8/9/2019 Practice Midterm 2 With Solutions

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8/9/2019 Practice Midterm 2 With Solutions

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Problem 4. Suppose you have

z= x2 +xy, x= sin t, y = et.

Computedz/dt.Solution:

Using the chain rule,dz

dt =

dz

dx

dx

dt +

dz

dy

dy

dt.

Butdz/dx= 2x+y,dz/dy= x, and dx/dt= cos t, dy/dt= et. Putting this together,

dz

dt = (2x+y)(cos t) + (x)(et).

It would be good form to now substitute in for x and y :

dz

dt = (2sin t+et)(cos t) + (sin t)(et).

Problem 5. Suppose you havef(x, y) =y2/x.

(a) Computef.(b) Evaluatefat (1, 2).(c) What is the slope in the direction = /4? How about the direction of the vector 2, 1?(d) What is the maximum slope at this point, and in what direction does it occur (note:

you can do this part without any dot products!).

Solution: We havef= y2/x2, 2y/x.

Evaluating at (1, 2), we get4, 4. Now notice that =/4 is the direction of1, 1, butthis dot product will be4 + 4 = 0, so 0 is the slope in that direction. In the direction of2, 1, normalized to 2/5, 1/5, we have 8/5 + 4/5 = 4/5. The maximum slopeis given by|f| = 16 + 16 = 42. It occurs in the direction off= 4, 4.Problem 6. Lets classify some critical points:

f(x, y) =x

2

+y

2

4xy.(a) Find all critical points off(well ok, theres only 1).

(b) Compute fxx,fxy =fyx, andfyy at this point.

(c) Use part (b) to determine whether the point is a maximum, minimum, or saddle point.

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Solution: Taking partials, a critical point would have x= 2y, and y = 2x. This gives only(0, 0). fxx = 2,fyy = 2, andfxy = 4 (for all (x, y)), so we get fxxfyy f2xy = 4 16 = 12,which is negative, meaning that (0, 0) is a saddle point.Problem 7. Lets find the absolute minimum and maximum of a function on a boundedregion: let

f(x, y) =xy2.

and consider the region

D= {(x, y)|x 0, y 0, x2 +y2 3}.

(a) Draw the region D.

(b) Check for critical points on the interior ofD; compute fat any ones that you find.

(c) Break the boundary ofD into three lines: L1, where y = 0, L2, where x = 0, and L3,wherex2 +y2 = 3. Write fas a function of one variable along each of these lines.

(d) Find the maximum and minimum values off along L1, L2,L3.

(e) Check the value offat the three corner points.

(f) What are the absolute minimum and maximum values offonD? Where do they occur?

Solution: Iffx = y2 = 0, then y = 0. Thus there are no critical points on the interior.

Along the line y = 0, we have f(x, 0) = 0. Along x = 0, we have f(0, y) = 0. Along thethird line, we have y =

3 x2, and we get f(x,3 x2 = 3x x2 for 0x 3. The

only critical point of this is x = 1, and f(1,

2) = 2, the maximum value. The minimumvalue of 0 occurs when x = 0, or when y = 0, or when x =

3 alongL3.

Problem 8. Find the points on the sphere

x2 +y2 +z2 = 4

which are closest to and farthest from the point (3, 1,1) via the following steps:(a) Write down the equation ford the distance between (x,y,z) and (3, 1,1).(b) Remember the trick for maximizing and minimizing distances: its easier to maximize

or minimized2!

(c) Write down your constraint equation,g, which comes from the fact that you want (x,y,z)

to be on the sphere.

(d) Write down the system of equations you get from(d2) =g.(e) Solve for , and use this to find the (x,y,z). If you feel like it, you could plug these

points intodto find which is the closest and which is the farthest.

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Solution: We haved=

(x 3)2 + (y 1)2 + (z+ 1)2.

Squaringd gets rid of the square root. Our equations are

2(x

3), 2(y

1), 2(z+ 1)

=

2x, 2y, 2z

x2 +y2 +z2 = 4.

Solving gives

x= 3

1 , y= 1

1 , z= 11

and from there

= 1

11

2 .

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