Question : How do you find gradient of modulus of position vector?
Solution :
We shall find gradient of modulus of position vector as follows
Position vector of an object is defined as \[\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \] So the modulus of position vector is \[ r=\left|\vec{r}\right|=\sqrt{x^2+y^2+z^2}\] Now gradient of modulus of position vector is \[\nabla\left(r\right)=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right)r\] \[\nabla\left(r\right)=\hat{i}\frac{\partial r}{\partial x}+\hat{j}\frac{\partial r}{\partial y}+\hat{k}\frac{\partial r}{\partial z}\] Now we shall find following partial derivatives of \[r=\sqrt{x^2+y^2+z^2}\] \[\frac{\partial r}{\partial x}=\frac{1}{2\sqrt{x^2+y^2+z^2}}\frac{\partial}{\partial x}\left(x^2+y^2+z^2\right)\] \[=\frac{1}{2\sqrt{x^2+y^2+z^2}}\left(2x\right)\] \[=\frac{x}{\sqrt{x^2+y^2+z^2}}=\frac{x}{r}\] \[\frac{\partial r}{\partial x}=\frac{x}{r}\] Similarly \[\frac{\partial r}{\partial y}=\frac{y}{r},\frac{\partial r}{\partial z}=\frac{z}{r}\] Substituting \[\nabla\left(r\right)=\hat{i}\frac{x}{r}+\hat{j}\frac{y}{r}+\hat{k}\frac{z}{r}\] \[\nabla\left(r\right)=\frac{\hat{i}x+\hat{j}y+\hat{k}z}{r}\] \[\nabla\left(r\right)=\frac{\ \vec{r}}{r}\] Thus we found gradient of modulus of position vector is \[\frac{\ \vec{r}}{r} \]