INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

March 23, 2020

Gradient, Divergence and Curl | IIT JAM 2014 | Problem 5

Try this problem from IIT JAM 2014 exam. It deals with calculating Gradient of a scalar point function, Divergence and curl of a vector point function.

$\nabla, \nabla ., \nabla \times$ Operators | IIT JAM 2014 | Problem 5

If $f(x,y,z)=x^2y+y^2z+z^2x,\quad \forall (x,y,z) \in \mathbb R$ and $\nabla=(\frac{\partial}{\partial x}\hat{i}+ \frac{\partial}{\partial y}\hat{j}+ \frac{\partial}{\partial z}\hat{k} )$ then the value of $\nabla.(\nabla \times \nabla f)+\nabla.(\nabla f)$ at $(1,1,1)$

  • $4$
  • $5$
  • $6$
  • $7$

Key Concepts

Vector Calculus

Scalar Point Function

Grad, Div , Curl

Check the Answer

Answer: $6$

IIT JAM 2014 , Problem 5

Try with Hints

Scalar Point Function : is a function which assigns a point$(x,y,z) \in \mathbb R^3$ to a scalar. Here $f$ is a scalar point function.

$\nabla=(\frac{\partial}{\partial x}\hat{i}+ \frac{\partial}{\partial y}\hat{j}+ \frac{\partial}{\partial z}\hat{k} ) $

Gradient of a function :($\nabla f) = (\frac{\partial f}{\partial x}\hat{i}+ \frac{\partial f}{\partial y}\hat{j}+ \frac{\partial f}{\partial z}\hat{k} ) $

Divergence of a function ($\nabla.\vec F$) =$ (\frac{\partial}{\partial x}\hat{i}+ \frac{\partial}{\partial y}\hat{j}+ \frac{\partial}{\partial z}\hat{k} ).\vec F $

Curl of a function ($\nabla\times \vec F$) =$ (\frac{\partial}{\partial x}\hat{i}+ \frac{\partial}{\partial y}\hat{j}+ \frac{\partial}{\partial z}\hat{k} )\times \vec F $

Now, $\nabla f = (2xy+z^2) \hat{i}+(2yz+x^2)\hat{j}+(2zx+y^2)\hat{k}$

Therefore $\nabla . (\nabla f)=\frac{\partial}{\partial x}(2xy+z^2)+ \frac{\partial}{\partial y}(2yz+z^2) + \frac{\partial}{\partial z}(2zx+y^2)$

$\quad= 2x+2y+2z $


$\nabla \times \nabla f = \begin{vmatrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2xy+z^2 & 2yz+x^2 & 2zx+y^2\end{vmatrix}= \vec 0$

Therefore $\nabla. (\nabla \times \nabla f)= 0$

then, $\nabla.(\nabla \times \nabla f)+\nabla.(\nabla f) = 2(x+y+z) \bigg|_{(1,1,1)}=6$

Subscribe to Cheenta at Youtube

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.