Question

For the following exercises, find the arc length of the curve on the indicated interval of the parameter.\(x=\frac{1}{3} t^{3}, \quad y=\frac{1}{2} t^{2}, \quad 0 \leq t \leq 1\)

Short answer

The arc length is approximately 0.9428.

Step-by-step solution

Set Up the Arc Length Formula

The formula for the arc length of a curve defined parametrically by \(x(t)\) and \(y(t)\) over an interval \([a, b]\) is: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt. \] For this problem, \(a = 0\) and \(b = 1\).

Compute the Derivatives

Find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). \[ \frac{dx}{dt} = \frac{d}{dt}\left(\frac{1}{3}t^3\right) = t^2, \]\[ \frac{dy}{dt} = \frac{d}{dt}\left(\frac{1}{2}t^2\right) = t. \]

Plug Derivatives into the Arc Length Formula

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

Simplify the Expression Inside the Integral

Factor the expression inside the square root.\[ t^4 + t^2 = t^2(t^2 + 1), \]which makes the integral simplify to:\[ L = \int_{0}^{1} t\sqrt{t^2 + 1} \, dt. \]

Use Substitution to Solve the Integral

Let \(u = t^2 + 1\), then \(du = 2t \, dt\) or \(t \, dt = \frac{1}{2} \, du\). When \(t = 0\), \(u = 1\), and when \(t = 1\), \(u = 2\). Substitute into the integral:\[ L = \int_{1}^{2} \sqrt{u} \frac{1}{2} \, du \]Which simplifies to:\[ L = \frac{1}{2} \int_{1}^{2} u^{1/2} \, du. \]

Integrate and Evaluate the Integral

Integrate \( u^{1/2} \) with respect to \( u \).\[ \int u^{1/2} \, du = \frac{2}{3} u^{3/2}. \]Now evaluate from 1 to 2:\[ L = \frac{1}{2} \left[ \frac{2}{3}(2^{3/2}) - \frac{2}{3}(1^{3/2}) \right]. \]

Final Calculation

Calculate the definite integral:\[ L = \frac{1}{2} \left[ \frac{2}{3}(\sqrt{8}) - \frac{2}{3}(1) \right], \]\[ L = \frac{1}{2} \left[ \frac{2}{3}(2\sqrt{2}) - \frac{2}{3} \right], \]Finally,\[ L = \frac{1}{3}(\sqrt{2}) + \frac{2}{3}, \]\[ L \approx 0.9428. \]

Conclude the Arc Length

The arc length of the curve from \(t = 0\) to \(t = 1\) is approximately 0.9428.

Key concepts

Integral Calculus

Integral calculus is one of the main branches of calculus, primarily focused on the concept of integration. Integration helps us find quantities like areas under curves, volumes of revolution, and, as in our exercise, the length of curves known as arc length.

Integral calculus revolves around the integral, which is a way of summing infinitely small quantities over an interval. We use the integral symbol \( \int \), and with definite integrals, we also specify the limits of integration (like \( a \) to \( b \) in our step-by-step solution).

For the arc length, the integral calculates the sum of tiny lengths along a curve between two points. This adds up to the total length of the curve. Here, we needed the definite integral because we were calculating something tangible and finite: the length of the curve from \( t=0 \) to \( t=1 \).

In practical terms, integral calculus enables us to solve problems that require the summation of infinite slices to find whole quantities. This process is essential in real-world applications across physics, engineering, economics, and beyond.

Parametric Equations

Parametric equations are a way of expressing mathematical functions that move beyond the simple \( y = f(x) \) format. Instead, both \( x \) and \( y \) are expressed in terms of a third variable, often \( t \), which we call the parameter.

This allows for more complex curves and paths to be drawn, which might not be possible with the standard function notation. In our exercise, the equations \( x=\frac{1}{3}t^3 \) and \( y=\frac{1}{2}t^2 \) define a curve in terms of the variable \( t \).

• Parametric equations are especially powerful when the relationship between \( x \) and \( y \) is not straightforward.
• They are widely used to describe trajectories, cycles, and anything involving movement in plane geometry.

To find the arc length using parametric equations, we establish the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) as seen in the step-by-step solution. These derivatives help denote how \( x \) and \( y \) change with respect to the parameter \( t \), setting the stage for our calculations.

Definite Integrals

In calculus, a definite integral calculates the net area under a curve and between two points on the x-axis, which are called the limits of integration. In our current exercise, it also plays a role in calculating the arc length of a parametric curve.

The expression \( \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt \) is a definite integral that finds the length of the curve from \( t=a \) to \( t=b \). Here, the actual or numerical limits indicate where the start (\( a=0 \)) and end (\( b=1 \)) of our calculation will be.

• Definite integrals provide both numeric and geometric interpretations.
• They're key in measuring areas, volumes, and in physics, real-life quantities like mass and charge on curves.

For the arc length, the definite integral collects all the tiny sections of the curve and sums them up, which then translates to the total arc length. This definite measure gives us a clear, finite result.

Mathematical Substitution

Mathematical substitution is a basic yet powerful technique employed to simplify the process of integrating complex functions. It involves replacing a complicated expression with a single variable, making the integral easier to solve.

In the step-by-step solution, substitution is used when changing the variable \( t \) to \( u \) with \( u = t^2 + 1 \). The derivative \( du \) is associated with the function of \( t \) to help in converting the entire integral into terms of \( u \).

• Substitution efficiently transforms integrals that are challenging to compute directly.
• This method is akin to the chain rule for differentiation but applied in reverse for integration.

Once substitution is complete, we integrate easier expressions and subsequently revert back to the original variable if needed. It allows solving integrals in a straightforward manner and is often indispensable in calculus problems, improving both understanding and execution.