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### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# (a). To show: The parametric equations $$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$ represent a hyperboloid of one sheet.(b). To draw: The graph of a hyperboloid of one sheet with the parametric equations $$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$ for $$a = 1,b = 2$$, and $$c = 3$$.(c). To find: The expression for surface area of the hyperboloid of one sheet with the parametric equations $$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$ for $$a = 1,b = 2$$, and $$c = 3$$.

(a). The parametric equations $$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$ are represents a hyperboloid of one sheet.

(c). The expression for surface area of the hyperboloid of one sheet with the parametric equations $$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$ for $$a = 1,b = 2$$, and $$c = 3$$ is

$$36{\cosh ^4}u{\cos ^2}v + 9{\cosh ^4}u{\sin ^2}v + 4{\sinh ^2}u{\cosh ^2}u$$

See the step by step solution

## Step 1: Formula used:

Write the expression for equation of hyperboloid of one sheet.

$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1(1)$$

Here,

$$a,b$$, and $$c$$ are scalar parameters (constant real values).

Write the parametric equations as follows.

$$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$

## Step 2: Rewrite the parametric equations as follows.

Take square and add the expressions as follows.

$${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} - {\left( {\frac{z}{c}} \right)^2} = {(\cosh u\cos v)^2} + {(\cosh u\sin v)^2} - {(\sinh u)^2}$$

Simplify the expression as follows.

\begin{aligned}&\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}\frac{{{z}^{2}}}{{{c}^{2}}}&={{\cosh }^{2}}u{{\cos }^{2}}v+{{\cosh }^{2}}u{{\sin }^{2}}v-{{\sinh }^{2}}u \\&={{\cosh }^{2}}u\left( {{\cos }^{2}}v+{{\sin }^{2}}v \right)-{{\sinh }^{2}}u\\&={{\cosh }^{2}}u(1)-{{\sinh }^{2}}u\left\{ \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right\}\\&={{\cosh}^{2}}u{{\sinh}^{2}}u\\&\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}\frac{{{z}^{2}}}{{{c}^{2}}}=1\quad \left\{ {{\cosh }^{2}}\theta -{{\sinh }^{2}}\theta =1 \right\} \\ & \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{{{z}^{2}}}{{{c}^{2}}}=1(2) \\\end{aligned}

As the equations (1) and (2) are similar, the parametric equations represent a hyperboloid of one sheet.

Thus, the parametric equations $$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$ are represents a hyperboloid of one sheet.

## Step 3: Consider the values of scalar parameters in the parametric equations

as follows.

$$a = 1,b = 2,{\rm{ and }}c = 3$$

Substitute 1 for $a, 2$ for $$b$$, and 3 for $$c$$ in the parametric equations and rewrite the parametric equations of the hyperboloid of one sheet as follows.

$$x = (1)\cosh u\cos v,y = (2)\cosh u\sin v,z =$$ (3) $$\sinh u$$

$$x = \cosh u\cos v,y = 2\cosh u\sin v,z = 3\sinh u(3)$$

From the analysis in Part (a) and equation (3), draw the graph of a hyperboloid of one sheet with the parametric equations $$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$ for $$a = 1,b = 2$$, and $$c = 3$$ as shown in figure 1

Thus, the graph of a hyperboloid of one sheet with the parametric equations $$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$ for $$a = 1,b = 2$$, and $$c = 3$$ is drawn.

## Step 4: Calculation of $${r_u}$$ :

Substitute $$\langle \cosh u\cos v,2\cosh u\sin v,3\sinh u\rangle$$ for $${\rm{r}}(u,v)$$ in equation (5),

$${{\rm{r}}_u} = \frac{\partial }{{\partial u}}\langle \cosh u\cos v,2\cosh u\sin v,3\sinh u\rangle$$

$$= \left\langle {\frac{\partial }{{\partial u}}(\cosh u\cos v),\frac{\partial }{{\partial u}}(2\cosh u\sin v),\frac{\partial }{{\partial u}}(3\sinh u)} \right\rangle$$

$$= \left\langle {\cos v\frac{\partial }{{\partial u}}(\cosh u),2\sin v\frac{\partial }{{\partial u}}(\cosh u),3\frac{\partial }{{\partial u}}(\sinh u)} \right\rangle$$

$$= \langle \sinh u\cos v,2\sinh u\sin v,3\cosh u\rangle$$

Calculation of $${{\rm{r}}_v}$$ :

