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Chapter 8. Further Applications of Integration

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/

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Chapter 8. Further Applications of Integration

8.1 Arc Length   http://youtu.be/7OVqI20z_Bw

문제풀이 by 남택현  http://youtu.be/A8N-mDD0ja8

8.2 Area of a Surface of Revolution   http://youtu.be/Eq4i2A8eKxA

문제풀이 by 정승찬  http://youtu.be/yZFJDJgTJfw

8.3 Applications of Integral Calculus    http://youtu.be/1ZAJeP16pAQ

8.4 Differential equations      http://youtu.be/uHfOjz8I4-s

8.1 Arc Length

We are well aware of the fact that the length of a polygon is found by adding the lengths of the line segments that form the polygon. A question arises: what about the length of a curve if it is not a polygon? To define the length of a general curve, we proceed as follows. First divide the curve into a large number of small parts, find the approximate lengths of the small parts and then add them up. This produces an approximation to the length of the curve. The larger the number of small parts, the better the approximation becomes. When, we take the limit as , where is the number of parts.

Consider a curve defined by the equation , . Assume that is continuously differentiable in the sense that the derivative is continuous for . We divide the interval into subintervals with endpoints , and, for the sake of convenience, of equal width . If , then the point lies on . These points divide the curve into smaller curves. Since the polygon with vertices , , ..., approximates (see Figure 1), we approximate the length of by the length of this polygon. As increases, the approximation gets better.

Figure 1

Therefore, the length of the curve is defined as follows:

[1]

Figure 2

provided the right hand limit exists.

We now try to deduce a more convenient expression for the computation of . Let in Figure 2, then

.

By the Mean Value Theorem applied to on the interval , there exists a number between and such that

and so

.

Therefore,

.

Since the right side of this equation is the limit of a Riemann sum, we recognize it as being a definite integral and so we have the following formula.

The Arc Length Formula

If is continuous on , then the length of the curve , , is

[2]                   .

If , denotes the equation for a curve , then the length of is given by

[3]

which is obtained by interchanging the roles of and in  [2].

Figure 3

EXAMPLE 1

For the curve , between the points and , determine the length of the curve.

For the upper branch of the curve we have   ,

and so the arc length formula gives

.

var('x')

integral (sqrt(x), x, 1, 4)

EXAMPLE 2

Calculate the length of the curve between and .

Figure 4

We have    and so, the length is

var('y')

integral (y^3 + 1/(4*y^3), y, 1, 2)

EXAMPLE 3

Determine the length of the curve , between the points and .

For the curve we have .  The length of the curve on the interval is

Figure 5

.

Let . Then and .

With the substitution, becomes and becomes

.

The Arc Length Function

Let , be the equation of a smooth curve . Let be the distance along from the initial point to the point . Then the function is called the arc length function, measuring the arc length of a part of the curve , is given by

[4]

Differentiating [4]  using FTC 1, we have

[5]

From [5], the rate of change of with respect to is always at least and is equal to when , the slope of the curve, is 0. The differential of the arc length is

[6]

or in the symmetric form

[7]

Figure 6

From the right triangle in Figure 6 we get [7]. Solving  [7] we have

Substituting the above in , we get the Arc Length Formula [2] or [37].

EXAMPLE 4

Determine the arc length function taking as the starting point.

Figure 7

Here, , then we have

,

,

.

Thus, the arc length function is given by

.

For example, the arc length along the curve from to is

.

8.1    EXERCISES (Arc Length)

http://matrix.skku.ac.kr/Cal-Book/part1/CS-Sec-8-1-Sol.html

1-6. Find the length of the curve.

1. ,

Since the length of the curve is

,

⇒ .

Substitute   and  .

So, .

⇒

2. ,

[CAS] 3. ,

4. from to

Since implies ,

Let , then and

.

5. , ,

6. ,

Since ,

.

7. Find the length of the curve as goes to .

,

var('k, x')

@interact

def _(y = input_box(tanh(k*x), label="y="), kappa=slider(0, 100, 0.01, default=1, label='k')):

plot(tanh(kappa*x), x, 0, 10, color='purple').show(aspect_ratio=1, xmin=0, xmax=2.5, ymin=0, ymax=1.2)

8. ,

[CAS] 9. Find the length of the curve when goes to 0.

