Problem 45 - Calculus Early Transcendentals

Question

Graph the curve and find its length. $$x=e^{t} \cos t, \quad y=e^{t} \sin t, \quad 0 \leqslant t \leqslant \pi$$

Step-by-Step Solution

Answer

The curve length is \( \sqrt{2} (e^{\pi} - 1) \).

Step 1: Understanding the Parametric Equations

We have two parametric equations given for a curve: \[ x = e^t \cos(t) \] \[ y = e^t \sin(t) \] These describe the x and y coordinates in terms of the parameter \( t \) ranging from 0 to \( \pi \). To graph the curve, plot these points for \( t \) within the given interval.

Step 2: Find the Derivatives of x and y

To find the length of the curve, we first need the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).Calculate \( \frac{dx}{dt} \) as follows:\[ \frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t \]Calculate \( \frac{dy}{dt} \) as follows:\[ \frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t \]

Step 3: Calculate the Expression Under the Square Root for Length

Using the derivatives found, we need to calculate:\[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \]Substitute and simplify:\[ \left(\frac{dx}{dt}\right)^2 = (e^t \cos t - e^t \sin t)^2 = e^{2t} (\cos^2 t - 2 \cos t \sin t + \sin^2 t) \]\[ \left(\frac{dy}{dt}\right)^2 = (e^t \sin t + e^t \cos t)^2 = e^{2t} (\sin^2 t + 2 \cos t \sin t + \cos^2 t) \]Sum them:\[ e^{2t}[\cos^2 t - 2 \cos t \sin t + \sin^2 t] + e^{2t}[\sin^2 t + 2 \cos t \sin t + \cos^2 t] = 2e^{2t} \] So we get:\[ \sqrt{e^{2t} \, 2} = \sqrt{2} \, e^t \]

Step 4: Set Up and Evaluate the Integral for Arc Length

To find the total arc length, integrate the expression from 0 to \( \pi \): \[ L = \int_{0}^{\pi} \sqrt{2} \, e^t \, dt \] Factor out \(\sqrt{2}\):\[ L = \sqrt{2} \int_{0}^{\pi} e^t \, dt \] Now evaluate \( \int e^t \, dt = e^t \): \[ L = \sqrt{2} \left[e^t \right]_{0}^{\pi} = \sqrt{2} (e^{\pi} - e^0) = \sqrt{2} (e^{\pi} - 1) \]

Step 5: Conclusion

The length of the curve, described parametrically by the given equations over the interval \( 0 \leq t \leq \pi \), is \( \sqrt{2} (e^{\pi} - 1) \). Graphically, this curve corresponds to a spiral from \( t = 0 \) to \( t = \pi \).

Key Concepts

Arc Length

Finding the arc length of a curve defined by parametric equations involves a specific process. With parametric equations, the curve's x and y coordinates are expressed in terms of a third variable, which is often denoted as \( t \). To calculate the arc length, you'll need to follow these steps:

• First, determine the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
• Next, you calculate the expression under the square root: \( \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \).
• Finally, set up an integral of this expression over the given range of \( t \) to find the total arc length.

For the equations \( x = e^t \cos(t) \) and \( y = e^t \sin(t) \), the arc length from \( t = 0 \) to \( t = \pi \) is found by integrating \( \sqrt{2} e^t \), leading to \( \sqrt{2} (e^{\pi} - 1) \). This method provides a way to measure the length of complex, non-linear paths defined via parametric forms.

Derivatives

When working with parametric equations, derivatives are crucial for understanding how the curve behaves. The derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) represent the rates of change of \( x \) and \( y \) with respect to the parameter \( t \).
Calculating these derivatives involves basic rules of differentiation. For instance:

• The derivative of the product \( e^t \cos(t) \) is calculated using the product rule: \( \frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t \).
• Similarly, \( \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t \).

These derivatives are then used to determine the rate at which the curve is traced out as \( t \) changes, and are vital for calculating arc length. Understanding and computing these derivatives accurately is essential as they provide insights into the geometric properties of the curve.

Integration

Integration is a fundamental process in calculus used to find the accumulated quantity, such as area under a curve or, in this case, the arc length. When it comes to parametric equations, integrating helps calculate the curve's length over a specified range of \( t \).
For example, once you determine the expression for the arc length of a parametric curve: \( \sqrt{2} e^t \), the task is to integrate it from \( t = 0 \) to \( t = \pi \):

• The integral \( \int_{0}^{\pi} e^t dt \) is solved as \( e^t \), evaluated from 0 to \( \pi \).
• Substituting the bounds gives \( e^{\pi} - e^0 \), simplifying to \( e^{\pi} - 1 \).

The integral is multiplied by \( \sqrt{2} \), resulting in the total arc length: \( \sqrt{2} (e^{\pi} - 1) \). Integration ties the process of calculating derivatives and understanding change to compute cumulative results like curve length neatly.

Curve Plotting

Plotting curves defined by parametric equations involves evaluating the equations at various intervals of \( t \) and then plotting the corresponding \( x \) and \( y \) values.
For the equations \( x = e^t \cos(t) \) and \( y = e^t \sin(t) \):

• Calculate sets of \( (x, y) \) by substituting different \( t \) values from 0 to \( \pi \).
• Plot these points on the coordinate plane to visualize the spiral-like curve.

This curve turns out to be a half spiral starting at a point when \( t = 0 \) and expanding outward as \( t \) increases to \( \pi \). Plotting helps provide a tangible view of how the parametric equations describe complex shapes and movements, demonstrating how analytical expressions translate to graphical representations.