Question

Find the length of the following polar curves. The spiral \(r=e^{\theta},\) where \(0 \leq \theta \leq 2 \pi n,\) for a positive integer \(n\)

Short answer

Answer: The length of the spiral is \(L = e^{2\pi n} - 1\), where n is a positive integer.

Step-by-step solution

Convert to Parametric Equations

Rewrite the polar equation \(r = e^{\theta}\) as parametric equations in terms of x and y: \(x = r \cos \theta = e^{\theta} \cos \theta\) and \(y = r \sin \theta = e^{\theta} \sin \theta\).

Compute Derivatives

Differentiate x and y with respect to \(\theta\) to obtain: \(\frac{dx}{d\theta} = e^{\theta} \cos \theta - e^{\theta} \sin \theta\) and \(\frac{dy}{d\theta} = e^{\theta} \sin \theta + e^{\theta} \cos \theta\).

Arc Length Formula

Use the arc length formula for parametric equations: \(L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} d\theta\), where \(\alpha = 0\) and \(\beta = 2 \pi n\).

Simplify the Expression

Compute the sum of squares and simplify: \(\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = \left(e^{\theta} \cos \theta - e^{\theta} \sin \theta\right)^2 + \left(e^{\theta} \sin \theta + e^{\theta} \cos \theta\right)^2 = e^{2\theta}\), since \((\cos^2 \theta +\sin^2 \theta)= 1\).

Evaluate the Integral

Evaluate the integral with its simplified form: \(L = \int_{0}^{2\pi n} \sqrt{e^{2\theta}} d\theta = \int_{0}^{2\pi n} e^{\theta} d\theta\). Applying the integral, we get \(L = \left[e^{\theta}\right]_{0}^{2\pi n} = e^{2\pi n} - 1\). The length of the spiral is given by \(L = e^{2\pi n} - 1\), where n is a positive integer.

Key concepts

Parametric Equations

Parametric equations allow us to express a curve in terms of a parameter, commonly denoted as \(t\) or \(\theta\), instead of using the traditional function notation \(y=f(x)\). This is particularly useful for curves that cannot be described as a function of \(x\) or \(y\) alone, such as loops or spirals.

For a given polar curve \(r=f(\theta)\), the set of parametric equations in Cartesian coordinates can be written as \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\), where \(r\) is the radial distance from the origin, and \(\theta\) is the angle measured from the positive x-axis. For example, the parametric equations of the spiral \(r=e^{\theta}\) are \(x=e^{\theta}\cos(\theta)\) and \(y=e^{\theta}\sin(\theta)\).

By expressing the curve parametrically, we can trace the path by varying \(\theta\) within a specified interval, capturing the entire geometry of the curve.

Arc Length Formula

The arc length of a curve in a plane is the distance you would travel if you were to walk along the path of the curve. For parametric curves, we use a specific formula to calculate this arc length. The formula for the arc length \(L\) of a curve defined by parametric equations \(x(t)\) and \(y(t)\) between the points \(\alpha\) and \(\beta\) on the parameter's interval is:
\[L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\]

This formula requires the derivatives of the parametric equations, \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), since these represent the change in \(x\) and \(y\) with respect to the parameter. Essentially, we are finding the hypotenuse of an infinitesimally small right triangle whose sides are \(dx\) and \(dy\), as the curve approaches a straight line for a very small interval of the parameter.

Derivatives

Derivatives are a fundamental tool in calculus that describe the rate at which a function is changing at any given point. In the context of parametric equations, we consider the derivatives of the separate functions with respect to the parameter. That is, if our parametric equations are \(x(t)\) and \(y(t)\), their derivatives are \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) respectively.

The derivatives are essential for computing the arc length of a curve because they give us the rate of displacement of each coordinate, which helps us measure how

Integral Calculus

Integral calculus, when dealing with arc lengths, involves finding the integral of a function over a certain interval. This is analogous to adding up an infinite number of infinitesimally small quantities – in the case of arc length, these are the lengths of an infinite number of infinitesimally small straight-line segments that make up the curve. For the spiral \(r=e^{\theta}\), once you have the simplified expression for the integrated function, you find the antiderivative and evaluate it at the upper and lower limits of the interval - from \(\theta = 0\) to \(\theta = 2\pi n\).

The fundamental theorem of calculus links the concept of differentiation with that of integration and is pivotal to solving problems involving arc lengths. Once the correct antiderivative is found, it gives us the total arc length over the interval, which is both a powerful and elegant application of integral calculus.