Question

Recall that for \(y=f(x)\), the are length from \(x=a\) to \(x=b\) is given by $$ \begin{array}{l} s=\int_{a}^{b} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \\ \text { (see }[5,87.4]) \end{array} $$ Let \(f(x)=x^{2} .\) Use symbolic functions to find an exact value for the are length from \(x=0\) to \(x=1\)

Short answer

The arc length from \( x=0 \) to \( x=1 \) is \( \frac{5^{3/2}}{6} \).

Step-by-step solution

Find the derivative

We begin by finding the derivative of the function \( f(x)=x^2 \). The derivative, using the power rule, is \( \frac{dy}{dx} = 2x \).

Set up the integrand

Substitute \( \frac{dy}{dx} = 2x \) into the arc length formula. The integrand becomes: \( \sqrt{1+(2x)^2} = \sqrt{1+4x^2} \).

Establish the integral limits

The problem sets the limits of integration from \( x=0 \) to \( x=1 \).

Compute the integral

Now compute the arc length integral \( s = \int_{0}^{1} \sqrt{1+4x^2} \, dx \). This integral requires a trigonometric substitution: namely, \( x = \frac{1}{2} \sinh(t) \), which simplifies the expression to \( \int \cosh(t) \, dt \).

Evaluate the integral

Evaluating \( \int \cosh(t) \, dt \) over the transformed bounds, we substitute back using the inverse hyperbolic function. The final result of the integral calculation gives the arc length \( \frac{5^{3/2}}{6} \) after simplifying the definite integral.

Key concepts

Arc Length Calculation

Arc length refers to the distance along a curved line, or the "arc," essentially measuring how long the curve is between two points. When we calculate arc length for a function like \( y = f(x) \), the formula to find the arc length \( s \) from \( x = a \) to \( x = b \) is given by: \[ s = \int_{a}^{b} \sqrt{1+\left(\frac{dy}{dx}\right)^{2}} \, dx \] This formula takes into account the slope of the curve by incorporating the derivative \( \frac{dy}{dx} \).

• The derivative helps us understand the change in the function, or the 'steepness' at a point.
• The expression \( \sqrt{1+(\frac{dy}{dx})^2} \) essentially ensures that we account for all directions in which the curve is moving.

For \( f(x) = x^2 \), the derivative is \( \frac{dy}{dx} = 2x \), thus leading to \( \sqrt{1 + (2x)^2} = \sqrt{1 + 4x^2} \). This showcases how derivatives play a pivotal role in determining the curve's characteristics.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals containing expressions like \( \sqrt{1 + x^2} \), \( \sqrt{1 - x^2} \), and similar forms. In our original problem, the expression \( \sqrt{1 + 4x^2} \) suggests using a hyperbolic substitution to simplify the integral. This is because the expression resembles a situation where hyperbolic identities can make calculations easier. By using the substitution \( x = \frac{1}{2} \sinh(t) \), we simplify \( \sqrt{1 + 4x^2} \) to \( \cosh(t) \).

• The hyperbolic sine function \( \sinh(t) \) and hyperbolic cosine function \( \cosh(t) \) relate closely to exponential functions, offering handy alternatives.
• This substitution transforms the integral into a simpler form \( \int \cosh(t) \, dt \), which is easier to handle.

Ultimately, the idea behind trigonometric substitution is to transform an integral into an easier one by switching to a more convenient variable.

Definite Integral Evaluation

Definite integrals provide the accumulated value across an interval on the x-axis from \( a \) to \( b \). This accumulated value is often associated with physical concepts like area under a curve or, in our case, the arc length of a curve. In the problem, after employing the trigonometric substitution, the integral \( \int \cosh(t) \, dt \) must be evaluated over the limits determined by the substitution. Evaluating this integral involves reversing the substitution to account for the original limits of \( x \).

• First, find the antiderivative \( \int \cosh(t) \, dt = \sinh(t) \).
• Then, translate the result back using the original limits of \( x \), which translate to specific \( t \) values.

For this specific problem, simplifying the definite integral yields a final arc length of \( \frac{5^{3/2}}{6} \). This showcases how definite integrals accumulate values, providing precise answers for real-world measurements.