Question

The temperature at a point (x, y, z) is given by t(x, y, z) = 200e−x2 − 3y2 − 7z2 where t is measured in °c and x, y, z in meters. (a) find the rate of change of temperature at the point p(4, −1, 4) in the direction towards the point (5, −5, 6).

Answer 1

Looks like the temperature is given by

$t(x,y,z)=200e^{-x^2-3y^2-7z^2}$

We have gradient at any point $(x,y,z)$

$\nabla t(x,y,z)=200e^{-x^2-3y^2-7z^2}(-2x,-6y,-14z)$

Then the rate of change of $t$ at $p$ in the direction of (5, -5, 6) is given by

$\nabla t(4,-1,4)\cdot\dfrac{(5,-5,6)}{\|(5,-5,6)\|}=\left(-\dfrac{400}{e^{131}}(4,-3,28)\right)\cdot\dfrac{(5,-5,6)}{\sqrt{86}}=-\dfrac{40600}{e^{131}}\sqrt{\dfrac2{43}}$

which is very nearly 0.