Problem 41 - Calculus Early Transcendentals

Question

Find the exact length of the curve. $$x=1+3 t^{2}, \quad y=4+2 t^{3}, \quad 0 \leqslant t \leqslant 1$$

Step-by-Step Solution

Answer

The exact length of the curve is \(4\sqrt{2} - 2\).

Step 1: Understand the Problem

We are given parametric equations for a curve with respect to the parameter \(t\) from 0 to 1. We need to find the length of this curve for the given interval of \(t\).

Step 2: Formula for Arc Length

The formula for the arc length of a parametric curve \((x(t), y(t))\) from \(t = a\) to \(t = b\) is given by: \[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\]This requires us to calculate the derivatives of \(x\) and \(y\) with respect to \(t\).

Step 3: Differentiate the Parametric Equations

Differentiate \(x = 1 + 3t^2\) with respect to \(t\) to get \(\frac{dx}{dt} = 6t\). Differentiate \(y = 4 + 2t^3\) with respect to \(t\) to get \(\frac{dy}{dt} = 6t^2\).

Step 4: Substitute into Arc Length Formula

Substitute \(\frac{dx}{dt} = 6t\) and \(\frac{dy}{dt} = 6t^2\) into the arc length formula: \[L = \int_{0}^{1} \sqrt{(6t)^2 + (6t^2)^2} \, dt\] Simplify to get:\[L = \int_{0}^{1} \sqrt{36t^2 + 36t^4} \, dt = \int_{0}^{1} \sqrt{36t^2(1 + t^2)} \, dt = 6 \int_{0}^{1} |t| \sqrt{1 + t^2} \, dt\] Since \(t\) is from 0 to 1, \(|t| = t\).

Step 5: Solve the Integral

Evaluate the integral: \[6 \int_{0}^{1} t \sqrt{1 + t^2} \, dt\]Use substitution: let \(u = 1 + t^2\), then \(du = 2t \, dt\) or \(t \, dt = \frac{1}{2} \, du\). When \(t = 0\), \(u = 1\) and when \(t = 1\), \(u = 2\).The integral becomes:\[6 \cdot \frac{1}{2} \int_{1}^{2} u^{1/2} \, du = 3 \int_{1}^{2} u^{1/2} \, du\]Integrate to find:\[3 \left[ \frac{2}{3} u^{3/2} \right]_{1}^{2} = 2 \left[ u^{3/2} \right]_{1}^{2}\]Evaluate it:\[2(2^{3/2} - 1^{3/2}) = 2(2\sqrt{2} - 1)\]Thus, the length of the curve is \(4\sqrt{2} - 2\).

Step 6: Validate the Calculation

Verify the final calculation to ensure accuracy and check the limits of integration. The final result is in the simplest form: \(4\sqrt{2} - 2\). This represents the exact length of the curve given the parametric equations.

Key Concepts

Parametric Equations

Parametric equations provide a way to describe a curve in the plane using a parameter, often denoted by \(t\). These equations separate the horizontal \(x\) and vertical \(y\) coordinates into functions of \(t\). This allows us to define more complex curves that can't easily be described using just \(y = f(x)\).
For example, in the exercise, the parametric equations are \(x = 1 + 3t^2\) and \(y = 4 + 2t^3\). These define a curve by expressing both \(x\) and \(y\) in terms of \(t\), where \(t\) ranges from 0 to 1.

• These equations can represent various types of paths, including loops and spirals, depending on their formulation.
• They are not limited by the constraints of functions where each \(x\) can only have one \(y\).

This flexibility is crucial for representing real-world scenarios such as the trajectory of an object or the design of curves in animation and graphics.

Integration Techniques

Integration is a central tool used to calculate the arc length of parametric curves. The key technique involves setting up the integral based on the derivatives of the parametric equations. We use the integration to sum the infinitesimal lengths along the curve from the start to the end values of \(t\).
In this exercise, you use the arc length formula: \[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\]After substituting the derivatives, you simplify the resulting expression before integrating.

• Techniques such as substitution are often used to handle complex integrands. In our case, by letting \(u = 1 + t^2\), we made the integration more manageable.
• It helps to recognize when substitution can simplify parts of the integral, as was done by converting \(t \, dt\) to \(\frac{1}{2} \, du\).
• Understanding and correctly applying limits is crucial as well, converting interval \(t = [0, 1]\) to \(u = [1, 2]\).

This process reflects the power of integration in solving complicated problems by breaking them down into simpler parts.

Differentiation

Differentiation is the process of finding the rate at which one quantity changes with respect to another. In the context of parametric curves, differentiating the functions for \(x(t)\) and \(y(t)\) gives the rates of change of \(x\) and \(y\) with respect to \(t\).
In the given problem:

• For \(x(t) = 1 + 3t^2\), the derivative \(\frac{dx}{dt} = 6t\) measures the rate of change of \(x\) as \(t\) changes.
• For \(y(t) = 4 + 2t^3\), the derivative \(\frac{dy}{dt} = 6t^2\) captures how \(y\) changes with \(t\).

These derivatives are essential because they form part of the integrand in the arc length formula. Differentiation gives the necessary components to calculate the magnitude of the velocity vector along the curve. This magnitude can be seen as the "speed" along the path, and integrating this "speed" over the interval of \(t\) results in the total length of the curve.
Differentiation in parametric equations allows the unraveling of complex relationships and is foundational for the accurate mathematical description of motion and change in various fields.