Problem 37 - Calculus Early Transcendentals

Question

Let \(f\) be differentiable on \([a, b]\) and suppose that \(g(x)=c f(x)\) and \(h(x)=f(c x),\) where \(c>0 .\) When the curve \(y=f(x)\) on \([a, b]\) is revolved about the \(x\) -axis, the area of the resulting surface is \(A\). Evaluate the following integrals in terms of \(A\) and \(c\). a. \(\int_{a}^{b} g(x) \sqrt{c^{2}+g^{\prime}(x)^{2}} d x\) b. \(\int_{a / c}^{b / c} h(x) \sqrt{c^{2}+h^{\prime}(x)^{2}} d x\)

Step-by-Step Solution

Answer

Answer: a. The value of the integral \(\int_{a}^{b} g(x) \sqrt{c^2 + g'(x)^2} dx\) is equal to \(c^2A\). b. The value of the integral \(\int_{a/c}^{b/c} h(x) \sqrt{c^2 + h'(x)^2} dx\) is equal to \(\frac{1}{c}A\).

Step 1: Derivatives of given functions

Since \(g(x) = cf(x)\), using the product rule and chain rule, we have: \(g'(x) = c f'(x)\). Similarly, for \(h(x) = f(cx)\), we have: \(h'(x) = f'(cx) \cdot c.\) Step 2: Express the given integrals in terms of \(A\) and \(c\)

Step 2: Relationship between surface area and the derivatives

We know that the surface area \(A\) of a curve revolved around the x-axis is given by: \(A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + f'(x)^2} dx\). a. First integral: \(\int_{a}^{b} g(x) \sqrt{c^2 + g'(x)^2} dx\) Using the definition of function \(g\), we can rewrite this integral as: \(A' = \int_{a}^{b} cf(x) \sqrt{c^2 + c^2 f'(x)^2} dx\). We can factor out \(c^2\) from the square root: \(A' = c^2 \int_{a}^{b} f(x) \sqrt{1 + f'(x)^2} dx\). Comparing this expression to the given formula for surface area \(A\), we can substitute \(A\) into the integral expression: \(A' = c^2 A\). b. Second integral: \(\int_{a/c}^{b/c} h(x) \sqrt{c^2 + h'(x)^2} dx\) Using the definition of function \(h\), we can rewrite this integral as: \(A'' = \int_{a/c}^{b/c} f(cx) \sqrt{c^2 + c^2f'(cx)^2} dx\). Next, we use substitution to simplify the integral. Let \(u = cx\), then \(du = cdx\) and when \(x = a/c\), \(u = a\), and when \(x = b/c\), \(u = b\): \(A'' = \int_{a}^{b} f(u) \sqrt{c^2 + c^2f'(u)^2} \frac{du}{c}\). Factor out \(c^2\) from the square root and simplify: \(A'' = \frac{1}{c} \int_{a}^{b} f(u) \sqrt{1 + f'(u)^2} du\). Comparing this expression to the given formula for surface area \(A\), we can substitute \(A\) into the integral expression: \(A'' = \frac{1}{c} A\). Thus, the two integrals are evaluated as: a. \(\int_{a}^{b} g(x) \sqrt{c^2 + g'(x)^2} dx = c^2A\) b. \(\int_{a/c}^{b/c} h(x) \sqrt{c^2 + h'(x)^2} dx = \frac{1}{c}A\)

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. It is the process of calculating the derivative—a measure of how a function's output value changes as its input value changes. The derivative of a function at a point represents the slope of the tangent line to the curve of the function at that point. In the given exercise, differentiation is used to determine the derivatives of functions \(g(x) = cf(x)\) and \(h(x) = f(cx)\). Here, calculating the derivatives involves:

• Product Rule: For a function that is the product of two functions, the derivative is calculated by the product rule formula: \((uv)' = u'v + uv'\). Though not directly used, awareness of such rules is essential when functions get complex.
• Chain Rule: This is crucial for functions involving compositions. For a composition of functions \(h(x) = f(cx)\), the chain rule allows finding the derivative \(h'(x) = c \cdot f'(cx)\). It helps compute the derivative by breaking it into simpler parts.

Through these rules, we deduce that for \(g(x)\), the derivative is \(g'(x) = c f'(x)\), and for \(h(x)\), it is \(h'(x) = c f'(cx)\). Differentiation provides the necessary derivatives needed to later evaluate surface areas in the problem by linking back to how curves behave when rotated.

Integration

Integration is the process of finding the integral, which is essentially the accumulation of quantities, such as areas under curves. In the context of surfaces of revolution, integration is used to sum up infinite thin strips, which forms the surface area when rotated around an axis. In this exercise,integrals are primarily linked to the formula for finding the surface area of a revolved curve:\[A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + f'(x)^2} dx\]The task is to express the integrals of modified functions \(g(x)\) and \(h(x)\) using their related expressions and constants like \(c\). Each integral is transformed using derivative and constant substitutions:

• For \(g(x)\), the substitution leads to \(c^2 A\), emphasizing how constants affect the surface area when scaling a function.
• For \(h(x)\), changes through integration bounds and substitution yield \(\frac{1}{c}A\), which illustrates how compression or expansion in one variable affects the result.

The concept of integration here highlights its versatility in calculating complex areas by modifying bounds, functions, and constants, which is crucial for evaluating surfaces of revolution.

Chain Rule

The chain rule is an essential differentiation tool, particularly when dealing with composite functions. It states that to differentiate a composite function \(u(v(x))\), you multiply the derivative of \(u\) with respect to \(v\) by the derivative of \(v\) with respect to \(x\). Symbolically, if \(y = u(v(x))\), then the derivative \(\frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}\). In the exercise, the chain rule is necessary for finding derivatives when the function’s argument is modified, for instance, substituting \(cx\) into the function \(f\).For example:

• In \(h(x) = f(cx)\), the modification of variable \(x\) by constant \(c\) requires using the chain rule. Thus, \(h'(x) = f'(cx) \cdot c\), where \(c\) is brought outside.
• This change impacts how the resulting function \(h(x)\) behaves, requiring adjustments in the computation of its derivative, crucial for integral transformation in this task.

Understanding and applying the chain rule is vital for navigating changes within a function’s structure due to variable substitutions—the heart of deriving the solution for calculating complex surfaces.