a spherical balloon is inflated and its volume increases at a rate of 19 in.³/min. what is the rate of…
a spherical balloon is inflated and its volume increases at a rate of 19 in.³/min. what is the rate of change of its radius when the radius is 12 in.?\nthe balloons radius is increasing at a rate of at the instant the radius is 12 in.\n(round to three decimal places as needed.)
Answer
Explanation:
Step1: Write the volume formula for a sphere
The volume formula for a sphere is ( V=\frac{4}{3}\pi r^{3}).
Step2: Differentiate both sides 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} = 19) and (r = 12). 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}}).
Step4: Substitute the given values
Substitute (\frac{dV}{dt}=19) and (r = 12) into the formula: (\frac{dr}{dt}=\frac{19}{4\pi\times(12)^{2}}). Calculate (4\pi\times(12)^{2}=4\pi\times144 = 576\pi). Then (\frac{dr}{dt}=\frac{19}{576\pi}). Using a calculator, (\frac{19}{576\pi}\approx\frac{19}{576\times3.14159}\approx\frac{19}{1809.557}\approx0.011).
Answer:
(0.011) in/min