# Find the rate of change of f at p in the direction of the vector u.

$f(x,y,z) = y^2e^{xyz}, P(0,1,-1), u = <\frac{3}{13},\frac{4}{13},\frac{12}{13}>$

This question aims to find the rate of change or gradient and projections of vector spaces onto a given vector.

Gradient of a vector can be found using following formula:

$\nabla f(x,y,z) = \bigg ( \frac{\partial f}{\partial x} (x,y,z),\frac{\partial f}{\partial y} (x,y,z),\frac{\partial f}{\partial z} (x,y,z) \bigg )$

Projection of a vector space can be found using dot product formula:

$D_uf(x,y,z) = \nabla f(x,y,z) \cdot u$

To solve the question, we will use the following steps:

1. Find partial derivatives.
3. Find the projection of gradient in the direction of the vector $u$.

Calculating partial derivative w.r.t $x$:

$\frac{\partial f}{\partial x} (x,y,z) = \frac{\partial}{\partial x}\bigg ( y^2e^{xyz} \bigg )= y^2e^{xyz}(yz) = y^3ze^{xyz}$

Calculating partial derivative w.r.t $y$:

$\frac{\partial f}{\partial y} (x,y,z) = \frac{\partial}{\partial y}\bigg ( y^2e^{xyz} \bigg )$

$\frac{\partial f}{\partial y} (x,y,z) = \frac{\partial}{\partial y} (y^2) e^{xyz} + y^2\frac{\partial}{\partial y} (e^{xyz})$

$\frac{\partial f}{\partial y}(x,y,z) = 2y^2e^{xyz}+y^2e^{xyz}(xz)$

$\frac{\partial f}{\partial y}(x,y,z) = 2y^2e^{xyz} +xy^2ze^{xyz}$

Calculating partial derivative w.r.t $z$:

$\frac{\partial f}{\partial z} (x,y,z) = \frac{\partial}{\partial z}\bigg ( y^2e^{xyz} \bigg )= y^2e^{xyz}(xy) = xy^3e^{xyz}$

Evaluating all partial derivatives at the given point $P$,

$\frac{\partial f}{\partial x} (0,1,-1) = (1)^3(-1)e^{(0)(1)(-1)} = -1$

$\frac{\partial f}{\partial y} (0,1,-1) = 2(1)^2e^{(0)(1)(-1)}+(0)(1)^2(-1)e^{(0)(1)(-1)} = 2$

$\frac{\partial f}{\partial z} (0,1,-1) = (0)(1)^3e^{(0)(1)(-1)} = 0$

Calculating the gradient of $f$ at point $P$:

$\nabla f(x,y,z) = \bigg ( \frac{\partial f}{\partial x} (x,y,z),\frac{\partial f}{\partial y} (x,y,z),\frac{\partial f}{\partial z} (x,y,z) \bigg )$

$\nabla f (0,1,-1) = \bigg ( \frac{\partial f}{\partial x} (0,1,-1),\frac{\partial f}{\partial y} (0,1,-1),\frac{\partial f}{\partial z} (0,1,-1) \bigg )$

$\nabla f (0,1,-1) = < -1 , 2 , 0 >$

Calculating the rate of change in the direction of $u$:

$D_uf(x,y,z) = \nabla f(x,y,z) \cdot u$

$D_uf (0,1,-1) = \nabla f (0,1,-1) \cdot <\frac{3}{13},\frac{4}{13},\frac{12}{13}>$

$D_uf (0,1,-1) = <-1 , 2 , 0 > \cdot <\frac{3}{13},\frac{4}{13},\frac{12}{13}>$

$D_uf (0,1,-1) = -1(\frac{3}{13}) + 2(\frac{4}{13}) + 0(\frac{12}{13})$

$D_uf (0,1,-1) = \frac{-1(3) + 2(4) + 0(12)}{13}$

$D_uf (0,1,-1) = \frac{-3 + 8 + 0}{13} = \frac{5}{13}$

The rate of change is calculated to be:

$D_uf (0,1,-1) = \frac{5}{13}$

## Example

We have the following vectors and we need to calculate the rate of change.

$f(x,y,z) = y^2e^{xyz}, P(0,1,-1), u = <\frac{1}{33},\frac{5}{33},\frac{7}{33}>$

Here, partial derivatives and the gradient values remain same, So:

$\frac{\partial f}{\partial x} (x,y,z) = y^3ze^{xyz}$

$\frac{\partial f}{\partial y} (x,y,z) = 2y^2e^{xyz}+xy^2ze^{xyz}$

$\frac{\partial f}{\partial z} (x,y,z) = xy^3e^{xyz}$

$\frac{\partial f}{\partial x} (0,1,-1) = -1$

$\frac{\partial f}{\partial y} (0,1,-1) = 2$

$\frac{\partial f}{\partial z} (0,1,-1) = 0$

$\nabla f (0,1,-1) = < -1 , 2 , 0 >$

Calculating the rate of change in the direction of $u$:

$D_uf(x,y,z) = \nabla f(x,y,z) \cdot u$

$D_uf (0,1,-1) = \nabla f (0,1,-1) \cdot <\frac{3}{13},\frac{4}{13},\frac{12}{13}>$

$D_uf (0,1,-1) = <-1 , 2 , 0 > \cdot <\frac{1}{33},\frac{5}{33},\frac{7}{33}>$

$D_uf (0,1,-1) = -1(\frac{1}{33}) + 2(\frac{5}{33}) + 0(\frac{7}{33})$

$D_uf (0,1,-1) = \frac{-1(1) + 2(5) + 0(7)}{33} = \frac{-1 + 10 + 0}{33} = \frac{5}{33}$