MATH 263 TEST I solutions

PART A: Answer all four problems in this section.

1. Find the directional derivative of g(x,y,z) = 3x²y² + 2yz at P = (-1,0,4) in the direction of –i + 3j + 3k.

First, normalize the direction to obtain Next, the gradient of g at (x,y,z),

grad g(x,y) = g_xi + g_yj + g_zk = 6xy²i + (6x²y+2z)j + 2yk.

Evaluating grad g at P yields: grad g(P) = 8j.

Now, to compute the directional derivative of g at P in the direction u, just take the dot product of grad g(P) with u, to obtain:

g_u(P) = , or approximately 5.5

2. A force F = 2i + j – 3k newtons is applied to a spacecraft with velocity vector

v = 3i – j. Express F as a sum of a vector parallel to v and a vector perpendicular to v.

We must resolve F into parallel and perpendicular components relative to v. Toward this end, we begin by normalizing v: .

Hence

Next, .

(You may wish to check that )

3. Let T = f(x,y) be the temperature in degrees Celsius at a point (x,y) in the xy-plane (where x and y are measured in cm.) Suppose that f has the following properties:

f(5,3) = 38, f_x(5,3) = 3, and f_y(5,3) = -2.

Charlotte the spider finds herself at the point R = (5,3).

(A) In which direction should she move to cool off as quickly as possible? (Express your answer in vector form.)

To cool off as quickly as possible, Charlotte should move in the direction of -grad f(5,3) = -(3i – 2j) = -3i + 2j.

(B) What is her rate of cooling in the direction that you found in part (A)? (Use appropriate units.)

Charlotte’s directional derivative in the direction of the gradient of f at (5,3) is just

||grad f(5,3)|| = . Thus her rate of cooling in the direction of the negative gradient is degrees Celsius / cm. This is approximately 3.6 degrees Celsius / cm.

(Her rate of heating equals - degrees Celsius / cm.)

(C) In which direction should she move to remain at the same temperature? (Express your answer in vector form.)

To remain at the same temperature, she should travel in a direction perpendicular to the gradient of f at (5,3). By inspection, we may choose either 2i + 3j, or –2i – 3j.

(D) If Charlotte were to move in the direction i + j, would she be warming up or cooling off? At what rate? (Use appropriate units.)

To compute her directional derivative in the direction of i + j, we first normalize the direction: Next,

f_u(5,3 )=== 0.707 degrees Celsius / cm.

Hence Charlotte would be warming up.

4. Consider the surface:

z = f(x,y) = x cos y – ye^x .

(A) Find the equation of the tangent plane to this surface at the point (0,0).

We begin by computing z_x and z_y, and then evaluating these derivatives at (0,0).

Since, z_x = cos y - ye^x and z_{y =} -x sin y – e^x, f_x(0,0) = 1, f_y(0,0) = -1. Also f(0,0) = 0.

Thus the equation of the tangent plane is: z – 0 = 1(x – 0) – 1(y-0).

Simplifying: z = x – y.

(B) Using your result from part (A), estimate the value of f(0.02, -0.03).

Since T(x,y) = x - y is the best linear approximation of f(x,y) near (0,0), we estimate f(0.02, -0.03) by computing T(0.02, -0.03) = 0.02 – (-0.03) = 0.05.

PART B: Answer any eight of the following nine problems. You may answer all nine for extra credit.

1. Consider the points P = (1,2,1), Q = (3,4,-2) and R = (-1,2,2).

(A) Find a vector of magnitude 5 perpendicular to each of PQ and PR.

Note that PQ = Q-P = 2i + 2j – 3k and PR = R-P = -2i + k.

To find a vector w perpendicular to each of PQ and PR, we compute the cross-product of PQ and PR:

w =

To find a vector of length 5 in the direction w,

(B) Find the equation of the plane which contains the three points P, Q, and R.

The equation of a plane with normal w = 2i + 4j + 4k passing through the point P = (1,2,1) is:

2(x-1) + 4(y-2) + 4(z-1) = 0.

