## Air is being pumped into a spherical balloon so that its volume increases at a rate of 150 cm3/s. How fast is the radius of the balloon incr

Air is being pumped into a spherical balloon so that its volume increases at a rate of 150 cm3/s. How fast is the radius of the balloon increasing when the diameter is 30 cm? SOLUTION We start by identifying two things: the given information: The rate of increase of the volume of air is 150 cm3/s. and the unknown: The rate of increase of the radius when the diameter is 30 cm. In order to express these quantities mathematically, we introduce some suggestive notation: Let V be the volume of the balloon and let r be its radius. The key thing to remember is that rates of change are derivatives. In this problem, the volume and the radius are both functions of the time. The rate of increase of the volume with respect to time is the derivative dV/dt, and the rate of increase of the radius is dr/dt. We can therefore restate the given and the unknown as follows: Given: dV dt = 150 cm3/s Unknown: dr dt when r = 30 Incorrect: Your answer is incorrect. seenKey 15 cm. In order to connect dV/dt and dr/dt, we first relate V and r by the formula for the volume of a sphere: V = 4 3 πr3. In order to use the given information, we differentiate each side of this equation with respect to t. To differentiate the right side, we need to use the Chain Rule:

## Answers ( )

Answer:

0.0159cm³/s

Step-by-step explanation:

Rate of change of tje volume, dV/dt=150cm³/s

Volume of a Sphere =⁴/₃πr³

Expressing dV/dt in terms of r and t using the Chain Rule

dV/dt=dV/dr X dr/dt ….. (i)

Since V=⁴/₃πr³

dV/dr=⁴/₃πX3r²=4πr²

From (i)

150=4πr² X dr/dt

Therefore, dr/dt = 150/4πr²

When r=30cm the rate of change of the sphere’s radius dr/dt will be:

dr/dt = 150/(4π X 30 X 30) =1/20π=0.0159cm³/s