Question

The temperature at a point (x, y, z) is given by:
T(x, y, z) = 400e^(−x^2 − 5y^2 − 7z^2)

Find the rate of change of temperature at the point
P(2, −1, 5) in the direction towards the point (6, −5, 6).

Answer 1

What is your guess?

Answer 2

I know I have to find the gradient of T and plug in P. I'm not sure if I'm taking the partial derivative right because my answer isn't right :(

Answer 3

so what did you get for
\[(T_x,T_y,T_z)\]

Answer 4

I will do two of them I will let you do T_z
\[T_x=(-x^2)'400 e^{-x^2-5y^2-7z^2}=-2x \cdot 400e^{-x^2-5y^2-7z^2} \\ T_x=-800xe^{-x^2-5y^2-7z^2} \\ T_y=(-5y^2)'400e^{-x^2-5y^2-7z^2}=-10y \cdot 400e^{-x^2-5y^2-7z^2} \\ T_y=-4000ye^{-x^2-5y^2-7x^2}\]

Answer 5

\[-800xe ^{-x^2-5y^2-7z^2},-4000ye ^{-x^2-5y^2-7z^2},-5600^{-x^2-5y^2-7z^2}>\]

Answer 6

That's what I got!

Answer 7

\[T_x|_{(2,-1,5)}=-800(2)e^{-2^2-5(-1)^2-7(5)^2}=-1600e^{-4-5-175} =-1600e^{-184}\]
eww that number is ugly...
I would do the same for T_y and T_z
plug in (2,-1,5)
\[T_y|_{(2,-1,5)}=-4000(-1)e^{-184}=4000e^{-184}\]

Answer 8

looks good except on the z part
but I think you just left ze^ thing out on accident

Answer 9

\[T_z|_{(2,-1,5)}=-5600(5)e^{-184}=-28000e^{-184}\]

Answer 10

Oh, calculated Tz wrong.

Now I need to.. Plug everything in? Is that it? Wow these numbers are just hideous.

Answer 11

Ok so so far we have:
\[\text{ directional derivative is } \\ <-1600e^{-184},4000e^{-184},-28000e^{-184}> \cdot \frac{<6,-5,6>}{||<6,-5,6>||}\]

Answer 12

do you agree with that?

Answer 13

\[||<6,-5,6>|| \text{ is meant to mean } \sqrt{6^2+(-5)^2+6^2}\]

Answer 14

@jpark are you there
I'm just wondering if our solution so far makes sense
we aren't entirely done

Answer 15

This website didn't load for a long time, I'm sorry about that!

Answer 16

But yes, that's what I got after I just did it.

Answer 17

lol your comment in between my comments didn't show up until now

Answer 18

The website is taking ages to load... :(

Answer 19

But what else needs to happen after that?

Answer 20

have you found the dot product of the vectors above?

Answer 21

\[\text{ directional derivative is } \\ <-1600e^{-184},4000e^{-184},-28000e^{-184}> \cdot \frac{<6,-5,6>}{||<6,-5,6>||} \\ =-1600e^{-184} \cdot \frac{6}{\sqrt{97}}+4000e^{-184} \cdot \frac{-5}{\sqrt{97}} +-28000e^{-184} \cdot \frac{6}{ \sqrt{97}}\]