Problem 3 - Calculus: Early Transcendentals

Question

Find the arc length of \(f(x)=(1 / 3)\left(x^{2}+2\right)^{3 / 2}\) on the interval \([0, a]\).

Step-by-Step Solution

Answer

The arc length is \( \frac{a^3}{3} + a \).

Step 1: Understand the formula

The formula for the arc length of the function \( f(x) \) from \( x = a \) to \( x = b \) is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} \, dx \]In this problem, \( a = 0 \) and our endpoint is going to be variable since we want to find the arc length from \( 0 \) to \( a \).

Step 2: Compute the derivative

First, we need to find the derivative \( f'(x) \) of the function \( f(x) = \frac{1}{3}(x^2 + 2)^{3/2} \).Using the chain rule:\[ f'(x) = \frac{1}{3} \cdot \frac{3}{2} (x^2 + 2)^{1/2} \cdot 2x = x(x^2 + 2)^{1/2} \]

Step 3: Plug into the arc length formula

Substitute \( f'(x) \) into the arc length formula:\[ L = \int_{0}^{a} \sqrt{1 + (x(x^2 + 2)^{1/2})^2} \, dx \]

Step 4: Simplify the integrand

Simplify the expression \( 1 + (x(x^2 + 2)^{1/2})^2 \):\[ (x(x^2 + 2)^{1/2})^2 = x^2(x^2 + 2) \]Thus, the integrand becomes:\[ 1 + x^2(x^2 + 2) = x^4 + 2x^2 + 1 \]

Step 5: Identify the perfect square

Notice that \( x^4 + 2x^2 + 1 \) is a perfect square:\[ x^4 + 2x^2 + 1 = (x^2 + 1)^2 \]

Step 6: Integrate to find arc length

The integrand now becomes:\[ \sqrt{(x^2 + 1)^2} = |x^2 + 1| \]Since \( x^2 + 1 \) is always positive over the interval \([0, a]\), we can rewrite this as \( x^2 + 1 \).Integrate from 0 to \( a \):\[ L = \int_{0}^{a} (x^2 + 1) \, dx \]\[ = \left[ \frac{x^3}{3} + x \right]_{0}^{a} \]\[ = \frac{a^3}{3} + a - \left(0 + 0\right) \]\[ = \frac{a^3}{3} + a \]

Step 7: Conclusion

The arc length of the function \( f(x) = \frac{1}{3}(x^2 + 2)^{3/2} \) from \( x = 0 \) to \( x = a \) is \( \frac{a^3}{3} + a \).

Key Concepts

Derivative using Chain Rule

When you want to find the derivative of a composite function, the chain rule is your best tool! It helps you differentiate a function that is nested inside another function.
To use the chain rule, follow these steps: identify the outer function and the inner function. In our exercise, the function is \( f(x) = \frac{1}{3}(x^2 + 2)^{3/2} \). Here's how it works:

• Outer function: \( u^{3/2} \), where \( u = x^2 + 2 \)
• Inner function: \( x^2 + 2 \)

Apply the chain rule by differentiating the outer function and multiplying by the derivative of the inner function:

• Derivative of the outer function with respect to \( u \): \( \frac{3}{2}u^{1/2} \)
• Derivative of the inner function: \( 2x \)

So, the derivative using the chain rule is calculated as:
\( f'(x) = \frac{1}{3} \cdot \frac{3}{2} (x^2 + 2)^{1/2} \cdot 2x = x(x^2 + 2)^{1/2} \).
This shows how the chain rule simplifies the process of finding derivatives for complex functions.

Integrals in Calculus

Integrals are essential in calculus, mainly when calculating the area under curves such as finding the arc length. Understanding integration helps transition from a list of numbers to continuous functions.
Generally, integrating consists of finding the anti-derivative or the accumulated sum of tiny bits under a curve over an interval. In our problem, we begin by integrating the function:
\[ L = \int_{0}^{a} (x^2 + 1) \, dx \]The result of this integral captures the arc length from 0 to \( a \). Integration involves finding the antiderivative:

• For the term \( x^2 \), the antiderivative is \( \frac{x^3}{3} \).
• For the term \( 1 \), the antiderivative is \( x \).

Combine them using the fundamental theorem of calculus:\[ \left[ \frac{x^3}{3} + x \right]_{0}^{a} = \frac{a^3}{3} + a \]This leads to understanding the arc length formula's outcome, which gives us concrete information through integration.

Perfect Square Identification

Recognizing perfect squares in algebraic expressions can greatly simplify calculations. This trick avoids overly complicated integration work.
In the example, the expression was transformed into a perfect square:
\[ 1 + x^2(x^2 + 2) = x^4 + 2x^2 + 1 \]This simplifies to:\[ (x^2 + 1)^2 \]Identifying this perfect square allows us to simplify further. Upon taking the square root to solve the problem:\( \sqrt{(x^2 + 1)^2} = |x^2 + 1| \)

In this situation, since \( x^2 + 1 \) is positive throughout the interval, absolute value simplifies to a straightforward \( x^2 + 1 \). That step ensures manageable integration and ultimately calculating arc lengths efficiently. Recognizing patterns and transformations like these is crucial in calculus and simplifies complex integrals.