvature for the parabola at .

Solution. Since and ,

and . Therefore the center is . ■

Example 7

Find the curvature of and sketch the graph of its curvature function .

Solution. Since and , gives .

Figure 6

The curvature at is . At the point , it is . Observe from the expression for or the graph of in Figure 6 that as . This corresponds to the fact that the graph appears to become flatter as . ■

http://interact.sagemath.org/taxonomy/term/13

Example 8

Sketch the graph of , and plot the curvature function of along with this curve. [This curve is called figure eight.]

http://matrix.skku.ac.kr/cal-lab/cal-12-3-Example-8.html

Solution.

var('t')

x=sin(t)

y=sin(2*t)

x1=diff(x,t)

y1=diff(y,t)

x2=diff(x1,t)

y2=diff(y1,t)

curvature=(x1(t)*y2(t)-x2(t)*y1(t))/((x1(t)^2+y1(t)^2)^(3/2))

plot(curvature,(t,0,2*pi),color='red',dpi=120)+parametric_plot((x,y),(t,0,2*pi),dpi=120)

Figure 7

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

■

The Normal Vector and Binormal Vector

At a given point on a smooth space curve , there are many vectors that are orthogonal to the unit tangent vector . We single out one that is orthogonal to by Theorem 5 in Section 12.2. Note that is not a unit vector. If we divide by its length , we obtain a unit vector that is orthogonal to .

DEFINITION 3

The principal unit normal vector (or simply unit normal) for a smooth curve in space is

.

DEFINITION 4

The binormal vector of a smooth curve in a space is given by

.

Since and are both unit vectors, the binormal vector is also a unit vector and is perpendicular to both and . (See Figure 8) The vectors , , and determine three mutually perpendicular planes that move along the curve as varies. Three mutually perpendicular unit vectors in a space are called the frame. It has important applications in differential geometry and in the study of the motion of spacecraft.

Figure 8

Example 9

Find the principal unit normal and binormal vectors for the circular helix

, , .

http://matrix.skku.ac.kr/cal-lab/cal-12-3-Example-9.html

Solution. The vector function for the helix is

.

Then we have

,

,

,

,

.

This shows that the normal vector at a point on the helix is horizontal and points toward the -axis. The binormal vector is

.

Plot the , and vector for helix

t=var('t')

r=vector((cos(t),sin(t),t))

T=diff(r,t)/norm(diff(r,t))

N=diff(T,t)/norm(diff(T,t))

B=T.cross_product(N)

p = parametric_plot(r,(t,0,2*pi),color='goldenrod',thickness=2)

p = p+sum([arrow3d(r(t=x), r(t=x)+T(t=x),width=1)+arrow3d( r(t=x), r(t=x)+N(t=x),color='red',width=1)+arrow3d(r(t=x), r( t=x)+B(t=x), color='green', width=1)+point3d((cos(x), sin(x), x), pointsize=20) for x in [-3..3,step=1]])

show(p)

Figure 9

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

■

The three unit vectors, , and form a right-handed set of mutually orthogonal vectors called the moving trihedral. The plane of and is called the osculating plane at a point on a curve , the plane of and is called the normal plane at a point on a curve , the plane of and is called rectifying plane at a point on a curve .

The three mutually orthogonal unit vectors , and can be thought of as a movable right-handed coordinate system. (See Figure 9.)

, , .

This movable coordinate system is referred to as the -frame.

The osculating circle or the circle of curvature of at is the circle with radius that lies in the osculating plane of at P, has the same tangent as at , lies on the concave side of towards which points. (See Figure 8) The behavior of the curve near is best described by osculating circle. Also it shares the same tangent, normal, and curvature at . The circle at is called circle of curvature and its center is the center of curvature. The circle of curvature has same tangent line at as the curve , and its center lies on the concave side of .