,

var('k, x')

@interact

def _(y = input_box(1/2*k*x^2, label="y="), kappa=slider(0, 100, 0.05, default=1,label='k')):

L=  lim(integral(sqrt(1+(k*x)^2), x, 0, 1), k=kappa)

html('<p style="text-align: left;">The Arc Length is = ' + str(L)+'</p>')

plot(1/2*kappa*x^2, x,0,1).show(aspect_ratio=1, xmin=0, xmax=1, ymin=0, ymax=1)

10. Find the length of the curve, , ,

var('t')

parametric_plot(((cos(t))^3, (sin(t))^3), 0, 2*pi, rgbcolor=hue(0.6))

var('t')

4*integral(3*sin(t)*cos(t), t, 0, pi/2)

[CAS] 11. (a) Graph the curve , .

(b)Compute the lengths of inscribed polygons with , , and sides. (Divide the interval into equal subintervals.) Illustrate by sketching these polygons.

(c) Set up an integral to find the length of the curve.

(d) Use any of your CAS tool to find the length of the curve to four decimal places. Compare with the approximations in part (b).

(a)

(b)

var('x')

S = plot( (1+x^2)^(2/3), x, 0, 4)

L = integral((1/3)*((9 +(16*x^2)/((1 + x^2 )^(2/3) ) )^(1/2)) , x, 0, 4)

show (L)             # 7.06022221757

print numerical_integral((1/3)*((9 +(16*x^2)/((1 + x^2 )^(2/3) ) )^(1/2)) , 0, 4)[0]

#P = plot(f,x, xmax=2.2, xmin=1.5, ymax=10, ymin=-10, color='blue')

show (S)

(c)

(d) Simpson’s Rule with gives .

12. Repeat Exercise 11 for the curve , .

(a)

var('x')

plot(x^2*cos(x), x, 0, 2*pi);

(b)

(c)

(d) Simpson’s Rule with gives .

8.2 Area of a Surface of Revolution

Figure 1

When a region bounded by curves is rotated about a line, a solid of revolution is generated. The lateral boundary of such a solid of revolution is the surface of revolution. The lateral surface area of a circular cylinder with radius and height is taken to be as in Figure 1. When we cut a circular cone with base radius , arc length and slant height along a dashed line as in Figure 2, and flatten it, it forms a sector of a circle with radius and central angle . Therefore the lateral surface area of this cone is

.

Figure 2

Observe that each of the lateral surfaces of a circular cylinder and a circular cone is generated when a simple polygon is rotated about a line. For later use in this section, let us find the lateral surface area of a portion  of a circular cone with slant height  and upper and lower radii and as shown in Figure 3. It is found by subtracting the areas of smaller cone from the larger cone as shown in the Figure 3;

Figure 3

.

The slant heights in Figure 3 are proportional to the radii.

so, .

The slant height of the frustum is

Substituting this into [1], using the last two equations

We obtain the surface as .

[2]           where is the average radius of the band.

Figure 4

We now consider finding the area of the surface of revolution obtained when a smooth curve , is rotated about the -axis. (See Figure 4.) Assume that is positive and has a continuous derivative. We divide the interval into subintervals with endpoints , , ..., and equal width . If , then the point lies on the curve. The part of the surface between and is approximated by the band obtained by taking the line segment and rotating it about the -axis. As increases, the approximation gets better. The band has slant height and average radius . So, by [2], its surface area is

[2]        .

We have

,   where is some number in . When is small, we have and also , since is a continuous function. Therefore, we have

.

So an approximation to the area of the total surface of revolution is

[3]            .

Figure 5

As the Riemann sum for the function , gives

.

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

[4]         or

[5]

If the curve is given by the equation , , then the area of the surface of revolution obtained by rotating the curve, about the -axis becomes

[6]        or

[7]

Or symbolically, for the rotation about the -axis the surface area is

[8]

whereas, for the rotation about the -axis, the surface area formula is

[9]

where we use for either  or .

Note that the circumference of a circle traced out by the point on the curve as it is rotated about the -axis or -axis, is or respectively. (See Figure 6.) Therefore, the surface area can be obtained by integrating of .

