I have done part a) of this question. I am confused about part b), as it doesn't say determine rate of change of temperature with respect to anything, so I am confused. Would it be ∂T/∂x + ∂T/dy ?
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They're asking about the directional derivative. – Hans Lundmark Oct 05 '18 at 09:16
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So v * ∂T/∂dx ? – J-Dorman Oct 05 '18 at 11:02
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No, not quite. The unit vector from $(0,-1)$ towards the origin is $\mathbf{v}=(0,1)$, so $\mathbf{v} \cdot \nabla T(0,-1)$ equals what? – Hans Lundmark Oct 05 '18 at 13:15
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You are missing the crucial piece of information: directional derivatives. Gradient tells you the direction of steepest increase, but if you move in some other direction, the change is less. If you move in some direction (let's you move with velocity vector $v$ when time $t$ changes), you can write down the chain rule: $$\frac{dT}{dt}=\frac{\partial T}{\partial x}\frac{dx}{dt}+\frac{\partial T}{\partial y}\frac{dy}{dt}$$ Notice that $dx/dt$ and $dy/dt$ are simply the components of velocity vector. You can rewrite this as:
$$\frac{dT}{dt}=\vec{v}\cdot \nabla T$$ So, dot product of velocity (so: direction in which you are looking for rate of change) with the gradient tells you the rate of change. In this case, you probably want a unit length $v$.
orion
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since it is moving towards the origin, direction vector would be: (0,0) - (0,-1) = (0,1)? @orion – J-Dorman Oct 06 '18 at 11:25
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@J-Dorman Yes, it is. So in your case you are just looking for derivative with respect to $y$. – orion Oct 06 '18 at 11:34
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