Question

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 2 cm/s (see figure). what is the rate of change of the volume of the cylinder when the piston is 3 cm from the base of the chamber? when the piston is 3 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.)

Explanation:

Step1: Write the formula for the volume of a cylinder

The volume \( V \) of a cylinder is given by \( V=\pi r^{2}h \), where \( r \) is the radius and \( h \) is the height. Here, \( r = 5\) cm (constant), so \( V = 25\pi h\).

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

Using the chain - rule \(\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}\). Since \( V = 25\pi h\), then \(\frac{dV}{dh}=25\pi\). We are given that \(\frac{dh}{dt}=- 2\) cm/s (negative because the height \( h \) is decreasing as the piston moves down).

Step3: Calculate \(\frac{dV}{dt}\)

Substitute \(\frac{dV}{dh}=25\pi\) and \(\frac{dh}{dt}=-2\) into the chain - rule formula \(\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}\). So \(\frac{dV}{dt}=25\pi\times(-2)=- 50\pi\approx - 157.08\) \(cm^{3}/s\). The negative sign indicates that the volume is decreasing.

Answer:

\(-157.08\) \(cm^{3}/s\)