Question

The temperature at a point (x, y, z) is given by t(x, y, z) = 200e−x2 − 5y2 − 9z2 where t is measured in °c and x, y, z in meters.

Answer 1

Part A:

Given that

$T(x, y, z) = 200e^{-x^2-5y^2-9z^2} \\ \\ \nabla T=\left( \frac{\partial}{\partial x} +\frac{\partial}{\partial y}+\frac{\partial}{\partial z}\right)T=200e^{-x^2-5y^2-9z^2}\left\ \textless \ -2x,-10y,-18z\right\ \textgreater \ \\ \\ \nabla T(4,-1,4)=200e^{-(4)^2-5(-1)^2-9(4)^2}\left\ \textless \ -2(4),-10(-1),-18(4)\right\ \textgreater \ \\ \\ =200e^{-165}\left\ \textless \ -8,10,-72\right\ \textgreater$

For the direction, let u be unit vector of point (6, -3, 6), then

$u= \frac{1}{||\left\ \textless \ 6,-3,6\right\ \textgreater \ ||} \left\ \textless \ 6,-3,6\right\ \textgreater \ \\ \\ = \frac{1}{\sqrt{6^2+(-3)^2+6^2}} \left\ \textless \ 6,-3,6\right\ \textgreater \ \\ \\ = \frac{1}{\sqrt{81}} \left\ \textless \ 6,-3,6\right\ \textgreater \ = \frac{1}{9} \left\ \textless \ 6,-3,6\right\ \textgreater \ \\ \\ =\frac{1}{3} \left\ \textless \ 2,-1,2\right\ \textgreater$

Thus, the directional derivative is given by:

$D_u(4,-1,4)=200e^{-165}\left\ \textless \ -8,10,-72\right\ \textgreater \ \cdot\frac{1}{3} \left\ \textless \ 2,-1,2\right\ \textgreater \ \\ \\ = \frac{200}{3} e^{-165}(-16-10-144)=-\frac{170(200)}{3} e^{-165}\approx-2.49\times10^{-68}$



Part B:

The gradient direction is the direction of fastest increase. The direction that the temperature increase at the fastest rate is at the direction of the gradient vector. 
From part A, the gradient vector is given by:
$\nabla T=200e^{-165}\left\ \textless \ -8,10,-72\right\ \textgreater$
Thus, the direction with the fastest rate is given by:

$\frac{1}{||\left\ \textless \ -8,10,-72\right\ \textgreater \ ||} \left\ \textless \ -8,10,-72\right\ \textgreater \ \\ \\ =\frac{1}{\sqrt{(-8)^2+10^2+(-72)^2}}} \left\ \textless \ -8,10,-72\right\ \textgreater \ \\ \\ = \frac{1}{\sqrt{5348}} \left\ \textless \ -8,10,-72\right\ \textgreater \ = \frac{1}{2\sqrt{1337}} \left\ \textless \ -8,10,-72\right\ \textgreater \ \\ \\ =\frac{1}{\sqrt{1337}} \left\ \textless \ -4,5,-36\right\ \textgreater$



Part C:

The magnitude of the maximum rate of increase is the length of the gradient vector. This is given by
$200e^{-165}\left(\sqrt{(-8)^2+10^2+(-72)^2}\right)=200\sqrt{5348}e^{-165} \\ \\ =100\sqrt{1337}e^{-165}$