Figure 10 Figure 11

Example 10

Find the equations of the normal plane, osculating plane and rectifying plane of the helix

at the point . (See Figure 12.)

http://matrix.skku.ac.kr/cal-lab/cal-12-3-Example-10.html

Figure 12

Solution. The normal plane at , the tangent vector is . Hence the equation of the normal plane is

or .

The osculating plane at is determined by the vectors and , hence its normal vector is .

,

.

Hence an equation of the osculating plane is

or

.

The rectifying plane at has normal vector

.

Hence an equation of the rectifying plane is

or

.

t=var('t')

r=vector((2*cos(t),2*sin(t),t))

T=diff(r,t)/norm(diff(r,t))

N=diff(T,t)/norm(diff(T,t))

B=T.cross_product(N)

x=pi/3

C=parametric_plot3d(r,(t,0,3*pi),thickness=2,color='goldenrod')

A=arrow3d(r(t=x), r(t=x)+T(t=x),thickness=1.1)+arrow3d(r(t=x),

r(t=x)+N(t=x),color='red',thickness=1.1)+arrow3d(r(t=x),

r(t=x)+B(t=x),color='green',thickness=1.1)+point3d((cos(x), sin(x),x),pointsize=20)

C+A

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

■

Example 11

Find the vectors , , and for at the given point .

http://matrix.skku.ac.kr/cal-lab/cal-12-3-Example-11.html

Solution. For point , the corresponding value is .

t=var('t')

x(t)=t^3

y(t)=t-2

z(t)=t^2

r(t)=(x(t), y(t), z(t))

dr(t)=(diff(x(t), t), diff(y(t), t), diff(z(t), t)) #

ndr=dr(t).norm() #

T=dr(t)/ndr #

dT=diff(T, t) #

ndT=dT.norm() #

N=dT/ndT #

print T(1)

print N(1)

print T(1).cross_product(N(1))

(3/14*sqrt(14), 1/14*sqrt(14), 1/7*sqrt(14))

(9/266 * sqrt(14) * sqrt(19), -11/266 * sqrt(14) * sqrt(19), -4/133 * sqrt(14) * sqrt(19))

(1/19*sqrt(19), 3/19*sqrt(19), -3/19*sqrt(19)

t=var('t')

r=vector((t^3, t-2, t^2))

T=diff(r,t)/norm(diff(r,t))

N=diff(T,t)/norm(diff(T,t))

B=T.cross_product(N)

x=1

C=parametric_plot3d(r,(t,0,pi/2),thickness=2,color='goldenrod')

A=arrow3d(r(t=x), r(t=x)+T(t=x), thickness=1.1)+arrow3d

(r(t=x), r(t=x)+N(t=x) ,color='red' ,thickness=1.1)+arrow3d

(r(t=x), r(t=x) +B(t=x), color='green', thickness=1.1) +

point3d((cos(x), sin(x), x), pointsize=20)

C+A

Figure 13

■

12.3 EXERCISES (Arc Length and Curvature)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-12-3-Sol.html

1-5. Find the length of the curve.

1. , .

Solution. . .

.

2. , .

http://matrix.skku.ac.kr/cal-lab/cal-13-3-2.html

Solution.

var('t')

r(t)=(2*sin(t)-t*cos(t), 2*t, 2*cos(t)+t*sin(t))

dr=diff(r(t), t);

s(t)=sqrt((dr[0]^2+dr[1]^2+dr[2]^2).simplify_trig());

length=integral(s(t), t, 0, sqrt(5))

print length

Answer : 1/2*sqrt(5)*sqrt(10) + 5/2*arcsinh(1)

3. , .

Solution. .

.

.

4. , .

http://matrix.skku.ac.kr/cal-lab/cal-13-3-4.html

Solution.

var('t')

x(t)=t

y(t)=2

z(t)=ln(t)

dx(t)=diff(x(t), t)

dy(t)=diff(y(t), t)

dz(t)=diff(z(t), t)

s(t)=sqrt(dx(t)^2+dy(t)^2+dz(t)^2)

length=integral(s(t), t, 1, sqrt(3))

print length

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer : -sqrt(2) - 1/2*log(sqrt(2) - 1) + 1/2*log(sqrt(2) + 1) - 1/2*log(3) + 2

5. , .

