Chapter 6.1, Problem 59E

a.

To determine

To find: the play area available to dog.

Answer to Problem 59E

  $530.1\text{ square feet}$

Explanation of Solution

Given information:

  $\begin{array}{c}\text{Length of the house}=33\text{ feet}\\ \text{Breadth of the house}=47\text{ feet}\\ \text{Length of the leash}=15\text{ feet}\end{array}$

Calculation:

The dog can move around in a circular path up to the length of the lease. So the radius is 15 feet. Since the dog is in the corner and he has access to ${\left(\frac{3}{4}\right)}^{th}$ of that area, so he can move by $270°$ . The play area of dog will be evaluated as follows:

  $\begin{array}{c}\text{Play area of dog}\left(A\right)=\frac{1}{2}{r}^2\theta \\ =\frac{1}{2}{\left(15\right)}^2\left[\left(270°\right)\left(\frac{\pi }{180°}\right)\right]\left[\text{Convert }\theta \text{ in radian measure}\right]\\ =\frac{1}{2}{\left(15\right)}^2\left[\frac{3\pi }{2}\right]\\ =530.1\text{ square feet}\end{array}$

Thus the play area available to dog is $530.1\text{ square feet}$ .

b.

To determine

To find: the length of the leash.

Answer to Problem 59E

  $17.8\text{ feet}$

Explanation of Solution

Given information:

  $\begin{array}{c}\text{Length of the house}=33\text{ feet}\\ \text{Breadth of the house}=47\text{ feet}\\ \text{Play area available for dog }\left(A\right)=750\text{ square feet}\end{array}$

Calculation:

Since dog can access only ${\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}^{th}$ area so he can move at a central angle $\left(\theta \right)$ of $270°$ or $\frac{3\pi }{4}$ radians. The length of leash $\left(r\right)$ will be evaluated as follows:

  $\begin{array}{c}A=\frac{1}{2}{r}^2\theta \\ 750=\frac{1}{2}{r}^2\left(\frac{3\pi }{4}\right)\left[\text{Put the values of }A\text{ and }\theta \right]\\ r^2=750\left(2\right)\left(\frac{4}{3\pi }\right)\left[\text{Rearrange in terms of }r\text{ and solve}\right]\\ r^2=636.6\\ r=\sqrt{636.6}\\ =17.8\text{ feet}\end{array}$

Thus the length of lease should be $17.8\text{ feet}$ .