Good ole hemispherical dome differentials problem

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Homework Help Overview

The problem involves estimating the amount of paint required to coat a hemispherical dome with a specified thickness using differentials. The subject area includes calculus and differential applications in geometry.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of diameter versus radius in the volume formula for a hemisphere, with some suggesting that the radius should be used since the paint increases the radius. There is also a mention of previous experiences with similar problems involving conversions between radius and diameter.

Discussion Status

Participants are exploring different approaches to the problem, including the validity of using diameter in the calculations. Some guidance has been offered regarding the application of the volume formula, and there is an acknowledgment of a potential error in the final answer calculation.

Contextual Notes

There is a specific thickness of paint to consider, and participants are navigating the implications of using different geometric parameters (diameter vs. radius) in their calculations. The original poster's previous experiences with homework problems are influencing their approach.

mlschiff
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Homework Statement


Use differentials to estimate the amount of paint needed to apply a coat of paint 0.03 cm thick to a hemispherical dome with diameter 54 m. (Round the answer to two decimal places.)

Homework Equations


V = 1/2(4/3(pi*r^3)) = 1/2(4/3(pi*(1/2D)^3)) = 1/2(4/3*pi(1/8D^3)) = 1/12pi*D^3


The Attempt at a Solution


How does V change if we change d from 54m to (54 = 0.03*10^-2)m?
dV/dD = 1/4*pi*D^2
dV = 1/4*pi*D^2*dD
= 1/4*pi*(54)^2(0.03*10^-2)

I got 0.679 as an awesome and got it wrong...
 
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Why did you switch to diameter? Since you are only painting a hemisphere, the paint only increases the radius, not the diameter. And use V= (2/3)\pi r^3.
 
Okay, using 2/3 seems a lot easier, but my professor gave an example in class of converting radius to diameter. Furthermore, I had a problem on a previous homework where I had to solve for the rate at which a sphere increased, and I got the right answer after converting the radius in the formula into diameter after attempting to solve for the rate at which the radius changed. If I do go about solving for the radius in this problem, what all needs to be done to address the problem of solving for diameter?
 
You can use the diameter if you want. But after applying the paint the diameter becomes 54m+2*(0.03cm).
 
okay. thanks for the help, all!
 
Your original differential seems to be correct, you used sound mathematical principles, the only problem I see is the final answer. When I put .25*pi*(54)^2*.0003 I get .687. I have tried to figure out what you might have miss entered, but the formula you used was good.
 

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