Solution. , ,

.

6-8. Reparametrize the curve with respect to arc length measured from the point in the direction of increasing .

6. .

Solution. Since , .

Thus , this implies . Substituting in , we have .

7.

Solution. Since

, .

Thus .

Substituting in , we have

8. .

http://matrix.skku.ac.kr/cal-lab/cal-13-3-8.html

Solution. Reparametrization

var('t')

x(t)=3*t*sin(t)

y(t)=3*t*cos(t)

var('t')

x(t)=3*t*sin(t)

y(t)=3*t*cos(t)

z(t)=--2*sqrt(2)*t^(3/2)

dx(t)=diff(x(t), t)

dy(t)=diff(y(t), t)

dz(t)=diff(z(t), t)

s(t)=sqrt(dx(t)^2+dy(t)^2+dz(t)^2)

length=integral(s(t), t, 1, sqrt(3))

print length

Answer : 1/2*(t+1)*sqrt19*t^2+18*t+9)-3/2

9-10. Find the unit tangent , unit normal vectors and the curvature .

9. .

Solution. , .

Hence

,

, .

Hence . .

10. , .

11-13. Use Theorem 2 to find the curvature.

11. .

http://matrix.skku.ac.kr/cal-lab/cal-12-3-Exercise-11.html

Solution.

,

, .

t=var('t')

r=vector((3*t,3+t,3-t^2))

rprime=diff(r,t)

rrprime=diff(rprime,t)

curvature = abs(rprime.cross_product (rrprime))/abs(rprime)^3

k=curvature.full_simplify()

show(k)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer :

12. .

http://matrix.skku.ac.kr/cal-lab/cal-13-3-12.html

Solution.

var('t')

r(t)=(3+4*t^(3/2), 6*t,3/2*t^2)

dr=diff(r(t), t)

ddr=diff(r(t), t, 2)

ANSWER=(dr.cross_product(ddr)).norm()/(dr.norm())^3

print ANSWER

Answer: 1/3(t^(5/2) +4*t^(3/2) + 4*sqrt(t))

13. .

Solution.

,

.

.

14. Find the curvature of at the point .

Solution.

,

,

,

, , .

15. Find the curvature of at the point .

Solution.

,

,

,

.

16-18. Use Formula 6 to find the curvature.

16. .

Solution. , .

17. .

Solution. , .

18. .

Solution. , .

19-20. Find the vectors , and at the given point, and plot them.

19. , .

http://matrix.skku.ac.kr/cal-lab/cal-13-3-19.html

Solution.

var('t')

r(t)=(t+1, 2*t, t^2)

dr=diff(r(t), t)

T=dr/dr.norm()

dT=diff(T, t)

N=dT/dT.norm()

B=T.cross_product(N)

print T.subs(t=1)

print N.subs(t=1)

print B.subs(t=1)

Answer : (1/3, 2/3, 2/3)

(-2/15*sqrt(5), -4/15*sqrt(5), 1/3*sqrt(5))

(2/5*sqrt(5), -1/5*sqrt(5), 0)

t=var('t')

r=vector((t+1, 2*t, t^2))

T=diff(r,t)/norm(diff(r,t))

N=diff(T,t)/norm(diff(T,t))

B=T.cross_product(N)

x=1

C=parametric_plot3d(r,(t,0,pi),thickness=2,color='goldenrod')

A=arrow3d(r(t=x), r(t=x)+T(t=x),thickness=1.1)+arrow3d(r(t=x),

r(t=x)+N(t=x),color='red',thickness=1.1)+arrow3d(r(t=x),

r(t=x)+B(t=x),color='green',thickness=1.1)+point3d((cos(x),sin(x),x),pointsize=20)

C+A

20. , .

