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 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.)

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.)

Answer

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)