Linear Approximation Part III

[Hckyplayer8]

How much plastic (in cubic feet) does it take to coat a sphere 20ft in radius if the plastic is .1" (1/120 ft) thick?

The volume of a sphere in relation to it's radius is 4/3pi³.

20ft = 240"

So subtracting the volume of the sphere plus plastic from the volume of the sphere will give me how much plastic.

V (240" + .1") - V(240") approx (.1") V'(240") = .1"(4)pi(240²) = (.4)pi(57600) = 23040/12 = 1920pi cubic feet

 

[Deleted member 4993]

How much plastic (in cubic feet) does it take to coat a sphere 20ft in radius if the plastic is .1" (1/120 ft) thick?

The volume of a sphere in relation to it's radius is 4/3pi³.

20ft = 240"

So subtracting the volume of the sphere plus plastic from the volume of the sphere will give me how much plastic.

V (240" + .1") - V(240") approx (.1") V'(240") = .1"(4)pi(240²) = (.4)pi(57600) = 23040/12 = 1920pi cubic feet

Another way to look at it:

The surface area of a sphere = 4 * pi * r^2 = 4 * pi * 400 = 1600 * pi

Volume of coating plastic = (The surface area of a sphere) * (thickness of coating) = 1600 * pi * 1/120 = 40/3 * pi = 41.9 cft

Your process is correct - there are "arithmetic" mistakes.

 

[Hckyplayer8]

Another way to look at it:

The surface area of a sphere = 4 * pi * r^2 = 4 * pi * 400 = 1600 * pi

Volume of coating plastic = (The surface area of a sphere) * (thickness of coating) = 1600 * pi * 1/120 = 40/3 * pi = 41.9 cft

Your process is correct - there are "arithmetic" mistakes.

Thanks for posting. For the final, did you just convert pi to the decimal form and round?

 

[Deleted member 4993]

Thanks for posting. For the final, did you just convert pi to the decimal form and round?

Yes ... rounded down to 3 significant digits.

 

[JeffM]

Another way to look at it:

The surface area of a sphere = 4 * pi * r^2 = 4 * pi * 400 = 1600 * pi

Volume of coating plastic = (The surface area of a sphere) * (thickness of coating) = 1600 * pi * 1/120 = 40/3 * pi = 41.9 cft

Your process is correct - there are "arithmetic" mistakes.

Dear SK

I do not think that your method is exact. Of course the error is small with this relative thickness, but let's change relative scale.

The surface area of the smaller sphere is 1 inch, but the thickness is still 0.1.

Your method says the plastic needed, in cubic inches, is

[MATH]4 \pi (1^2) * 0.1 = 0.4 pi.[/MATH]
The difference of volume method gives, in cubic inches,

[MATH]4 \pi (1.1^2) - 4 \pi (1^2) = 4\pi(0.21) = 0.84 \pi.[/MATH]
This is a relative error of over 100%.

Back to the problem at hand.

[MATH]\dfrac{4 \pi}{3} * \left (20 + \dfrac{1}{120} \right )^3 - \dfrac{4 \pi}{3} * 20^3 = \dfrac{4 \pi}{3} * \left ( 3 * 20^2 * \dfrac{1}{120} + 3 * 20 * \dfrac{1}{120^2} + \dfrac{1}{120^3} \right ) =[/MATH]
So your method is good enough for all practical purposes.

[MATH]\dfrac{4 \pi}{3} * \left ( 10 + \dfrac{60}{14400} + \dfrac{1}{1728000} \right ) \approx \dfrac{4 \pi}{3} * 10.004 = 41.905.[/MATH]
But I am not sure that either method is what is expected for this problem on linearization.

 

[MarkFL]

As with all such problems, I would begin with:

[MATH]f(x+\Delta x)-f(x)\approx\d{f}{x}\Delta x[/MATH]
Here, we have:

[MATH]f(x)=\frac{4}{3}\pi x^3\implies f'(x)=4\pi x^2[/MATH]
[MATH]x=20\text{ ft}[/MATH]
[MATH]\Delta x=\frac{1}{120}\text{ ft}[/MATH]
Hence:

[MATH]\Delta V=f\left(\frac{2401}{120}\text{ ft}\right)-f(20\text{ ft})\approx4\pi \left(20\text{ ft}\right)^2\left(\frac{1}{120}\text{ ft}\right)=\frac{40\pi}{3}\text{ ft}^3\approx41.8879\text{ ft}^3[/MATH]
For comparison, the exact answer is:

[MATH]\Delta V=\frac{4}{3}\pi\left(\left(\frac{2401}{120}\right)^3-20^3\right)\text{ ft}^3=\frac{17287201\pi}{1296000}\approx41.9054\text{ ft}^3[/MATH]