http://matrix.skku.ac.kr/cal-lab/cal-13-3-20.html

Solution.

var('t')

r(t)=(exp(t)*cos(t), exp(t)*sin(t), sqrt(2)*exp(t))

dr=diff(r(t), t)

T=dr/dr.norm()

dT=diff(T, t)

N=dT/dT.norm()

B=T.cross_product(N)

print T.subs(t=0)

print N.subs(t=0)

print B.subs(t=0)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer : (1/2, 1/2, 1/2*sqrt(2))

(-sqrt(1/2), sqrt(1/2), 0)

(-1/2*sqrt(1/2)*sqrt(2),-1/2*sqrt(1/2)*sqrt(2), qrt(1/2))

t=var('t')

r=vector([exp(t)*cos(t), exp(t)*sin(t), sqrt(2)*exp(t)])

T=diff(r,t)/norm(diff(r,t))

N=diff(T,t)/norm(diff(T,t))

B=T.cross_product(N)

x=0

C=parametric_plot3d(r,(t,-pi/2,pi/2),thickness=2,color='goldenrod')

A=arrow3d(r(t=x), r(t=x)+T(t=x),thickness=1.1)+arrow3d(r(t=x), r(t=x)+N(t=x), color='red', thickness=1.1) + arrow3d(r(t=x), r(t=x) + B(t=x), color='green', thickness=1.1) + point3d((cos(x), sin(x), x), pointsize=20)

C+A

21-22. Find equations of the normal plane and osculating plane of the curve at the given point, and plot the graphs.

21. .

http://matrix.skku.ac.kr/cal-lab/cal-13-3-21.html

Solution. Normal plane

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

r(t)=(2*sin(t), 5*t, 2*cos(t))

dr=diff(r(t), t)

T=dr/dr.norm()

dT=diff(T, t)

N=dT/dT.norm()

B=T.cross_product(N)

N=N.subs(t=pi)

expand(N[0]*(x-0)+N[1]*(y-5*pi)+N[2]*(z+2)==0)

Answer : sqrt(1/29)*sqrt(29)*z + 2*sqrt(1/29)*sqrt(29)==0

# Osculating plane

B=B.subs(t=pi)

expand(B[0]*(x-0)+B[1]*(y-5*pi)+B[2]*(z+2)==0)

Answer: -10*pi*sqrt(1/29) + 5*sqrt(1/29)*x + 2*sqrt(1/29)*y == 0

# Graph

t=var('t')

r=vector([2*sin(t), 5*t, 2*cos(t)])

T=diff(r,t)/norm(diff(r,t))

N=diff(T,t)/norm(diff(T,t))

B=T.cross_product(N)

x=pi

C=parametric_plot3d(r,(t,0,2*pi),thickness=2,color='goldenrod')

A=arrow3d(r(t=x), r(t=x) + T(t=x), thickness=1.1) + arrow3d(r(t=x), r(t=x) + N(t=x), color='red', thickness=1.1) + arrow3d(r(t=x), r(t=x) + B(t=x), color='green', thickness=1.1) + point3d((cos(x), sin(x), x), pointsize=20)

C+A

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

22. .

http://matrix.skku.ac.kr/cal-lab/cal-13-3-22.html

Solution.

# Normal plane

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

r(t)=(t^2, 2/3*t^3, t)

dr=diff(r(t), t)

T=dr/dr.norm()

dT=diff(T, t)

N=dT/dT.norm()

B=T.cross_product(N)

N=N.subs(t=1)

expand(N[0]*(x-1)+N[1]*(y-2/3)+N[2]*(z-1)==0)

Answer : -1/3*x + 2/3*y - 2/3*z + 5/9 == 0

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

# Osculating plane

B=B.subs(t=1)

expand(B[0]*(x-1)+B[1]*(y-2/3)+B[2]*(z-1)==0)

Answer : -2/3*x + 1/3*y + 2/3*z - 2/9 == 0

# Graph

t=var('t')

r=vector([t^2, 2/3*t^3, t])

