Solving Related Rates Prob: Check Work w/ Hemispherical Bowl

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Homework Help Overview

The discussion revolves around a related rates problem involving a hemispherical bowl being filled with water. The original poster seeks verification of their calculations related to the rate at which water is flowing into the bowl, given the water level's rise over time.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the height of the water and the radius of the bowl, questioning the derivation of the volume formula used by the original poster. There is a focus on understanding how the variables are connected and ensuring clarity in the reasoning behind the formulas presented.

Discussion Status

Some participants have expressed understanding of the original poster's approach after reviewing a provided drawing. There appears to be a productive exchange regarding the derivation of the volume formula, with no explicit consensus reached on the correctness of the calculations, but clarification has been offered.

Contextual Notes

The original poster's calculations are based on a specific scenario involving a hemispherical bowl of fixed radius and a given rate of water level increase. The discussion highlights the importance of visual aids in conveying relationships between variables in related rates problems.

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Related rates prob. [solved]

A hemispherical bowl of radius 8 in. is being filled with water at a constant rate. If the water level is rising at the rate of 1/3 in./s at the instant when the water is 6 in. deep, find how fast the water is flowing in by using the fact that if V is the volume of the water at time t, then dV/dt = (pi)(r)^2(dh/dt)

I just need someone to check my work. From my picture, I have:

dV/dt = (pi)(64-(8-h)^2) dh/dt.

just plug in 6 for h and 1/3 for dh/dt and solve right? Thanks.
 
Last edited:
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The first formula you show looks fine, but can you tell us how you arrived at your second formula, namely dV/dt = pi (64 - (8 - h)^2) dh/dt?
You mentioned that it comes from your picture, but since I can't see your picture, it's not clear to me how h is related to the radius of the disk you are using for your incremental volume. I'm not saying your formula is wrong, but you haven't provided enough information so that I can tell.
 
Mark44 said:
The first formula you show looks fine, but can you tell us how you arrived at your second formula, namely dV/dt = pi (64 - (8 - h)^2) dh/dt?
You mentioned that it comes from your picture, but since I can't see your picture, it's not clear to me how h is related to the radius of the disk you are using for your incremental volume. I'm not saying your formula is wrong, but you haven't provided enough information so that I can tell.

http://img363.imageshack.us/img363/3018/problemab0.jpg http://g.imageshack.us/img363/problemab0.jpg/1/

I suck at drawing on paint but this is kind of how my pic looks. y + h is radius of the bowl which is 8. 8 - h = y and then i just used c^2 - a^2 = b^2. And thanks for your help.
 
Last edited by a moderator:
elitespart said:
http://img363.imageshack.us/img363/3018/problemab0.jpg http://g.imageshack.us/img363/problemab0.jpg/1/

I suck at drawing on paint but this is kind of how my pic looks. y + h is radius of the bowl which is 8. 8 - h = y and then i just used c^2 - a^2 = b^2. And thanks for your help.


OK, thanks for including the drawing. Now I know how h and y are related. I don't see anything wrong in your second formula.
 
Last edited by a moderator:
Thanks again.
 

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