Question

find the gradient of function f(x, y, z) =x2+y3+z4​

Answer 1

Answer:

Gradient of a function is represented as the partial derivatives of the given scalar function using arithmetic operators.

In terms of formula, it is represented as:

$\boxed{ \bf{\nabla \:F = \dfrac{\partial f}{\partial x} + \dfrac{\partial f}{\partial y} + \dfrac{\partial f}{\partial z} }}$

Now it is given that, scalar function f(x,y,z) is equal to x² + y³ + z⁴.

Using the formula given above we get:

$\implies \dfrac{ \partial f}{ \partial x} = \dfrac{ \partial (x^2 + y^3 + z^4)}{ \partial x}\\\\\\\implies \dfrac{ \partial f}{ \partial x} = 2x + 0 + 0 = 2x$

$\implies \dfrac{ \partial f}{ \partial y} = \dfrac{ \partial (x^2 + y^3 + z^4)}{ \partial y}\\\\\\\implies \dfrac{ \partial f}{ \partial y} = 0 + 3y^2 + 0 = 3y^2$

$\implies \dfrac{ \partial f}{ \partial z} = \dfrac{ \partial (x^2 + y^3 + z^4)}{ \partial z}\\\\\\\implies \dfrac{ \partial f}{ \partial z} = 0 + 0 + 4z^3 = 4z^3$

Combining all the three in the formula we get:

$\implies \boxed{ \bf{\nabla\:F = 2x + 3y^2 + 4z^3}}$

This is the required answer.

Answer 2

Answer:

The gradient of the function $f(x,y,z)=x^{2} +y^{3} +z^{4}$ is $i(2x)+j(3y^{2} )+k(4z^{3} )$

Explanation:

Del is an operator used in vector calculus as a vector differential operator

To convert a scalar quantity to a vector quantity del operator is used as a gradient

For the cartesian coordinate system, the gradient is given by the formula

$id/dx+jd/dy+kd/dz$

From the question, it is given that the function is

$f(x,y,z)=x^{2} +y^{3} +z^{4}$

now the gradient of the operator is

$id/dx+jd/dy+kd/dz(x^{2} +y^{3} +z^{4})$

$i\frac{d(x^{2} +y^{3} +z^{4})}{dx}+j\frac{d(x^{2} +y^{3} +z^{4})}{dy}+k\frac{d(x^{2} +y^{3} +z^{4})}{dz}$

differentiate the given function with respect to x, y, and z

$i(2x)+j(3y^{2} )+k(4z^{3} )$

now the scalar function is converted to a vector