Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Christy plans to paint both sides of a fence whose base is in the \(x y\)-plane with shape \(x=30 \cos ^{3} t, y=30 \sin ^{3} t\), \(0 \leq t \leq \pi / 2\), and whose height at \((x, y)\) is \(1+\frac{1}{3} y\), all measured in feet. Sketch a picture of the fence and decide how much paint she will need if a gallon covers 200 square feet.

Step-by-Step Solution

Answer

Christy needs approximately 3.03 gallons of paint.

Step 1: Understand the Fence Shape

The fence base is described by parametric equations: \(x=30 \cos^3 t\) and \(y=30 \sin^3 t\) for \(0 \leq t \leq \frac{\pi}{2}\). This implies it is one-quarter of an astroid since both equations are described in terms of \(\cos^3 t\) and \(\sin^3 t\). The base lies in the first quadrant.

Step 2: Determine Height Formula

The height of the fence at any point \((x, y)\) is given by \(h(x, y) = 1 + \frac{1}{3}y\). This means the height at any given \(t\) is \(1 + \frac{1}{3} \cdot 30 \sin^3 t = 1 + 10 \sin^3 t\).

Step 3: Calculate Differential Segment Length

Use the parametric equations to find the length of a small segment \(dL\) along the curve. The formula is given by \(dL = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). Calculate \(\frac{dx}{dt} = -90 \cos^2 t \sin t\) and \(\frac{dy}{dt} = 90 \sin^2 t \cos t\).

Step 4: Setup and Solve the Integral for Length

Plug the derivatives into the \(dL\) formula and simplify: \[dL = \sqrt{(-90 \cos^2 t \sin t)^2 + (90 \sin^2 t \cos t)^2}\, dt = 90 |\sin t \cos t|\, dt\]. Compute total length by integrating: \[L = \int_0^{\pi/2} 90 \cos t \sin t\, dt = \int_0^{\pi/2} 45 \sin(2t)\, dt\]. Evaluating gives \[L = 45 \left[-\frac{1}{2} \cos(2t)\right]_0^{\pi/2} = 45\].

Step 5: Calculate Total Surface Area

The surface area is the length of the fence base times its height, doubled because both sides are painted: \[ \text{Total Surface Area} = 2 \times \int_0^{\pi/2} (1 + 10 \sin^3 t) \cdot 90 \cos t \sin t \, dt \].This simplifies to: \[ 2 \times 90 \left[ \frac{1}{2} t + \frac{10}{3} \cdot \frac{3}{8} \sin^4 t \right]_0^{\pi/2} = 2 \times 90 \left[\frac{\pi}{4} + \frac{10}{8}\right] = 135\frac{\pi}{2} + 225.\]

Step 6: Determine Amount of Paint

Amount of paint needed is the total surface area over the coverage per gallon:\[\text{Paint Needed} = \frac{135 \pi/2 + 225}{200} \approx 3.03 \text{ gallons}.\]

Key Concepts

Parametric Equations

Parametric equations allow us to define a set of equations that describe a curve by defining both the x and y coordinates in terms of a third parameter, often denoted as \( t \).
Instead of expressing \( x \) directly as a function of \( y \) or vice versa, we use functions like \( x = f(t) \) and \( y = g(t) \). This becomes particularly useful when dealing with complex curves that are not easily described using standard equations.
In this exercise, the base of the fence is described by:

• \( x = 30 \cos^3 t \)
• \( y = 30 \sin^3 t \)

These equations sketch one-quarter of an astroid, a curve shaped like a squished star, lying in the first quadrant of the coordinate plane. Parametric equations offer a robust way to model such intricate paths or shapes.

Surface Area Calculation

Calculating the surface area of a 3D shape often involves integrating over the shape's dimensions. In this problem, we're considering a fence with a 2D base and varying height.
The surface area is computed by determining the perimeter of the base curve and multiplying it by the height at each point along this curve.
Given that the height of the fence at any point \((x, y)\) is \( h(x, y) = 1 + \frac{1}{3} y \), the expression for height incorporates the parametric equations introduced earlier.
The overall surface area will include areas on both sides of the fence, which necessitates doubling the computed integral:

• \( \text{Total Surface Area} = 2 \times \int (1 + 10 \sin^3 t) \cdot 90 \cos t \sin t \, dt \)

This approach leverages integration to account for variations in height across the entire fence.

Integration of Trigonometric Functions

The integration of trigonometric functions is a common procedure in calculus, especially when dealing with periodic functions or curves expressed in terms of trigonometry.
In computing the fence length, the integral

• \( \int_0^{\pi/2} 45 \sin(2t)\, dt \)

involves recognizing the double angle identity for sine. Transforming the product \( \sin t \cos t \) into a sin of double angle \( \sin(2t) \) simplifies the integration process significantly.
Evaluating such integrals often involves substitution methods or using known antiderivatives like

• \( \int \sin x \, dx = -\cos x + C \)

Practice with integrating these expressions builds a strong foundation for solving real-world problems involving periodic behavior or cyclic patterns.

Application of Differentiation

Differentiation, a fundamental tool in calculus, is crucial for understanding the instantaneous rate of change. It is vital for finding derivative functions that describe how one variable changes with respect to another.
In the context of the exercise, we find derivatives of parametric functions to calculate small length segments \( dL \) along a curve:

• \( \frac{dx}{dt} = -90 \cos^2 t \sin t \)
• \( \frac{dy}{dt} = 90 \sin^2 t \cos t \)

These derivatives quantify how changes in \( t \), affect the changes in \( x \) and \( y \), which are then plugged into the formula for \( dL \):

• \( dL = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \)

This reveals the total differential segment length along the path, facilitating the calculation of the overall length by integrating \( dL \) over the given interval for \( t \). Understanding differentiation and its use in breaking down curves into manageable segments is vital for analyzing shapes beyond simple lines or circles.