Equivalently: x + 2y + 2z = 7.

(C) Find the angle between the vectors PQ and PR.

We may use either the dot product or the cross product here.

Solution using dot product: and thus degrees.

Solution using cross product: . Solving for T,

We find q = 40.6 deg or 139.4 deg. We choose the obtuse angle since the dot product of PQ and PR is negative.

2. Find the distance between the two planes x + 2y + 6z = 1 and x + 2y + 6z = 10.

Choose a point Q = (1,0,0) on the first plane and a point P = (10,0,0) on the second plane. Noting that the planes are parallel, we have only to compute the distance from P to the first plane. Let n = i + 2j + 6k be a normal to the plane. We must find the length of the projection of PQ onto n. To simplify the calculation, we compute the unit normal

. Note that PQ = Q – P = -9 i. Now the distance from P to the first plane is simply . This is approximately equal to 1.41.

3. Consider the surface cos(x+y) = exp(xz+2). Check that the point Q = (-1,1,2) lies on the surface and, viewing the surface as a level surface of a function of three variables, find the equation of the tangent plane to the surface at Q.

Note that Q satisfies our equation since cos(-1+1) = 1 and exp(-2+2) = 1.

Let f(x,y,z) = cos(x+y) – exp(xz+2). Then grad f (Q) is normal to the tangent plane to our surface cos(x+y)=exp(xz+2) at Q, since our surface is just a level surface of f.

Now grad f = f_x i + f_y j + f_z k =

(-sin(x+y) – z exp(xz + 2)) i + (-sin(x+y)) j + (-x exp(xz+2)) k

Evaluating at Q, we have grad f (Q) = (-sin 0 – 2 exp(0)) i + (-sin(0)) j + (-(-1)exp(0)) k

= -2 i + k.

Thus the equation of the tangent plane to our level surface at Q is given by:

-2(x-(-1)) + 1(z-2) = 0. Simplifying, we have 2x - z + 4 = 0.

4. Given z = f(x,y), x = g(u,v), y = h(u,v), and g(1,2) = 5, h(1,2) = 3, calculate z_u(1,2) in terms of some of the numbers a, b, c, d, e, k, p, q where

f_x(1,2) = a, f_y(1,2) = c, g_u(1,2) = e, h_u(1,2) = p, f_x(5,3) = b, f_y(5,3) = d, g_v(1,2) = k, h_v(1,2) = q.

This is an application of the Chain Rule (version 2): f may be viewed as a function of x and y (namely, f(x,y)) or as a function of u and v (namely, f(g(u,v), h(u,v)) ).

. Since z is evaluated at u=1 and v=2, z_x and z_y should be evaluated at g(1,2)=5 and h(1,2)=3. Thus = be + dp.

5. Let g(x,y) = x(x+y)³. Show that g_xy = g_yx.

This is a computational problem. Some students chose to expand the polynomial in x and y using the binomial theorem, but this is not as easy as simply applying the product rule:

g_x = x 3(x+y)² + 1(x+y)³ = 3x(x+y)² + (x+y)³

g_xy = 3x 2(x+y) + 3(x+y)² = 6x(x+y) + 3(x+y)²

Also:

g_y = x 3(x+y)² = 3x(x+y)²

g_yx = 3[ x 2(x+y) + 1(x+y)²] = 6x(x+y) + 3(x+y)²

Hence g_xy = g_yx

6. The lengths x, y, z of the edges of a closed rectangular box are changing with time.

(A) What is the equation of the volume V of the box? Of the surface area S of the box?

V = xyz

S = 2(xy + yz + xz)

(B) At time = 5 minutes, x = 1 m, y = 2 m, z = 3 m, dx/dt = 1 m/sec, dy/dt = 1 m/sec,

and dz/dt = -3 m/sec. At what rates are V and S changing at t = 5 minutes? (Show your work.)

Using the chain rule (version 1):

= (2)(3)(1) + (1)(3)(1) + (1)(2)(-3)

= 3 m³ / sec.

