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Chapter 3: Problem 36

Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05 \(\mathrm{cm}\) thick to a hemispherical dome with diameter 50 \(\mathrm{m} .\)

Short Answer

Expert verified
About 1.96 cubic meters of paint is needed.

Step by step solution

01

Understand the Problem

We need to find the additional volume created when a coat of paint is applied on the outer surface of a hemisphere. The thickness of the paint layer is given as 0.05 cm (converted to 0.0005 m for consistency in units), and the diameter of the dome is 50 m, so its radius is 25 m.
02

Calculate Surface Area of the Hemisphere

The surface area of a hemisphere is given by the formula \( S = 2\pi r^2 \). Substituting the radius \( r = 25 \) m into the formula, we get \( S = 2\pi (25)^2 = 1250\pi \) square meters.
03

Apply the Concept of Differentials

To estimate the volume \( \Delta V \) of the paint, we can use the formula for small changes: \( \Delta V = S \cdot d \), where \( S \) is the surface area of the hemisphere and \( d = 0.0005 \) m is the thickness of the paint layer. Substitute the values into the formula: \( \Delta V = 1250\pi \times 0.0005 \).
04

Calculate the Volume of Paint Needed

Compute \( \Delta V = 1250\pi \times 0.0005 = 0.625\pi \) cubic meters. Evaluating \( \pi \) as approximately 3.1416, the volume of paint needed is \( 0.625 \times 3.1416 \approx 1.9635 \) cubic meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hemisphere Surface Area
When dealing with geometric shapes, like a hemisphere, it's essential to understand how to calculate the surface area. A hemisphere is half of a sphere. The surface area of a sphere is given by the formula \(4\pi r^2\). Thus, a hemisphere, being half of a sphere, has a surface area equal to half of this formula, which is \(2\pi r^2\). This calculation includes the curved surface area only.

In practical terms, if you have a hemisphere with a radius \(r\), plugging this value into the formula \(2\pi r^2\) gives you the total surface area that you have to cover, for example, with paint or any coating. Understanding and correctly calculating this is crucial, as it sets the stage for further estimations related to the object, especially when additional layers are added.

In our example, the radius of 25 meters gives a surface area of \(1250\pi\) square meters, ready for the next steps in estimating additional layers on this surface.
Volume Estimation
Volume estimation comes into play when we want to figure out how much "space" an object or a modification takes up. In this exercise, we aim to estimate the added volume resulting from a layer of paint.

The key is to multiply the surface area by the thickness of the layer. However, rather than recalculating the volume of a whole new larger hemisphere, we use differentials鈥攁 powerful calculus tool that helps us approximate changes without exhaustive recalculations.
  • We identify the main surface: the hemisphere with a surface area of \(1250\pi\).
  • The layer being added is 0.0005 meters thick.
Thus, the differential formula \(\Delta V = S \cdot d\) helps calculate the small volume change, making it suitable for thin layers. That's why the estimated volume of paint is \(0.625\pi\) cubic meters, allowing us to move forward with precision.
Paint Layer Thickness
Understanding the implications of layer thickness is crucial in many practical applications. Here, the thickness of the paint layer is given as 0.05 cm, or 0.0005 meters.

In construction and manufacturing, knowing how to convert and work with different thickness measurements ensures you work within the required specifications. This is especially important because small variations can lead to significant changes in the material used or structural properties.

When a thin, uniform layer is added, such as our paint, this simplification makes calculations manageable and reliable. By using the thickness in meters as part of our differential formula \(\Delta V = S \cdot d\), we efficiently and accurately determine the volume of additional material needed to apply the layer, ensuring proper order and complete coverage.

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Most popular questions from this chapter

On page 431 of Physics: Calculus, 2\(d\) ed, by Eugene Hecht (Pacific Grove, CA: Brooks/Cole, \(2000 ),\) in the course of deriving the formula \(\mathrm{T}=2 \pi \sqrt{\mathrm{L} / \mathrm{g}}\) for the period of a $$ \begin{array}{l}{\text { pendulum of length } L, \text { the author obtains the equation }} \\ {\mathrm{a}_{\mathrm{T}}=-g \sin \theta \text { for the tangential acceleration of the bob of the }}\end{array} $$ pendulum, He then says, "for small angles, the value of \(\theta\) in radians is very nearly the value of \(\sin \theta\) ; they differ by less than 2\(\%\) out to about \(20^{\circ} .\) $$ \begin{array}{c}{\text { (a) Verify the linear approximation at } 0 \text { for the sine function: }} \\ {\sin x=x}\end{array} $$ (b) Use a graphing device to determine the values of \(x\) for which sin \(x\) and \(x\) differ by less than 2\(\%\) . Then verify Hecht's statement by converting from radians to degrees.

Suppose \(y = f ( x )\) is a curve that always lies above the \(x\) -axis and never has a horizontal tangent, where \(f\) is differentiable everywhere. For what value of \(y\) is the rate of change of \(y ^ { 5 }\) with respect to \(x\) eighty times the rate of change of \(y\) with respect to \(x ?\)

$$ \begin{array}{l}{29-31 \text { Explain, in terms of linear approximations or differentials, }} \\ {\text { why the approximation is reasonable. }}\end{array} $$ $$ \ln 1.05 \approx 0.05 $$

At noon, ship \(A\) is 100 \(\mathrm{km}\) west of ship \(\mathrm{B} .\) Ship \(\mathrm{A}\) is sailing south at 35 \(\mathrm{km} / \mathrm{h}\) and ship \(\mathrm{B}\) is sailing north at 25 \(\mathrm{km} / \mathrm{h}\) . How fast is the distance between the ships changing at \(4 : 00 \mathrm{PM}\) ?

The gas law for an ideal gas at absolute temperature T (in kelvins), pressure \(P\) (in atmospheres), and volume \(V\) ( in liters) is \(P V=n R T\) , where \(n\) is the number of moles of the gas and \(R=0.0821\) is the gas constant. Suppose that, at a certain instant, \(\mathrm{P}=8.0\) atm and is increasing at a rate of 0.10 atm/min and \(\mathrm{V}=10 \mathrm{L}\) and is decreasing at a rate of 0.15 \(\mathrm{L} / \mathrm{min}\) . Find the rate of change of \(\mathrm{T}\) with respect to time at that instant if \(\mathrm{n}=10 \mathrm{mol} .\)
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