Q. 31 - Calculus

Question

In Exercises $29–34$, find the equation of the line tangent to the surface at the given point$\mathrm{P}$ and in the direction of the given unit vector $\mathrm{u}$. Note that these are the same functions, points,

and vectors as in Exercises $21–26$.

$\mathrm{f}(\mathrm{x},\mathrm{y})=\frac{\mathrm{x}}{{\mathrm{y}}^2}$at$\mathrm{P}=(-2,1),\mathbf{u}=\left(\frac{\sqrt{10}}{10},-\frac{3\sqrt{10}}{10}\right)$

Step-by-Step Solution

Answer

The equation of the line tangent is $\mathrm{x}=-2+\frac{\sqrt{10}}{10}\mathrm{t},\mathrm{y}=1-\frac{3\sqrt{10}}{10}\mathrm{t},\mathrm{z}=-2-11\frac{\sqrt{10}}{10}\mathrm{t}$.

Step 1: Equation for line of tangent

Given function is,

$\mathrm{f}(\mathrm{x},\mathrm{y})=\frac{\mathrm{x}}{{\mathrm{y}}^2}$

$\mathrm{P}=\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)=(-2,1)$ and$\mathrm{u}=(\mathrm{\alpha },\mathrm{\beta })=\left(\frac{\sqrt{10}}{10},-\frac{3\sqrt{10}}{10}\right)$

Equation of line of tangent is,

${\mathrm{f}}_{\mathrm{x}}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\left(\mathrm{x}-{\mathrm{x}}_0\right)+{\mathrm{f}}_{\mathrm{y}}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\left(\mathrm{y}-{\mathrm{y}}_0\right)=\mathrm{z}-\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$

${\mathrm{f}}_{\mathrm{x}}(-2,1)(\mathrm{x}+2)+{\mathrm{f}}_{\mathrm{y}}(-2,1)(\mathrm{y}-1)=\mathrm{z}-\mathrm{f}(-2,1) .....1$

Step 2: Values of function

Functions are,

${\mathrm{f}}_{\mathrm{x}}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)={\left.\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{f}(\mathrm{x},\mathrm{y})\right|}_{\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}={\left.\frac{\mathrm{d}}{\mathrm{dx}}\frac{\mathrm{x}}{{\mathrm{y}}^2}\right|}_{\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}$

${\mathrm{f}}_{\mathrm{x}}(-2,1)={\left.\frac{1}{{\mathrm{y}}^2}\right|}_{(-2,1)}$

${\mathrm{f}}_{\mathrm{x}}(-2,1)=1 ......2$

${\mathrm{f}}_{\mathrm{y}}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)={\left.\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{f}(\mathrm{x},\mathrm{y})\right|}_{\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}={\left.\frac{\mathrm{d}}{\mathrm{dy}}\frac{\mathrm{x}}{{\mathrm{y}}^2}\right|}_{\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}$

${\mathrm{f}}_{\mathrm{y}}(-2,1)={\left.\frac{-2\mathrm{x}}{{\mathrm{y}}^3}\right|}_{(-2,1)}$

${\mathrm{f}}_{\mathrm{y}}(-2,1)=4 .........3$

$\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)=\mathrm{f}(-2,1)=\frac{-2}{1}$

$\mathrm{f}(-2,1)=-2 ....4$

Substitute $2,3,4$ equation in $1$

we get,

$1(\mathrm{x}+2)+4(\mathrm{y}-1)=\mathrm{z}+2$

$\mathrm{x}+4 \mathrm{y}-\mathrm{z}=2-2+4$

$\mathrm{x}+4 \mathrm{y}-\mathrm{z}=4$

Step 3: Directional derivative

Regarded the equation of the normal line is,

$\mathrm{x}={\mathrm{x}}_0+\mathrm{\alpha t},\mathrm{y}={\mathrm{y}}_0+\mathrm{\beta t},\mathrm{z}={\mathrm{z}}_0+\mathrm{\gamma t}$

${\mathrm{z}}_0=\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$ and $\mathrm{\gamma }={\mathrm{D}}_{\mathrm{w}}\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$

The directional derivative is,

${\mathrm{D}}_{\mathrm{u}}\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)={Lim}_{\mathrm{h}\to 0}\frac{\mathrm{f}\left({\mathrm{x}}_0+\mathrm{\alpha h},{\mathrm{y}}_0+\mathrm{\beta h}\right)-\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}{\mathrm{h}}$

${\mathrm{D}}_{\mathrm{u}}\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)={Lim}_{\mathrm{h}\to 0}\frac{\mathrm{f}\left({\mathrm{x}}_0+\mathrm{\alpha h},{\mathrm{y}}_0+\mathrm{\beta h}\right)-\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}{\mathrm{h}} .......5$

$\mathrm{f}\left(-2+\frac{\sqrt{10}}{10}\mathrm{h},1-\frac{3\sqrt{10}}{10}\mathrm{h}\right)=\frac{-2+\frac{\sqrt{10}}{10}\mathrm{h}}{{\left(1-\frac{3\sqrt{10}}{10}\mathrm{h}\right)}^2}$

$=\frac{-2+\frac{\mathrm{h}}{\sqrt{10}}}{{\left(1-\frac{3}{\sqrt{10}}\mathrm{h}\right)}^2}$

$=\frac{\frac{-2\sqrt{10}}{\sqrt{10}}+\mathrm{h}}{\left(\frac{10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2}{10}\right)}$

$=\frac{-2^{}\times 10+\mathrm{h}\sqrt{10}}{10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2} .........6$

And $\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)=\mathrm{f}(-2,1)=\frac{-2}{1^2}=-2 ......7$

Step 4: Calculation for equation of line tangent

Substitute $6,7$ equation in $5$ equation,

we get,

${\mathrm{D}}_{\mathrm{u}}\mathrm{f}(-2,1)={Lim}_{\mathrm{h}\to 0}\frac{\frac{-2\times 10+\mathrm{h}\sqrt{10}}{10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2}-(-2)}{\mathrm{h}}$

$={Lim}_{\mathrm{h}\to 0}\frac{-20+\mathrm{h}\sqrt{10}+20-12\sqrt{10}\mathrm{h}+18{\mathrm{h}}^2}{\mathrm{h}\left(10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2\right)}$

$={Lim}_{\mathrm{h}\to 0}\frac{-11\mathrm{h}\sqrt{10}+18{\mathrm{h}}^2}{\mathrm{h}\left(10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2\right)}$

$={Lim}_{\mathrm{h}\to 0}\frac{(-11\sqrt{10}+18\mathrm{h})}{\left(10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2\right)}$

${\mathrm{D}}_{\mathrm{u}}\mathrm{f}(-2,1)=\frac{-11\sqrt{10}}{10}$