T=diff(r,t)/norm(diff(r,t))

N=diff(T,t)/norm(diff(T,t))

B=T.cross_product(N)

x=1

C=parametric_plot3d(r,(t,0,pi/2),thickness=2,color='goldenrod')

A=arrow3d(r(t=x), r(t=x) + T(t=x), thickness=1.1) + arrow3d(r(t=x), r(t=x) + N(t=x),

color='red', thickness=1.1) + arrow3d(r(t=x), r(t=x) + B(t=x), color='green',

thickness=1.1) + point3d((cos(x), sin(x), x), pointsize=20)

C+A

23. At what point on the curve is the tangent plane parallel to the plane ?

http://matrix.skku.ac.kr/cal-lab/cal-13-3-23.html

Solution.

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

r(t)=(t^4, 3*t, t^2)

dr=diff(r(t), t)

T=dr/dr.norm()

T=T.subs(t=1)

expand(T[0]*(x-r(1)[0])+T[1]*(y-r(1)[1])+T[2]*(z-r(1)[2])==0)

Answer: 4/29*sqrt(29)*x + 3/29*sqrt(29)*y + 2/29*sqrt (29)*z - 15/29*sqrt(29) == 0.

24. The curvature at a point of a curve is defined as , where is the angle of inclination of the tangent line at , as shown in the figure. Thus, the curvature is the absolute value of the rate of change of with respect to arc length. It can be regarded as a measure of the rate of change of direction of the curve at .

(a)For a parametric curve , derive the formula

.

where the dots indicate derivatives with respect to ; that is, . [Hint: Use and Equation to find . Then use the Chain Rule to find .]

(b)For a curve as the parametric curve , , show that the formula in part (a) becomes .

Solution.

(a) Here

,

,

and

. Thus .

(b) Here .

By applying the result of (a) to this case, we obtain that

.

25. (a) Show that the curvature at each point of a straight line is .

(b) Show that the curvature at each point of a circle of radius is .

Solution. (a) For a straight line, we parametrize . Since is a straight line,

is a constant, and hence is zero. .

(b)

Then, and . .

*12.4 Motion Along a Space Curve: Velocity and Acceleration

In this section, we will show how tangent and normal vectors can be used in physics to study the motion of a particle moving along a space curve, including its velocity and acceleration.

Suppose that a particle is moving through space along a smooth curve defined by the vector function . From Figure 1, for small values of , consider the vector

.

The vector gives the average velocity over a time interval of length . Then the velocity vector at time is defined by

.

Hence the velocity vector is also the tangent vector and points in the direction of the tangent line. We have also defined the speed at time to be the magnitude of the velocity vector, that is, . Notice that from and in Section 12.3, we have

rate of change of distance with respect to time.

The acceleration vector at time is defined as the derivative of the velocity:

.

Example 1

The position vector of an object moving in a plane is given by . Find its velocity, speed, and acceleration when and illustrate geometrically.

Solution. The components of give the parametric equations

and .

Hence .

In particular, the object is moving along the ellipse. (See Figure 2.)

The velocity and acceleration at time are

,

,

and the speed is

.

When , we have

, , .

These velocity and acceleration vectors are shown in Figure 2. ■

Figure 2

Since the velocity vector of a particle moving in space can be obtained by differentiating the position vector , it follows that can be obtained by integrating . Similarly, can be obtained by integrating .

Example 2

A moving particle starts at an initial position with initial velocity . Its acceleration is . Find its position at time .

Solution. Since we have

.

To find the value of the constant vector , we use the fact that . The preceding equation gives , and hence . Thus

.

Since we have

To find , we use the initial condition , and so .

Hence the position at time is given by

. ■

The Newton's Second Law gives us an exact relationship between force, mass and acceleration. It can be expressed as a mathematical vector version:

where is a force, a mass and an acceleration.

