Math 2350 Fall 2004 Exam 2 Problem Set with Answers - Prof. Lance Douglas Drager, Assignments of Advanced Calculus

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PROBLEM SET

Practice Problems for Exam

Math 2350, Fall 2004

Nov. 7, 2004

Corrected Nov. 10

ANSWERS

i

Answer : We have ∂r ∂u = v − 2 u ∂r ∂v

= u − 2 v ∂u ∂x = 2xy ∂u ∂y

= x^2

∂v ∂x

∂v ∂y

By the chain rule, ∂r ∂x

∂r ∂u

∂u ∂x

∂r ∂v

∂v ∂x = (v − 2 u)2xy + (u − 2 v) = (2x − y − 2 x^2 y)2xy + (x^2 y − 4 x + 2y) = 4x^2 y − 2 xy^3 − 4 x^3 y^2 + 2x^2 y − 8 x + 4y = 6x^2 y − 2 xy^3 − 4 x^3 y^2 − 8 x + 4y

Similarly, we have

∂r ∂y

∂r ∂u

∂u ∂y

∂r ∂v

∂v ∂y = (v − 2 u)x^2 + (u − 2 v)(−1) = (2x − y − 2 x^2 y)x^2 − (x^2 y − 4 x + 2y) = 2x^3 − x^2 y − 2 x^4 y − x^2 y + 4x − 2 y = 2x^3 − 2 x^2 y − 2 x^4 y + 4x − 2 y.

Problem 3. Suppose that p = f (u, v, w), u = g(x, y), v = h(y, z) and w = k(x, y, z). Write the chain rule formals for the partial derivatives of p with respect to x, y and z.

Answer : ∂p ∂x

∂p ∂u

∂u ∂x

∂p ∂w

∂w ∂x ∂p ∂y

∂p ∂u

∂u ∂y

∂p ∂v

∂v ∂y

∂p ∂w

∂w ∂y ∂p ∂z

∂p ∂v

∂v ∂z

∂p ∂w

∂w ∂z

Problem 4. A box is measured to have a length of 15 inches, a width of 10 inches and a depth of 5 inches. From these measurements, we would calculate the volume to be 750 cubic inches. If there is a possible error of up to 0.25 inch in the measurements, use differentials to estimate the maximum possible error in the calculated volume.

Answer : Let x, y and z denote the length, width and depth of the box. Then the volume is given by V = xyz. The total differential of the volume is

dV =

∂V

∂x

dx +

∂V

∂y

dy +

∂V

∂z

dz

= yz dx + xz dy + xy dz.

To approximate the error in the volume, we plug in x = 15, y = 10, z = 5 and dx = dy = dz = 0.25. Thus,

Error in Volume ≈ dV = (10)(5)(0.25) + (15)(5)(0.25) + (15)(10)(0.25) = 68. 75

cubic inches.

Problem 5. A rectangular block of ice is melting in the sun. At a certain instant the block has a height of 5 inches, a length of 10 inches, and a width of 12 inches, and each of the dimensions is decreasing at a rate of 2 inches per hour. How fast is the volume of the block changing?

Answer : Let x be the height of the box, y the length and z the width so the volume of the box is V = xyz. By the chain rule

dV dt

∂V

∂x

dx dt

∂V

∂y

dy dt

∂V

∂z

dz dt

= yz

dx dt

• xz

dy dt

• xy

dz dt

At the given instant, we have

x = 5, y = 10, z = 12, dx dt

dy dt

dz dt

Answer :

If the bug’s position at time t is (x(t), y(t)), the bug’s temperature at time t is T (t) = f (x(t), y(t)). The rate of change of the bug’s temperature is

dT dt

(t) = ∇f (x(t), y(t)) · v(t),

where v(t) is the bug’s velocity vector. We have

∇f (x, y) = (2x + y, x + 2y).

At our instant, call it t = t 0 , we have (x(t 0 ), y(t 0 )) = (2, 1) and v(t 0 ) = (3, −4). Thus, we have

dT dt

(t 0 ) = ∇f (x(t 0 ), y(t 0 )) · v(t 0 )

= ∇f (2, 1) · (3, −4) = (5, 4) · (3, −4) = 15 − 16 = − 1.

B. Suppose the bug is at (2, 1). If the bug wants to cool off, what direction should he travel (at speed 5) to get fastest decrease in temperature? What is this rate of change of temperature?

Answer : The direction of fastest decrease is the direction of −∇f (2, 1). The vector of length 5 in this direction is

v = −

‖∇f (2, 1)‖

∇f (2, 1),

so this should be the bug’s velocity vector. The rate of change of temperature will be dT dt

= ∇f (2, 1) · v

= ∇f (2, 1) ·

[

‖∇f (2, 1)‖

∇f (2, 1)

]

‖∇f (2, 1)‖^2 ‖∇f (2, 1)‖ = − 5 ‖∇f (2, 1)‖ = − 5

Problem 8. Find the equation of the tangent plane to the cone z^2 = x^2 + y^2 at the point P = (4, − 3 , 5). Find parametric equations for the normal line to the surface at this point.

Answer : The cone is the level surface

g(x, y, z) = x^2 + y^2 − z^2.

The gradient of this function is

∇g(x, y, z) = (2x, 2 y, − 2 z).

