Problem 38 - Calculus Early Transcendentals

Question

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places. $$x=1+e^{t}, \quad y=t^{2}, \quad-3 \leq t \leqslant 3$$

Step-by-Step Solution

Answer

The length of the curve is approximately 49.3186 units.

Step 1: Understand the problem

We need to find the length of a curve described parametrically by the equations \(x=1+e^{t}\) and \(y=t^{2}\) for \(-3 \leq t \leq 3\). The length of the curve from \(t=a\) to \(t=b\) is given by \(L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\).

Step 2: Find the derivatives

First, we calculate \(\frac{dx}{dt}\). Since \(x = 1 + e^{t}\), we derive \(\frac{dx}{dt} = e^{t}\). Next, for \(y = t^2\), we derive \(\frac{dy}{dt} = 2t\).

Step 3: Set up the integral

Substitute \(\frac{dx}{dt} = e^{t}\) and \(\frac{dy}{dt} = 2t\) into the formula for the arc length: \[L = \int_{-3}^{3} \sqrt{(e^t)^2 + (2t)^2} \, dt = \int_{-3}^{3} \sqrt{e^{2t} + 4t^2} \, dt\]

Step 4: Evaluate the integral

Using a calculator, evaluate the integral \(\int_{-3}^{3} \sqrt{e^{2t} + 4t^2} \, dt\) to find the length of the curve.

Key Concepts

Parametric Equations

Parametric equations are a way to express mathematical curves by using parameters instead of explicit functions. This helps in describing more complex and intricate curves. In the original exercise, the curve is defined by the parametric equations

• \( x = 1 + e^{t} \)
• \( y = t^{2} \)

This means that instead of describing \( y \) as a function of \( x \) directly, both \( x \) and \( y \) are described in terms of a third variable, \( t \). This method is beneficial because it allows us to trace curves in a more flexible manner.
The parameter \( t \) typically represents time or another variable that affects the position of a point on the curve. In this exercise, it varies from \( -3 \) to \( 3 \), ensuring we cover the entire length of the curve. Parametric equations are particularly useful in physics and engineering where the path of an object is often described in terms of time or along a parameterized path.

Arc Length

Arc length is the measurement of the distance along a curve. It's a natural extension of the concept of length to curved forms.
In parametric form, the length \( L \) of a curve defined by equations \( x(t) \) and \( y(t) \) from \( t = a \) to \( t = b \) is calculated using the formula:

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

This formula comes from the Pythagorean theorem, considering infinitesimal segments of the curve as hypotenuses of tiny right triangles. Each segment contributes a small length which is aggregated by the integral over the specified interval.
In our exercise, this formula helps us find the curve's length over the parameter \( t \) from \( -3 \) to \( 3 \). The expression under the square root accounts for the changes in both the \( x \) and \( y \) directions, providing an overall measure of the path's length.

Derivative Calculation

Derivatives in parametric equations are essential to create formulas for arc length. Derivatives reflect how much a variable is changing in respect to another. For arc length, we calculate the derivative of each parametric equation with respect to \( t \).
For the given equations:

• The derivative \( \frac{dx}{dt} \) of \( x = 1 + e^{t} \) is \( e^{t} \).
• The derivative \( \frac{dy}{dt} \) of \( y = t^{2} \) is \( 2t \).

These derivative functions are essential because they tell us how rapidly \( x \) and \( y \) change with respect to \( t \). Their squares add up under the square root in the arc length formula, capturing how quickly and in what direction the curve moves as \( t \) changes.
Understanding derivatives here is important for handling complex curves as they dissect the movement into more manageable components, which are easier to analyze using calculus.