Math 126 Homework hints
Note: As I’m sure you’ve all seen in section, I make mistakes too! I can’t guarantee that everything
here will be right 100% of the time. If you think there’s an error in something, please e-mail me
and I’ll try to get a correction up. Same goes for any clarification or further questions; if I’ve made
something even more confusing, please e-mail or come to office hours to clear it up!
14.1
Problem 22: If V (x, y) is the electric potential at a point (x, y) in the xy-plane, then the level
curves of V are called equipotential curves because at all points on such a curve the electric potential
is the same. Sketch some equipotential curves if
V (x, y) = p
c
r2
− x2 − y 2
,
where c and r are positive constants.
First off, ignore the physics language. All you need to know is that it’s asking you to sketch
something called the equipotential curves of V , and that equipotential curves are the level sets of
V . So, these curves are just what you get if you set V to be some constant. You can see that doing
this for any fixed constant creates some function in x and y. Say we have some number d and we
want to see what values of x and y satisfy V (x, y) = d. Then we get an equation of the form:
c
d= p
2
r − x2 − y 2
p
d r 2 − x2 − y 2 = c
d2 (r2 − x2 − y 2 ) = c2
Is this type of equation starting to look familiar? They’ll all be the same type of equation depending
on what we plug in for d. What kind of shapes will you be getting if you plot this equation? Match
them up to one of the graphs in the problem statement.
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14.3 Part I
Problem 9: Find the first partial derivatives of the function u = 5xy/z .
Recall the most fundamental aspect of finding partial derivatives: when you’re differentiating with
respect to any given variable, treat the others as constants. I’m going to do this explicitly by
setting x, y, z equal to constants when I compute the derivatives below, but that’s only to illustrate
how you should be visualizing the equations when you’re taking derivatives. Don’t make these
substitutions when you’re actually solving a problem on an exam!
Let’s start with the derivative with respect to x. In this case, we think of y/z just being some
number. In order to illustrate this, let’s pretend that y/z is just some constant c. Now think of u
as the function
u = 5xc
How would you differentiate this normally? It’s the straightforward power rule:
∂u
= 5cx1−c
∂x
Plug our value of c back in and you’re good to go.
The other two are similar, but I’ll just do the trickier case of the partial derivative with respect to
z. Once again, we just need to pretend that x and y are constants. Let’s say x = c and y = d to
illustrate this. Now recall the following derivative:
d z
c = cz log z
dz
We’ll need to combine this with the chain rule to find the answer. First let’s re-write u with the
constants we chose:
u = 4cd/z
We’ll use the derivative formula above to get a factor of 4cd/z log(d/z) in the answer, but through
the chain rule we also need to multiply by the derivative of d/z with respect to z. Do this, plug
x = c and y = d back in, and you’ve got your answer. The derivative with respect to y is the same
as this one, but with an easier chain rule step.
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