## Second Midterm-Practice test

You know the drill: work these problems and let me know if you have any trouble. I will try to answer some questions until Monday evening (Oct 10). This is a good opportunity to compare notes and work with other students. Enjoy!

1. Find three positive numbers whose sum is 12 and the sum of whose squares is as small as possible.
2. The base of an aquarium with volume 9,000cm3 is made of slate and the sides are made of glass. If slate costs five times as much as glass (per square centimeter), find the dimensions of the aquarium that minimizes the cost of the materials.
3. If $u=f(x,y),$ where $x=e^s \cos t$ and $e^s \sin t,$ show that
$\displaystyle{\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = e^{-2s} \Big( \frac{\partial^2 u}{\partial s^2} + \frac{\partial^2 u}{\partial t^2} \Big)}$
4. If $z=f(x,y),$ where $x=s+t$ and $y =s-t,$ show that
$\displaystyle{ \Big( \frac{\partial z}{\partial x} \Big)^2 - \Big( \frac{\partial z}{\partial y} \Big)^2 = \frac{\partial z}{\partial s}\, \frac{\partial z}{\partial t} }$
5. Suppose that over a certain region of space the electrical potential $V$ is given by $V(x,y,z) = 5x^2 - 3xy + xyz.$
• Find the rate of change of the potential at the point $P=(3,4,5)$ in the direction of the vector $\boldsymbol{v} = \boldsymbol{i} + \boldsymbol{j} - \boldsymbol{k}.$
• In which direction does $V$ change most rapidly at $P?$
• What is the maximum rate of change at $P?$
6. Find the directional derivative of $f(x,y) = \sqrt{xy\strut}$ at $P=(2,8)$ in the direction of $Q=(5,4).$
7. Find the local maximum and minimum values and saddle point of the function $f(x,y) = x^2 y e^{-x^2-y^2}.$
8. Find the local maximum and minimum values and saddle point of the function $f(x,y) = \sin x \sin y$ in the square $-\pi \leq x \leq \pi, -\pi \leq y \leq \pi.$
9. Find the absolute maximum and minimum values of $f(x,y)=xy^2$ in the set $D= \big\{ (x,y) : x\geq 0, y\geq 0, x^2+y^2 \leq 3 \big\}.$
10. Find and sketch the domain of the following functions
$f(x,y) = \ln (9-x^2-9y^2), f(x,y) = \sqrt{y} + \sqrt{25-x^2-y^2}, f(x,y) =\displaystyle{\frac{\sqrt{y-x^2}}{1-x^2}}$
11. Find the following limits if they exist, or show that they do not exist:
$\displaystyle{\lim_{(x,y)\to(0,0)} \frac{x^2ye^y}{x^4+4y^2}, \lim_{(x,y)\to(0,0)} \frac{6x^3y}{2x^4+y^4}. }$
12. Find all the second partial derivatives of $v = \displaystyle{\frac{xy}{x-y}}.$
13. Verify that the function $u=(x^2+y^2+z^2)^{-1/2}$ is a solution of the three-dimensional Laplace equation
$\displaystyle{\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = 0}$
14. Find an equation of the tangent plane to the surface $z=y\cos(x-y)$ at the point $(2,2,2).$
15. Use differentials to estimate the amount of tin in a closed tin can with diameter 8cm and height 12cm if the tin is 0.04cm thick.
16. if $z=x^2-xy+3y^2$ and $(x,y)$ changes from $(-3,-1)$ to $(2.96, -0.95),$ compare the values of $\Delta z$ and $dz.$
17. Four positive numbers, each less than 50, are rounded to the first decimal place and then multiplied together. Use differentials to estimate the maximum possible error in the computed product that might result from the rounding.
1. October 7, 2011 at 11:52 am

I need help with number 3 and number 4. How do you get the second partials with the s and t variables?

