Problem 47 - Calculus Early Transcendentals

Question

Graph the curve and find its length. $$x=e^{t}-t, \quad y=4 e^{t / 2}, \quad-8 \leq t \leqslant 3$$

Step-by-Step Solution

Answer

The length of the curve is \(e^3 - e^{-8} + 11\).

Step 1: Parametric Equations

The given problem involves parametric equations where \(x\) and \(y\) are functions of the parameter \(t\). We have \(x = e^t - t\) and \(y = 4e^{t/2}\). The task is to graph this curve and find its length over the interval \(-8 \leq t \leq 3\).

Step 2: Graphing the Curve

To graph the curve, evaluate the parametric equations at a series of values within the interval \(-8 \leq t \leq 3\). For each \(t\), calculate \(x\) and \(y\) and plot the resulting points on the coordinate plane. Connect these points smoothly to visualize the curve. A graphing calculator or software can facilitate this step.

Step 3: Arc Length Formula

To find the length of the curve, we use the parametric arc length formula: \[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]. This requires finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Step 4: Derivatives of x and y

Calculate the derivatives: \( \frac{dx}{dt} = e^t - 1 \) and \( \frac{dy}{dt} = 2e^{t/2} \). These are found by differentiating \(x = e^t - t\) and \(y = 4e^{t/2}\) with respect to \(t\).

Step 5: Integrate the Arc Length Formula

Substitute \( \frac{dx}{dt} = e^t - 1 \) and \( \frac{dy}{dt} = 2e^{t/2} \) into the arc length formula to get \[L = \int_{-8}^{3} \sqrt{(e^t - 1)^2 + (2e^{t/2})^2} \, dt\]. Simplifying, this becomes:\[L = \int_{-8}^{3} \sqrt{e^{2t} - 2e^t + 1 + 4e^t} \, dt = \int_{-8}^{3} \sqrt{e^{2t} + 2e^t + 1} \, dt\].

Step 6: Simplify the Integrand

Recognize that the integrand \(\sqrt{e^{2t} + 2e^t + 1}\) simplifies to \(\sqrt{(e^t + 1)^2}\). This is simply \(e^t + 1\) as it is the positive root, thus:\[L = \int_{-8}^{3} (e^t + 1) \, dt\].

Step 7: Compute the Integral

Break the integral into simpler parts:\[L = \int_{-8}^{3} e^t \, dt + \int_{-8}^{3} 1 \, dt\]. These integrals evaluate to \[L = \left[e^t\right]_{-8}^{3} + \left[t\right]_{-8}^{3}\].

Step 8: Evaluate the Integral

First evaluate \( \left[e^t\right]_{-8}^{3} = e^3 - e^{-8} \) and \( \left[t\right]_{-8}^{3} = 3 - (-8) = 11 \). Combine these results to find the length of the curve: \[L = (e^3 - e^{-8}) + 11\]. Compute these for the final length.

Key Concepts

Arc Length Formula

In calculus, the arc length formula is a powerful tool for finding the length of a curve defined by parametric equations. It allows you to calculate the distance along a curve from one point to another. The general formula for the arc length, \( L \), of a parametric curve defined by \( x(t) \) and \( y(t) \) 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 formula combines the derivatives of the curve’s parametric equations. Here's how to use it:

• Start by differentiating the given parametric equations to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
• Plug these derivatives into the formula under the square root sign, sum their squares, and take the square root.
• Integrate this result from the beginning parameter \(a\) to the endpoint \(b\).

This approach helps us understand the continuous nature of a curve, providing a measure of its actual path length between specified bounds.

Parametric Curve Graphing

Graphing a parametric curve involves plotting points calculated from the parametric equations over a specific range of the parameter, \( t \). This process allows us to visualize the trajectory of a moving point, which is particularly useful for curves that are not functions but still represent paths through the plane.
To graph these curves:

• Select a range for \( t \), such as \(-8 \leq t \leq 3\) for our example.
• Calculate the corresponding \(x\) and \(y\) values for several points within this range.
• Plot these \((x, y)\) coordinate points on the graph.
• Connect the plotted points smoothly to form the curve.

Graphing parametric equations can be effectively done using graphing calculators or computer software, allowing you to see how the curve behaves over the specified interval.

Derivatives of Parametric Functions

Derivatives of parametric functions play a crucial role in various calculations such as finding tangents, rates of change, and lengths of curves. When dealing with parametric equations, \(x(t)\) and \(y(t)\), we compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) with respect to the parameter \(t\).
This process involves differentiating each of the parametric equations individually:

• Differentiate \(x(t)\) to find \( \frac{dx}{dt} \).
• Differentiate \(y(t)\) to find \( \frac{dy}{dt} \).

These derivatives provide pivotal information:

• They are used in calculating the arc length of the parametric curve.
• They help in analyzing the curve's slope and understanding changes in direction.
• Moreover, combined derivatives can offer insight into the curve's trajectory and speed.

Mastering these concepts allows us to perform deeper analysis and make informed conclusions about the geometrical and physical properties of parametric curves.