Question

Find the arc length of the function on the given interval. \(f(x)=2 x^{3 / 2}-\frac{1}{6} \sqrt{x}\) on [0,9] .

Short answer

The arc length is found by evaluating the integral \( \int_0^9 \sqrt{10x + \frac{37}{72x}} \, dx \).

Step-by-step solution

Recall the Arc Length Formula

The arc length of a function on an interval \([a, b]\) is given by the integral \( \int_a^b \sqrt{1 + (f'(x))^2} \, dx \). First, we need to compute the derivative \( f'(x) \) of the function \( f(x) = 2x^{3/2} - \frac{1}{6}\sqrt{x} \).

Find the Derivative

To find \( f'(x) \), differentiate each term of \( f(x) = 2x^{3/2} - \frac{1}{6}\sqrt{x} \). The derivative of \( 2x^{3/2} \) is \( 3x^{1/2} \), and the derivative of \( -\frac{1}{6}\sqrt{x} \) is \( -\frac{1}{12}x^{-1/2} \). Therefore, \( f'(x) = 3x^{1/2} - \frac{1}{12}x^{-1/2} \).

Set up the Integral

Next, substitute \( f'(x) = 3x^{1/2} - \frac{1}{12}x^{-1/2} \) into the arc length integral formula. This gives us the integral \( \int_0^9 \sqrt{1 + \left(3x^{1/2} - \frac{1}{12}x^{-1/2}\right)^2} \, dx \). Simplify and solve this integral to find the arc length.

Simplify the Expression Inside the Integral

Simplify the expression inside the square root: \( \left(3x^{1/2} - \frac{1}{12}x^{-1/2}\right)^2 \). Expand this to get \( 9x + \frac{1}{144x} - \frac{1}{2} \). Add 1, leading to \( 10x + \frac{37}{72x} \).

Evaluate the Integral

Evaluate the integral \( \int_0^9 \sqrt{10x + \frac{37}{72x}} \, dx \). You can perform this integral using numerical methods, or by applying a suitable substitution to simplify it. Solve to find the arc length.

Key concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It focuses on slightly different principles, primarily represented in calculus as differentiation and integration. These two major concepts allow us to not only compute quantities like slopes and areas but also to model changing systems in science and engineering.

In the arc length problem, calculus helps determine the length of a curve. This curve is defined by a function within a particular interval. The use of derivatives and integrals in this problem shows calculus in action. Here, they help identify how a function behaves and how to accumulate that behavior throughout an interval to find length.

Integral Calculus

Integral calculus is concerned with accumulation of quantities and the areas under and between curves. When we want to find the arc length, integral calculus comes into play as it allows us to add up an infinite number of infinitesimally small lengths to find a total length.

This is done using an integral. In this case, \(\int_a^b \sqrt{1 + (f'(x))^2} \, dx \) is used for computation of arc length between two points. Integral calculus thus provides solutions for real-world problems by bringing summation into play, whether it's finding a total length, area, or volume.

Derivative

A derivative describes the rate at which a function is changing at any given point. In simple terms, it helps us know how steep a curve is at any point. To find curve lengths, it's crucial to first understand how the curve itself changes; that's where derivatives are important.

In our example, we differentiate \( f(x) = 2x^{3/2} - \frac{1}{6}\sqrt{x} \) into \( f'(x) = 3x^{1/2} - \frac{1}{12}x^{-1/2} \). The resultant expression \( 3x^{1/2} - \frac{1}{12}x^{-1/2}\) tells us how the function behaves, thereby making it possible to move toward integration.

Numerical Integration

The expression inside the integral for arc length can often become complex, as seen here with \( \sqrt{10x + \frac{37}{72x}\). When evaluating such integrals analytically is difficult or impractical, numerical integration methods become helpful.

Techniques like Simpson’s Rule or the Trapezoidal Rule approximate the value of integrals. While they do not provide an exact answer, they give highly accurate estimates. In the arc length context, numerical methods help when a substitution that simplifies the integral is not apparent.