A piston is seated at the top of a cylindrical chamber with radius 3 cm when it starts moving into the chamber at a constant speed of 4 cm/s (see figure). What is the rate of change of the volume of the cylinder when the piston is 15 cm from the base of the chamber? a) Let V, r, and h be thevolume, radius, and height of a cylinder, respectively. Write an equation relating V, r, and h.
b) Find the related rates equation.
c) When the piston is 15cm from the base of the chamber, the volume of the cylinder is changing at a rate of about

1. V = πr²h

2. dv/dt = πr²dh/dt

3. dv/dt = 113.04

Step-by-step explanation:

We have these information

Radius R = 3cm

Height = h

Volume = V

We have been given dh/dt = 4cm

1.

An equation relating V, r, h

Volume of cylinder = πr²h

2.

The related rates equation

dv/dt = πr²dh/dt

dv/dt = π3²dh/dt

3. Continuing from number 2 above, we have

dv/dt = π3²dh/dt

= π9x4

= 3.14x9x4

= 113.04

These are the answers to the questions

518,400

step-by-step explanation:

the area is multiplying all the given measurements together.

40 * 18 * 30 * 24 = 518,400

a² + b²= c²

24² + 7² = c²

c= 25