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The temperature of any given point on a plate is given by: T(x,y)=180e^(−(x^2/4)−(y^2/3)) I need to find the rate of change in temperature along the direction from (2,1) to (-1,-3). I have tried taking the dot product of the gradient and unit vector, but this has not worked. I have correctly used the same technique to find the direction and magnitude of greatest increase in temperature. I have found the gradient of T to be <-180e^(-4/3), -120e^(-4/3). How do I find the rate of change from (2,1) to (-1, -3)?

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Let $A = (2,1), B = (-1,-3)\implies \vec{u} = \dfrac{1}{|\vec{AB}|}\vec{AB}= \left(\dfrac{-4}{5}, \dfrac{-3}{5}\right)$ is the unit vector which is also the direction in which the derivative being taken along. Thus the directional derivative you are looking for is: $\nabla_{\vec{u}} f= \nabla f\cdot \vec{u}= (f_x,f_y)\cdot \vec{u}= \dfrac{-4}{5}f_x - \dfrac{3}{5}f_y$. You can find the partial derivatives, can't you ....? Also, you need to specify the point that the directional derivative is calculated at...

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  • $\begingroup$ Wouldn't $\vec{AB}=\langle (-1-2), (-3-1) \rangle = \langle -3, -4 \rangle$? $\endgroup$ – Andrew Tawfeek Oct 21 '17 at 6:15
  • $\begingroup$ Directional derivatives are located at (2,1), yielding Tx = -180e^(-4/3) and Ty = -120e^(-4/3). When plugging those in to the equation multiplying gradient by unit vector, I found the answer to be 36e^(-4/3), which was incorrect. $\endgroup$ – Julia97 Oct 21 '17 at 6:15
  • $\begingroup$ @Julia97 $\left\langle-\frac35,-\frac45\right\rangle\cdot\langle-180,-120\rangle\ne36$. $\endgroup$ – amd Oct 21 '17 at 6:22
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The directional derivative along a vector $\vec{v}$ is given by $$\vec{\nabla}_v f( \vec{x} ) = \vec{\nabla} f( \vec{x} ) \cdot \vec{v} $$

But since the question asks merely in the direction of $\vec{v}$, you must instead dot with $\hat{v}=\frac{\vec{v}}{|\vec{v}|}$.

Since you were given $f(\vec{x})=T(x,y)$ and $\vec{v}$ (found through the two points), all there's left to do is compute $$\vec{\nabla} f( \vec{x} ) \cdot \hat{v} = \vec{\nabla} T(x,y) \cdot \hat{v} = \left< \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right> \cdot \hat{v}$$


Complete solution: We have $T(x,y)=180 e^{-\frac{x^2}{4} - \frac{y^2}{3}}$ thus $$\vec{\nabla}T(x,y)= \left< -90 x e^{-\frac{-x^2}{4}-\frac{y^2}{3}}, -120 y e^{-\frac{x^2}{4} - \frac{y^2}{3}} \right>$$ and $\vec{v}= \left< -1-2, -3-1 \right> = \left<-3,-4 \right> \implies \hat{v}=\left< -\frac{3}{5}, -\frac{4}{5} \right>$. Thus

$$\begin{align*} \left< \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right> \cdot \hat{v} &= \left< -90 x e^{-\frac{-x^2}{4}-\frac{y^2}{3}}, -120 y e^{-\frac{x^2}{4} - \frac{y^2}{3}} \right> \cdot \left< -\frac{3}{5}, -\frac{4}{5} \right> \\ &= 54 x e^{-\frac{-x^2}{4}-\frac{y^2}{3}} + 96y e^{-\frac{x^2}{4} - \frac{y^2}{3}} \end{align*}$$

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