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Let's suppose that the temperature of a particle at a given position $(x,y,z)$ is given by $ T(x,y,z)=e^{-(x^2+2y^2+3z^2)}$ where $x,y,z$ let's consider that $x,y,z$ are given in meters=m$

If the particle moves at a speed of $ 8 m$ per second, compute the velocity of decreasing of the temperature.

I don't understand, since the problem talk about time, then the position is a function of the form $ r(t)= (x(t),y(t),z(t)) $ and the temperature also depends on time. I thought that I would need to use the gradient, but if I derivate with respect to $t$ I can not find the derivative explicitly.

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  • $\begingroup$ You need to know not only the speed of the particle, but also the direction of its motion, i.e., its velocity vector. Then the chain rule will appear with the gradient and the velocity vector. $\endgroup$ – Ted Shifrin May 11 '13 at 21:42
  • $\begingroup$ And How can I use it? $\endgroup$ – Trafalgaw May 11 '13 at 22:44
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The chain rule says that $$\frac d{dt} T(r(t)) = \frac d{dt}T(x(t),y(t),z(t)) = \nabla T(r(t))\cdot \frac{dr}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt} + \frac{\partial T}{\partial z}\frac{dz}{dt}\,.$$ To complete the problem, we need to know more than the speed (which is the magnitude of the velocity vector). We need to know both the position and the velocity vector at the desired instant $t_0$. Then we can evaluate the gradient vector at the desired point ($r(t_0)$) and dot it with the velocity vector.

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