a spherical balloon is inflated and its volume increases at a rate of 23 in.³/min. what is the rate of…

a spherical balloon is inflated and its volume increases at a rate of 23 in.³/min. what is the rate of change of its radius when the radius is 8 in.? the balloons radius is increasing at a rate of at the instant the radius is 8 in. (round to three decimal places as needed.)

a spherical balloon is inflated and its volume increases at a rate of 23 in.³/min. what is the rate of change of its radius when the radius is 8 in.? the balloons radius is increasing at a rate of at the instant the radius is 8 in. (round to three decimal places as needed.)

Answer

Explanation:

Step1: Recall the volume formula for a sphere

The volume formula for a sphere is (V=\frac{4}{3}\pi r^{3}).

Step2: Differentiate the volume formula with respect to time (t)

Using the chain - rule, (\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}).

Step3: Solve for (\frac{dr}{dt})

We know that (\frac{dV}{dt} = 23) in³/min and (r = 8) in. Rearranging the equation (\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}) for (\frac{dr}{dt}), we get (\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^{2}}). Substitute (\frac{dV}{dt}=23) and (r = 8) into the formula: (\frac{dr}{dt}=\frac{23}{4\pi(8)^{2}}).

Step4: Calculate the value

First, calculate (4\pi(8)^{2}=4\pi\times64 = 256\pi). Then (\frac{dr}{dt}=\frac{23}{256\pi}\approx\frac{23}{256\times3.14159}\approx\frac{23}{804.247}\approx0.029) in/min.

Answer:

(0.029) in/min