Use derivative of volume to find weight of a shpere

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Homework Help Overview

The discussion revolves around calculating the weight of a spherical soccer ball made of leather, utilizing the derivative of volume for a linear approximation. The problem involves understanding the relationship between the volume, surface area, and weight based on given dimensions and material density.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the derivative of volume and its connection to surface area, questioning how these concepts relate to estimating the weight of the sphere. There is uncertainty about the next steps after identifying the surface area and its implications for weight calculation.

Discussion Status

Some participants have provided hints regarding the relationship between the derivative, surface area, and weight, while others express confusion about how to proceed with the calculations. Multiple interpretations of the problem are being explored, indicating a productive dialogue without a clear consensus.

Contextual Notes

Participants note the thickness of the leather and the density of the material as important factors in the calculations. There is an acknowledgment of the challenge in applying linear approximation to the problem.

rburt
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Suppose a soccer ball is made of leather 1/8 in thick. If the outside diameter is 9 in, and the density of leather is assumed to be 0.64 oz/cubic in. Use the derivative of volume to get a linear approximation to figure the weight of the ball (assume the ball is spherical).
Work already done: dV=4piR^2
I don't know where to go from here.
 
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rburt said:
Suppose a soccer ball is made of leather 1/8 in thick. If the outside diameter is 9 in, and the density of leather is assumed to be 0.64 oz/cubic in. Use the derivative of volume to get a linear approximation to figure the weight of the ball (assume the ball is spherical).
Work already done: dV=4piR^2
I don't know where to go from here.

You left out an important part of that expression rburt.

dV=4 \pi r^2 \ \mathbf{dr}Can you think of how this expression is hinting to us about how to estimate the volume of a thin shell. (Hint. What is the expression for the surface area?)
 
I understand that the derivative is the surface area but I do not understand how that relates to the weight of the sphere. I also know that the rate at which the volume of a sphere increases when the radius r increases is the measure of the surface area if the sphere but I do not understand what step is next once you have the surface area and how it relates to the weight.
 
rburt said:
I understand that the derivative is the surface area but I do not understand how that relates to the weight of the sphere.

Find out how much material you have, and use the density to find the weight.
 
How would i do it using linear approximation?
 
If you cut the soccer ball so you could lay it out flat, its volume would be its area times its thickness. Of course, it won't actually lay flat, so that's just an approximation. But look at the formula in uart's post #2. That is a linear approximation.
 

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