M408D Final Exam Solutions, December 14, 2002

1. Evaluate the following limits and improper integrals, or write ``DNE'' if the limit does not exist. EXPLAIN YOUR REASONING.


tex2html_wrap_inline125 , which diverges as tex2html_wrap_inline127 , so the answer is DNE.


Since tex2html_wrap_inline131 , this is a 0/0 indeterminate form. By L'Hôpital's rule, the limit is tex2html_wrap_inline133 .


The numerator is bounded while the denominator goes to tex2html_wrap_inline137 , so the limit is zero.

tex2html_wrap_inline139 .

Note that tex2html_wrap_inline141 , and tex2html_wrap_inline143 , which goes to zero.

2. Indicate which of these series converge absolutely, which converge conditionally, and which diverge. EXPLAIN YOUR REASONING.

a) tex2html_wrap_inline145

This is an alternating series that converges. However, tex2html_wrap_inline147 diverges by comparison to tex2html_wrap_inline149 , so the answer is CONVERGES CONDITIONALLY.

b) tex2html_wrap_inline151

This CONVERGES ABSOLUTELY by the ratio test (or by the root test).

c) tex2html_wrap_inline153

This DIVERGES by the integral test (compare to problem 1a).

d) tex2html_wrap_inline155

This CONVERGES ABSOLUTELY by the ratio (or root) test.

3. Consider the infinite power series tex2html_wrap_inline157 .

a) Find the radius of convergence.

By the root test, the radius of convergence is tex2html_wrap_inline159 .

b) Use the series to estimate f(0.2). Your answer should be good to 3 decimal places.

tex2html_wrap_inline163 .

c) Estimate tex2html_wrap_inline165 to 3 decimal places.

Two terms are enough: tex2html_wrap_inline167 .

d) Estimate f'(0.1) to 3 decimal places.


4. Let L be the line through the points (0,1,2) and (3,2,3). Let P be the plane containing the origin and the line L. Let Q=(5,1,2).

a) Find the equation for the line L.

d = (3,2,2)-(0,1,2) = (3,1,1), so our line is tex2html_wrap_inline189 .

b) Find the equation for the plane P.

We want the plane through (0,0,0), (0,1,2) and (3,2,3). The normal vector is tex2html_wrap_inline193 , so the plane is -x + 6y - 3z=0.

c) Find the distance from Q to L.


d) Find the distance from Q to P.


5. Consider the parametrized curve tex2html_wrap_inline209

a) Find the velocity (vector) at time t=0.

tex2html_wrap_inline213 , so tex2html_wrap_inline215 .

b) Find the speed as a function of time.

The speed is tex2html_wrap_inline217 .

c) Find the distance traveled (i.e., arclength) from t=-1 to t=1.

Distance = tex2html_wrap_inline223 speed tex2html_wrap_inline225 .

6. a) The width of a rectangle is increasing at a rate of 0.2 meters/second, while the height is increasing at a rate of 1 meter/second. At what rate is the area increasing when the height is 10 meters and the width is 0.5 meters?

A = wh, where A is area, w is width and h is height. tex2html_wrap_inline229 square meters per second.

b) A bird is flying through a forest fire at 6 meters/second, in the direction (2,2,1). The temperature as a function of position is tex2html_wrap_inline231 . At what rate is the temperature changing when the bird passes the point (5,20,10)?

Velocity is a multiple of (2,2,1), so v=k(2,2,1) for some constant k. Since tex2html_wrap_inline237 , we have k=2, so v=(4,4,2).


7. Consider the surface tex2html_wrap_inline245 .

a) Find the equation of the plane tangent to this surface at the point (1,1,1).

The normal vector is tex2html_wrap_inline249 , so our plane is 2x+2y+5z=9.

b) Find the equation of the line normal to this surface at the point (1,1,1).

Since d=(2,2,5), our line is tex2html_wrap_inline257 .

8. Consider the function of two variables: tex2html_wrap_inline259 .

a) Find all critical points of this function.

tex2html_wrap_inline261 and tex2html_wrap_inline263 . Setting these equal to zero gives x=y=0. This is the only critical point.

b) Use the second derivative test to determine which critical points are local maxima, which are local minima, and which are saddle points.

tex2html_wrap_inline267tex2html_wrap_inline269 , and tex2html_wrap_inline271 , so we are at a local MINIMUM.

9. Find the maximum and minimum values of the function tex2html_wrap_inline273 on the unit sphere tex2html_wrap_inline275 . Where do these maximum and minimum values occur?

tex2html_wrap_inline277 and tex2html_wrap_inline279 , so our equations are


The solutions are tex2html_wrap_inline283 , (-1, 0, 0, 1), tex2html_wrap_inline287tex2html_wrap_inline289tex2html_wrap_inline291 , and tex2html_wrap_inline293 . The first two have f=1, the next two have f=2, and the last two have f=-2, so the last two are minima, the middle two are maxima, and the first two are neither. That is, the maximum value is 2, and is achieved at tex2html_wrap_inline301 , while the minimum value is 2, and is achieved at tex2html_wrap_inline303 .

10. Let T be the triangle in the x-y plane with vertices (0,0), (2,0) and (1,1). We are interested in the double integral tex2html_wrap_inline317 .

a) Write tex2html_wrap_inline317 as an iterated integral, where you integrate first over x and then over y. (Do not evaluate yet)


b) Break T up into two smaller triangles, tex2html_wrap_inline329 with vertices at (0,0), (1,0) and (1,1), and tex2html_wrap_inline337 with vertices at (1,0), (2,0) and (1,1). Write tex2html_wrap_inline345 as an iterated integral, where you integrate first over y and then over x. Do the same for tex2html_wrap_inline351 .

tex2html_wrap_inline353 and tex2html_wrap_inline355 .

c) Compute tex2html_wrap_inline317 , either by doing the iterated integral you wrote down in (a) or the ones you wrote down in (b).

In (a), tex2html_wrap_inline359 , so our outer integral is tex2html_wrap_inline361 .

Lorenzo Sadun

Sat Dec 14 09:45:28 CST 2002