The temperature $T:\mathbb{R}^3\setminus\{(0,0,0)\}\to \mathbb{R}$ at any point $P(x,y,z)$ is inversely proportional to the square of the distance of $P$ from the origin. If the value of the temperature $T$ at the point $R(0,0,1)$ is $\sqrt{3}$, then the rate of change of $T$ at the point $Q(1,1,2)$ in the direction of $\vec{QR}$ is equal to _______ (round off to $2$ places of decimal).
This is a rather simple problem, at first we define $$T(x,y,z)=\frac{K}{x^2+y^2+z^2}$$ Then using $T(0,0,1)=\sqrt{3}$ we have $K=\sqrt{3}$. So $$T(x,y,z)=\frac{\sqrt{3}}{x^2+y^2+z^2}$$ Now since we need rate of change of $T$ at $(1,1,2)$, we calculate $|\nabla T|_{(1,1,2)}$. Therefore $$\nabla T=\bigg(-\frac{2x\sqrt{3}}{(x^2+y^2+z^2)^2},-\frac{2y\sqrt{3}}{(x^2+y^2+z^2)^2},-\frac{2z\sqrt{3}}{(x^2+y^2+z^2)^2}\bigg)$$ So required rate of change at $(1,1,2)$ is $$|\nabla T|_{(1,1,2)}=\sqrt{\bigg(\frac{2\sqrt{3}}{36}\bigg)^2+\bigg(\frac{2\sqrt{3}}{36}\bigg)^2+\bigg(\frac{4\sqrt{3}}{36}\bigg)^2}=\sqrt{\frac{72}{36^2}}=\frac{1}{3\sqrt{2}}=0.2357\simeq0.24$$ whereas the answer is given $0.22$. I can't understand where it is going wrong. Any help is appreciated.
It is easier to use spherical polars to calculate the magnitude of grad$(T)$. Since there is no angle dependence the magnitude is just $T'(r)$.
We have $T(r)=\frac{\sqrt3}{r^2}$. Hence $T'(r)=-\frac{2\sqrt3}{r^3}$ and $T'(Q)=-\frac{\sqrt2}{6}$. From here on we use ordinary Cartesian coordinates. The unit vector in the direction $OQ$ is $\frac{1}{\sqrt6}(1,1,2)$. So grad$(T)$ is $-\frac{1}{6\sqrt3}(1,1,2)$.
The unit vector in the direction $QR$ is $\frac{1}{\sqrt3}(-1,-1,-1)$. So to get the rate of change in that direction we take the dot product to get $$\frac{1}{18}(1+1+2)=\frac{2}{9}=0.2222\dots$$
Your answer was wrong because you took the rate of change in the direction $QO$ instead of $QR$. If you multiply yours by $\frac{2\sqrt2}{3}$ you would get the right answer.