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Suppose that the temperature at a point $(x,y,z)$ in space is given by $T(x,y,z)=\frac{80}{1+x^2+2y^2+3z^2}$ where $T$ is measured in degrees celsius and $x$,$y$ and $z$ in meters.

In which direction does the temperature increase fastest at the point $(1,1,-2)$? What is the maximum rate of increase?

I first had to find the vector field which is $\operatorname{grad}T=-5i/8-5j/4+15k/4$ and assume it be the direction and then determine the magnitude as the rate. I need a quick critic on this and please help to make a critical decision on how make corrections. Thank you in advance

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  • $\begingroup$ Hint: use the gradient operator $\endgroup$ – danimal Jun 7 '15 at 21:52
  • $\begingroup$ Given a scalar field $T$ in your case, the direction of fastest increase at the point $P(1,1,-2)$ will be $\nabla T(P)$ and the maximum rate of increased will be $|\nabla T(P)|$. $\endgroup$ – MathNewbie Jun 7 '15 at 21:53
  • $\begingroup$ Is my assumption accurate? $\endgroup$ – DOCTOR NGILAZI BANDA JOSHUA Jun 7 '15 at 22:08
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You compute the partial derivatives at the point you are given. You get

$$(\partial_{x}T,\partial_{y}T,\partial_{z}T)=-160\left(\frac{x}{(1+x^{2}+2y^{2}+3z^{2})^{2}},\frac{2y}{(1+x^{2}+2y^{2}+3z^{2})^{2}},\frac{3z}{(1+x^{2}+2y^{2}+3z^{2})^{2}}\right)$$

that in $(1,1,-2)$ is equal to

$$-\frac{5}{8}(1,2,-6)=\left(-\frac{5}{8},-\frac{5}{4},\frac{15}{4}\right)$$

Then the direction is that given by the above vector. After normalization it is approximately $\left(-\frac{5}{32},-\frac{5}{16},\frac{15}{16}\right)$. The modulus of this vector is $1.00098$, so the approximation is pretty precise. Your reasoning is totally correct.

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  • $\begingroup$ why have you used the unit vector of gradT(1,1,-2) $\endgroup$ – DOCTOR NGILAZI BANDA JOSHUA Jun 7 '15 at 22:20
  • $\begingroup$ Because if you're asked a direction, a direction is by definition a unit vector. To leave the vector unnormalized is not a mistake, but the normalized version looks more correct. In the end, it depends on the reader's taste and on the future use you make of it. $\endgroup$ – Giorgio Comitini Jun 7 '15 at 23:01

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