Since ∇g is orthogonal to the level surfaces of g, the vector

N = ∇g(4, − 3 , 5) = (8, − 6 , −10)

is perpendicular to the surface (i.e., perpendicular to the tangent plane of the surface) at P. Hence the tangent plane is the plane that passes through the point with position vector x 0 = (4, − 3 , 5) and is perpendicular to N. The vector form of the equation is

N · (x − x 0 ) = 0

where x = (x, y, z). Thus, we have

0 = N · (x − x 0 ) = N · [(x, y, z) − (4, − 3 , 5)] = N · (x − 4 , y + 3, z − 5) = (8, − 6 , −10) · (x − 4 , y + 3, z − 5) = 8(x − 4) − 6(y + 3) − 10(z − 5).

Thus, the equation of the tangent plane is

8(x − 4) − 6(y + 3) − 10(z − 5) = 0,

which can be simplified to standard form as

4 x − 3 y − 5 z = 0.

The normal line passes through P and is parallel to ∇g(P ) = N, so the vector form of the parametrization is

x = x 0 + tN = (4, − 3 , 5) + t(8, − 6 , −10) = (4 + 8t, − 3 − 6 t, 5 − 10 t),

so the parametric equations are

x = 4 + 8t y = − 3 − 6 t z = 5 − 10 t

Setting these equal to zero gives the system of equations

y = 0 x = 0.

Thus, we get one critical point (0, 0) of f which is in the interior of the region. We have f (0, 0) = 0. Now we have to find the max and min of f on the boundary circle x^2 +y^2 = 2. You could do this by parameterizing the circle by x =

√ 2 cos(θ) and^ y^ = 2 sin(θ). I’ll do it by Lagrange multipliers. Thus we want to maximize and minimize f (x, y) = xy subject to the con- straint g(x, y) = x^2 + y^2 = 2. The Lagrange multiplier equations are ∇f = λ∇g and g = 2 in vector form. In order words

fx = λgx fy = λgy g = 2.

Plugging in the formulas we get this system

(10.1) y = 2λx

(10.2) x = 2λy

(10.3) x^2 + y^2 = 2.

Plugging (10.1) into (10.2) gives x = 4λ^2 x or

(10.4) (4λ^2 − 1)x = 0.

First, consider the case where 4λ^2 − 1 6 = 0. Then (10.4) implies that x =

1. But then (10.1) implies that y = 0. The point (0, 0) does not satisfy the constraint equation (10.3), so there is no solution of the system with 4λ^2 − 1 6 = 0. Thus, we must have 4λ^2 − 1 = 0, i.e., λ = ± 1 /2. Consider first the case λ = 1/2. Then equation (10.1) and equation (10.2) become y = x. Substituting this into the constraint (10.3) gives 2x^2 = 2, so x = ±1. Since y = x, we get two critical points (1, 1) and (− 1 , −1). The value of the function at these points is f = 1. In the case λ = − 1 /2, (10.1) gives y = −x. Plugging this into the constraint equation again gives 2x^2 = 2, so x = ±1. Thus, we get two critical points (1, −1) and (− 1 , 1). The value of the function at these points is f = −1. At the points that are candidates for the absolute max and min of f , we have values −1, 0, 1. Thus, the absolute min is of f on the disc is −1, which occurs at the points (1, −1) and (− 1 , 1). The absolute max of f on the disk is 1, which occurs at the points (1, 1) and (− 1 , −1).

Problem 11. Find the absolute max and min of the function f (x, y, z) = xyz^2 on the spherical surface x^2 + y^2 + z^2 = 128. (The absolute max and min exist because the surface is closed and bounded.)

Answer : The problem is to extermize the function f (x, y, z) = xyz^2 subject to the con- straint g(x, y, z) = x^2 + y^2 + z^2 = 128. The Lagrange multiplier equations are ∇f = λ∇g and g = 128. Plugging in the formulas for the partial derivatives this is equivalent to

(11.1) yz^2 = 2λx

(11.2) xz^2 = 2λy

(11.3) 2 xyz = 2λz =⇒ xyz = λz.

(11.4) x^2 + y^2 + z^2 = 128

Now multiply (11.1) by x, (11.2) by y, and (11.3) by z to get

(11.5) xyz^2 = 2λx^2

(11.6) xyz^2 = 2λy^2

(11.7) xyz^2 = λz^2.

Thus, we have

(11.8) 2 λx^2 = 2λy^2 = λz^2.

Consider first the case λ = 0. Then, from (11.1) through (11.3) we must have xz^2 = 0, yz^2 = 0 and xyz = 0. These equations are satisfied if z = 0. Thus, all the points on the circle x^2 + y^2 = 128 in the plane z = 0 are critical points. The value of the function f at these points is zero. If z 6 = 0, we must have x = y = 0. From the constraint, this implies that z^2 = 128, so z = ± 8

The function is zero at these points. If λ 6 = 0, the we can divide (11.8) by λ to get

(11.9) 2 x^2 = 2y^2 = z^2.

Multiplying the constraint equation by 2 gives

2 x^2 + 2y^2 + 2z^2 = 256

Substituting z^2 for 2z^2 and 2y^2 gives 4z^2 = 256, so z = ±8. We then have 2 x^2 = z^2 = 64, so x = ±

2. Similarly, y = ± 4

2. Thus, we get 8 critical points (± 4

2 , ±8). The values of the function at these points are ±256. Thus, the absolute min is −256 and the absolute max is 256. I’ll leave it to the reader to list the points at which these occur.

The smallest value of f at a critical point is 1. Thus, the points on C closest to the origin are (1, 0 , 0) and (0, 1 , 0) at a distance of 1. The largest value of f at a critical point is 1 + (1 +

2)^2. Thus, the point on C that is farthest from the origin is (−

2. at a distance

of

2)^2 ≈ 2 .6131.