Question

Air is blown into a spherical balloon. If its diameter d is increasing at the rate of 3 cm/min, then the rate at which the volume of the balloon is increasing when d = 10 cm, is

• A. 120πcm³/min
• B. 150πcm³/min
• C. 100πcm³/min
• D. 180πcm³/min
• E. 210πcm³/min

Answer 1

The Correct Option is B

The volume of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3, \] where \( r \) is the radius of the sphere. We are given that the diameter \( d \) is increasing at a rate of 3 cm/min. Since the radius \( r \) is half of the diameter, the rate of change of the radius is: \[ \frac{dr}{dt} = \frac{1}{2} \frac{dd}{dt} = \frac{1}{2} \times 3 = 1.5 \, \text{cm/min}. \] Now, differentiate the volume equation with respect to time \( t \): \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}. \] Substitute the given values: when \( d = 10 \) cm, the radius \( r = 5 \) cm. We already know that \( \frac{dr}{dt} = 1.5 \) cm/min. Thus, the rate of change of the volume is: \[ \frac{dV}{dt} = 4 \pi (5)^2 \times 1.5 = 4 \pi \times 25 \times 1.5 = 150 \pi \, \text{cm}^3/\text{min}. \]

So, the correct option is (B) : \(150\pi \ cm^3/min\)

Answer 2

• We are given that the diameter \(d\) of a spherical balloon is increasing at a rate of 3 cm/min, i.e., \(\frac{dd}{dt} = 3\) cm/min. We want to find the rate at which the volume \(V\) is increasing, \(\frac{dV}{dt}\), when \(d = 10\) cm.
• The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\), where \(r\) is the radius. Since \(d = 2r\), we have \(r = \frac{d}{2}\). Substitute this into the volume formula: \[V = \frac{4}{3}\pi \left(\frac{d}{2}\right)^3 = \frac{4}{3}\pi \frac{d^3}{8} = \frac{\pi}{6} d^3\]
• Now, differentiate both sides with respect to time \(t\): \[\frac{dV}{dt} = \frac{\pi}{6} \cdot 3d^2 \cdot \frac{dd}{dt} = \frac{\pi}{2} d^2 \frac{dd}{dt}\]
• We are given that \(\frac{dd}{dt} = 3\) cm/min and we want to find \(\frac{dV}{dt}\) when \(d = 10\) cm. Substitute these values into the equation: \[\frac{dV}{dt} = \frac{\pi}{2} (10)^2 (3) = \frac{\pi}{2} (100)(3) = 150\pi\]
• Therefore, the rate at which the volume of the balloon is increasing when \(d = 10\) cm is \(150\pi\) cm³/min.