Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-46, find the length of the parametric curve defined over the given interval. $$ x=3 t^{2}, y=t^{3} ; 0 \leq t \leq 2 $$

Step-by-Step Solution

Answer

The length of the parametric curve is 8.

Step 1: Understand the Formula for Arc Length

The arc length of a parametric curve defined by \( x(t) \) and \( y(t) \) from \( t = a \) to \( t = b \) is given by the formula:\[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]In this problem, \( a = 0 \) and \( b = 2 \). We need to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Step 2: Compute Derivatives

Given \( x(t) = 3t^2 \) and \( y(t) = t^3 \), compute their derivatives with respect to \( t \):\[\frac{dx}{dt} = \frac{d}{dt}(3t^2) = 6t\]\[\frac{dy}{dt} = \frac{d}{dt}(t^3) = 3t^2\]

Step 3: Substitute Derivatives into the Arc Length Formula

Substitute \( \frac{dx}{dt} = 6t \) and \( \frac{dy}{dt} = 3t^2 \) into the arc length formula:\[L = \int_{0}^{2} \sqrt{(6t)^2 + (3t^2)^2} \, dt\]Simplifying, we get:\[L = \int_{0}^{2} \sqrt{36t^2 + 9t^4} \, dt\]Further simplification gives:\[L = \int_{0}^{2} \sqrt{9t^2(4 + t^2)} \, dt\]\[L = \int_{0}^{2} 3t\sqrt{4 + t^2} \, dt\]

Step 4: Solve the Integral

We'll substitute \( u = 4 + t^2 \) with \( du = 2t \, dt \). Rearrange to solve for \( dt \):\[t\, dt = \frac{1}{2} \, du\]Change the limits for \( t =0 \) to 0 and for \( t =2 \) to 8:\[L = \int_{0}^{8} \frac{3}{2} \cdot \sqrt{u} \, du\]Integrate to obtain:\[\int \frac{3}{2}u^{1/2} \, du = \frac{3}{2} \cdot \frac{2}{3}u^{3/2} = u^{3/2}\]Evaluate from 0 to 8:\[L = [u^{3/2}]_{0}^{8} = (8)^{3/2} - (0)^{3/2} = 8\]

Step 5: State the Final Answer

The length of the parametric curve is 8.

Key Concepts

Derivatives of Parametric Equations

When dealing with parametric equations, an important step in finding the arc length is computing the derivatives of the given functions with respect to the parameter. In our example, we have the parametric equations \(x = 3t^2\) and \(y = t^3\). To find the arc length of this curve, we first need to determine \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

• For \(x(t) = 3t^2\), the derivative \(\frac{dx}{dt} = 6t\). This tells us how the x-coordinate changes with respect to time.
• For \(y(t) = t^3\), the derivative \(\frac{dy}{dt} = 3t^2\). This tells us how the y-coordinate changes over time.

These derivatives will be used in the arc length formula to assess the change in direction of the curve as you move along it. Calculating these accurately is crucial, as they are foundational to finding the overall arc length across a given interval.

Integral Calculus

Integral calculus is a fundamental tool used to calculate various aspects of geometry and motion through accumulation. In our problem, we utilize integral calculus to compute the arc length of a curve.
The goal of integration, in this context, is to sum up infinitely small line segments that make up the curve to find its total length. The arc length formula requires integration over a specific interval, ensuring that the entire segment of the curve is considered.
In practice, this involves setting up a definite integral where the limits reflect the range over which the parameter, \(t\), is defined. In our scenario, the limits are from 0 to 2. Integral calculus allows us to accumulate small stretches of the curve, given by \(\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\), to compute the full arc length.

Arc Length Integration

Calculating the length of a parametric curve involves using a specific integration formula for arc length. This utilizes the derivatives of the parametric equations. For parametric equations \(x(t)\) and \(y(t)\), the arc length from \(t = a\) to \(t = b\) is given by:\[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]After substituting the derivatives into this formula, it simplifies our problem considerably. In our example:

• Substituted derivatives: \(\frac{dx}{dt} = 6t\) and \(\frac{dy}{dt} = 3t^2\).
• The arc length integral becomes \(\int_{0}^{2} \sqrt{36t^2 + 9t^4} \, dt\), which further simplifies to \(\int_{0}^{2} 3t\sqrt{4 + t^2} \, dt\).

Utilizing the arc length formula helps transform the curve into length calculations, allowing even complex curves to be broken down into manageable pieces via integration.

Change of Variables in Integration

To solve integrals involving complex expressions, the method known as 'change of variables' or 'substitution' is often used. This helps transform an integral into a simpler one, making it easier to solve.
In our example, we use substitution to handle the integral \(\int_{0}^{2} 3t\sqrt{4 + t^2} \, dt\). We let \(u = 4 + t^2\), then calculate the differential \(du = 2t \, dt\).

• This implies \(t \, dt = \frac{1}{2} \, du\), which we can substitute back into the integral.
• We also adjust the limits of integration: when \(t = 0\), \(u = 4\); when \(t = 2\), \(u = 8\).

The substitution simplifies our integral to \(\int_{4}^{8} \frac{3}{2} \sqrt{u} \, du\), which can be integrated easily. Evaluating it yields the arc length of the curve. Substitution is a powerful integration technique for simplifying complex integrals by changing variables, ultimately leading to a solution.