A projectile is some object thrown in air or space. The curved path along which the projectile travels is what is known as trajectory. Projectile motion is the free fall motion of any body in a horizontal path with constant velocity.

Example 3

Suppose that a projectile is fired from the origin at time at angle of above the horizontal and initial velocity . (See Figure 3.) Assume that air resistance is negligible and the only external force is due to the gravity. What is the position vector of the projectile?

Figure 3

Solution. By Newton's Second Law and the gravity, the force is

where . Integrating ,

. Since , .

Integrating again, the position vector is

.

Since , . Hence

.

Let be the initial speed of the projectile. Then

.

Substituting from , the position vector of the projectile is

. ■

Example 4

A foot ball player kicks a football at an angle of above the horizontal in the field. The initial speed of the ball is 18 m/s. How far away from the kick off point will the football land, and what speed on the ground?

Solution. With , and , we get the components of position vector at time

and .

Impact occurs when on the ground, that is, . Solving this equation, we get .

Then which means the football hits the field about 28.63m away.

The position vector of this football at time is

.

Then the velocity of this football at time is

.

Hence its speed on the ground is

. ■

Tangential and Normal components of Acceleration

We are going to think about acceleration of the motion of an object along a curve as the second derivative of position, and we can further break down acceleration into two components called the tangential and normal components. (See Figure 4.)

Figure 4

The tangential acceleration allows us to know how much of the acceleration acts in the direction of motion. The normal acceleration is how much of the acceleration is orthogonal to the tangential acceleration. The tangential acceleration is a measure of the rate of change in the magnitude of the velocity vector, that is, speed, and the normal acceleration is a measure of the rate of change of the direction of the velocity vector.

Let be a position vector along a curve at time . We can find the tangential acceleration by the Chain Rule to rewrite the velocity vector

using properties of Section 12.3.

Now, since acceleration is simply the derivative of velocity, we can find acceleration

.

Note that .

Let the speed of the object be . Since and , we define the tangential acceleration and normal acceleration of acceleration by

and ,

respectively.

Then the acceleration of the motion of an object is

.

Since and are unit vectors and orthogonal to each other,. The magnitude of the acceleration is .

Although provides useful expressions for the tangential and normal components of acceleration, it is often desirable to have expressions in terms of and . To get this, we take the dot product of and as given by :

(since and .

Thus

.

By Theorem 2 in Section 12.3, we have

.

Example 5

A particle moves with position function . Find the tangential and normal components of acceleration.

http://buzzard.pugetsound.edu/talks/beezer-2012-kms-math-ed.pdf

Solution. We have

,, ,

.

Hence by , the tangential component is

.

Since

.

By , the normal component is

. ■

Example 6

Find the velocity, acceleration, and speed of a particle with the given position function:

http://matrix.skku.ac.kr/cal-lab/cal-13-4-Exm6.html

Solution.

t=var('t')

x(t)=cos(t)

y(t)=t^2

z(t)=3*sin(t)

r(t)=(x(t), y(t), z(t))

v(t)=(diff(x(t), t), diff(y(t), t), diff(z(t), t)) # velocity

a(t)=(diff(x(t), t, 2), diff(y(t), t, 2), diff(z(t), t, 2)) # acceleration

s(t)=v(t).norm() # speed

v(t)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer : (-sin(t), 2*t, 3*cos(t)) # velocity

a(t)

Answer : (-cos(t), 2, -3*sin(t)) # acceleration

s(t)

Answer : sqrt(abs(2*t)^2 + abs(-sin(t))^2 + abs(3*cos(t))^2) ■

*12.4 EXERCISES (Motion Along a Space Curve: Velocity and Acceleration)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-12-4-Sol.html

1-5. Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for specified value of .

1. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-1.html

Solution.

var('t')

r(t)=(t^3+1, t)

v=diff(r(t), t)

a=diff(v, t)

s=v.norm()

print v, a, s

v2=v.subs(t=2)

a2=a.subs(t=2)

print v2, a2

Answer : (3*t^2, 1) (6*t, 0) sqrt(abs(3*t^2)^2 + 1) (12, 1) (12, 0)

p1=parametric_plot(r(t), (t, 0, 3))

p2=line([r(2), r(2)+v2], color='red')

p3=line([r(2), r(2)+a2], color='green')

show(p1+p2+p3)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

2. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-2.html

Solution.

var('t')

r(t)=(2+2*t, 4*sqrt(t))

v=diff(r(t), t)

a=diff(v, t)

v1=v.subs(t=1)

a1=a.subs(t=1)

s=v.norm()

print v, a, s

Answer : (2, 2/sqrt(t)) (0, -1/t^(3/2)) sqrt(abs(2/sqrt(t))^2 + 4)

(2, 2), (0, -1)

p1=parametric_plot(r(t), (t, 0, 3))

p2=line([r(1), r(1)+v1], color='red')

p3=line([r(1), r(1)+a1], color='green')

show(p1+p2+p3)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

3. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-3.html

Solution.

var('t')

r(t)=(exp(2*t), exp(-t))

v=diff(r(t), t)

a=diff(v, t)

v0=v.subs(t=0)

a0=a.subs(t=0)

s=v.norm()

print v, a, s

Answer : (2*e^(2*t), -e^(-t)) (4*e^(2*t), e^(-t)) sqrt(abs (-e^(-t))^2 + abs(2*e^(2*t))^2)

(2, -1), (4, 1)

p1=parametric_plot(r(t), (t, -1, 1))

p2=line([r(0), r(0)+v0], color='red')

p3=line([r(0), r(0)+a0], color='green')

show(p1+p2+p3)

4. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-4.html

Solution.

var('t')

r(t)=(2*sin(t), cos(t))

v=diff(r(t), t)

a=diff(v, t)

vt=v.subs(t=pi/3)

at=a.subs(t=pi/3)

s=v.norm()

print v, a, s

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer : (2*cos(t), -sin(t)) (-2*sin(t), -cos(t)) sqrt(abs (-sin(t))^2 + abs(2*cos(t))^2)

,

p1=parametric_plot(r(t), (t, 0, 3))

p2=line([r(pi/3), r(pi/3)+vt], color='red')

p3=line([r(pi/3), r(pi/3)+at], color='green')

show(p1+p2+p3)

5. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-5.html

Solution.

var('t')

r(t)=(3*sin(t), t, 3*cos(t))

v=diff(r(t), t)

a=diff(v, t)

v0=v.subs(t=0)

a0=a.subs(t=0)

s=v.norm()

print v, a, s

Answer : (3*cos(t), 1, -3*sin(t)) (-3*sin(t), 0, -3*cos(t)) sqrt(abs(-3*sin(t))^2 + abs(3*cos(t))^2 + 1)

p1=parametric_plot3d(r(t), (t, -1, 2))

p2=line([r(0), r(0)+v0], color='red')

p3=line([r(0), r(0)+a0], color='green')

show(p1+p2+p3)

6-10. Find the velocity, acceleration and speed of a particle with the given position function.

6. .

Solution. ,

,

.

7. .

Solution. ,

,

.

8. .

Solution. ,

,

.

9. .

Solution. ,

,

.

10. .

Solution. ,

,

.

11-12. Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.

11. .

Solution. .

Since , we have

.

.

Since , we have .

12. .

Solution. .

Since , we have

.

.

Since , we have

.

13-14. Find the position vector of a particle that has the given acceleration and the specified initial velocity and position.

13. .

Solution. .

Since , we have .

.

Since , we have

.

14.

.

Solution. .

Since , we have

.

Since , we have . Hence,

15. The position function of a particle is given by . When is the speed a minimum?

http://matrix.skku.ac.kr/cal-lab/cal-13-4-15.html

Solution.

var('t')

r(t)=(t^2, t, t^2-4*t)

v=diff(r(t), t)

s=v.norm()

s0=diff(s, t)

solve(s0==0, t)

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer : t=1

plot(s, (t, 0, 3))

16-20. Find the tangential and normal components of the acceleration vector.

16. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-16.html

Solution.

var('t')

r(t)=(t-t^3, t^2, 0)

dr=diff(r(t), t)

ddr=diff(r(t), t, 2)

T=dr.dot_product(ddr)/dr.norm()

print T

N=(dr.cross_product(ddr)).norm() / dr.norm()

print N

Answer : 2*(3*(3*t^2 - 1)*t + 2*t)/sqrt(abs(-3*t^2 + 1)^2 + abs(2*t)^2) sqrt(abs(6*t^2 + 2)^2)/sqrt(abs (-3*t^2 + 1)^2 + abs(2*t)^2)

17. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-17.html

Solution.

var('t')

r(t)=(cos(t), sin(t), 2*t)

dr=diff(r(t), t)

ddr=diff(r(t), t, 2)

T=dr.dot_product(ddr)/dr.norm()

print T

N=(dr.cross_product(ddr)).norm() / dr.norm()

print N

Answer : 0 sqrt(abs(sin(t)^2 + cos(t)^2)^2 + abs(2*sin (t))^2 + abs(-2*cos(t))^2)/sqrt(abs(-sin(t))^2 + abs(cos(t))^2 + 4)

18. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-18.html

Solution.

var('t')

r(t)=(2*t, t^2, t^3)

dr=diff(r(t), t)

ddr=diff(r(t), t, 2)

T=dr.dot_product(ddr)/dr.norm()

print T

N=(dr.cross_product(ddr)).norm() / dr.norm()

print N

Answer : 2*(9*t^3 + 2*t)/sqrt(abs(3*t^2)^2 + abs(2*t)^2 + 4) sqrt(abs(6*t^2)^2 + abs(-12*t)^2 + 16)/sqrt (abs(3*t^2)^2 + abs(2*t)^2 + 4)

19. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-19.html

Solution.

var('t')

r(t)=(exp(2*t), t, exp(-2*t))

dr=diff(r(t), t)

ddr=diff(r(t), t, 2)

T=dr.dot_product(ddr)/dr.norm()

print T

N=(dr.cross_product(ddr)).norm() / dr.norm()

print N

Answer : -8*(e^(-4*t) - e^(4*t))/sqrt(abs(-2*e^(-2*t))^2 + abs(2*e^(2*t))^2 + 1) sqrt(abs(4*e^(-2*t))^2 + abs(-4*e^(2*t))^2 + 256)/sqrt(abs(-2*e^(-2*t))^2 + abs(2*e^(2*t))^2 + 1)

20. .

http://matrix.skku.ac.kr/cal-lab/cal-13-4-20.html

Solution.

var('t')

r(t)=(t, 1/2*(cos(t))^2, 1/2*(sin(t))^2)

dr=diff(r(t), t)

ddr=diff(r(t), t, 2)

T=dr.dot_product(ddr)/dr.norm()

print T

N=(dr.cross_product(ddr)).norm() / dr.norm()

print N

[CAS] http://sage.skku.edu/ 또는 https://sagecell.sagemath.org/

Answer : -2*(sin(t)^2 - cos(t)^2)*sin(t)*cos(t)/sqrt(abs(-sin (t)*cos(t))^2 + abs(sin(t)*cos(t))^2 + 1)

sqrt(2)*sqrt(abs(sin(t)^2 - cos(t)^2)^2)/sqrt(abs (-sin(t)*cos (t))^2 + abs(sin(t)*cos(t))^2 + 1)

Calculus

About this book : http://matrix.skku.ac.kr/Cal-Book/

Copyright @ 2019 SKKU Matrix Lab. All rights reserved.

Made by Manager: Prof. Sang-Gu Lee and Dr. Jae Hwa Lee http://matrix.skku.ac.kr/sglee/ and http://matrix.skku.ac.kr/cal-book/

*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).