Problem 24 - Calculus Single Variable

Question

Calculate the length of the given parametric curve. $$ x=\cos ^{3}(t) \quad y=\sin ^{3}(t) \quad 0 \leq t \leq \pi / 2 $$

Step-by-Step Solution

Answer

The length of the parametric curve is \( \frac{3}{2} \).

Step 1: Understanding the Parametric Curve

To find the length of a parametric curve defined by \( x = f(t) \) and \( y = g(t) \) over an interval \([a, b]\), we use the formula: \[ L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt. \] This formula calculates the integral of the speed function for the given range of \( t \).

Step 2: Differentiate the Functions

We need to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).For \( x = \cos^3(t) \), we apply the chain rule: \[ \frac{dx}{dt} = 3\cos^2(t)(-\sin(t)) = -3\cos^2(t)\sin(t). \]For \( y = \sin^3(t) \), applying the chain rule gives:\[ \frac{dy}{dt} = 3\sin^2(t)\cos(t). \]

Step 3: Setting Up the Integral

Substitute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) into the length formula:\[ L = \int_0^{\pi/2} \sqrt{(-3\cos^2(t)\sin(t))^2 + (3\sin^2(t)\cos(t))^2} \, dt. \]

Step 4: Simplifying the Expression

Simplify inside the square root:\[ (3\cos^2(t)\sin(t))^2 + (3\sin^2(t)\cos(t))^2 = 9\cos^4(t)\sin^2(t) + 9\sin^4(t)\cos^2(t). \]Factor out common terms:\[ = 9\cos^2(t)\sin^2(t)(\cos^2(t) + \sin^2(t)). \]Since \( \cos^2(t) + \sin^2(t) = 1 \), it simplifies to:\[ 9\cos^2(t)\sin^2(t). \]

Step 5: Compute the Integral

Simplify further:\[ L = \int_0^{\pi/2} \sqrt{9\cos^2(t)\sin^2(t)} \, dt \]\[ = \int_0^{\pi/2} 3|\cos(t)\sin(t)| \, dt. \]Since \( 0 \leq t \leq \pi/2 \), \( \cos(t), \sin(t) \geq 0 \), so we can drop the absolute value:\[ = 3\int_0^{\pi/2} \cos(t)\sin(t) \, dt. \]

Step 6: Evaluate the Integral

Use the identity \( \cos(t)\sin(t) = \frac{1}{2}\sin(2t) \):\[ = \frac{3}{2}\int_0^{\pi/2} \sin(2t) \, dt. \]Evaluate the integral:\[ = \frac{3}{2} \left[-\frac{1}{2}\cos(2t)\right]_0^{\pi/2} \]\[ = -\frac{3}{4}\left[\cos(\pi) - \cos(0)\right] \]\[ = -\frac{3}{4}[-1 - 1] \]\[ = \frac{3}{2} \].

Step 7: State the Final Result

The length of the parametric curve from \( t = 0 \) to \( t = \pi/2 \) is \( \frac{3}{2} \).

Key Concepts

Arc Length

The concept of arc length is crucial when dealing with the geometry of curves described by parametric equations. It represents the actual distance traveled along the curve. For a curve defined as \( x = f(t) \) and \( y = g(t) \), to find its arc length over an interval \([a, b]\), we use the integral formula: \[ L = \int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt. \] This computation involves integrating the derivative of the curve's position functions—essentially, calculating the length of the hypotenuse for infinitesimally small segments of the curve, then summing those up over the specified range of \(t\). For example, in our exercise, we calculate the length of the curve \( x = \cos^3(t) \) and \( y = \sin^3(t) \) from \( t=0 \) to \( t=\pi/2 \). You differentiate these position functions and substitute into the arc length formula, through integration, to find the distance the curve makes between these two points.

Calculus Integration

When finding the length of a curve or solving similar calculus problems, integration serves as a powerful tool. For instance, once you determine the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) for your parametric equations, the task involves integrating the formula: \[ L = \int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt. \] Integration helps accrue small changes across a continuous curve into a single value, which is essential for exact measurement in calculus. It's fundamental to solve these integrals correctly, often requiring simplification and sometimes even substitution, to arrive at a result. This allows us to find the total arc length or evaluate other properties like areas under or between curves.

Chain Rule

The chain rule in calculus is especially useful for finding derivatives of composite functions, which appear frequently in parametric equations. Essentially, if you have a function \( h(x) = f(g(x)) \), you apply the chain rule by differentiating the outer function and multiplying by the derivative of the inner function: \[ h'(x) = f'(g(x)) \cdot g'(x). \] In our example, the parametric equations \( x = \cos^3(t) \) and \( y = \sin^3(t) \) require the use of the chain rule for differentiation. This involves differentiating \( \cos^3(t) \) and \( \sin^3(t) \) by treating these expressions as composites of trigonometric and power functions. You'll differentiate the trigonometric part first and then multiply by the derivative of the power function, simplifying to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Successfully applying the chain rule is crucial in setting up the integral for the calculation of arc length.