Chapter 11.2, Problem 66E

a.

To determine

Determine the domain of the function.

Answer to Problem 66E

Domain: $L>18$

Explanation of Solution

Formula used:

  $\sin \theta =\frac{Perpendicular}{Hypotenuse}$

Pythagoras theorem:

  ${\left(Hypotenus\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2$

Given information:

Weight of light in auditorium= 24 pound

Radius of the light= 18 Inches

By symmetry, the vertical component of the tension on each cable are same so that:

  $T_j=\frac{W}{3}=\frac{24}{3}=8lbs$

Using the right triangle formed,

  

  $\begin{array}{l}\sin \theta =\frac{\sqrt{L^2-{18}^2}}{L}\\ Also,\\ T_j=T\sin \theta \\ 8=T\left(\frac{\sqrt{L^2-{18}^2}}{L}\right)\\ \boxed{T=\frac{8L}{\sqrt{L^2-{18}^2}}}\end{array}$

Domain: $L>18$

b.

To determine

The asymptotes of the graph.

Answer to Problem 66E

Tension on each rope will be 8 lbs.

Explanation of Solution

Given information:

Weight of light in auditorium= 24 pound

Radius of the light= 18 Inches

  

There is a vertical asymptote at $L=18$ which means the maximum length is 18 inches and a horizontal asymptote at $T=8$ , which means the maximum tension on each rope is 8 lbs.

c.

To determine

Determine the minimum length of each cable.

Answer to Problem 66E

Length of each cable will be 30 inches.

Explanation of Solution

Formula used:

  $\sin \theta =\frac{Perpendicular}{Hypotenuse}$

Pythagoras theorem:

  ${\left(Hypotenus\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2$

Given information:

Weight of light in auditorium= 24 pound

Radius of the light= 18 Inches

Now finding the length of cable using $T=\frac{8L}{\sqrt{L^2-{18}^2}}$ :

At $T=10lbs$ :

  $10=\frac{8L}{\sqrt{L^2-{18}^2}}$

  $10\left(\sqrt{L^2-{18}^2}\right)=8L$ Squaring both the sides

  $\begin{array}{l}100\left(L^2-{18}^2\right)=64L^2\\ 100L^2-32400=64L^2\\ 36L^2=32400\\ L^2=900\\ \boxed{L=30inches}\end{array}$