=

2(2+3)(1)+2(1+3)(1)+2(1+2)(-3) = 0 m³ / sec.

(C) Is the major diagonal of the box increasing or decreasing at t = 5 minutes? (Justify your answer.)

To simplify the computation we will examine the square of the length of the diagonal rather than the diagonal. (Note that the square of the diagonal’s length is increasing if and only if the diagonal is increasing.)

Let f(x,y,z) = x² + y² +z². So

= 2(1)(1) + 2(2)(1) + 2(3)(-3) = -12 < 0. Hence the diagonal is decreasing in length.

7. Let .

(A) Find the domain of g. Explain.

g is defined as long as x² + y² is not zero. Thus the domain of g consists of all points (x,y) in the xy-plane excluding the origin.

(B) Find the range of g. Explain.

Since cos (t) varies from –1 to +1, the range of g consists of the closed interval [0,2].

(C) Describe or draw the level curves of g.

If g is constant, then x² + y² is constant. Thus the level curves consist of circles centered at the origin.

(D) Does the limit of g exist as (x,y) approaches (0,0)? Explain.

No: Consider what happens to g as we approach the origin along the x-axis (y=0).

In this case, . This function does not approach a limit as x tends toward zero, as the following graph illustrates. Notice that cos oscillates ever more wildly as one moves closer to the origin.

8. An experiment to measure the toxicity of formaldehyde yielded the data shown in the table below. The values show the percent, P = f(t,c), of rats surviving an exposure to formaldehyde at a concentration of c (in parts per million, ppm) after t months. Estimate f_t(18,6) and f_c(18,6). Using complete sentences, interpret each answer in terms of formaldehyde toxicity.

Time t (months)

c (ppm) \ t (months)	14	16	18	20	22	24
0	100	100	100	99	97	95
2	100	99	98	97	95	92
6	96	95	93	90	86	80
15	96	93	82	70	58	36

(A) f_t(18,6) = percent per month, or, taking a difference quotient from the left, percent per month. Taking the average of the two quotients yields a better estimate: percent per month.

Meaning: For a fixed formaldehyde concentration of 6 ppm, the percentage survival of a rat population drops by about 1.25% for each additional month of exposure beyond 18 months.

(B) f_c(18,6) = percent per ppm. Taking a difference quotient from the left, percent per month. Taking the average of the two quotients yields a better estimate: percent per ppm.

Meaning: For a fixed exposure time of 18 months, the percentage survival rate of a rat population drops by about 1.23 % for each additional ppm increase in formaldehyde concentration.

9. The energy, E, of a body of mass m moving with speed v is given by the formula

The speed, v, is non-negative and less than the speed of light, c, which is a constant.

(A) Find E_m. What would you expect the sign of E_m to be? Explain!

> 0. As the mass increases for fixed velocity, so does energy.

(B) Find E_v. What would you expect the sign of E_v to be? Explain!

Extra Credit:

To make different people comparable in studies of cardiac output, researchers divide the measured cardiac output by the body surface area to find the cardiac index C;

C = (cardiac output) / (body surface area).

The body surface area B of a person with weight w and height h is approximated by the formula

B = 71.84w^0.425h^0.725,

which gives B in square centimeters when w is measured in kilograms and h in centimeters. You are about to calculate the cardiac index of a person with the following measurements:

Cardiac output: 7 liters / min

Weight: 70 kg

Height: 180 cm.

Which will have a greater effect on the calculation: a 1 kg error in measuring the weight or a 1 cm error in measuring the height? Explain your reasoning! (Use calculus in your solution.)

Since C is proportional to the reciprocal of B, let us first analyze the errors in B resulting from errors in w or h. To simplify computation, let F(w,h) = w^0.425h^0.725_. Using a tangent plane approximation to F = F(w,h) at the point Q = (70, 180) we find:

=

=

Thus in computing B, a 1 kg error in weight will result in a larger error than a 1 cm error in height. Hence, a 1 kg error in weight will have a greater effect upon C than would a 1 cm error in height.