Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the length of the indicated curve. \(x=e^{t} \cos t, y=e^{t} \sin t\) for \(0 \leq t \leq 1\)

Step-by-Step Solution

Answer

Length of curve is \(\sqrt{2} (e - 1)\).

Step 1: Understand the Curve Representation

The problem provides parametric equations of a curve: \(x=e^{t} \cos t\) and \(y=e^{t} \sin t\). We need to find the length of this curve from \(t=0\) to \(t=1\). The parameter \(t\) traces out the curve as it varies from 0 to 1.

Step 2: Determine the Parametric Derivatives

Find the derivatives of \(x\) and \(y\) with respect to \(t\). The derivatives are needed for the arc length formula:\[\frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t)\]\[\frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t)\]

Step 3: Use the Arc Length Formula for Parametric Equations

The length \(L\) of a curve defined by parametric equations \(x(t)\) and \(y(t)\) from \(t=a\) to \(t=b\) is given by:\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]Substituting the derivatives, the formula becomes:\[ L = \int_{0}^{1} \sqrt{(e^t (\cos t - \sin t))^2 + (e^t (\sin t + \cos t))^2} \, dt \]

Step 4: Simplify the Expression Inside the Integral

Simplify the expression under the square root:\[(e^t (\cos t - \sin t))^2 = e^{2t} (\cos t - \sin t)^2 = e^{2t} (\cos^2 t - 2 \cos t \sin t + \sin^2 t) = e^{2t} (1 - 2 \cos t \sin t)\]\[(e^t (\sin t + \cos t))^2 = e^{2t} (\sin^2 t + 2 \sin t \cos t + \cos^2 t) = e^{2t} (1 + 2 \cos t \sin t)\]\[\sqrt{e^{2t} (1 - 2 \cos t \sin t) + e^{2t} (1 + 2 \cos t \sin t)} = \sqrt{2 e^{2t}} = \sqrt{2} e^t\]

Step 5: Integrate the Simplified Function

Now integrate the simplified function from 0 to 1:\[L = \int_{0}^{1} \sqrt{2} e^t \, dt = \sqrt{2} \int_{0}^{1} e^t \, dt\]Perform the integration:\[\int e^t \, dt = e^t\]Evaluating from 0 to 1 gives:\[\sqrt{2} (e^1 - e^0) = \sqrt{2} (e - 1)\]

Step 6: Final Result

The length of the curve from \(t = 0\) to \(t = 1\) is \(\sqrt{2} (e - 1)\).

Key Concepts

Parametric Equations

Parametric equations are a way of defining a curve through a set of parameters, rather than using a direct relationship between x and y. In the given problem, the parametric equations are defined as \(x = e^t \cos t\) and \(y = e^t \sin t\).
These two equations use the parameter \(t\) to describe the position on the curve. As \(t\) varies, you get different points on the curve. This parameterization can be particularly useful for curves that loop or turn, where defining y directly in terms of x might be complex or impossible.
In this example:

• The parameter \(t\) ranges from 0 to 1. As \(t\) changes within this range, we trace the linked motion between \(x\) and \(y\).
• The exponential function \(e^t\) affects how both x and y increase exponentially, giving an idea of the shape as a spiral starting at \(t = 0\).

Using parametric equations allows for great flexibility in both defining and analyzing complex curves.

Differentiation

Differentiation is a crucial process in calculus, used to find how quickly something changes at any point along a curve. In this problem, we differentiated the parametric equations for x and y with respect to \(t\). Doing so allows us to apply the formula for arc length.

• For \(x = e^t \cos t\), the derivative \(\frac{dx}{dt}\) is calculated as \(e^t (\cos t - \sin t)\).
• For \(y = e^t \sin t\), the derivative \(\frac{dy}{dt}\) is found to be \(e^t (\sin t + \cos t)\).

Differentiation helps us understand the rate at which the x and y coordinates change with respect to the parameter \(t\). This information is necessary to compute the length of the curve using the arc length formula.

Integration

Integration in calculus enables us to accumulate small increments over a continuous range. For this exercise, after computing the parametric derivatives, integration was used to calculate the entire length of the curve.
The arc length formula we use is:
\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \] where \(a = 0\) and \(b = 1\) for this particular curve.

• The expression under the square root simplifies to \(\sqrt{2} e^t\).
• This leads us to integrate \(\sqrt{2} e^t\) from 0 to 1.
• The integration results in \(\sqrt{2} (e^1 - e^0)\), which simplifies to \(\sqrt{2} (e - 1)\).

Thus, integration plays the role of adding up infinite tiny lengths to find the total curve length.

Calculus

Calculus is the fundamental mathematical framework used to study change, and it involves two primary operations: differentiation and integration. By mastering these, we can analyze and solve a variety of complex problems involving curves, such as finding the arc length.
In this exercise, both differentiation and integration were necessary to compute the arc length of a parametric curve. Calculus helps us:

• Understand how different parts of a curve behave by using derivatives.
• Calculate the total distance along a curve by integrating those behaviors.

These concepts combined allow us to solve for intricate qualities of curves that are specified parametrically. Calculus not only helps in theoretical mathematics but also finds application in fields such as physics, engineering, and even economics.