Related rates problem for the dimensions of a cylinder

[nikk33213]

TL;DR Summary: related rates problem

The radius of a cylinder is increasing at a constant rate of 3 meters per minute. The volume remains a constant 1115 cubic meters. At the instant when the height of the cylinder is 44 meters, what is the rate of change of the height?

 

[erobz]

TL;DR Summary: related rates problem

The radius of a cylinder is increasing at a constant rate of 3 meters per minute. The volume remains a constant 1115 cubic meters. At the instant when the height of the cylinder is 44 meters, what is the rate of change of the height?

You must show an honest attempt at a solution to receive help via forum rules.

 

[nikk33213]

given: dr/dt=3, v=1115, h = 44, dh/dt = ?
v= (pi)r^2(h)
dv/dt = 2(pi)r(dr/dt)(h) + (pi)r^2(dh/dt)
0 = 2(pi)r(3)(44) + (pi)r^2(dh/dt)

1115 = (pi)r^2(44)
r = 2.84012

0 = 2(pi)*3*2.84012*44 + (pi)(8.066262)(dh/dt)
(-2355.540034)/(25.340909) = dh/dt

-92.954 = dh/dt

Is this correct?

 

[erobz]

given: dr/dt=3, v=1115, h = 44, dh/dt = ?
v= (pi)r^2(h)
dv/dt = 2(pi)r(dr/dt)(h) + (pi)r^2(dh/dt)
0 = 2(pi)r(3)(44) + (pi)r^2(dh/dt)

1115 = (pi)r^2(44)
r = 2.84012

0 = 2(pi)*3*2.84012*44 + (pi)(8.066262)(dh/dt)
(-2355.540034)/(25.340909) = dh/dt

-92.954 = dh/dt

Is this correct?

I don't get that result.

Use latex to present your solution, It's not going to take you long to learn ( it's a very natural extension to what you are already doing) see LaTeX Guide Please, leave it all in variables, plug in values only as the very last step.

 

[erobz]

Never mind, I missed a factor of $h$ in there; it checks out(other than the lack of presented units and extra sig figs).

$$ \frac{dh}{dt} = -2 h \sqrt{ \frac{\pi h }{V}} \frac{dr}{dt} $$

 

[erobz]

You’re welcome.

 

[nikk33213]

thank you

 

[Mark44]

-92.954 = dh/dt
Is this correct?

No, as it does not include units. All of the dimensions and rates are given with units. The solution should also include the relevant units.

 

[naveed]

TL;DR Summary: related rates problem

The radius of a cylinder is increasing at a constant rate of 3 meters per minute. The volume remains a constant 1115 cubic meters. At the instant when the height of the cylinder is 44 meters, what is the rate of change of the height?

When the radius of the cylinder is increasing at a constant rate and the volume remains constant, the rate of change of the height can be found using the formula:

Rate of change of height = (constant volume) / (π * radius^2)

So, at the instant when the height of the cylinder is 44 meters, the rate of change of the height is:

Rate of change of height = 1115 / (π * radius^2)

We need to find the radius at this instant. Since the volume of the cylinder is constant, we can use the formula for the volume of a cylinder:

Volume = π * radius^2 * height

Given that the volume is 1115 cubic meters and the height is 44 meters, we can rearrange the formula to solve for the radius:

Radius^2 = Volume / (π * height)

Now, we can plug in the values:

Radius^2 = 1115 / (π * 44)

Then, we find the square root of the result to get the radius:

Radius ≈ √(1115 / (π * 44))

Once we have the radius, we can substitute it into the formula for the rate of change of height to find the rate of change of the height.

This is the simple explanation of how to find the rate of change of the height of the cylinder when its volume is constant and the radius is increasing at a constant rate.

 

[Mark44]

When the radius of the cylinder is increasing at a constant rate and the volume remains constant, the rate of change of the height can be found using the formula:

Rate of change of height = (constant volume) / (π * radius^2)

Your formula doesn't take into account that the radius is changing, so this really is just a formula for the height, not $\frac{dh}{dt}$.

 

[naveed]

Your formula doesn't take into account that the radius is changing, so this really is just a formula for the height, not dhdt.

Apologies for any confusion. To incorporate changing radius, you'll need to use a formula that considers both height and radius variation.

 

[Mark44]

Apologies for any confusion. To incorporate changing radius, you'll need to use a formula that considers both height and radius variation.

In other words, $\frac{dh}{dt}$ and $\frac{dr}{dt}$. To get these you need to use some calculus.