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Let $T(x,y,z)$ be the temperature, in Celsius, inside a room at the point $(x,y,z)$, such that $x,y,z$ are given in meters.

It's given that $$T(5,4,2)=30º,\quad\frac{\partial T}{\partial x}(5,4,2)=3º/m,\quad\frac{\partial T}{\partial y}(5,4,2)=-1º/m,\quad\frac{\partial T}{\partial z}(5,4,2)=1º/m$$

A fly flies in the room and its path is given by $$x=t^2+1,\quad y=2t,\quad z=10-t^3$$ at the instant $t$ given in seconds.

I have to find the rate of the temperature subject to this path at $t=2$ seconds.

What I made was: the position of the fly is given by the function $$s(t)=(t^2+1,2t,10-t^3)$$ and then, $$s(2)=(5,4,2)$$ and in this point, the partial derivatives are given.

Since $$s'(t)=(2t,2,-3t^2)$$ so $$s'(2)=(4,2,-12)$$ Therefore, the rate I'm looking for is $$s'(2)\cdot\nabla T(5,4,2)=-2$$

My question is: is it right or should I normalize the vector $\vec{u}=(4,2,-12)$?

mvfs314
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1 Answers1

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Related: Why isn't the directional derivative generally scaled down to the unit vector?

In short, by not scaling, you are incorporating the fly's velocity into the answer. If the fly were moving in the same direction, but super slowly the instantaneous rate of change would be smaller, whereas if the fly were moving super quickly in the same direction, the instantaneous rate of change of temperature would be larger.

Math Helper
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