Problem 109 - Calculus Volume 3

Question

Find the arc length of the curve on the indicated interval of the parameter. $$ x=\frac{1}{3} t^{3}, \quad y=\frac{1}{2} t^{2}, \quad 0 \leq t \leq 1 $$

Step-by-Step Solution

Answer

The arc length is \( \frac{1}{3} (2\sqrt{2} - 1) \).

Step 1: Write the Arc Length Formula

The arc length of a parametric curve defined by \( x = f(t) \) and \( y = g(t) \) from \( t = a \) to \( t = b \) is given by the integral:\[L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt.\]

Step 2: Differentiate x(t) and y(t) with respect to t

Calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) for the given functions:\[ x = \frac{1}{3} t^3 \Rightarrow \frac{dx}{dt} = t^2, \quad y = \frac{1}{2} t^2 \Rightarrow \frac{dy}{dt} = t. \]

Step 3: Substitute into the Arc Length Formula

Substitute \( \frac{dx}{dt} = t^2 \) and \( \frac{dy}{dt} = t \) into the arc length formula:\[L = \int_{0}^{1} \sqrt{(t^2)^2 + t^2} \, dt = \int_{0}^{1} \sqrt{t^4 + t^2} \, dt.\]

Step 4: Simplify the Integrand

Factor \( t^2 \) out of the square root: \[L = \int_{0}^{1} \sqrt{t^2(t^2 + 1)} \, dt = \int_{0}^{1} t \sqrt{t^2 + 1} \, dt.\]

Step 5: Use Substitution to Evaluate the Integral

Use the substitution \( u = t^2 + 1 \), which gives \( du = 2t \, dt \) or \( t \, dt = \frac{1}{2} du \). Change the limits of integration: when \( t = 0 \), \( u = 1 \); when \( t = 1 \), \( u = 2 \). Substitute and evaluate the integral:\[L = \frac{1}{2} \int_{1}^{2} \sqrt{u} \, du.\]

Step 6: Integrate the Function

Find the antiderivative of \( \sqrt{u} \):\[\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}.\]So, evaluate from 1 to 2:\[L = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_1^2 = \frac{1}{2} \left( \frac{2}{3} \cdot 2^{3/2} - \frac{2}{3} \cdot 1^{3/2} \right).\]

Step 7: Compute the Final Answer

Evaluate the expression:\[L = \frac{1}{2} \left( \frac{2}{3} \left( 2\sqrt{2} - 1 \right) \right) = \frac{1}{3} \left( 2\sqrt{2} - 1 \right).\]

Key Concepts

Parametric Curves

In mathematics, parametric curves offer a versatile way to describe a wide range of complex shapes that cannot be captured using standard Cartesian equations. Instead of expressing one variable directly in terms of another, parametric equations use a third parameter, often denoted as \( t \), to independently determine both \( x \) and \( y \).
For instance, in the exercise problem, the equations \( x = \frac{1}{3}t^3 \) and \( y = \frac{1}{2}t^2 \) describe the curve's path as \( t \) varies from 0 to 1. This parameterization allows us to find not just the location of points on the curve, but also the rate at which the curve is traversed. By setting \( 0 \leq t \leq 1 \), the entire segment of the curve can be examined systematically.

• This approach offers a more intuitive grasp of motion and speed along the path of the curve, which is especially useful in physics and engineering applications.
• Parametric equations are also key to computing arc lengths, as they provide the necessary functions to differentiate and integrate.

Integration Techniques

Integration is a fundamental mathematical tool used to find quantities like areas, volumes, and arc lengths. In the context of arc length, integration allows one to sum an infinite number of infinitesimally small line segments to measure a curve's actual length.
The arc length formula for parametric curves, \( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \), involves integrating the square root of the sum of squares of derivatives, highlighting the role of calculus in measuring curves.

• Differentiating \( x(t) \) and \( y(t) \) is crucial to set up the integrand correctly. For the given functions \( x = \frac{1}{3}t^3 \) and \( y = \frac{1}{2}t^2 \), these derivatives enable the formulation of a proper integral.
• Simplifying the integrand involves mathematical manipulation to make the integral easier to solve, as in factoring terms for clarity.
• Proper evaluation of such integrals frequently relies on a variety of techniques, such as substitution or partial fraction decomposition, to address different types of expressions.

Each step demands careful attention, underscoring the technique's importance in rendering problems like arc length computations manageable and solvable.

Substitution Method

The substitution method is a powerful tool in calculus for evaluating integrals that are difficult to solve in their original form. It involves a change of variable to simplify the integral. For instance, in our exercise, the function \( L = \int_{0}^{1} t \sqrt{t^2 + 1} \, dt \) can be more easily integrated by substitution.
Here, setting \( u = t^2 + 1 \) transforms the integral, as it leads to \( du = 2t \, dt \) or rearranged \( t \, dt = \frac{1}{2}du \). This substitution simplifies the square root expression, easing integration.

• The limits of integration must be adjusted to reflect the new variable. When \( t = 0 \), \( u = 1 \), and when \( t = 1 \), \( u = 2 \).
• This method streamlines the process, turning what can be an otherwise difficult integral into a simpler problem.
• After integration, switching back to the original variable is often a straightforward step, allowing one to apply the calculated \( u \) integral values to obtain the final solution.

For many calculus problems, particularly those involving complex parametric curves, substitution can turn intricate integrals into more approachable forms, paving the way for efficient and accurate solutions.