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

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

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:

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

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

$$D_{u}f(x,y,z)=\mathrm{\nabla }f(x,y,z)\cdot u$$

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

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

Expert Answer

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$D_{u}f(x,y,z)=\mathrm{\nabla }f(x,y,z)\cdot u$$

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

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

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

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

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

Numerical Answer

The rate of change is calculated to be:

$$D_{u}f(0,1,-1)=\frac{5}{13}$$

Example

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

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

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

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

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

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

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

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

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

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

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

$$D_{u}f(x,y,z)=\mathrm{\nabla }f(x,y,z)\cdot u$$

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

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

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

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

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