Physics 1 related rates homework question.

[fletchea]

A child loves to watch as you fill a transparent plastic bottle with shampoo. Every horizontal cross section of the bottle is circular, but the diameters of the circles have different values. You pour the brightly colored shampoo into the bottle at a constant rate of 17.0 cm3/s.

At what rate is its level in the bottle rising at a point where the diameter of the bottle is = 6.80 cm?

At what rate is its level in the bottle rising at a point where the diameter of the bottle is = 1.39 cm?

no height was given for the problem. I would show attempts at solving but I am pretty lost on the whole thing. any help would be appreciated.

 

[tiny-tim]

welcome to pf!

hi fletchea! welcome to pf!

forget the varying diameter, just answer the question as if the bottle was an ordinary cylinder with diameter 6.80 cm

 

[HallsofIvy]

tiny-tim's answer is for the first question, of course. For the second question treat as a cylinder of diameter 1.39 cm.

 

[tiny-tim]

tiny-tim's answer is for the first question, of course. For the second question treat as a cylinder of diameter 1.39 cm.

awww … you're making it too easy!

 

[asteriatic]

Hello! It seems like you have been given a related rates problem in your Physics 1 homework. These types of problems involve finding the rate of change of one quantity with respect to another, while both are changing. In this case, we are looking at the rate of change of the level of shampoo in the bottle with respect to the diameter of the bottle.

To solve this problem, we will use the formula for the volume of a cylinder, V = πr^2h, where r is the radius and h is the height of the cylinder. Since we are only concerned with the rate of change of the level of shampoo, we can ignore the height and focus on the radius (which is half of the diameter).

Let's start by finding the radius of the bottle at the given diameters. At 6.80 cm, the radius would be 6.80/2 = 3.40 cm. At 1.39 cm, the radius would be 1.39/2 = 0.695 cm.

Now, we can find the rate of change of the level of shampoo by taking the derivative of the volume formula with respect to time. This gives us dV/dt = 2πrh(dr/dt), where r is the radius, h is the height (which we can ignore), and (dr/dt) is the rate of change of the radius with respect to time.

We know that the rate of change of the volume (dV/dt) is given as 17.0 cm^3/s, so we can plug that in and solve for (dr/dt) at each diameter.

At 6.80 cm: 17.0 cm^3/s = 2π(3.40 cm)(dr/dt)
Solving for (dr/dt), we get (dr/dt) = 17.0 cm^3/s / (2π(3.40 cm)) = 0.796 cm/s

At 1.39 cm: 17.0 cm^3/s = 2π(0.695 cm)(dr/dt)
Solving for (dr/dt), we get (dr/dt) = 17.0 cm^3/s / (2π(0.695 cm)) = 12.3 cm/s

Therefore, the rate at which the level of shampoo is rising at a point with a diameter of 6.80 cm