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

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MATH200 April 2012

• Q1 (a) • Q1 (b) • Q2 (a) • Q2 (b) • Q3 • Q4 (a) • Q4 (b) • Q4 (c) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 • Q8 • Q9 • Q10 •

Other MATH200 Exams

• 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
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.