Question

An open hemispherical storage tank has radius $13$ m. Oil flows into the tank such that the depth $h$ of oil in the tank changes at the rate of $3$ m/hr. When $h=1$ m, the rate of change of the area of the top surface of the oil is:

• A. $24\pi \text{ m}^2/\text{hr}$
• B. $72\pi \text{ m}^2/\text{hr}$
• C. $26\pi \text{ m}^2/\text{hr}$
• D. $75\pi \text{ m}^2/\text{hr}$

Hint:

For spherical liquid problems, the standard relation \[ r^2=2Rh-h^2 \] is extremely useful and appears frequently in related rates questions.

Answer 1

The Correct Option is B

Concept: This is a related rates problem involving a hemisphere. The oil surface forms a circle whose radius changes as the depth changes. The area of the circular surface is: \[ A=\pi r^2 \] We first establish a relation between the radius $r$ of the oil surface and the depth $h$.

Step 1: Relating $r$ and $h$ using geometry. Let the radius of the hemisphere be: \[ R=13 \text{ m} \] Consider the vertical cross-section through the center. Using the right triangle formed: \[ r^2+(R-h)^2=R^2 \] Substituting $R=13$: \[ r^2+(13-h)^2=13^2 \] \[ r^2+169-26h+h^2=169 \] \[ r^2=26h-h^2 \]

Step 2: Expressing area in terms of $h$. Area of oil surface: \[ A=\pi r^2 \] Substituting $r^2$: \[ A=\pi(26h-h^2) \]

Step 3: Differentiating with respect to time. Differentiate both sides with respect to $t$: \[ \frac{dA}{dt}=\pi\left(26\frac{dh}{dt}-2h\frac{dh}{dt}\right) \] \[ \frac{dA}{dt}=\pi(26-2h)\frac{dh}{dt} \]

Step 4: Substituting the given values. Given: \[ h=1 \] and \[ \frac{dh}{dt}=3 \text{ m/hr} \] Therefore, \[ \frac{dA}{dt}=\pi(26-2)(3) \] \[ \frac{dA}{dt}=\pi(24)(3) \] \[ \frac{dA}{dt}=72\pi \text{ m}^2/\text{hr} \] Hence, \[ \boxed{72\pi \text{ m}^2/\text{hr}} \]