Question

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 by 0.7 cm per second, find the rate at which the radius is changing when the radius is 2 cm and the length is 9 cm.

Answer 1

Answer:

The radius is decreatsing at a rate 0.78 cm per second.

Step-by-step explanation:

We are given the following in the question:

$\dfrac{dl}{dt} = 0.7\text{ cm per second}$

Instant radius,r = 2 cm

Instant length,l = 9cm

The radius of the cylinder decreases.

Volume of cylinder =

$V =\pi r^2l$

Since the volume of the cylinder does not changes, we can write:

$\dfrac{dV}{dt}=0$

Rate of change of volume:

$\dfrac{dV}{dt} = \pi(2r\dfrac{dr}{dt}l + r^2\dfrac{dl}{dt})$

Puttiung values, we get

$0 = \pi(2(2)\dfrac{dr}{dt}(9) + (2)^2(0.7))\\\\\dfrac{dr}{dt}=-\dfrac{4\times 0.7}{4\times 9} \approx -0.78$

Thus, the radius is decreatsing at a rate 0.78 cm per second.

Answer 2

Using differentiation, it is found that the radius is increasing at a rate of 0.35 cm/second.

The length of a cylinder is the diameter, which is double the radius, hence:

$l = 2r \rightarrow r = \frac{l}{2}$

Applying differentiation, the rate of change of the radius is given by:

$\frac{dr}{dt} = \frac{1}{2}\frac{dl}{dt}$

We are given that the rate of change of the length is $\frac{dl}{dt} = 0.7$, hence:

$\frac{dr}{dt} = \frac{1}{2}\frac{dl}{dt} = \frac{1}{2}(0.7) = 0.35$

The radius is increasing at a rate of 0.35 cm/second.

A similar problem is given at https://brainly.com/question/9543179