## Precalculus (6th Edition) Blitzer

The required solution is $\text{157}\ \text{ft/min}$
The angle covered in one revolution is $2\pi \ \text{radians}$. The carousel is rotating at 2.5 revolutions per minute. The angle covered in one minute is $2.5\left( 2\pi \ \text{radians} \right)=5\pi \ \text{radians}$ Angular speed $\omega$ is $5\pi \ \text{radians per minute}$. And the outer row of the animals is $20\ \text{feet}$ from the center denoted by ${{r}_{1}}$. The linear speed ${{v}_{1}}$ of the outer row of animals is given by ${{v}_{1}}={{r}_{1}}\omega$ Put $20\ \text{feet}$ for ${{r}_{1}}$ and $5\pi \ \text{radians per minute}$ for $\omega$: \begin{align} & {{v}_{1}}=\left( 20\ \text{feet} \right)\left( 5\pi \ \text{radians per minute} \right) \\ & =100\pi \ \text{feet per minute} \end{align} The inner row of animals is $10\ \text{feet}$ from the center denoted by ${{r}_{2}}$. The linear speed ${{v}_{2}}$ of the inner row of animals is given by ${{v}_{2}}={{r}_{2}}\omega$ put $10\ \text{feet}$ for ${{r}_{2}}$ and $5\pi \ \text{radians per minute}$ for $\omega$: \begin{align} & {{v}_{2}}=\left( 10\ \text{feet} \right)\left( 5\pi \ \text{radians per minute} \right) \\ & =50\pi \ \text{feet per minute} \end{align} And the difference in linear speeds of outer and inner row of animals is given by $\Delta v={{v}_{1}}-{{v}_{2}}$ Put $100\pi \ \text{feet per minute}$ for ${{v}_{1}}$ and $50\pi \ \text{feet per minute}$ for ${{v}_{2}}$: \begin{align} & \Delta v=100\pi \ \text{feet per minute}-50\pi \ \text{feet per minute} \\ & =50\pi \ \text{feet per minute} \end{align} Put $\pi =3.14159$: \begin{align} & \Delta v=50\left( 3.14159 \right)\ \text{feet per minute} \\ & =\text{157}\text{.07}\ \text{feet per minute}\approx \text{157}\ \text{feet per minute} \end{align}