Question

If water is poured into a cylindrical tank of radius 3.5 ft at the rate of 1 cubic ft/min, then the rate at which the level of the water in the tank increases (in ft/min) is:

• A. \( \frac{1}{154} \)
• B. Missing option
• C. \( \frac{2}{77} \)
• D. \( \frac{1}{11} \)

Hint:

For cylindrical volume rate problems, differentiate the volume equation with respect to time and solve for the rate of height increase.

Answer 1

The Correct Option is C

Step 1: Formula for Volume of a Cylinder The volume of a cylinder is given by: \[ V = \pi r^2 h \] Differentiating both sides with respect to time \( t \): \[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]

Step 2: Substituting the Given Values Given: \[ \frac{dV}{dt} = 1 \text{ cubic ft/min}, \quad r = 3.5 \text{ ft} \] \[ 1 = \pi (3.5)^2 \frac{dh}{dt} \] \[ 1 = \pi (12.25) \frac{dh}{dt} \]

Step 3: Solving for \( \frac{dh}{dt} \) \[ \frac{dh}{dt} = \frac{1}{12.25\pi} \] Approximating \( \pi \approx 3.14 \), \[ \frac{dh}{dt} = \frac{1}{12.25 \times 3.14} = \frac{1}{38.465} \] Converting to a fraction, \[ \frac{dh}{dt} = \frac{2}{77} \]