MATH 263 TEST II solutions

1. Let f(x,y) = e^y cos x + x³y. Compute the quadratic Taylor polynomial, Q(x,y), approximating f near (0,0).

Solution: f_x(x,y) = - e^y sin x + 3x²y

f_y(x,y) = e^y cos x + x³

f_xx(x,y) = - e^y cos x + 6xy

f_xy(x,y) = - e^y sin x +3x²

f_yy(x,y) = e^y cos x

Thus f(0,0) = 1, f_x(0,0) = 0, f_y(0,0) = 1, f_xx(0,0) = -1, f_xy(0,0) = 0, f_yy(0,0) = 1.

So Q(x,y) = f(0,0) + f_x(0,0)x + f_y(0,0)y + (1/2)f_xx(0,0)x² + f_xy(0,0)xy + (1/2)f_yy(0,0)y²

= 1 + y – x²/2 + y²/2

2. Reverse the order of integration of the following iterated integral. Begin by sketching the region of integration.

Solution:

3. Consider the region in the first quadrant bounded by the curve and the x-axis between x = 0 and x = p. (Assume that x and y are measured in cm.) Suppose that the density function of this region is given by d(x,y) = 9y³ grams/sq cm. Find the total mass of the region.

Let R denote the region bounded by the curve and the x-axis between x = 0 and x = p. Then the area of R equals

4. Let F(x,y) = ax² – 2axy + y² – bx, where a, b are constants, and a ≠ 0, a ≠ 1.

(A) Find the critical point of F. (Express your answer in terms of the constants a and b.)

First, compute the gradient of F: grad F = (2ax-2ay-b) i + (-2ax+2y) j

Setting grad F = 0, we obtain: 2ax-2ay-b = 0 and -2ax+2y = 0.

The first equation yields x = since a ≠ 0 and a ≠ 1. The second equation yields y = ax =

Thus the critical point is . Let us call this point Q.

(B) Find the condition on a such that the critical point will be a saddle point.

Computing: f_xx(Q) = 2a, f_xy (Q) = -2a, and f_yy(Q) = 2.

Computing the discriminant, D(Q) = f_xx(Q)f_yy(Q)-(f_xy(Q))² = 4a-4a² = 4a(1-a).

Now, Q is a saddle point when D(Q)<0. Solving 4a(1-a)<0 yields a<0 or a>1.

(C) Find the condition on a such that the critical point will be a local maximum point.

Q will be a local maximum when D>0 and f_xx(Q) < 0. Now D>0 implies that 0<a<1. But f_xx(Q) < 0 implies that a<0. Hence no value of a will produce a local maximum at Q.

5.    Using spherical coordinates, evaluate

 where R is the unit ball centered at the origin.

Solution: Since r² = x² + y² + z², the integrand is just r. The region of integration is given by: 0 < r < 1, 0 < f < p, 0 < q < 2p. Thus, using the appropriate volume element for spherical coordinates:

6.     The path of a particle moving in xyz-space is given by:

s(t) = 6 sin (2t) i + 6 cos (2t) j + 5t k where time is measured in minutes and distance in centimeters.

(A)   Find the velocity, v, of the particle at time t = 1 minute.

(B)   Find the speed of the particle at time t = 1 minute.

(C)   Compute the arc length of this curve from t = 0 to t = p minutes.

Solution: (A) v(t) = s¢(t) = 12 cos (2t) i - 12 sin (2t) j + 5 k

When t = 1, v(1) = 12 cos (2) i - 12 sin (2) j + 5 k centimeters/minute

(B) The speed at time t is given by ||v(t)|| = centimeters/minute

Thus at time t = 1, the speed is 13 cm/min.

(C) The arc length is given by: centimeters.

7. Consider the function g(x,y) = x⁴ + 2x³y – 6x²y² + 7xy³ – y⁴ – 19.

(A) Show that (0,0) is a critical point of g.

grad g = (4x³+6x²y-12xy²+7y³) i + (2x³-12x²y+21xy²-4y³)

Now, at the point (0,0), grad g = 0 i + 0 j = 0. Hence (0,0) is a critical point of g.

(B) Find the cross sections of g(x,y) with x fixed at x = 0, and with y fixed at y = 0.

g(0,y) = – y⁴ – 5 and g(x,0) = x⁴ – 5.

(C) Classify the critical point (0,0) of g.

The second derivative test fails, since D(0,0)=0. However, examining the cross sections near (0,0) reveal that (0,0) is a saddle point. This is the case since along the x-axis g achieves a minimum at (0,0) whereas along the y-axis g achieves a maximum at (0,0).

8. A open rectangular box has volume 108 cm³. What are the lengths of the edges of the box that yield the minimum surface area?

Solution: Let x, y, and z denote the three dimensions of the box, assuming that z is the height. Then the surface area of such an open box is xy + 2yz + 2xz.

Since we are told that xyz=108, we may solve for z to obtain z = 108/(xy).

Thus the surface area may now be expressed as a function of two variables:

S(x,y) = xy + 2y(108/(xy)) + 2x(108/(xy)) = xy + 216/x + 216/y.

To find critical points, we set S_x and S_y equal to 0:

S_x = y – 216/x² = 0

S_y = x – 216/y² = 0.

Thus the first equation yields y = 216/x² and the second equation yields x = 216/y². Plugging the former into the latter equation yields:

x = 216/(216/x²)² = 216 (x²/216)² = x⁴/216.

So 216x = x⁴ from which we obtain: x(x³-216)=0. Hence either x=0 (which we reject) or x = 216^1/4 = 6.

