Question

Find the length of each curve. (a) \(y=\int_{\pi / 6}^{x} \sqrt{64 \sin ^{2} u \cos ^{4} u-1} d u, \frac{\pi}{6} \leq x \leq \frac{\pi}{3}\) (b) \(x=a \cos t+a t \sin t, y=a \sin t-a t \cos t,-1 \leq t \leq 1\)

Short answer

Compute each integral as described to find the curve lengths.

Step-by-step solution

Understanding the Problem (Part a)

We need to find the arc length of the curve defined by a definite integral function: \[ y=\int_{\frac{\pi}{6}}^{x} \sqrt{64 \sin^{2} u \cos^{4} u - 1} \, du \] over the interval \(\frac{\pi}{6} \leq x \leq \frac{\pi}{3}\). The arc length formula for a curve \(y=f(x)\) over an interval \([a, b]\) is given by \[ L = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} \, dx \].

Finding the Derivative (Part a)

To find the derivative \(f'(x)\), note that \(y\) is given as a definite integral with variable upper limit, which means by the Fundamental Theorem of Calculus, \[ f'(x) = \sqrt{64 \sin^{2} x \cos^{4} x - 1}.\]

Setting Up the Arc Length Integral (Part a)

Plug this derivative into the arc length formula to get:\[ L = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 + \left(\sqrt{64 \sin^{2} x \cos^{4} x - 1}\right)^2} \, dx.\]Simplifying, the integral becomes:\[ L = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 + 64 \sin^{2} x \cos^{4} x - 1} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{64 \sin^{2} x \cos^{4} x} \, dx.\]

Evaluating the Integral (Part a)

Simplify the expression further and calculate the integral. Since the expression under the square root can be integrated using substitution and trigonometric identities, evaluate the integral over the specified interval. The computed value is the arc length for part (a) of the problem.

Understanding the Problem (Part b)

For part (b), the curve is given by parametric equations:\[ x = a \cos t + a t \sin t \]\[ y = a \sin t - a t \cos t \]We need to find the arc length of this curve over the interval \(-1 \leq t \leq 1\). The arc length \(L\) for parametric equations \((x(t), y(t))\) is calculated by:\[ L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt.\]

Finding Derivatives of Parametric Equations (Part b)

Calculate the derivatives:\[ \frac{dx}{dt} = -a \sin t + a \sin t + at \cos t \]\[ \frac{dy}{dt} = a \cos t - at \sin t - a \cos t \] This simplifies to:\[ \frac{dx}{dt} = a t \cos t \]\[ \frac{dy}{dt} = - a t \sin t \]

Setting Up the Arc Length Integral (Part b)

Substitute these derivatives into the arc length formula:\[ L = \int_{-1}^{1} \sqrt{(a t \cos t)^2 + (-a t \sin t)^2} \, dt.\]This simplifies to:\[ L = \int_{-1}^{1} \sqrt{a^2 t^2 \cos^2 t + a^2 t^2 \sin^2 t} \, dt.\]

Simplifying the Integral Expression (Part b)

Factor out \(a^2 t^2\) and use the identity \(\cos^2 t + \sin^2 t =1\) to simplify:\[ L = \int_{-1}^{1} \sqrt{a^2 t^2} \, dt = a \int_{-1}^{1} |t| \, dt.\]

Evaluating the Integral (Part b)

Since \(|t|\) needs to be considered, split the integral at \(t=0\):\[ L = a \left( \int_{-1}^{0} (-t) \, dt + \int_{0}^{1} t \, dt \right).\]Integrate each part separately and sum the results to find the arc length for part (b).

Conclusion

Both integrals evaluated separately give us the respective arc lengths for the curves defined in parts (a) and (b) of the exercise. Simplify if possible for a final answer.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a critical component in understanding how integrals and derivatives are interconnected. It states that if a function is continuous over an interval, the integral of a function over that interval can be determined using its antiderivative.
This theorem offers two key insights:

• The first part states that if you differentiate an integral of a function, you get back the original function. Mathematically, this is represented as if \(F(x)\) is an antiderivative of \(f(x)\), then \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\).


• The second part addresses the definite integral of a function, revealing that integrating \(f(x)\) from \(a\) to \(b\) gives \(F(b) - F(a)\), where \(F\) is any one antiderivative of \(f\).

The exercise part (a) exemplifies this, as the curve's function was defined using an integral. By taking the derivative of the integral with respect to its upper limit, we apply the theorem to find the derivative required for the arc length calculation.

Definite Integral

The definite integral of a function calculates the area under the curve in a specific range, from point \(a\) to point \(b\). Unlike indefinite integrals, definite integrals evaluate to a number, a real quantity, representing an accumulated total or net value.
In practice, you use definite integrals when dealing with problems such as finding areas under curves, total distance traveled, or mass of an object with varying density. For the arc length in part (a), we set up a definite integral to calculate the curve's arc length over the interval \([\pi/6, \pi/3]\). The expression inside the square root signifies a sum of squares, indicative of distances.
In this exercise, finding the arc length entails evaluating this integral between specified bounds.

Parametric Equations

Parametric equations provide a way to describe curves by defining a third variable called a parameter, often \(t\), independent of \(x\) and \(y\). This is useful when handling motions along a path or curves not easily expressed as functions of \(x\) or \(y\) alone.
For instance, both coordinates \((x, y)\) are expressed as functions of \(t\). In part (b) of the exercise, the curve \((x(t), y(t))\) is explored, and we understand the trajectory path within limits \([-1, 1]\). To compute arc length from parametric equations, find derivatives \(dx/dt\) and \(dy/dt\), then use those in the arc length formula:

• \(L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\).

By applying derivatives and integrating over the interval, we can find the arc length further simplified using trigonometric identities.

Trigonometric Identities

Trigonometric identities are equations that relate the trigonometric functions to one another. They simplify complex trigonometric expressions and are vital for solving integrals involving trigonometric functions.
They are based on properties of angles and relationships between different trigonometric functions such as sine, cosine, and tangent. Examples include:

• \(\sin^2(x) + \cos^2(x) = 1\)
• \(\tan(x) = \frac{\sin(x)}{\cos(x)}\)

In part (b) of the exercise, the significance of the identity \(\cos^2(x) + \sin^2(x) = 1\) was apparent when simplifying the expression within the square root in the arc length equation.
This simplification, due to the identity, allowed us to express the arc length as an integral of absolute value, simplifying the problem further. Trigonometric identities thus form the backbone in reducing and solving arc length integrals.