• October 7, 2011 at 1:18 pm

Look at example 7 in page 905

• October 8, 2011 at 6:16 pm

Hey, Nate. For number 4 this is what did:

The partial derivative of z with respect to s= the partial derivative of z with respect to x times the partial derivative of x with repsect to s plus the partial derviative of z with respect to y times the partial derivative of y with respect to s. The partial derivative of x with respect to s = 1, and the partial derivative of y with respect to s =1. Therefore, the partial derivative of z with repsect to s = the partial dervative of z with respect to x + the partial derivative of z with respect to y. Next, the partial derivative of z with respect to t =the partial derivative of z with repsect to x time the partial derivative of x with repsect to t plus the partial derivative of z with respect to y times the partial derivative of y with respect to t. The partial derivative of x with repsect to t equals 1 and the partial derivative of y with repsect to t= -1. Thus, the partial derivative of z with respect to t = the partial derivative of z with respect to x – the partial derivitive of z with repsect to y. Therefore the partial derivative of z with respect to s times the partial derivative of t with respect to t= (the partial derivative of z with respect to x + the partial derivative of z with repsect to y) times (the partial derivative of z with respect to x – the partial derivative of z with repsect to y). When you foil it, you get (the partial derivative of z with respect to x)^2 – (the partial derivative of z with respect to y)^2. Hence, you have shown what the question wants you to show.

p.s. sorry i had to write everything out in words, but i have no clue how to type it to make it look nice.

• October 10, 2011 at 10:14 am

Lauretta, thanks a lot for the help. Something that might make solutions easier to read is using for example Maple’s notation instead of words. Something like “diff(f,x,x)” is easier to interpret, for example.

Y’all are doing a good job, keep it up!

2. October 8, 2011 at 5:09 pm

Hello. For number one I got x=4, y=4, and z=4, which I think is correct. Is that correct? Also, I used the Lagrange method. How would you go about solving the problem if it said that the sum of the squares was a large as possible?

• October 9, 2011 at 6:28 am

Did you take into account all the different constraints?
$x+y+z=12,$
$x,y,z\geq 0.$

With only the first constraint you may apply the method of Lagrange multipliers. Note how it gave you only one solution for this particular problem. This is very significative, right? It means that, with only one of the constraints of this problem, there is only one critical point that satisfies the solution. This point might be either a max or a min—did you check that it is a min? (it is)

It also means that (with only one constraint) there is no max: you can find three numbers that add up to 12 and the sum of whose squares is as big as you want.

We need to use the other constraints of the problem: if all of the variables are to be non-negative, then the maximum is attained at $(12,0,0), (0,12,0)$ or $(0,0,12).$

We did in class a problem where the method of Lagrange multipliers gave us four critical points: two were min and two were max. If the method gives you only one, then you only have one kind.

Good question!

3. October 8, 2011 at 5:12 pm

Also, in number two we are assuming the aquarium is in the shape of a rectangular prism, correct?

• October 9, 2011 at 6:10 am

Yes

4. October 8, 2011 at 5:40 pm

For #2, if x and y represent the dimension of the base of the aquarium and z represents the height of the aquarium, I got that x= (3600)^(1/3), y=(3600)^(1/3), and z= (5/2)(3600)^(1/3). Are those the correct dimensions?

• October 9, 2011 at 5:11 pm

For #2 I’m having problems with solving the system of equations. My constraint function is g(x,y,z) = xyz – 9000 and my function to minimize is f(x,y,z) = 2xz + 2yz + 5xy

Are these correct functions to start with? The reason I ask is that I get some crazy system of equations that is really difficult to solve. For example, my first equation is yz / (2z + 5y) = lamda. The other three look the same. I’m having trouble solving for x, y, and z with these equations.

• October 9, 2011 at 5:38 pm

It looks good, except that it should be, for example the first equation, $2z+5y=yz\lambda.$ (it looks like you placed the lambda in the wrong side of the equation). If you consider now the first two equations and equal the lambdas, you should get (after simplifying) something like $x=y.$ Go from there.

• October 9, 2011 at 6:08 pm

Aaah, you are right. I switched my two gradient vectors. Thanks!

• October 10, 2011 at 12:01 pm

I got the exact same answer…

5. October 8, 2011 at 6:48 pm

For #5, I got
1) 32/(3^(1/3))
2)
3) 1624^(1/2)

• October 9, 2011 at 5:50 pm

For this one I actually got
1) 50 / 3^(1/2)
2)
3) 2164^(1/2)

• October 9, 2011 at 5:51 pm

Hah, I think putting vectors in angle brackets is making the blog think that we’re using html tags… Anyway, for part 2 my vector is
(38, 24, 12)

• October 10, 2011 at 9:57 am

I got

1) 32/(3^(1/2))
2) <38,6,12
3) 1624^(1/2)

6. October 8, 2011 at 6:49 pm

Also, is there an answer key anywhere so I don’t have to keep asking you if my answers are right?

