9 a conical tank has height 10 ft and radius 5 ft at the top. water flows into the tank at a rate of 2…

Explanation:

Step1: Recall volume formula for cone

The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. For a similar - shaped cone (the cone of water in the tank), the ratio of radius to height is constant. Given the large cone has $h = 10$ ft and $r = 5$ ft, so $\frac{r}{h}=\frac{5}{10}=\frac{1}{2}$, then $r=\frac{h}{2}$. Substitute $r=\frac{h}{2}$ into the volume formula: $V=\frac{1}{3}\pi(\frac{h}{2})^{2}h=\frac{1}{12}\pi h^{3}$.

Step2: Differentiate with respect to time

Differentiate both sides of $V=\frac{1}{12}\pi h^{3}$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=\frac{1}{12}\pi\times3h^{2}\frac{dh}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}$.

Step3: Solve for $\frac{dh}{dt}$

We know that $\frac{dV}{dt}=2$ ft³/min and $h = 4$ ft. Substitute these values into the equation $\frac{dV}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}$: [ \begin{align*} 2&=\frac{\pi}{4}(4)^{2}\frac{dh}{dt}\ 2&=4\pi\frac{dh}{dt}\ \frac{dh}{dt}&=\frac{1}{2\pi}\text{ ft/min} \end{align*} ]

Answer:

$\frac{1}{2\pi}$ ft/min

Explanation for problem 10:

Step1: Establish a relationship between angle and distance

Let $\theta$ be the angle that the light rotates, and $x$ be the distance along the shore from the closest point on the shore to the observer. We know that $\tan\theta=\frac{x}{1}$ (since the distance from the island to the shore is 1 mile), so $x = \tan\theta$.

Step2: Differentiate with respect to time

Differentiate both sides of $x=\tan\theta$ with respect to time $t$. Using the chain - rule, $\frac{dx}{dt}=\sec^{2}\theta\frac{d\theta}{dt}$.

Step3: Find $\frac{d\theta}{dt}$ and $\sec\theta$

The light rotates at 10 revolutions per minute. Since 1 revolution is $2\pi$ radians, $\frac{d\theta}{dt}=10\times2\pi = 20\pi$ radians per minute. The observer is 2 miles away from the closest point on the shore, and the distance from the island to the shore is 1 mile. By the Pythagorean theorem, the hypotenuse of the right - triangle formed by the island, the closest point on the shore, and the observer is $\sqrt{1 + 4}=\sqrt{5}$ miles. And $\sec\theta=\frac{\sqrt{5}}{1}=\sqrt{5}$ (since $\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}$ in the right - triangle with respect to $\theta$).

Step4: Calculate $\frac{dx}{dt}$

Substitute $\frac{d\theta}{dt}=20\pi$ and $\sec\theta=\sqrt{5}$ into $\frac{dx}{dt}=\sec^{2}\theta\frac{d\theta}{dt}$: [ \begin{align*} \frac{dx}{dt}&=(\sqrt{5})^{2}\times20\pi\ &= 100\pi\text{ miles/min} \end{align*} ]

Answer for problem 10:

$100\pi$ miles/min