a piston is seated at the top of a cylindrical chamber with radius 5 cm when it starts moving into the…

a piston is seated at the top of a cylindrical chamber with radius 5 cm when it starts moving into the chamber at a constant speed of 6 cm/s (see figure). what is the rate of change of the volume of the cylinder when the piston is 11 cm from the base of the chamber? when the piston is 11 cm from the base of the chamber, the volume of the cylinder is changing at a rate of about (round to the nearest hundredth as needed.)
Answer
Explanation:
Step1: Recall the volume formula for a cylinder
The volume formula for a cylinder is (V=\pi r^{2}h). Here, the radius (r = 5) cm (constant), and (h) is the height (distance of the piston from the base of the chamber). So (V=\pi\times(5)^{2}\times h=25\pi h).
Step2: Differentiate the volume with respect to time
Differentiate (V = 25\pi h) with respect to time (t) using the chain - rule. (\frac{dV}{dt}=25\pi\frac{dh}{dt}). We are given that (\frac{dh}{dt}=6) cm/s (the speed of the piston).
Step3: Calculate the rate of change of volume
Substitute (\frac{dh}{dt}=6) into the equation (\frac{dV}{dt}=25\pi\frac{dh}{dt}). (\frac{dV}{dt}=25\pi\times6). [ \begin{align*} \frac{dV}{dt}&=150\pi\ &\approx150\times 3.14159\ & = 471.24 \end{align*} ]
Answer:
(471.24) cm³/s