Question

Calc 3 derivative question. Please help!
I think it's pretty simple, I'm just confusing myself...

Answer 1

The temperature at a point (x,y,z) is given by\[T(x,y,z) = 2048*2^{-2x^2-3y^2-z^2}\]where T is in degrees Kelvin and x,y,z are in meters.

Answer 2

a) Find the rate of change of the temperature at the point P := (1,-1,2) in the direction towards the point (3,-3,2).

Answer 3

b) In which direction does the temperature increase the fastest at P?

c) Find the maximum rate of increase at P.

Answer 4

So, essentially we want to find the directional derivative which is defined as

\[D _{u}f = (Gradient) . u\]

I would start as rewriting the function as so

T(x,y,z) = 2^(11 - 3x^2 - 3x^y -z^2)

Then take the partials of the function as it will make it much easier.

Is this what you were having trouble on?

Answer 5

That helps a little actually.

Also, just in case you're solving it, the coefficient of the x^2 in the exponent is 2 not 3

Answer 6

So, when you take the partials of the T function, do you multiply the gradient vector by the vector of the 2 points that were given in part A? is that what you meant by (Gradient) * u?

Answer 7

The first partial with respect to x would as follows

There was a typo in the equation above it should be 3y^2 not 3x^y

let u = 11 - 3x^2 - 3y^2 -z^2

then d/du(2^u) = 2^u log(2)

From here we have

log(2)(2^(11 - 3x^2 - 3y^2 -z^2))d/dx(11 - 3x^2 - 3x^y -z^2)


d/dx(11 - 3x^2 - 3y^2 -z^2) = -6x -0 -0

So, you would get

log(2)(2^(11 - 3x^2 - 3y^2 -z^2)(-6x)

From here you just repeat the same process for the other two variables.

Answer 8

No, it is the gradient dotted with the unit vector of (3,-3,2).

Answer 9

So not from point P to (3,-3,2)? Why?

Answer 10

Because you want to evaluate the derivative at (1,-1,2) in the direction of (3,-3,2).

Answer 11

Oh I see, thank you!

Answer 12

No, problem I hope this helps you out :)

Answer 13

Just remember that U is the unit vector of (3,-3,2) so remember to take this vector and divide it by its magnitude.

Answer 14

Ah, I see. Thanks!