Calculating Rate of Change of Water in a Semi-Circular Bathtub

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The discussion revolves around calculating the rate of change of water height in a semi-circular bathtub with a radius of 5 m, filled at a rate of 2 m³/min. The volume of water is described by the formula V = πy²(R - y/3), where R is the radius of the bathtub and y is the height of the water. Participants clarify that R refers to the bathtub's radius, and the problem's interpretation as a vertical cross-section is confirmed. The key steps involve differentiating the volume formula with respect to time and applying the chain rule to find dy/dt. The conversation emphasizes the need for clarity in problem statements and the realistic dimensions of bathtubs versus pools.
songoku
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Homework Statement


A bathtub with shape semi-circle has radius 5 m is filled with water with rate 2 m^3/minute. How fast will the height of water in the bathtub change when it is 2 m from the bottom? Given that the volume of water when the radius = R and height = y is πy^2(R - y/3)


Homework Equations


differential


The Attempt at a Solution


The volume of bathtub (constant) = 2/3 πr^3

The rate of change of volume of water :
\frac{dV}{dt}=2\frac{m^3}{minute}

I have to find dy/dt. I think I have to find the relation between R and y, but I don't know how...

Thanks
 
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R doesn't change. R=5 m. Only V and y change. And they already gave you the relation between them. V=πy^2(R - y/3). That part of the job is done. You just have to start with the differentiation.
 
Hi Dick

Oh I Think R is the radius of the water, not the bathtub. I just realized that's not possible.

Thanks a lot
 
Can you provide the exact statement of the problem? Saying that "A bathtub with shape semi-circle" doesn't clearly describe it. Is a vertical cross-section of the tub semicircular? That would probably be the most reasonable interpretation, but I suppose it could have a horizontal cross-section that is semicircular. In either case, a radius of 5 m. seems very large for a bathtub. A radius of .5 m seems more realistic.
 
Hi Mark44

I got this question from my friend and that's all what he told me. I interpreted it as the vertical cross-section. Will it be different if it's horizontal cross-section? I think it will be the same.
About the radius, yeah 5 m is ridiculous for a bathtub. Maybe it's a pool :smile:
 
You are given a formula for the volume of water in the pool when the water is y feet deep, and the radius of the pool is R feet:
V = \pi y^2(R - y/3)

Differentiate both sides with respect to t, using the chain rule. That will give you
dV/dt = (something involving y)*dy/dt

Solve this equation for dy/dt, and substitute in all the other known quantities at the time when y = 2 feet.
 
Hi Mark44

Ok Thanks a lot Mark :smile:
 

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