Figure 6

EXAMPLE 1

Determine the area of the surface of revolution obtained by rotating about the -axis the curve , .

var('u')

revolution_plot3d(sin(u), (u, 0, pi/2), opacity=0.5, color='blue', show_curve=True, parallel_axis='x')

Figure 7

Let and .  We have

.

var('x')

integral ( 2*pi* (sec(x))^3 , x, 0, pi/4)

EXAMPLE 2

Determine the area of the surface of revolution obtained by rotating about the -axis the arc of the parabola from to .

var('u')

revolution_plot3d(u^3, (u, 0, 2), opacity=0.5, color='blue', show_curve=True, parallel_axis='x')

Figure 8

Since and from (8) we have,

(where )

var('u')

pi/9*integral (u*(1+u^2)^1/2, u, 0, 12)

EXAMPLE 2

Find the area of the surface of a sphere of radius , .

The equation of the upper circle with radius and center at is .

We have

.

Therefore, the desired area of the surface is

.

var('x, R')

integral ( 2*pi * R , x, -R, R)   # 4*pi*R^2

[CAS] EXAMPLE 1

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

[Solution]

var('x, u')

y=x+cos(x)

P=plot(y, (x, 0, pi))

Q=revolution_plot3d(u+cos(u), (u, 0, pi), opacity=0.5, color='blue', show_curve=True, parallel_axis='x')

show(Q+P)

print numerical_integral(2*pi*y*sqrt(1+(y.diff(x))^2), 0, pi)[0]

Figure 9                                       Figure 10

8.2 EXERCISES (Area of a Surface of Revolution)

1-2. Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about the given axis.

1. , ; -axis

[CAS] 2. , ; -axis

(range from to )

Let,

var('x,y')

implicit_plot(4*y==2*x^2-ln(x),(1,4),(0,8))

revolution_plot3d(y==0.5*x^2-0.25*ln(x),(x,1,4),(y,0,10),axis=(1,0,0),show_curve=False,opacity=0.7).show(aspect_ratio=(1,1,1))

3-7. Find the area of the surface obtained by rotating the curve about the -axis.

3. Find the area of the surface when goes to 0.

,

var('k, x')

@interact

def _(y = input_box(1/2*k*x^2, label="y="), kappa=slider(0, 1, 0.05, default=1, label='k')):

html(AREA '\$\$=%s\$\$'%( lim(integral(pi*2*1/2*k*x^2*sqrt(1+(k*x)^2), x, 0, 1), k=kappa) ))

plot(1/2*kappa*x^2, x, 0, 1, color='purple', fill=true).show(aspect_ratio=1, xmin=0, xmax=1, ymin=0, ymax=0.5)

4. Find the area of the surface when is a positive even integer.

,

k, x=var('k, x');

@interact

def _(y = input_box(sin(k*x), label="y="), kappa=slider(0, 100, 1, default=1, label=' k')):

html(AREA '\$\$=%s\$\$'%(  lim(integral(2*pi*sin(k*x)*sqrt(1+diff(sin(k*x), x)), x, 0, pi), k=kappa) ))

plot(sin(kappa*x), x, 0, pi, color='purple', fill=true).show(aspect_ratio=1, xmin=0, xmax=pi, ymin=-1, ymax=1)

5. ,

6. ,

The surface obtained by rotating the curve about the -axis is .

Substitute ,

⇒ .

Let and ,

then .

Calculate and simplify to have

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

,

p1=plot(sqrt(1-(x)^2), (0,1),

rgbcolor=hue(0.4))

p2=plot(-2*log(x)*x, (0, 1),

rgbcolor=hue(0.6))

From the above figure, it is clear that the maximum is circle of unit radius in the first quadrant. When we rotate the unit circle in the first quadrant about the x-axis we obtain half unit sphere. Hence the required area is .

8-10. The given curve is rotated about the -axis. Find the area of the resulting surface.

8. ,

9. ,

10. ,

The surface obtained by rotating the curve about the -axis becomes

(Range form to )

Then substitute to above equation.

(, then (because of ) and so )

11. Find the area of the surface of the solid of revolution obtained by rotating about the -axis the circle , .

The right semicircle and left semicircle equations are given by and

, respectively.

.

[CAS] 12. Use a CAS to find the area of the surface obtained by rotating the curve about the given axis. Use Simpson’s Rule with .

