Q. 38 - Calculus

Question

In exercise $35-38$, find the directional derivative of the given function at the specified point $P$ and in the direction of the given vector $v$.

$f(x,y,z)=x^2+y^2-z^3 at P=(0,-2,5),v=-i+3j+5k$

Step-by-Step Solution

Answer

The directional derivative of the given function $D_{w}f(0,-2,5)=-\frac{137}{\sqrt{35}}$

Step 1: Directional of derivative.

For given $P=(x_0,y_0,z_0)=(0,-2,5)$ and $v=-i+3j=5k$, we must compute the directional derivative of function $f(x,y,z)=x^2+y^2-z^2$.

Think this as directional derivative

$D_{u}f\left(x_0,y_0,z_0\right)={\mathrm{Lim}}_{h\to 0}\frac{f\left(x_0+\alpha h,y_0+\beta h,z_0+\gamma h\right)-f\left(x_0,y_0,z_0\right)}{h} ....\left(1\right)$

$||v||=\sqrt{{(-1)}^2+3^2+5^2}=\sqrt{1+9+25}$

$||v||=\sqrt{35}$

$u=(\alpha ,\beta ,\gamma )=\left(\frac{-1}{\sqrt{35}},\frac{3}{\sqrt{35}},\frac{5}{\sqrt{35}}\right)$

Step 2: Equation.

Therefore,

$f\left(x_0+\alpha h,y_0+\beta h,z_0+\gamma h\right)=f\left(0-\frac{1}{\sqrt{35}}h,-2+\frac{3}{\sqrt{35}}h,5+\frac{5}{\sqrt{35}}h\right)$

$={\left(0-\frac{1}{\sqrt{35}}h\right)}^2+{\left(-2+\frac{3}{\sqrt{35}}h\right)}^2-{\left(5+\frac{5}{\sqrt{35}}h\right)}^3$

$=\frac{1}{35}{h}^2+\left(4-\frac{12}{\sqrt{35}}h+\frac{9}{35}{h}^2\right)-\left(125+\frac{375}{\sqrt{35}}h+\frac{375}{35}{h}^2+\frac{125}{{\left(\sqrt{35}\right)}^3}{h}^3\right)$

$=\frac{1}{35}{h}^2+4-\frac{12}{\sqrt{35}}h+\frac{9}{35}{h}^2-125-\frac{375}{\sqrt{35}}h-\frac{375}{35}{h}^2-\frac{125}{{\left(\sqrt{35}\right)}^3}{h}^3$

$=-121-\frac{365}{35}{h}^2-\frac{137}{\sqrt{35}}h-\frac{125}{{\left(\sqrt{35}\right)}^3}{h}^3 .....\left(2\right)$

$f(x_0,y_0,z_0)=0^2+{(-2)}^2-5^3=-121 ......\left(3\right)$

Step 3: Directional derivatives

When equations $2$ and  $3$are substituted for equation$1$ , we get

${\mathrm{D}}_{\mathrm{w}}\mathrm{f}(0,-2,5)={\mathrm{Lim}}_{\mathrm{h}\to 0}\frac{-121-\frac{137}{\sqrt{35}}\mathrm{h}-\frac{365}{35}{\mathrm{h}}^2-\frac{125}{(\sqrt{35}{)}^3}{\mathrm{h}}^3-121}{\mathrm{h}}$

$={\mathrm{Lim}}_{h\to 0}-\frac{137}{\sqrt{35}}-\frac{365}{35}h-\frac{125}{(\sqrt{35}{)}^3}{h}^2$

$D_{w}f(0,-2,5)=-\frac{137}{\sqrt{35}}$