Linear approximation and errors

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Homework Help Overview

The discussion revolves around estimating the maximum possible error in the volume of a spherical ball based on a measured circumference. The original poster, Juan, provides a circumference measurement of 40 cm with a potential error of up to 2 cm, and references the formulas for circumference and volume of a sphere.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore methods for calculating the maximum error in volume based on the given circumference and its associated error. There are discussions on expressing volume as a function of circumference and considering maximum error scenarios.

Discussion Status

Some participants have shared their calculations and results, while others have provided guidance on approaching the problem. There is an ongoing exploration of different methods to verify the maximum error in volume, but no consensus has been reached on the final answer.

Contextual Notes

Participants note the importance of understanding the relationship between circumference and radius, as well as the implications of the error in circumference on the calculated volume. The original poster expresses uncertainty about the computation process, and there are repeated references to the need for clarification on derivatives and their application in this context.

anthonym44
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[SOLVED] Linear approximation

Homework Statement


Juan measures the circumference C of a spherical ball at 40cm and computes the ball's volume V. Estimate the maximum possible error in V if the error in C is as most 2cm. Recall that C=2(pi)r and V=(4/3)pi(r)


Homework Equations


deltaf - f'(a)h, a= 40, V=(4/3)pi(C/2pi)^3


The Attempt at a Solution



im not sure exactly how to compute to find the final answer but i believe that a = 40cm + or - 2cm and that somehow solving for r using circumference and then plugging it into the volume equation. if you can point me in the right direction i would appreciate it, thanks.
 
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The easiest way is to assume maximum error (plus, then minus). Then take the difference between the calculated volume and true volume.

Or you could try expressing V as a function of C.
 
thanks, that actually helps me a lot.
 
after solving for the volume when the radius is (21/pi) which is a +2cm error i got the answer 1251.1cm^3. when i solved for the volume when the radius was (19/pi) a -2cm error i got 926.6cm^3. last i solved for the actual measured raius which was 40 and found the radius to be (20/pi) when i solved for this i got 1080.8. i then subtracted 926.6 (-2cm error) from this answer and got 154.2. when i subtracted 1080.8 from 1251.1 (+2cm error minus actual) i got the answer 170.3. I'm guessing this means that the maximum error is 170.3cm^3. can anyone verify this?
 
anthonym44 said:

Homework Statement


Juan measures the circumference C of a spherical ball at 40cm and computes the ball's volume V. Estimate the maximum possible error in V if the error in C is as most 2cm. Recall that C=2(pi)r and V=(4/3)pi(r)


Homework Equations


deltaf - f'(a)h, a= 40, V=(4/3)pi(C/2pi)^3


The Attempt at a Solution



im not sure exactly how to compute to find the final answer but i believe that a = 40cm + or - 2cm and that somehow solving for r using circumference and then plugging it into the volume equation. if you can point me in the right direction i would appreciate it, thanks.

V=(4/3)pi(C/2pi)^3= C^3/(6pi^2) is exactly what you want. What is the derivative of V with respect to C?
 
4pi(c/2pi)^2
 
i think i finally got the answer to be 324.22pi by plugging in 40 to the above equation for C and then multipliing by delta x which in this case is 2
 

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