Problem 5 - Essential Calculus

Question

In Exercises \(3-14,\) find the arc length of the graph of the function over the indicated interval. $$ y=\frac{3}{2} x^{2 / 3}, \quad[1,8] $$

Step-by-Step Solution

Answer

The method described can be used to compute the exact arc length, which involves performing an integral operation. The exact operations to be performed with this integral are outside this current scope, and they will result to a numerical value that will represent the arc length of the given function in the interval [1,8].

Step 1: Find the derivative

To find the derivative of y = 3/2 * x^(2/3), apply the power rule for differentiation. The power rule states that the derivative of x^n is n*x^(n-1), where n is a real number. Here, n = 2/3, so (d/dx)(x^(2/3)) = (2/3)*x^((2/3)-1) = (2/3)*x^(-1/3). Therefore, dy/dx of y = 3/2 * x^(2/3) becomes (3/2)*(2/3)*x^(-1/3) = x^(-1/3).

Step 2: Substitute the derivative into the arc length formula

Substitute the derivative into the arc length formula as follows: L = \int_1^8 sqrt(1+(dy/dx)^2) dx = \int_1^8 sqrt(1+(x^(-1/3))^2) dx.

Step 3: Simplify the integral

Simplify the integral by factoring out the square root sign: \[ L = \int_1^8 dx * sqrt(1+x^{(-2/3)}) \].

Step 4: Evaluate the integral

The final step is to evaluate this integral using the method of choosing the correct form for a standard integral. The integral can be calculated to obtain an exact numerical value for L.

Key Concepts

Arc Length

Finding the arc length of a curve involves understanding the distance along a continuous function between two points. It's like measuring the length of a path on a graph. Arc length can be found using a specific formula that incorporates an integral. The general formula for arc length of a function, represented by \( y = f(x) \) from \( a \) to \( b \), is given by:

• \( L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \)

This formula tells us to take the integral of the square root of \(1\) plus the square of the derivative of the function, across the interval from \(a\) to \(b\). This helps account for how the curve rises and falls between those points.
Finding the arc length can be essential in various fields such as physics, engineering, and even computer graphics, making it vital to grasp this concept well!

Derivative

The derivative of a function provides crucial information about the rate of change of that function. In simple terms, it indicates how steep a function is at any given point. For the function given in the exercise, \( y = \frac{3}{2} x^{2/3} \), we find the derivative to understand how the height of the graph changes with respect to \(x\).

• The derivative is found using the power rule, turning \(y\) into its slope formula \(dy/dx\).

The power rule for differentiation states that the derivative of \(x^n\) is \(n \cdot x^{n-1}\). Here, applying the differentiation process step-by-step, we obtain \(x^{-1/3}\) as the derivative. This calculation is integral to proceeding with solving for arc length as it measures the change.
The derivative helps us not only in finding arc lengths but also in various aspects of calculus, such as identifying maximum and minimum points of functions.

Power Rule

The power rule is one of the most fundamental techniques in calculus for differentiating polynomial functions. It simplifies the process of finding derivatives for expressions raised to a power.
The expression \(x^n\) when differentiated becomes \(n \times x^{n-1}\). This quick formula is a core tool in calculus that helps us transform complex expressions into more manageable derivatives.
Applying the power rule is straightforward:

• Identify the power \(n\) in the term \(x^n\).
• Multiply \(n\) by the coefficient of \(x\).
• Decrease the exponent \(n\) by 1.

In the solution of the exercise, the power rule allows us to efficiently determine the derivative of \(x^{2/3}\), making it easier to conduct further calculations for the arc length.

Integral

An integral computes the area under a curve within a certain interval, giving us important information about the overall trend and accumulation of values within that space. For the arc length formula, integrals help account for every tiny segment that constitutes the curve.
Integrals can be thought of as the reverse process of differentiation, essentially "adding up" infinitesimal data points to find a total. In the exercise, the integral \( \int_1^8 \sqrt{1 + (x^{-1/3})^2} \, dx \) calculates the total arc length over the interval \([1,8]\).
This calculation is vital as it provides the definitive "length" derived from the continuous path of the function. Integrals, along with derivatives, form the twin pillars of calculus, and understanding integrals enables us to unlock numerous real-world problem-solving opportunities in science and engineering.