A hemispherical dome of radius 40 feet is to be given 7 coats of paint, each of which is 1/100 inch thick. How do you use linear approximation to estimate the volume of paint needed for the job?

Answer 1

#approx " 117.3 ft"^3 #

I would go with the (brilliant) result that, for a sphere:

#V = 4/3 pi r^3# and #(dV)/(dr) = 4 pi r^2#
[ie the surface area of a sphere of radius #r# is the derivative wrt #r# of its volume]

To first order we can say that:

#delta V = (dV)/(dr) delta r = 4 pi r^2 delta r#
and so with #delta r = 7 * 1/(12*100)# (7 layers, and adjusting to Imperial ft measurements), we have
#delta V = 4 pi ( 40)^2 * 7/100 = 112/3 pi approx " 117.3 ft"^3 #

Actual increase is

#DeltaV = 4/3 pi ( (40+ 7/1200)^3- 40^3) = " 117.3 ft"^3 #

I did those to 1 dp, so you'll find a slightly bigger discrepancy if you plug the numbers into a calculator

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Answer 2

To estimate the volume of paint needed for the job using linear approximation, you can approximate the dome's surface area as a cylinder. Here's how to do it:

  1. Calculate the surface area of the hemisphere using the formula: ( A = 2\pi r^2 ), where ( r ) is the radius of the hemisphere.

  2. Since each coat of paint is ( \frac{1}{100} ) inch thick, multiply the surface area calculated in step 1 by ( \frac{7}{100} ) to get the approximate volume of paint needed.

  3. The linear approximation assumes that the thickness of the coats is small relative to the size of the dome, so the error introduced by this approximation should be minimal.

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Answer 3

To estimate the volume of paint needed for the job using linear approximation, we can approximate the increase in volume of the dome after each coat of paint and then sum up these increases.

First, let's find the volume of the dome without any paint, which is given by the formula for the volume of a hemisphere: V = (2/3)πr^3, where 'r' is the radius of the dome. Substituting the given radius of 40 feet into this formula, we get the initial volume of the dome.

After each coat of paint, the thickness of the paint layer adds to the radius of the dome. Since each coat is 1/100 inch thick, after the first coat, the new radius will be 40 + 1/100 feet. After the second coat, it will be 40 + 2/100 feet, and so on.

Using linear approximation, we can estimate the change in volume of the dome after each coat of paint by approximating it with a cylinder. The change in volume ΔV after each coat is approximately the product of the surface area of the dome's base (which is a circle) and the thickness of the paint layer.

The surface area of the base of the dome is πr^2, where 'r' is the new radius after each coat of paint. Therefore, the change in volume ΔV after each coat is approximately πr^2 * (1/100), where 'r' is the new radius after each coat.

We can then sum up these changes in volume for all 7 coats of paint to estimate the total volume of paint needed for the job.

By using this linear approximation method, we can estimate the volume of paint needed for the job without having to calculate the exact volume increase after each coat of paint, making the estimation process more manageable.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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