Science:Math Exam Resources/Courses/MATH200/April 2012/Question 04 (b)

MATH200 April 2012

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  •  April 2012  •  December 2011  •  

Question 04 (b)
The temperature at a point (x,y,z) is given by $\displaystyle {}T(x,y,z)=5\exp(-2x^{2}-y^{2}-3z^{2})$ where T is measured in centigrade and x,y,z in metres. In which direction does the temperature decrease most rapidly?

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Hint
Recall that for two vectors $\mathbf {a} \cdot \mathbf {b} =|\mathbf {a} ||\mathbf {b} |\cos \theta .$ How would we choose $\theta$ to make this dot product the smallest? How can we relate this idea to the directional derivative?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For any given unit vector $\displaystyle {}{\widehat {\mathbf {u} }}$, the rate of change of temperature at P in direction $\displaystyle {}{\widehat {\mathbf {u} }}$ is the scalar ${\begin{aligned}D_{\widehat {\mathbf {u} }}T(1,2,-1)&=\nabla T(1,2,-1)\cdot {\widehat {\mathbf {u} }}\\&=\left\vert \nabla T(1,2,-1)\right\vert \,\left\vert {\widehat {\mathbf {u} }}\right\vert \cos(\theta )\\&=\left\vert \nabla T(1,2,-1)\right\vert \cos(\theta )\end{aligned}}$ where θ is the angle between the given vectors. (Recall that $\displaystyle {}\vert {\widehat {\mathbf {u} }}\vert =1$.) For optimum decrease, we should make the directional derivative as small as possible. Since our only degree of freedom is the angle above and we know that $\displaystyle {}-1\leq \cos \theta \leq 1$ we should arrange $\displaystyle {}\cos \theta =-1$ by choosing θ=π, i.e., by making $\displaystyle {}{\widehat {\mathbf {u} }}$ parallel to $-\nabla T(1,2,-1)$: ${\widehat {\mathbf {u} }}={\frac {-\nabla T(1,2,-1)}{\left\vert -\nabla T(1,2,-1)\right\vert }}={\frac {\left\langle 2,2,-3\right\rangle }{\sqrt {17}}}$ This is the direction of most rapid temperature decrease.