Related Rates Clay Pot Problem

[Burjam]

Homework Statement

A potter forms a piece of clay into a cylinder. As he rolls it, the length, L, of the cylinder increases and the radius, r decreases. If the length of the cylinder is increasing at 0.1 cm per second, find the rate at which the radius is changing when the radius is 1 cm and the length is 5 cm.

Homework Equations

N/A

The Attempt at a Solution

So I know that dL/ds=0.1. But I don't know exactly how the radius changes as L changes. I'm having trouble setting up a function for this. If I had that I could do the rest of the problem.

 

[Dick]

Homework Statement

A potter forms a piece of clay into a cylinder. As he rolls it, the length, L, of the cylinder increases and the radius, r decreases. If the length of the cylinder is increasing at 0.1 cm per second, find the rate at which the radius is changing when the radius is 1 cm and the length is 5 cm.

Homework Equations

N/A

The Attempt at a Solution

So I know that dL/ds=0.1. But I don't know exactly how the radius changes as L changes. I'm having trouble setting up a function for this. If I had that I could do the rest of the problem.

The volume is constant as the potter rolls. What's an equation for the volume of a cylinder in terms of the radius and length?

 

[Burjam]

The volume is constant as the potter rolls. What's an equation for the volume of a cylinder in terms of the radius and length?

V=Lπr²

But how can I connect this with time?

 

[Dick]

V=Lπr²

But how can I connect this with time?

Differentiate both sides d/dt.

 

[Burjam]

How can I differentiate this function with respect to time?

 

[Dick]

How can I differentiate this function with respect to time?

V is a constant. r and L are both functions of time. Write V=πr(t)^2*L(t). Now differentiate it.

 

[Burjam]

V is a constant. r and L are both functions of time. Write V=πr(t)^2*L(t). Now differentiate it.

Thank you I think I figured it out:

dV/dt=2L(t)πr(t)r'(t)+L'(t)πr(t)²
0=2L(t)πr(t)r'(t)+L'(t)πr(t)²
0=2L(t)r'(t)+L'(t)r(t)
2L(t)r'(t)=-L'(t)r(t)
r'(t)=-L'(t)r(t)/2L(t)r'(t)
r'(t)=-0.1(1)/2(5)
r'(t)=-0.1/10=-0.01cm/s

 

[Dick]

Thank you I think I figured it out:

dV/dt=2L(t)πr(t)r'(t)+L'(t)πr(t)²
0=2L(t)πr(t)r'(t)+L'(t)πr(t)²
0=2L(t)r'(t)+L'(t)r(t)
2L(t)r'(t)=-L'(t)r(t)
r'(t)=-L'(t)r(t)/2L(t)r'(t)
r'(t)=-0.1(1)/2(5)
r'(t)=-0.1/10=-0.01cm/s

Right. Well done.