(a) , ; -axis

(b) , ; -axis

(a)

var('x');

y=x^5;

d=diff(y);

f(x)=y*(1+d^2)^(1/2);

q=n(2*pi*(0.1/3*(f(0)+4*f(0.1)+2*f(0.2)+4*f(0.3)+2*f(0.4)+4*f(0.5)+2*f(0.6)+4*f(0.7)+2*f(0.8)+4*f(0.9)+f(1))));

show(q)

(b)

var('y');

x=ln(y);

d=diff(x);

f(y)=x*(1+d^2)^(1/2);

k=4;

q=n(2*pi*(0.1*k/3*(f(1)+4*f(1+0.1*k)+2*f(1+0.2*k)+4*f(1+0.3*k)+2*f(1+0.4*k)+4*f(1+0.5*k)+2*f(1+0.6*k)+4*f(1+0.7*k)+2*f(1+0.8*k)+4*f(1+0.9*k)+f(1+1*k))));

show(q)

8.3 Center of Mass

Figure 1 the center of mass

Center of Mass

In physics, the center of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero. Calculations in mechanics are simplified when formulated with respect to the center of mass. In the case of a single rigid body, the center of mass is fixed in relation to the body, and if the body has uniform density, it will be located at the centroid. The center of mass may be located outside the physical body.

The center of mass is the mean location of all the mass in a system. We consider first two bodies, with masses and , and positions and respectively. We denote by the center of mass for the bodies. Then, due to the Law of the Lever (i.e., the rod is balanced at the center of mass), we note that .

Therefore, the center of mass in this case is given as

[1]             .

In general, for bodies, with masses and their positions from the origin, respectively, the center of mass, denoted by , is located at

[2]     ,

where is the total mass of the bodies. The numerator in the formula [2] , is called the moment of the masses.

EXAMPLE 1

Find the center of mass of an object that has masses 3, 4, and 8 at the points , and , respectively.

We use the formula  [2] :

.

Thus, the center of mass is .

Figure 2   Discrete Masses

Centroids

We consider a flat plate, denoted by , with uniform density in the plane. The centroid of a plane region is geometrically the intersection of all straight lines that divide into two parts of equal moment about the line. Roughly speaking, the centroid is the average of all points of . The concept of the centroid can be extended to any domain in as follows: the centroid is the intersection of all hyperplanes that divide into two parts of equal moment. The centroid, denoted by of a domain in can be obtained as an integral

[3]                   ,

where is a characteristic function, which is if and otherwise. Another formula for each coordinate for the centroid is given as follows:

[4]             ,

where is the measure of the intersection of and the hyperplane defined by the equation . The denominator is nothing but the measure of . For the plane, due to the formula [4], the coordinates of the centroid become

[5]             ,

where is the length of the intersection of the flat plate and the line parallel to -axis passing through and is similarly the length of the intersection of and the line parallel to -axis passing through .

We remark that if the domain under consideration is symmetric with respect to a hyperplane, then the centroid lies on the hyperplane. Therefore, for a flat plane if is symmetric about a line , then the center of mass is on the line . This is intuitively obvious, and is often referred as the “symmetric principle”, from which a computation may become simpler, as long as symmetry is observed for the domains under consideration.

EXAMPLE 2

Figure 3

Consider a flat plate , which is enclosed by , , -axis, and the graph of , where is a continuous function. Suppose that the density is , which is uniformly distributed on the plate. Show that the centroid (center of mass) of the plate is

[6]     .

is given in the form of [6] directly from [5]. It remains to show [6]

for the case of .

We divide the interval into subintervals with endpoints , , , , and equal width . We denote , which is the midpoint of th subinterval. Note that the centroid of the th approximating rectangle, denoted by , is and its mass is . Furthermore, the moment of about the -axis is . The total moment of approximating the plate is the sum of each moment, that is, . Now passing it to the limit, total moment about -axis becomes the following integral: . Therefore, -coordinate of centroid is just the above integral divided by the total mass. The formula [6] is deduced.

We remark that if a flat plate is enclosed by and , where for , then the formula [6] becomes

[7]     , .

Its verification is the same as the case of [6] and thus the details are omitted.