• October 9, 2011 at 6:10 am

Nope. I want y’all to obtain the correct answers through “networking.” It is part of the learning process. I will only offer hints in the right direction.

7. October 8, 2011 at 6:59 pm

Is the answer to #6 2/5 ?

• October 10, 2011 at 10:05 am

That’s what I got.

8. October 8, 2011 at 7:49 pm

When you are finding local maxima and minima and saddle points, what do you do if the second partial derivative of the function with respect to x twice of a critical point = 0?

• October 9, 2011 at 6:07 am

That’s a good question. If $\tfrac{\partial^2 z}{\partial x^2}(a,b)=0,$ then most probably your Hessian $D(a,b)$ is negative (substitute in the formula to realize this fact). This means that you have a saddle point.

The other possibility is that the Hessian is zero, and you have no means to decide the kind of critical point with this method. We use then other methods.

9. October 9, 2011 at 12:13 pm

For number 8 I found 11 critical points. Am I on the right track?

• October 9, 2011 at 5:33 pm

There are many critical points, yes. Which ones did you find?

• October 9, 2011 at 6:07 pm

Yeah, I actually found 13 critical points. I got
(pi/2, pi/2) (pi/2, -pi/2)
(-pi/2, pi/2) (-pi/2, -pi/2)
(pi, pi) (pi, -pi) (pi, 0)
(-pi, pi) (-pi, -pi) (-pi, 0)
(0, pi) (0, -pi) (0, 0)

10. October 9, 2011 at 7:47 pm

for number 5, what exactly is it asking in part 2? im kind of confused

11. October 9, 2011 at 9:47 pm

Guys…i have a bunch of problems…i cannot figure this out…the second partials on numbers 3 and 4…i also dont understand d^2x/dxdy or d^2y/dydx…i need some major help

12. October 10, 2011 at 9:44 am

On number 9, how do we find the boundary given by x^2+ y^2 ≤ 3? I got two of the boundaries since we can say that the max y value will happen when x is 0 and the max x value will happen when y is 0, so our two boundaries are (0,0) to (root(3), 0) and (0,0) to (0, root(3)), however since the last boundary occurs on a curve and not a straight line, how do we account for this?

Also, when we find critical points where all points of the form (0, a) are critical points and the Hessian comes out to 0 such as in number 7, how do we conclude this? If you look at the function for number 7 you can see that the points (0, a) are minima, but they all occur in a straight line, so are they still local minima or are they not critical points at all?

• October 10, 2011 at 9:48 am

Good questions too!

About the first one: That part of the boundary is a piece of a circle, so you can parametrize it as $h(\theta) = \big( \cos \theta, \sin \theta \big).$ You just have to find what values of $\theta$ give you that piece of the circle. Go from there.

About the second part: good job at realizing that the points of the form $(0,a)$ are indeed critical points. Since the second derivative test doesn’t give you information, we need to do something different. This one is specially tricky. Some of the points are going to be maxima, and some others are going to be minima! Graph the function (with WolframAlpha, or Maple or whatever) to realize it, and see how you can prove it algebraically.

13. October 10, 2011 at 10:13 am

For #7 I keep getting that y(x^2) = 0. Wouldn’t this yield an infinite number of critical points when either x, y, or both are 0?

• October 10, 2011 at 10:19 am

Remember to combine the “zeros” with both equations. The first equation will give you three possibilities (something like $x=0, y=0$ or $x=\pm 1.)$. The second equation will give you some other possibilities… but then you have to combine them all: (1) the first possibility from the first equation, with the first possibility of the second equation; (2) the first from the first, with the second of the second; (3) first from first, third from second; etc.

Note that at most, you will have to make eight different combinations, since the “degrees” of your expressions are 4 and 4.

14. October 10, 2011 at 10:36 am

Are both of the limits in #11 DNE?