Substitute $$\langle \cosh u\cos v,2\cosh u\sin v,3\sinh u\rangle$$ for $${\rm{r}}(u,v)$$ in equation (6),

$${{\bf{r}}_v} = \frac{\partial }{{\partial v}}\langle \cosh u\cos v,2\cosh u\sin v,3\sinh u\rangle$$

$$= \left\langle {\frac{\partial }{{\partial v}}(\cosh u\cos v),\frac{\partial }{{\partial v}}(2\cosh u\sin v),\frac{\partial }{{\partial v}}(3\sinh u)} \right\rangle$$

$$= \left\langle {\cosh u\frac{\partial }{{\partial v}}(\cos v),2\cosh u\frac{\partial }{{\partial v}}(\sin v),3\sinh u\frac{\partial }{{\partial v}}(1)} \right\rangle$$

$$= \langle - \cosh u\sin v,2\cosh u\cos v,0\rangle$$

## Step 5: Calculation of $${{\rm{r}}_u} \times {{\rm{r}}_v}$$ :

Substitute $$\langle \sinh u\cos v,2\sinh u\sin v,3\cosh u\rangle$$ for $${{\rm{r}}_u}$$ and $$\langle - \cosh u\sin v,2\cosh u\cos v,0\rangle$$ for $${{\rm{r}}_v}$$ in the expression $${{\rm{r}}_u} \times {{\rm{r}}_v}$$,

$$${{\rm{r}}_u} \times {{\rm{r}}_v} = \left\langle {\begin{array}{*{20}{l}}{\sinh u\cos v,}\\{2\sinh u\sin v,3\cosh u}\end{array}} \right\rangle \times \left\langle {\begin{array}{*{20}{l}}{ - \cosh u\sin v,}\\{2\cosh u\cos v,0}\end{array}} \right\rangle$$$

Rewrite and compute the expression as follows.

$${{\rm{r}}_u} \times {{\rm{r}}_v} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{\sinh u\cos v}&{2\sinh u\sin v}&{3\cosh u}\\{ - \cosh u\sin v}&{2\cosh u\cos v}&0\end{array}} \right|$$

$$= \left( {\begin{array}{*{20}{c}}{\left| {\begin{array}{*{20}{c}}{2\sinh u\sin v}&{3\cosh u}\\{2\cosh u\cos v}&0\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}{\sinh u\cos v}&{3\cosh u}\\{ - \cosh u\sin v}&0\end{array}} \right|{\rm{j}}}\\{ + \left| {\begin{array}{*{20}{c}}{\sinh u\cos v}&{2\sinh u\sin v}\\{ - \cosh u\sin v}&{2\cosh u\cos v}\end{array}} \right|{\rm{k}}}\end{array}} \right)$$

$$= \left\{ {\begin{array}{*{20}{l}}{\left( {0 - \left( {6{{\cosh }^2}u\cos v} \right)} \right){\rm{i}} - \left( {0 - \left( { - 3{{\cosh }^2}u\sin v} \right)} \right){\rm{j}}}\\{ + \left( {2\sinh u\cosh u{{\cos }^2}v - \left( { - 2\sinh u\cosh u{{\sin }^2}v} \right)} \right){\rm{k}}}\end{array}} \right\}$$

$$= - 6{\cosh ^2}u\cos v{\rm{i}} - 3{\cosh ^2}u\sin v{\rm{j}} + 2\sinh u\cosh u\left( {{{\cos }^2}v + {{\sin }^2}v} \right){\rm{k}}$$

Simplify the expression as follows.

$${{\rm{r}}_u} \times {{\rm{r}}_v} = \left\langle { - 6{{\cosh }^2}u\cos v, - 3{{\cosh }^2}u\sin v,2\sinh u\cosh u} \right\rangle$$

Substitute $$\left\langle { - 6{{\cosh }^2}u\cos v, - 3{{\cosh }^2}u\sin v,2\sinh u\cosh u} \right\rangle$$ for $$_u \times {{\rm{r}}_v}$$ in equation (4),

Consider the limits of $$v$$ as follows.

$$0 \le v \le 2\pi$$

As the $$z$$ lies between $$( - 3)$$ to $$3$$ , find the limits of $$u$$ as follows.

$$\begin{array}{l}{u_{{\rm{lower }}}} = - {\sinh ^{ - 1}}( - 1) = - \ln (1 + \sqrt 2 )\\{u_{{\rm{upper }}}} = {\sinh ^{ - 1}}(1) = \ln (1 + \sqrt 2 )\end{array}$$

The limits of $$u$$ are written as follows.

$$- \ln (1 + \sqrt 2 ) \le u \le \ln (1 + \sqrt 2 )$$

Apply the limits of \$u, v$$,andrewritetheexpressioninequation(7)asfollows.$$$$A(S) = \int_0^{2\pi } {\int_{ - \ln (1 + \sqrt 2 )}^{\ln (1 + \sqrt 2 )} {\sqrt {36{{\cosh }^4}u{{\cos }^2}v + 9{{\cosh }^4}u{{\sin }^2}v + 4{{\sinh }^2}u{{\cosh }^2}u} } } dudv$$

Thus, the expression for surface area of the hyperboloid of one sheet with the parametric equations $$x = a\cosh u\cos v,y = b\cosh u\sin v,z = c\sinh u$$ for $$a = 1,b = 2$$, and $$c = 3$$ is

$$A(S) = \int_0^{2\pi } {\int_{ - \ln (1 + \sqrt 2 )}^{\ln (1 + \sqrt 2 )} {\sqrt {36{{\cosh }^4}u{{\cos }^2}v + 9{{\cosh }^4}u{{\sin }^2}v + 4{{\sinh }^2}u{{\cosh }^2}u} } } dudv.$$