It is easy to check that this is a local and a global minimum. Hence the optimal values for the dimension of the box are: 6 cm x 6 cm x 3 cm.

9. Set up but do not evaluate a triple iterated integral that represents the volume of the tetrahedron having vertices A=(0,0,0), B=(0,0,1), C=(0,1,0), and D=(1,1,0).

Hint: First draw a picture of the solid and then find the equation of the plane that forms the top of the solid.

Solution: The top of the solid is the plane that passes through the points B, C, and D. To find the slope in the x-direction use the points C and D to obtain Dz/Dx = (0-0)/(1-0) = 0.

To find the slope in the y-direction use the points B and C to obtain Dz/Dy = (0-1)/(1-0) = -1.

Hence the top of the solid is given by: z = 1-y

In the xy-plane, the base of the solid is bounded by s=0, y=x, and y=1.

Thus the volume of the tetrahedron is given by

V =

10. Albertine’s Bakery in BetaVille produces two types of chocolate cakes: ordinary and crafted. Each Ordinary Cake requires 0.1 lb of Swiss chocolate, while each Crafted Cake requires 0.2 lb of Swiss chocolate. Currently there are only 233 lb of Swiss chocolate available each month. Suppose that the profit function is given by:

p(x,y) = 150x – 0.2x² + 200y – 0.1y²,

where x is the number of Ordinary cakes and y is the number of Crafted cakes that the bakery produces each month.

How many of each type of chocolate cake should Albertine’s bakery produce each month to maximize profit?

Solution:

The constraint equation is: 0.1x + 0.2y < 233.

Let g(x,y) = x + 2y be the constraint function.

Compute grad p to obtain: grad p = (150-0.4x) i + (200-0.2y) j. Note that if grad p = 0, then x = 150/0.4 = 375 and y = 200/0.2 = 1000. But (375, 1000) is not within our constraint: namely: 0.1(375) + 0.2(1000) = 237.5 > 233. Hence there are no critical points of p within the interior of our constraint triangle. Thus the maximum value of p must occur on the boundary of the triangle. Clearly p = 0 when either x or y equals 0. Hence the maximum must occur on the hypotenuse: 0.1x + 0.2y = 233.

To find this maximum we use Lagrange multipliers:

grad p = λ grad g.

Thus: (150-0.4x) i + (200-0.2y) j = λ(i + 2j)

So: 150-0.4x = λ and 200-0.2y = 2λ.

This yields: 150-0.4x = (200-0.2y)/2. That is: 150-0.4x=100-0.1y. So y = 4x-500.

But, since x+2y=2330, we obtain x+2(4x-500)=2330. Thus x=3330/9 = 370 and y = 980.
Thus optimal profit occurs when Albertine produces 370 ordinary cakes and 980 crafted cakes each month.

11. The function G(x,y) = ax² + 4y has an average value of 5 on the triangle with vertices at (0,0), (1,0) and (0,1). What can you say about the constant a?

Solution:

Let R denote the triangular region above. Note that the area of R is ½.

Since the average value of G over R is 5, we have:

5 =

=

= = 2(a/3 – a/4 + 2 – 2 + 2/3) = a/6 + 4/3 =

Solving for a: a = 6(5) - 8 = 22

12. Consider the iterated integral .
(A) Sketch the region of integration.

(B) Convert this integral to polar coordinates, but do not evaluate.

Solution: Since dA = r dr dθ and r² = x² + y², we have:

13. Consider the integral .

(A) Evaluate the integral.

=

(B) Sketch the region of integration.

(C) Reverse the order of integration (but do not try to evaluate again).

Extra Credit:

Is the point P = (1,2,4) visible from the point Q = (-3,-2,1) if there is an opaque ball of radius 3 centered at the origin?  Justify your answer!

Solution: We may view this as an optimization problem. P is visible from Q if no point (x,y,z) between P and Q is within distance 3 or less from the origin. Note that a point (x,y,z) on the line joining P and Q is between P and Q if and only if x is between -3 and 1. Thus we will attempt to determine the minimum value of the square of the distance from the origin to a point (x,y,z) on the line segment PQ. (Note that (x,y,z) is within distance 3 of the origin if and only if the square of the distance from (x,y,z) to the origin is no greater than 9.

Let f(x,y,z) = square of distance from R = (x,y,z) to the origin = x² + y² + z². The equation of the line passing though P and Q is given by L(t) = Q + t(P-Q) = (-3,-2,1) + t(4,4,3) = (-3+4t, -2+4t, 1+3t). Note that L(0) = Q and L(1) = P. Thus L(t) traces out the line segment from P to Q as t varies from 0 to 1.

Now, as a function of t, f = (-3+4t)² + (-2+4t)² + (1+3t)². This is just a function of one variable. To find its minimum, we search for a critical point:

df/dt = 2(-3+4t)4 + 2(-2+4t)4 + 2(1+3t)3 = 2[(-3+4t)4 + (-2+4t)4 + (1+3t)3] = 2(-12+16t-8+16t+3+9t) = 2(41t – 17).

Thus a minimum value of f occurs when t = 17/41. Since 17/41 lies between 0 and 1, L(17/41) lies on the line segment joining P and Q. Now we must compute

f(17/41) = (-3+68/41)² + (-2+68/41)² + (1+51/41)² < 9. Hence the line segment PQ does intersect the opaque ball of radius 3 centered at the origin. In other words, P is not visible from Q.

Alternative solutions:

A1. Saudur found the distance from the origin to the line joining the points P and Q. Since this distance is 2.6, P is not visible from Q.

A2. Nidhi wrote the line joining P and Q parametrically, and asked if this line intersects the sphere centered at the origin of radius 3.