Hints for Sample Final problem #2
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Problem #2 on the sample final is a tricky problem, but you can do it in
at least three seemingly different ways. (Maybe this is why it's so
tricky!)
(A) You can do it using the directional derivative. This is probably the
method most people are trying, but there's a tricky step at the beginning.
(B) You can do it using the chain rule, but there's a tricky step to set
it up this way.
(C) Once you've figured out the tricky step for (B), you can actually
solve this problem using single variable calculus!
Here are a few comments on each approach.
METHOD A
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Using the given information, you should be able to find the directional
derivative of T(x,y,z) at the point (2,3,1) in the direction of the vector
which goes from (2,3,1) to (3,4,3). BE CAREFUL - you're not done. This
gives you the (instantaneous) change in temperature per change in
kilometer in that direction. You're supposed to give an answer in terms
of change in temperature per second. So you need to figure out how to use
the rest of the given information to get your final answer.
METHOD B
--------
You're trying to find a partial derivative of T, which is a function of x,
y, and z. But these three variables are themselves functions of t (time).
This sounds like the chain rule! [Read the discussion from the bottom of
page 203 through 204, but realize that the notation is a little deceiving;
whenever they say "du/dx" for example, they really mean "du/dx evaluated
at (x(s,t),y(s,t))."] To use the chain rule in our case, you need to do
two things:
(1) Find a parametrization f(t)=(x(t),y(t),z(t)) for the path -- i.e. the
line -- which goes straight from (2,3,1) to (3,4,3) at a constant speed of
5. (Remember, the "speed" is the length of the tangent vector f'(t).)
This is actually pretty tricky.
(2) Set up the chain rule, where you're interested in T(f(t)), so the
derivative is JT(f(t))Jf(t). You're going to want to evaluate this at
whatever value of t gives you f(t)=(2,3,1). Note that T has three inputs
and one output, while f has one input and three outputs, so the chain rule
looks like this:
[dT/dx dT/dy dT/dz ] [ dx/dt ]
[ dy/dt ]
[ dz/dt ]
Which multiplies out to:
dT dT dx dT dy dT dz
-- = -- -- + -- -- + -- --
dt dx dt dy dt dz dt
Remember, the partial derivatives like dT/dx here need to be evaluated at
f(t) here.
METHOD C
--------
Suppose you've found the parametrization f(t) mentioned above. Let's say
f(0)=(2,3,1), although you could have it set up a little differently.
f(t) tells us what x, y, and z are in terms of t, so you could plug that
into T(x,y,z) and get a new function Temp(t) which expresses temperature
in terms of time. Then this is a single variable calculus problem, and
you're looking for Temp'(0), the derivative at 0.
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This sounds like a complicated problem, but in fact you can do it using
method A in a few minutes once you realize the trick at the end. I've
included the other approaches here since a few people have tried them.
The answer, incidentally, is 205/Sqrt[6].
Hope this helps!
Jonathan Rogness (rogness@math.umn.edu)