Science:Math Exam Resources/Courses/MATH200/April 2012/Question 04 (c)/Solution 1

When T is decreasing most rapidly from its value at P, the actual rate of change (a scalar) is based on the direction chosen in part (b):

${\begin{aligned}D_{\widehat {\mathbf {u} }}T(1,2,-1)&=\nabla T(1,2,-1)\cdot {\frac {-\nabla T(1,2,-1)}{\left\vert -\nabla T(1,2,-1)\right\vert }}\\&={\frac {-\left\vert {\nabla T(1,2,-1)}\right\vert ^{2}}{\left\vert \nabla T(1,2,-1)\right\vert }}\\&=-\left\vert {\nabla T(1,2,-1)}\right\vert \\&=-10e^{-9}{\sqrt {17}}.\end{aligned}}$

This has units of $\displaystyle {}{\text{C}}/{\text{m}}$.

In plain English, the words can absorb the minus sign: The maximum rate of decrease of temperature at point P is

$10e^{-9}{\sqrt {17}}{\text{C}}/{\text{m}}$

obtained in the direction $\left\langle 2,2,-3\right\rangle /{\sqrt {17}}$.