• October 10, 2011 at 4:05 pm

I got zero for the first limit and DNE for the second limit.

15. October 10, 2011 at 12:42 pm

Would the last problem be equal to 50^4 – 49.95^4 ?

16. October 10, 2011 at 1:32 pm

In number 17, I’m not sure how to set up the problem. Since there are constraints, it seems like it should be a LaGrange Multipliers problem but I don’t see anyway to relate all of the variables in a single equation. Also, for the equation that we need to maximize, I set it up as

F(w,x,y,z) = (w+0.05)(x+0.05)(y+0.05)(z+0.05) – wxyz

I did this because 0.05 is the maximum error per number that could be caused by rounding to the first decimal place but I’m not sure if this is correct.

• October 10, 2011 at 5:03 pm

You’re going about number 17 the wrong way. You use differentials to solve this one.

Your equation is f(w,x,y,z) = wxyz So the first thing to do is calculate the differential. The differential is
df = xyz(dw) + wyz(dx) + wxz(dy) + wxy(dz)
When you’re using differentials to estimate df you simply use dx = delta(x), etc…

The tricky part to this problem is what to do with the actually parameters he gave you. You’re definitely right about the maximum error. It would be 0.05. But what about the “less than 50” part? Well… the question just asks what the maximum error could be…

Think about it this way… Four numbers multiplied together… would your error be very big if, say, all the numbers were really small (like less than 1)? No, you’d have a really small error. The difference between multiplying 0.90^4 and 0.95^4 is not that big… So… what about if the four numbers were really big? The difference between 49.90^4 and 49.95^4 is pretty significant. So if we want to find the MAXIMUM POSSIBLE ERROR, then we should consider a number as close to 50 as we can get.

So then you just take your differential equation (above) and substitute in w = x = y = z = 50 and dw = dx = dy = dz = 0.05 .

The answer I got was 25000 as the maximum possible error.

• October 10, 2011 at 6:12 pm

Thanks, that makes a lot more sense! I got the same answer once I worked it out.

Also, for number 11 you said that you got 0 for the first limit and DNE for the second limit.
I got DNE for both of them. I plugged in the polar coordinates and got 0/[4*sin^2(theta)]. So in most cases the limit is 0, however in the direction of theta = 0, pi, -pi and so forth wouldn’t you get 0/0? Maybe I interpreted this wrong but I just wanted to see what your thoughts were on it.

• October 10, 2011 at 8:36 pm

Yeah, you’re exactly right. I totally forgot to take into account when sin^2theta was zero. Good catch.

17. October 10, 2011 at 2:58 pm

Can anyone help me with number 13?

• October 10, 2011 at 5:58 pm

Hey Nate, I figured this one out earlier after I left the tutoring center. The best way to work this one out is to make sure you always keep the (x^2 + y^2 + z^2)^(-1/2) and anything similar to it as just a negative exponent and not try to move it around.

For starters, just look at the partial of u for x. You should get something like this:

(-1/2) * (x^2 + y^2 + z^2)^(-3/2) * (2x)

Now take the second partial of u for x of this using the product rule. Leave the (x^2 + y^2 + z^2) stuff alone for the moment but simplify the rest of it and you should get something like this:

3x^2 * (x^2 + y^2 + z^2)^(-5/2) – (x^2 + y^2 + z^2)^(-3/2)

Notice at this point that all of the second partials will look exactly like this. The only difference is that the second differential for y will have 3y^2 at the beginning instead of 3x^2 and the z will have 3z^2. Put all three of these second partials together and you will notice that each of the parts of this whole equation has (x^2 + y^2 + z^2) to some sort of fractional exponent. Since the base is the same for all of them, you can go ahead and pull out (x^2 + y^2 + z^2)^(-5/2). What this means for the ones that are raised to the (-3/2) is that they are now raised to the 1st power since if you were to multiply them out and add exponents then (-5/2) + 1 = (-3/2). So now you have two sets of equations multiplied by each other and only one of them needs to equal 0, so you can ignore (x^2 + y^2 + z^2)^(-5/2). Now if you combine everything together in the other equation and simplify as much as possible you should be able to cancel out the rest of the equation.