We close this section by recalling Pappus’s centroid theorem (also known as the Guldinus theorem, Pappus–Guldinus theorem or Pappus’ theorem), which shows a connection between the surface areas and volumes enclosed by surfaces and solids of revolution.

THEOREM 1   Pappus’s Centroid Theorem (PARTⅠ)

The surface area of a surface of revolution generated by rotating a plane curve about an axis external to and on the same plane is equal to the product of the arc length of and the distance traveled by its geometric centroid.

.

THEOREM 2   Pappus’s Centroid Theorem (PART II)

The volume of a solid of revolution generated by rotating a plane figure about an external axis is equal to the product of the area of and the distance d traveled by its geometric centroid.

[8]              .

EXAMPLE 3

(a) Find the centroid of the region bounded by the line and the parabola (we assume that the density of the enclosed region is 1).

(b) Use Pappus’s centroid theorem to find the volume of the solid obtained by rotating about the -axis the region enclosed by the line and the parabola .

(a) The area of the enclosed region is

Due to formula [8], we have

So the centroid is .

(b) Using result (a), we know that  -coordinate of the centroid for the enclosed region is . Therefore, with the aid of Pappus’ centroid theorem, the volume of rotating the region is .                             ■

8.3 EXERCISES (Center of Mass)

1. Find the center of mass of the region bounded by the cycloid and axis (we assume that the density of the enclosed region is 1).

, ,

(Hint:

)

By the symmetry principle, the center of mass must lie on , so . The area of the region is computed as follows:

Recall the formula for -coordinate of the center of mass is given as follows:

Summing up, the center of mass is

2. By using graphical tools, find the approximate center of mass of the region which is bounded by the , .

theta=var('theta');

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

By looking at the figure it is clear that the center of mass of the region is approximately .

* 8.4 Differential Equations

Differential equations is one of the most important applications of differentiation and integration in Calculus.1) In this section we introduce several models of ordinary differential equations, which seem relatively simple but quite informative.2) A method of finding solutions is also presented for a separable differential equation although in general, it is not an easy task to solve differential equations explicitly.

by  Joyner and Hampton

Models of Ordinary Differential Equations

One very simple model for population growth assumes that the rate of growth of a population is proportional to the size of the population. This assumption is ideal and may be observable in certain situations. For convenience, we denote by the number of inhabitants in the population, where indicates the time. The hypothesis is then written as the following equation:

[1]                     ,

where is the proportionality constant. Suppose that the population size at a certain time (here without loss of generality such time is ) can be measured, let us denote it as . Then the question is what the population is at time . This type of question is called an initial-value problem. It turns out that the solution for [1] is (you can check this directly). In this case, finding the solution is not difficult and details of the procedure will be presented subsequently in example 2.

A modified version of the above population model3) is the "logistic model", which is based on the following assumption: There is a number , the carrying capacity, so that the population approaches as time passes by. More precisely, if size of population is small, the rate of the growth of population is assumed to be proportional to the size of population as in the previous case. However, we suppose that if decreases if it ever exceeds ). A typical example satisfying such assumptions would be

[2]    .

This seems more realistic compared to the previous case in the sense that it does not allow never-ending growth (but this does not mean that the logistic model reflects reality).

So far, we have considered some models for a single population, next we consider models for two populations that interact with one another. One simple example is predator-prey models, which can be written in the form of differential equations shown below: (predator-prey models4) have a variety of descriptions but here we introduce only a simple case).

In the predator-prey model, there are two species, one is the prey and the other is the predator. For convenience, we denote by and the number of the prey and the predator, respectively. In this case, the hypothesis is that the rate of predation on the prey is proportional to the rate at which the predators and the prey meet, and vice versa, possibly with a different proportional constant. The differential equations taking these assumptions into account is described as follows:

[3]    ,

[4]    ,

where , , and are positive constants. The system of this prey-predator model is sometimes referred to the Lotka-Volterra equations.5)

Compared to the previous two examples, the solutions are a pair of functions to a coupled system of differential equations.

Separable Differential Equations

For first-order ordinary differential equations, a separable equation is of the form .

If , we separate variables in the following manner:

.

Integrating both sides of the equation we get

.

Summing up, if a differential equation is separable, it can be eventually separated into a pair of integrals.

EXAMPLE 1

Solve with initial condition .

is separable variable equation.

By integration on both sides of the equations

,

or .

