A metal plate is placed on the $xy$-plane in such a way that the temperature $T$ in degrees Celsius at any point $P=(x,y)$ is inversely proportional to the distance of $P$ from $(0,0)$. Suppose the temperature of the plate at the point $(-3,4)$ equals ${50^{\circ}{\rm C}}$.

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Solution (a) The temperature $T$ at any point $(x,y)$ is inversely proportional to the distance of $( x,y)$ from the origin $(0,0)$. We can model $T=T( x,y)$ as $$T(x,y)=\dfrac{k}{\sqrt{x^{2}+y^{2}}}$$

where $k$ is the constant of proportionality. Since $T=50^{\circ}{\rm C}$ when $(x,y)=(-3,4)$, then $$k=T\sqrt{x^{2}+y^{2}}=50\sqrt{( -3) ^{2}+4^{2}}=50(5)=250$$

So, $T=T( x,y) =\dfrac{250}{\sqrt{x^{2}+y^{2}}}$.

Figure 5 $T=T( x,y) =\dfrac{250}{\sqrt{x^{2}+y^{2}}}$

(b) The gradient of $T=T( x,y) =\dfrac{250}{\sqrt{ x^{2}+y^{2}}}$ is $${\bf\nabla }T(x,y)=T_{x}(x,y)\mathbf{i}+T_{y}(x,y)\mathbf{j}=-\frac{250x }{(x^{2}+y^{2})^{3/2}}\mathbf{i}-\frac{250y}{(x^{2}+y^{2})^{3/2}}\mathbf{ j}$$

At $( -3,4)$, $${\bf\nabla }T(-3,4)=-\frac{250 ( -3) }{ [ ( -3) ^{2}+4^{2}] ^{3/2}}\mathbf{i}-\frac{250 ( 4) }{ [ ( -3) ^{2}+4^{2}] ^{3/2}}\mathbf{j}=\frac{750}{125} \mathbf{i}-\frac{1000}{125}\mathbf{j}=6\mathbf{i}-8\mathbf{j}$$

(c) The temperature increases most rapidly in the direction ${\bf\nabla} T(-3,4)=6\mathbf{i}-8\mathbf{j}$.

(d) The temperature decreases most rapidly in the direction $-{\bf\nabla }T(-3,4)=-6\mathbf{i}+8\mathbf{j}$.

(e) The rate of change in $T$ at $(-3,4)$ equals $0$ for directions orthogonal to ${\bf\nabla} T(-3,4)=6\mathbf{i}-8\mathbf{j}$. That is, the rate of change in $T$ is $0$ in either of the two directions orthogonal to $6\mathbf{i}-8\mathbf{j}$, namely $\pm ( 8\mathbf{i}+6 \mathbf{j})$.