Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

, 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 units.

Step 1: Determine the Derivatives

First, we need to find the derivatives of the parametric equations. For \(x = 3t^2\), the derivative with respect to \(t\) is \(\frac{dx}{dt} = 6t\). For \(y = t^3\), the derivative with respect to \(t\) is \(\frac{dy}{dt} = 3t^2\).

Step 2: Substitute Derivatives into the Arc Length Formula

The arc length \(L\) of a parametric curve from \(t = a\) to \(t = b\) is given by \(L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). In this case, \(a = 0\) and \(b = 2\). Using the derivatives found, the formula becomes \(L = \int_{0}^{2} \sqrt{(6t)^2 + (3t^2)^2} \, dt\).

Step 3: Simplify the Expression Under the Square Root

Simplify the expression under the square root: \(L = \int_{0}^{2} \sqrt{36t^2 + 9t^4} \, dt\). This simplifies to \(L = \int_{0}^{2} \sqrt{9t^2(4 + t^2)} \, dt\).

Step 4: Simplify Further

Factor out the 9 from under the square root: \(L = \int_{0}^{2} 3t \sqrt{4 + t^2} \, dt\).

Step 5: Use Substitution to Solve the Integral

Let \(u = 4 + t^2\), then \(du = 2t \, dt\) or \(t \, dt = \frac{1}{2} du\). Change the limits: when \(t = 0\), \(u = 4\); when \(t = 2\), \(u = 8\). Thus, the integral becomes \(L = \int_{4}^{8} \frac{3}{2} \sqrt{u} \, du\).

Step 6: Integrate and Substitute the Limits

The integral \(\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}\). Thus, \(L = \frac{3}{2} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{4}^{8} = \left[ u^{3/2} \right]_{4}^{8}\). Substitute the bounds: \(L = 8^{3/2} - 4^{3/2}\).

Step 7: Calculate the Final Answer

Calculate \(8^{3/2}\) and \(4^{3/2}\): \(8^{3/2} = (2^3)^{3/2} = 2^4 = 16\) and \(4^{3/2} = (2^2)^{3/2} = 2^3 = 8\). Therefore, \(L = 16 - 8 = 8\).

Key Concepts

Derivatives of Parametric Equations

In order to find the arc length of a parametric curve, it's necessary to compute the derivatives of the equations describing the curve. Consider the parametric equations given by \(x=3t^2\) and \(y=t^3\). To find the derivative of \(x\) with respect to \(t\), apply the power rule. This gives \(\frac{dx}{dt} = 6t\), as the derivative of \(3t^2\) comes out to be \(6t\).
Similarly, differentiate \(y = t^3\) with respect to \(t\), yielding \(\frac{dy}{dt} = 3t^2\). These derivatives are crucial because they help determine how fast each component of the curve is changing with \(t\), and are integral to computing the arc length.

Arc Length Formula

The arc length formula for a parametric curve quantifies the total distance traced by a point moving along the curve. Specifically, if a curve is characterized by the parametric equations \(x(t)\) and \(y(t)\), the arc length \(L\) between \(t=a\) and \(t=b\) is calculated as:

• \(L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\)

For our specific curve, use \(\frac{dx}{dt} = 6t\) and \(\frac{dy}{dt} = 3t^2\) within this formula over the interval \([0, 2]\).
The resulting integral becomes \(L = \int_{0}^{2} \sqrt{(6t)^2 + (3t^2)^2} \, dt\), which symbolizes the process of accumulating the infinitesimal lengths of all segments of the curve from \(t=0\) to \(t=2\).

Substitution in Integrals

Integrals often need simplification for ease of calculation, and substitution is a powerful technique to achieve this simplification. In our integral, simplify by letting \(u = 4 + t^2\). Consequently, differentiate \(u\) to get \(du = 2t \, dt\), or \(t \, dt = \frac{1}{2} \, du\).
Change the integration limits according to the new variable \(u\):

• When \(t = 0\), \(u = 4\).
• When \(t = 2\), \(u = 8\).

Substituting these transformations turns the integral into \(L = \int_{4}^{8} \frac{3}{2} \sqrt{u} \, du\), significantly easing the computation process.

Simplifying Expressions

Simplifying expressions, especially under a square root, can make subsequent calculations tractable and prevent errors. Beginning with \(L = \int_{0}^{2} \sqrt{36t^2 + 9t^4} \, dt\), factor out the greatest common factor under the square root.
The expression simplifies to \(L = \int_{0}^{2} \sqrt{9t^2(4 + t^2)} \, dt\), and further to \(L = \int_{0}^{2} 3t \sqrt{4 + t^2} \, dt\) by extracting the constant \(9\) as \(3\) from the square root.
This step is crucial as it transforms a complex integral into a form that allows for linear substitution methods, ultimately turning the integral into a solvable form \(L = \int_{4}^{8} \frac{3}{2} \sqrt{u} \, du\), ready for evaluation.