Problem 23 - Thomas Calculus in SI Units

Question

Find the length of the curve $$ y=\int_{0}^{x} \sqrt{\cos 2 t} d t $$ from \(x=0\) to \(x=\pi / 4\)

Step-by-Step Solution

Answer

The length of the curve is 1.

Step 1: Understand the problem

We need to find the length of the curve defined by \( y = \int_{0}^{x} \sqrt{\cos 2 t} \, dt \) from \(x=0\) to \(x=\pi/4\). This requires using the curve length formula for a function \(y=f(x)\).

Step 2: Recall the curve length formula

The formula for the length \(L\) of a curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by:\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] Here, \(f(x)\) is our integral function and we need to find its derivative first.

Step 3: Find the derivative \( \frac{dy}{dx} \)

The function is \( y(x) = \int_{0}^{x} \sqrt{\cos 2t} \, dt \). By the Fundamental Theorem of Calculus, \( \frac{dy}{dx} = \sqrt{\cos 2x} \).

Step 4: Substitute into the curve length formula

Substitute \( \frac{dy}{dx} = \sqrt{\cos 2x}\) into the length formula:\[ L = \int_{0}^{\pi/4} \sqrt{1 + ( \sqrt{\cos 2x} )^2} \, dx \] which simplifies to:\[ L = \int_{0}^{\pi/4} \sqrt{1 + \cos 2x} \, dx \]

Step 5: Simplify and evaluate the integral

We use the trigonometric identity \(1 + \cos 2x = 2\cos^2 x\) to get:\[ L = \int_{0}^{\pi/4} \sqrt{2}\cos x \, dx = \sqrt{2} \int_{0}^{\pi/4} \cos x \, dx \]Now, evaluate the integral \( \int \cos x \, dx = \sin x \):\[\sqrt{2}\left[ \sin x \right]_{0}^{\pi/4} = \sqrt{2}\left(\sin \frac{\pi}{4} - \sin 0\right) = \sqrt{2}\left(\frac{\sqrt{2}}{2} - 0\right) = 1 \]

Step 6: Conclusion

The length of the curve \( y = \int_{0}^{x} \sqrt{\cos 2 t} \, dt \) from \(x=0\) to \(x=\pi/4\) is 1.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key principle that bridges differential and integral calculus. It consists of two main parts:

• The first part ensures that if you have a continuous function integrated from a constant to a variable, the derivative of this integral is simply the original function.
• The second part connects definite integrals and antiderivatives. It states that the definite integral of a function over an interval can be evaluated using any one of its infinitely many antiderivatives.

In the exercise, we use the first part of the theorem. For the integral function \(y = \int_{0}^{x} \sqrt{\cos 2t} \, dt\), the derivative \(\frac{dy}{dx}\) is simply \(\sqrt{\cos 2x}\). This simplifies solving problems involving integrals by converting the integral into a more straightforward differentiation problem.

Trigonometric Identity

Trigonometric identities are equations involving trigonometric functions that are true for all values of the involved variables. They are essential for simplifying expressions and solving trigonometric equations. In this exercise, we employ the identity:

• \(1 + \cos 2x = 2\cos^2 x\)

This identity helps to simplify the expression under the square root in the curve length formula. By rewriting \(1 + \cos 2x\) as \(2\cos^2 x\), we can transform \(\sqrt{1 + \cos 2x}\) into \(\sqrt{2}\cos x\). Simplifying expressions in this way often makes integration more manageable, leading to an exact solution.

Definite Integral

Definite integrals are used to find the net area under a curve between two bounds. In contrast to indefinite integrals, definite integrals have limits and result in a numerical value. The definite integral of a function \(f(x)\) from \(x = a\) to \(x = b\) is denoted by:\[ \int_{a}^{b} f(x) \, dx \]In the given exercise, we evaluated a definite integral to find the curve's length from \(x = 0\) to \(x = \pi/4\). By integrating the expression \(\sqrt{2}\cos x\), the definite integral \(\int_{0}^{\pi/4} \cos x \, dx\) is computed using the antiderivative of \(\cos x\), which is \(\sin x\). Thus, the length is found by calculating \(\sqrt{2}\left[ \sin x \right]_{0}^{\pi/4}\), elegantly producing a result of 1.