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

MATH200 April 2012

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  •  April 2012  •  December 2011  •  

Question 04 (a)
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. Find the rate of change of temperature at the point P(1,2,-1) in the direction toward the point (1,1,0).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that the directional derivative of a scalar function f in the direction $\mathbf {u}$ is $D_{u}f=\nabla {f}\cdot {\hat {\mathbf {u} }}$ where ${\hat {\mathbf {u} }}$ is a unit vector.

Hint 2
What is the direction vector $\mathbf {u}$ that we need?

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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 $\displaystyle {}T(x,y,z)=5e^{-2x^{2}-y^{2}-3z^{2}}$, the gradient is ${\begin{aligned}\nabla T(x,y,z)&=5e^{-2x^{2}-y^{2}-3z^{2}}\left\langle {-4x,-2y,-6z}\right\rangle \\&=10e^{-2x^{2}-y^{2}-3z^{2}}\left\langle {-2x,-y,-3z}\right\rangle \end{aligned}}$ At the point $\displaystyle {}P(1,2,-1)$, we have $\nabla T(1,2,-1)=10e^{-9}\left\langle {-2,-2,3}\right\rangle .$ Each component has units of $\displaystyle {}{\text{C}}/{\text{m}}$. The vector from $\displaystyle {}P(1,2,-1)$ to $\displaystyle {}Q(1,1,0)$ is $\displaystyle {}\left\langle {0,-1,1}\right\rangle$ A unit vector in this direction is ${\widehat {\mathbf {u} }}={\frac {\left\langle {0,-1,1}\right\rangle }{\left\vert \left\langle {0,-1,1}\right\rangle \right\vert }}={\frac {1}{\sqrt {2}}}\left\langle {0,-1,1}\right\rangle$ Each component is dimensionless because the units of the vector $\mathbf {u}$ and its magnitude are the same. The rate of change of T at P in direction $\displaystyle {}{\widehat {\mathbf {u} }}$ is ${\begin{aligned}D_{\widehat {\mathbf {u} }}T(1,2,-1)=\nabla T(1,2,-1)\cdot {\widehat {\mathbf {u} }}=10e^{-9}\left({\frac {5}{\sqrt {2}}}\right)=25{\sqrt {2}}e^{-9}.\end{aligned}}$ Its units are $\displaystyle {}{\text{C}}/{\text{m}}$.