Given the problem, we know the following information:

• The rate at which the tank is being filled: $\frac{dV}{dt} = 88$ cubic cm per hour.
• The rate at which the water level rises: $\frac{dh}{dt} = 7$ cm per hour.

We need to find the radius of the tank. Let's denote the radius by $r$ cm and the height of the water level by $h$ cm.

Step 1: Formula for the volume of a cylinder

The volume $V$ of a cylinder is given by the formula:

$V = \pi r^2 h$

Step 2: Differentiate with respect to time

To relate the rate of volume increase to the rate of height increase, differentiate both sides of the equation with respect to time $t$:

$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$

Step 3: Substitute the known values

Now, substitute the given rates:

$88 = \pi r^2 \times 7$

Step 4: Solve for $r$

Simplify the equation to solve for $r$:

$r^2 = \frac{88}{7\pi}$

$r^2 = \frac{88}{7 \times 3.1416} \approx 4$

$r = \sqrt{4} = 2$

Final Answer:

The radius of the tank is $2$ cm.

Would you like more details, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. How would the formula change if the tank were a cone instead of a cylinder?
2. What happens to the radius if the rate of water filling doubles?
3. How would you calculate the height of the water after 5 hours?
4. If the radius is doubled, how does it affect the volume of the tank?
5. How would you calculate the volume if the height of the tank were fixed?

Tip: Always make sure units are consistent when dealing with rates of change in geometry problems.