A.P. Calculus - Practice with Related Rates: Problems on volume and surface area changes in geometric shapes.

Here are the step-by-step solutions for the three related rates problems shown in your work.

Problem 1

Goal: Find the surface area ($S$) when $\frac{dV}{dt} = 9 \frac{dr}{dt}$.

1. Start with the Volume formula:
$$V = \frac{4}{3}\pi r^3$$

2. Differentiate with respect to time ($t$):
Using the chain rule, take the derivative of both sides:
$$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$$
$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$

3. Use the given condition:
The problem states that $\frac{dV}{dt} = 9 \frac{dr}{dt}$. Substitute this into our derived equation:
$$9 \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$$

4. Solve for $r$:
Divide both sides by $\frac{dr}{dt}$ (assuming the radius is changing, so $\frac{dr}{dt} \neq 0$):
$$9 = 4\pi r^2$$

5. Find the Surface Area:
Recall the formula for the surface area of a sphere: $S = 4\pi r^2$.
Looking at our equation from Step 4, we see that $4\pi r^2$ is exactly equal to 9.
Therefore, $S = 9$.

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Problem 2

Goal: Find how fast the water level is rising ($\frac{dh}{dt}$) when the depth $h = 5$ meters.
*Given:* Height of tank $H = 10$, Diameter of base $= 8$ (so Radius $R = 4$), Water pumped in at $\frac{dV}{dt} = 2$.

1. Relate Radius ($r$) and Height ($h$):
Use similar triangles. The ratio of the radius to the height is constant for the cone.
$$\frac{r}{h} = \frac{R}{H} = \frac{4}{10} = \frac{2}{5}$$
So, $r = \frac{2}{5}h$.

2. Substitute into the Volume formula:
We want an equation with only $V$ and $h$ because we are looking for $\frac{dh}{dt}$.
$$V = \frac{1}{3}\pi r^2 h$$
Substitute $r = \frac{2}{5}h$:
$$V = \frac{1}{3}\pi \left(\frac{2}{5}h\right)^2 h$$
$$V = \frac{1}{3}\pi \left(\frac{4}{25}h^2\right) h$$
$$V = \frac{4\pi}{75} h^3$$

3. Differentiate with respect to time ($t$):
$$\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \frac{dh}{dt}$$
Simplify ($3/75$ becomes $1/25$):
$$\frac{dV}{dt} = \frac{4\pi}{25} h^2 \frac{dh}{dt}$$

4. Plug in known values and solve:
We know $\frac{dV}{dt} = 2$ and $h = 5$.
$$2 = \frac{4\pi}{25} (5)^2 \frac{dh}{dt}$$
$$2 = \frac{4\pi}{25} (25) \frac{dh}{dt}$$
The 25s cancel out:
$$2 = 4\pi \frac{dh}{dt}$$
$$\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi}$$

***

Problem 3

Goal: Find the Volume ($V$) when $\frac{dV}{dt} = 12 \frac{dr}{dt}$.
*Given:* Height is always three times the radius ($h = 3r$).

1. Substitute into the Volume formula:
We want an equation with only $V$ and $r$.
$$V = \frac{1}{3}\pi r^2 h$$
Substitute $h = 3r$:
$$V = \frac{1}{3}\pi r^2 (3r)$$
The $\frac{1}{3}$ and $3$ cancel out:
$$V = \pi r^3$$

2. Differentiate with respect to time ($t$):
$$\frac{dV}{dt} = 3\pi r^2 \frac{dr}{dt}$$

3. Use the given condition:
The problem states $\frac{dV}{dt} = 12 \frac{dr}{dt}$. Substitute this in:
$$12 \frac{dr}{dt} = 3\pi r^2 \frac{dr}{dt}$$

4. Solve for $r$:
Divide by $\frac{dr}{dt}$:
$$12 = 3\pi r^2$$
Divide by 3:
$$4 = \pi r^2$$
$$r^2 = \frac{4}{\pi} \implies r = \frac{2}{\sqrt{\pi}}$$

5. Calculate the Volume:
Use the simplified volume formula from Step 1: $V = \pi r^3$.
$$V = \pi \left( \frac{2}{\sqrt{\pi}} \right)^3$$
$$V = \pi \left( \frac{8}{\pi\sqrt{\pi}} \right)$$
Cancel one $\pi$ from the numerator and denominator:
$$V = \frac{8}{\sqrt{\pi}}$$

Final Answer:
1. Surface Area = 9
2. Rate of rise = $\frac{1}{2\pi}$ m/min
3. Volume = $\frac{8}{\sqrt{\pi}}$