The general solution is , where is arbitrary constant.

From the given initial condition , we obtain .

Thus the solution of the initial-value problem is .

sage: x = var('x')

sage: y = function('y', x)

sage: de = lambda y: diff(y,x) + y/x

sage: desolve(de(y),[y,x],[1,1])    # 1/x

EXAMPLE 2

Find the general solution to either “by hand” or using Sage.

The equation is with separable variables. By integration of both sides we have

,

or  .

The solution is . ( is arbitrary constant.)

Next we recall the model of population growth , which can also be solved explicitly by the method of separable variables. We note that the solution is in fact growing if and is decaying in case that .

EXAMPLE 3

Find the general solution of the equation [1], that is , where is a constant.

We can rewrite the equation as follows: , so integrating both sides,

where is a constant. Simplifying it, we obtain .

We obtained

.

Hence is the initial value of the given equation.

The next example is for logistic models such as [2], which are again separable differential equations. This means that such equations are explicitly solvable (as long as we can find antiderivatives).

EXAMPLE 4

Solve , with an initial condition .

Using the method of separable variables, we have

.

Due to the method of partial fractions, the left side of the integrals can be rewritten and computed as follows:

where

Since , the solution is

.

EXAMPLE 5

Find the general solution to either “by hand” or using Sage.

http://matrix.skku.ac.kr/cal-lab/cal-8-4-Exm-8.html

We rewrite this as . Now compute   so the formula gives

sage: t = var('t')

sage: x = function('x', t)

sage: de = lambda y: diff(y,t) + (1/t)*y - exp(t)/t

sage: desolve(de(x),[x,t])    # (c + e^t)/t

Orthogonal Trajectory

Figure 1 Orthogonal trajectories

We close this section by discussing orthogonal trajectories. Orthogonal trajectories are two families of curves in the plane that intersect a given family of curves at right angles. For example, the family of straight lines parallel to -axis, that is for any , is an orthogonal trajectory for the family of straight line parallel to -axis, that is for any . We say that two families of curves (curves including the straight lines) are orthogonal trajectories of each other.

var('y')

L=[]

for i in srange(-2, 2, 0.2):

p=implicit_plot(x^2+y^2==i*x, (x, -2, 2), (y, -2, 2))

L.append(p)

M=[]

for i in srange(-2, 2, 0.4):

p=implicit_plot(x^2+y^2==i*y, (x, -2, 2), (y, -2, 2), cmap=['red'])

M.append(p)

sum(L)+sum(M)

Figure 2

EXAMPLE 6

Find the orthogonal trajectories of the family of straight lines through the origin, that is for any .

If we differentiate , we get . To eliminate , we note that and so the differential equation can be rewritten as . Since an orthogonal trajectory must have the slope of its tangent line that is negative reciprocal of this slope. Thus, the orthogonal trajectories must satisfy . Since it is separable, we compute

,

where is an arbitrary positive constant. Therefore, the orthogonal trajectory is the family of concentric circles centered at the origin.

var('y')

L=[]

for i in srange(-2, 2, 0.4):

p=implicit_plot(y==i*x, (x, -2, 2), (y, -2, 2))

L.append(p)

M=[]

for i in srange(0, 2, 0.2):

p=implicit_plot(x^2+y^2==i, (x, -2, 2), (y, -2, 2), cmap=['red'])

M.append(p)

sum(L)+sum(M)

Figure 3

There are many other applications of first order differential equations,  such as decay of radioactive materials, mixing problems involving two or more fluids, and so on (some examples are in the exercises).

8.4 EXERCISES (Differential Equations)

1. Find the solutions of the given initial-value problem  , .

The given equation is separable, and therefore, it can be rewritten as

Using the initial condition, we know that , and therefore,

2. Find the orthogonal trajectories of the family of the following curves: .

Since , the orthogonal trajectories must have their derivative perpendicular to the given family of curves, namely

(1)

Note that the given curves never touch x-axis, which means that , and so the above differential equation (1) makes sense. Now noting that (1) is of separable type, we get

3. Solve the differential equation .

Separation of variables gives By the integration, we have So, by simplification, we obtain

4. Solve the differential equation that satisfies the initial condition .

We first note that differential equation is separable. Therefore, after splitting it into variables and using the integration in each variable, we  have

Since , we have . Summing up, we

obtain

5. A tank contains of brine with 15kg of dissolved salt. Brine that contains 0.07kg of salt per liter of water enters the tank at a rate of and another brine that contains 0.02kg of salt per liter of water enters the tank at a rate of .

