Chapter 10.4, Problem 50E

$\left(a\right)$

To determine

To calculate: The equation that represent the cross section of base of clock,

  

Answer to Problem 50E

The equation that represent the cross section of base of clock is $x^2-\frac{y^2}{27}=1$ .

Explanation of Solution

Given information:

The cross section of base of clock is provided below,

  

Formula used:

The equation of horizontal hyperbola is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , here $\left(h,k\right)$ is the center, vertices are a units from center and foci are c units from center and $c^2=a^2+b^2$ .

Calculation:

Consider cross section of base of a clock that is provided below,

  

Observe that it is in shape of hyperbola whose transverse axis is horizontal.

Recall that the equation of horizontal hyperbola is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , here $\left(h,k\right)$ is the center, vertices are a units from center and foci are c units from center and $c^2=a^2+b^2$ .

Here center is origin so $\left(h,k\right)=\left(0,0\right)$ . Next the vertices of hyperbola are at the point $\left(-1,0\right)\text{ and }\left(1,0\right)$ . Since a is distance between vertex and center therefore, $a=1$ .

Also, the hyperbola passes through the point $\left(2,9\right)$ .

Substitute x as 2, y as 9, a as 1, h as 0 and k as 0 in the equation $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ to compute value of b .

  $\begin{array}{c}\frac{{\left(2-0\right)}^2}{1^2}-\frac{{\left(9-0\right)}^2}{b^2}=1\\ 4-1=\frac{81}{b^2}\\ b^2=\frac{81}{3}\\ b^2=27\end{array}$

Now, substitute $a^2=1,b^2=27,h=0\text{ and }k=0$ in the equation $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

  $x^2-\frac{y^2}{27}=1$

Thus, the equation that represent the curved sides of the sculpture is $x^2-\frac{y^2}{27}=1$ .

$\left(b\right)$

To determine

To calculate: The width of the base 4 inches from the bottom of the clock,

  

Answer to Problem 50E

The width of the base 4 inches from the bottom of the clock is $1.89\text{ feet}$ .

Explanation of Solution

Given information:

The cross section of a sculpture is provided below,

  

Formula used:

The equation of horizontal hyperbola is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , here $\left(h,k\right)$ is the center, vertices are a units from center and foci are c units from center and $c^2=a^2+b^2$ .

Calculation:

Consider cross section of base of a clock that is provided below,

  

Observe that it is in shape of hyperbola whose transverse axis is horizontal.

Recall that the equation of horizontal hyperbola is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , here $\left(h,k\right)$ is the center, vertices are a units from center and foci are c units from center and $c^2=a^2+b^2$ .

Here center is origin so $\left(h,k\right)=\left(0,0\right)$ . Next the vertices of hyperbola are at the point $\left(-1,0\right)\text{ and }\left(1,0\right)$ . Since a is distance between vertex and center therefore, $a=1$ .

Also, the hyperbola passes through the point $\left(2,9\right)$ .

Substitute x as 2, y as 9, a as 1, h as 0 and k as 0 in the equation $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ to compute value of b .

  $\begin{array}{c}\frac{{\left(2-0\right)}^2}{1^2}-\frac{{\left(9-0\right)}^2}{b^2}=1\\ 4-1=\frac{81}{b^2}\\ b^2=\frac{81}{3}\\ b^2=27\end{array}$

Now, substitute $a^2=1,b^2=27,h=0\text{ and }k=0$ in the equation $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

  $x^2-\frac{y^2}{27}=1$

Therefore, the equation that represent the curved sides of the sculpture is $x^2-\frac{y^2}{27}=1$ .

Now, it is provided that each unit in coordinate plane represents $\frac{1}{2}$ foot. So width of the width of the base 4 inches from the bottom of the clock is evaluated below.

Now 4 inches from the bottom is equivalent to $\frac{1}{3}$ foot which is further equal to $\frac{2}{3}$ units on Cartesian plane.

Substitute $y=-\frac{25}{3}$ in the above equation to evaluate the width of clock,

  $\begin{array}{c}{x}^2-\frac{{\left(\frac{-25}{3}\right)}^2}{27}=1\\ x=1.89\end{array}$

Because each unit in coordinate plane represents $\frac{1}{2}$ foot, width is twice of x that is,

  $\begin{array}{c}2x=2\left(\frac{1}{2}\right)\left(1.89\right)\\ =1.89\end{array}$

Thus, the width of the base 4 inches from the bottom of the clock is $1.89\text{ feet}$ .