Consider the surface $$z=(\frac{x}{2}{)}^{2}-(\frac{y}{3}{)}^{2}$$. Compute the tangent plane at each point. Find the point such that the normal vector is vertical at that point.

Deborah Proctor
2022-10-17
Answered

Consider the surface $$z=(\frac{x}{2}{)}^{2}-(\frac{y}{3}{)}^{2}$$. Compute the tangent plane at each point. Find the point such that the normal vector is vertical at that point.

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Plutbantonavv

Answered 2022-10-18
Author has **15** answers

Given surface

$$z=(\frac{x}{2}{)}^{2}-(\frac{y}{3}{)}^{2}$$

tangent plane at each point

Here $$f(x,y,z)=(\frac{x}{2}{)}^{2}+(\frac{y}{3}{)}^{2}-z$$

let a point in given surface, whose posiion vector is

$${r}_{0}=ai+bj+ck$$

let p be any point of tangent plane

$$(r-{r}_{0})\mathrm{\u25b3}f=0\phantom{\rule{0ex}{0ex}}\Rightarrow ((x-a)u+(y-b)j+(z-jk))\phantom{\rule{0ex}{0ex}}\Rightarrow (x-a)\frac{x}{2}+\frac{2(y-b)y}{9}-(z-c)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x(x-a)}{2}+\frac{2y}{9}(y-b)-(z-c)$$

This is the equation of tangent plane at each point.

$$z=(\frac{x}{2}{)}^{2}-(\frac{y}{3}{)}^{2}$$

tangent plane at each point

Here $$f(x,y,z)=(\frac{x}{2}{)}^{2}+(\frac{y}{3}{)}^{2}-z$$

let a point in given surface, whose posiion vector is

$${r}_{0}=ai+bj+ck$$

let p be any point of tangent plane

$$(r-{r}_{0})\mathrm{\u25b3}f=0\phantom{\rule{0ex}{0ex}}\Rightarrow ((x-a)u+(y-b)j+(z-jk))\phantom{\rule{0ex}{0ex}}\Rightarrow (x-a)\frac{x}{2}+\frac{2(y-b)y}{9}-(z-c)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x(x-a)}{2}+\frac{2y}{9}(y-b)-(z-c)$$

This is the equation of tangent plane at each point.

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a)

b)

c)

d)

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The derivation the following equation is given as a statement of D'Alambert's principle:

${\overrightarrow{F}}_{\text{net}}\cdot \delta \overrightarrow{r}=m\frac{{d}^{2}\overrightarrow{r}}{d{t}^{2}}\cdot \delta \overrightarrow{r}$

After which the following equation is used: $(xy{)}^{\prime}={x}^{\prime}y+x{y}^{\prime}$ to rewrite the right hand side so that:

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What is x, what is y and how is the equation applied?

${\overrightarrow{F}}_{\text{net}}\cdot \delta \overrightarrow{r}=m\frac{{d}^{2}\overrightarrow{r}}{d{t}^{2}}\cdot \delta \overrightarrow{r}$

After which the following equation is used: $(xy{)}^{\prime}={x}^{\prime}y+x{y}^{\prime}$ to rewrite the right hand side so that:

$m\frac{{d}^{2}\overrightarrow{r}}{d{t}^{2}}\cdot \delta \overrightarrow{r}=m[\frac{d}{dt}(\frac{d\overrightarrow{r}}{dt}\cdot \delta \overrightarrow{r})-\frac{d\overrightarrow{r}}{dt}\cdot \frac{d\delta \overrightarrow{r}}{dt}]$

What is x, what is y and how is the equation applied?