Problem 28 - Thomas Calculus

Question

Find the lengths of the curves. $$ x=(2 t+3)^{3 / 2} / 3, \quad y=t+t^{2} / 2, \quad 0 \leq t \leq 3 $$

Step-by-Step Solution

Answer

The length of the curve is 10.5.

Step 1: Understand the Arc Length Formula

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

Step 2: Compute Derivatives

First, we calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = \frac{(2t+3)^{3/2}}{3} \), use the chain rule to find \( \frac{dx}{dt} = \frac{1}{2} (2t+3)^{1/2} \times 2 = (2t+3)^{1/2} \). For \( y = t + \frac{t^2}{2} \), simply derive to find \( \frac{dy}{dt} = 1 + t \).

Step 3: Prepare the Arc Length Integral

Substitute the derivatives into the arc length formula: \( L = \int_{0}^{3} \sqrt{((2t+3)^{1/2})^2 + (1+t)^2} \, dt \). Simplify the expression under the square root: \( (2t+3) + (1+t)^2 = 2t+3 + 1 + 2t + t^2 = t^2 + 4t + 4 \). It becomes \( \sqrt{(t+2)^2} \).

Step 4: Calculate the Integral

Since \( \sqrt{(t+2)^2} = |t+2| \), and \( t+2 \) is always positive over the interval \([0, 3]\), this simplifies to \( t+2 \). Integrate \( \int_{0}^{3} (t+2) \, dt = \left[ \frac{t^2}{2} + 2t \right]_{0}^{3} \). Evaluate this to find \( \frac{3^2}{2} + 2 \times 3 = \frac{9}{2} + 6 \).

Step 5: Evaluate the Integral Result

Evaluate \( \left[ \frac{t^2}{2} + 2t \right] \) from 0 to 3: \( \left( \frac{9}{2} + 6 \right) - \left( \frac{0}{2} + 0 \right) = \frac{9}{2} + 6 = \frac{21}{2} \). Thus, the length of the curve is \( \frac{21}{2} = 10.5 \).

Key Concepts

Parametric Equations

In mathematics, parametric equations are a way to express a set of related variables as functions of one or more independent parameters. Instead of defining a curve by directly relating x and y, we use a parameter, typically t, to define both coordinates separately. This approach is especially useful for describing complex curves and motion in a more intuitive and flexible manner.
For example, consider our parametric equations in the exercise:

• \( x = \frac{(2t+3)^{3/2}}{3} \)
• \( y = t + \frac{t^2}{2} \)

Here, both x and y are functions of t. As t varies, the pair \((x, y)\) traces out a path in the plane. This is why understanding parametric equations is key when you want to analyze curves described in terms of parameters rather than direct relationships between x and y.
Using parametric equations provides a powerful tool for describing paths and figures that are otherwise difficult to represent in standard Cartesian form.

Chain Rule

The chain rule is a fundamental concept in calculus used for deriving composite functions. It helps us find the derivative of a function that is composed of other functions. This is essential when working with parametric equations, as we must differentiate each component—a task that often involves nested functions.
For instance, look at our previous equation:

• \( x = \frac{(2t+3)^{3/2}}{3} \)

To differentiate \( x \) with respect to \( t \), we utilize the chain rule. It lets us first find the derivative of the outer function and multiply it by the derivative of the inner function. Specifically for our case:

• First, the derivative of \((2t + 3)^{3/2}\) with respect to \((2t+3)\) is \( \frac{3}{2}(2t+3)^{1/2} \)
• Then, we multiply by the derivative of \( (2t+3) \), which is 2
• This finally gives us \( \frac{3}{2} \times (2) \times (2t + 3)^{1/2} \)

The chain rule thus simplifies the process of differentiation, keeping track of how each part of the composite function contributes to the overall derivative.

Integration of Curves

Understanding integration is crucial for calculating the length of a curve defined by parametric equations. The integration process allows us to compute the accumulated sum of quantities over a region, which in our case, is used to determine the arc length of a curve.
To find the arc length, we use the formula \[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \ dt\]This integral sums up small segments of a curve, calculated by evaluating both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) over the interval from \( t = a \) to \( t = b \).
In our exercise:

• \( \frac{dx}{dt} = (2t+3)^{1/2} \)
• \( \frac{dy}{dt} = 1 + t \)

These derivatives are plugged into the arc length formula to form the integrand:
\[L = \int_{0}^{3} \sqrt{(2t+3) + (1+t)^2} \, dt = \int_{0}^{3} \sqrt{(t+2)^2} \, dt\]Upon simplification, since \( t+2 \) is always positive within our bounds, the square root simplifies to \( |t+2| = t+2 \). Integrating \( t+2\) from 0 to 3 gives us the final arc length of the curve, which is 10.5 units.
Integration, in this context, not only calculates area under a curve but can also derive various other physical properties like work, distance, and in our case, the complete length of a path on a plane.