The solution is kept thoroughly mixed and drains from the tank at a rate of . How much salt is in the tank as tends to ?

Let be the amount of salt (in kilograms) after minutes. The rate at which enters the tank is

The tank always contains of liquid and so the rate at which leaves the tank is

Thus, the differential equation is

Solving this separable differential equation, we obtain

Since , we have . As tends to , converges to .

6. Let and be the number of prey and predators at time , respectively. We suppose that  these two species satisfy the following system of differential equations:

, .

(a) Find all equilibrium solutions and the expression for .

(b)Solve with given initial conditions

and by the method of separable variables.

(a) The equilibrium are steady state solutions, and so they must satisfy

Using  the chain rule, we obtain

(1)

(b) Note that the initial condition when and by solving the  equation (1), we get

It  remains to specify the constant . Indeed, due to initial condition, we can  see

Hence, the solution satisfies

7. The half-life of Cesium-137 is 30 years. Suppose that we have a 100mg sample. After how long will only 1mg remain?

Let be the mass of Cesium-137 (in milligrams) that remains after years.  Then and , so we have . In order to  determine the value of , we use the fact that . Thus we get

We want to find the value of such that . Solving the following equation for and taking the natural  logarithm of both sides, we have

8. Let be a positive constant. We consider the following differential equation .

, , .       (1)

For what value of does the solution of (1) exist for all time (global existence or blow up in a finite time)? Verify your answer.

Since the equation is separable, we get . We consider first the case . Keeping in mind that , we obtain

In this  case, a solution exists globally for all time. Next we consider the case .  Since and , we obtain , which again

exists globally. Finally, in case that , as in the previous case of , we have

The  above solution is well-defined as long as but its limit, goes  to as approaches . Therefore, the solution blows up in a finite time  for the case .

[CAS] 9. Solve the differential equation , with the initial condition and determine the interval in which the solution exists.

(Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent)

(You may do it with Sage at

Since

and , we obtain the following integral curve.

f = lambda x,y: y^2-2*y-2*x^3-2*x^2-2*x-3

contour_plot(f, (-5, 5), (-5, 5), fill=0, contours=20,plot_points=200)

Let us enlarge it.

f = lambda x,y: y^2-2*y-2*x^3-2*x^2-2*x-3

contour_plot(f, (-5, 5), (-5, 5), fill=0, contours=2, plot_points=200)

The integral curve has a vertical tangent around . Therefore, the desired interval is .

Now, we can consider another solution. Since  and , we obtain . To find the interval of definition, look for points where the integral curve has a vertical tangent.

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

p1=implicit_plot(y^2-2*y-2*x^3-2*x^2-2*x-3, (x, -3, 3), (y, -3, 3))

show(p1)

The integral curve has a vertical tangent around . Therefore, the desired interval is .

[CAS] 10. Solve the differential equation , with the initial condition . Hence plot the solution curve. Also plot the solution curves for a range of initial conditions.

reset()

var('t')

y = function('y', x)

de = diff(y, x) == 3*x^2/(6*y+sin(y))

sol=desolve(de, y, [0,-1])

print "The required solution is given by"

show(sol)

s=sol.subs_expr(y(x)==t)

p=implicit_plot(s, (x,-3,3), (t,-3,3));

p

The required solution is given by

.

We plot the solutions curves for a range of initial values.

reset()

var('t')

y = function('y', x)

de = diff(y,x) == 3*x^2/(6*y+sin(y))

sol=desolve(de, y, [0,-1])

s=sol.subs_expr(y(x)==t)

p=implicit_plot(s, (x, -3, 3), (t, -3, 3))

x0 = srange(-1, 2, 0.2)

soln = [desolve(de, y, [0,float(c)]) for c in x0]

for i in range(len(soln)):

g=soln[i].subs_expr(y(x)==t)

p=p+implicit_plot(g, (x, -3, 3), (t, -3, 3))

p

<Abacus>

-  The End of Ch. 8 -

Calculus