Question

Consider the spiral \(r=4 \theta,\) for \(\theta \geq 0\) a. Use a trigonometric substitution to find the length of the spiral, for \(0 \leq \theta \leq \sqrt{8}\). b. Find \(L(\theta),\) the length of the spiral on the interval \([0, \theta],\) for any \(\theta \geq 0\). c. Show that \(L^{\prime}(\theta)>0 .\) Is \(L^{\prime \prime}(\theta)\) positive or negative? Interpret your answer.

Short answer

In summary, given the polar equation \(r = 4\theta\), we found the following: a. The length of the spiral for the interval \([0, \sqrt{8}]\) is given by the following integral and has to be solved numerically: \[L = \int\limits_0^{\sqrt{8}} \sqrt{16 + 8\theta^2 - 16\theta \sin(2\theta)} d\theta\] b. The length of the spiral for any interval \([0, \theta]\) is given by: \[L(\theta) = \int\limits_0^{\theta} \sqrt{16 + 8t^2 - 16t \sin(2t)} dt\] c. The first derivative of the length function, \(L'(\theta)\), is: \[L'(\theta) = \sqrt{16 + 8\theta^2 - 16\theta \sin(2\theta)}\] Since \(L'(\theta) > 0\), the spiral is always increasing in length as \(\theta\) increases. The second derivative, \(L''(\theta)\), determines the curvature of the spiral and can be found using numerical methods. The sign of \(L''(\theta)\) indicates whether the curvature increases or decreases along the spiral.

Step-by-step solution

a. Finding the length of the spiral for \(0 \leq \theta \leq \sqrt{8}\)

We are given \(r = 4\theta\), a polar equation. We can convert this to Cartesian coordinates by using the conversion equations: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). This gives us: \(x = 4\theta \cos(\theta)\) \(y = 4\theta \sin(\theta)\) Now we need to find the arc length of the curve for \(\theta \in [0, \sqrt{8}]\). The arc length formula for a curve is defined as: \(L = \int\limits_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt\). To use this formula, first, we need to find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\): \(\frac{dx}{d\theta} = 4\cos(\theta) - 4\theta \sin(\theta)\) \(\frac{dy}{d\theta} = 4\sin(\theta) + 4\theta \cos(\theta)\) Now, we can find the arc length for \(\theta \in [0, \sqrt{8}]\): \(L = \int\limits_0^{\sqrt{8}} \sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} d\theta\)

Finding the integral

Now we need to calculate the following integral: \(L = \int\limits_0^{\sqrt{8}} \sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} d\theta\). First, let's compute the sum of their squares: \((\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2 = 16 + 8\theta^2 - 32\theta \sin(\theta) \cos(\theta)\) Now let's make the integrand easier to compute using a trigonometric identity: \((\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2 = 16 + 8\theta^2 - 16\theta \sin(2\theta)\) Now we can compute the integral: \(L = \int\limits_0^{\sqrt{8}} \sqrt{16 + 8\theta^2 - 16\theta \sin(2\theta)} d\theta\) Unfortunately, this integral is too complex to solve analytically. However, it can be solved using numerical methods (e.g., with a calculator).

b. Finding the length of the spiral on the interval \([0, \theta]\)

We have found that the length of the spiral for \(\theta \in [0, \sqrt{8}]\) is given by: \(L = \int\limits_0^{\sqrt{8}} \sqrt{16 + 8\theta^2 - 16\theta \sin(2\theta)} d\theta\). In order to find the length of the spiral on the interval \([0, \theta]\), we simply need to replace the upper limit in the integral with \(\theta\): \(L(\theta) = \int\limits_0^{\theta} \sqrt{16 + 8t^2 - 16t \sin(2t)} dt\) Again, this integral is too complex to solve analytically, but it can be solved using numerical methods.

c. Finding \(L'(\theta)\) and \(L''(\theta)\)

Before we find \(L'(\theta)\) and \(L''(\theta)\), let's rewrite the function \(L(\theta)\) in a more compact form. From part (b): \(L(\theta) = \int\limits_0^{\theta} \sqrt{16 + 8t^2 - 16t \sin(2t)} dt\) Now calculate the first derivative of \(L(\theta)\) with respect to \(\theta\) using the Fundamental Theorem of Calculus: \(L'(\theta) = \sqrt{16 + 8\theta^2 - 16\theta \sin(2\theta)}\) Now for the second derivative, which can be found using the chain rule: \(L''(\theta) = \frac{d}{d\theta}(\sqrt{16 + 8\theta^2 - 16\theta \sin(2\theta)})\) To find \(L''(\theta)\), we can use numerical methods (e.g., a calculator). Since \(L'(\theta)\) is the square root of a function that has positive values for the domain we are interested in, we can say that \(L'(\theta) > 0\). This means that the spiral is always increasing in length as \(\theta\) increases in the given domain. The sign of \(L''(\theta)\) determines the curvature of the spiral. If \(L''(\theta)\) is positive, then the spiral's curvature increases; and if it is negative, the spiral's curvature decreases.

Key concepts

Polar Coordinates

Polar coordinates provide us with a unique way to represent a point in a plane, which is especially useful when dealing with curves like spirals. Unlike Cartesian coordinates, which use a pair of \(x\) and \(y\) coordinates, polar coordinates employ a radius \(r\) and an angle \(\theta\). This makes it simpler to describe circular and spiral patterns.
For the given spiral \(r = 4\theta\), the radius is dependent on the angle, meaning as the angle increases, the distance from the origin also increases, creating a spiral pattern. This relationship is pivotal, as it allows us to express complex curves in simple forms that are easy to interpret in a visual and mathematical sense.

• When you move counterclockwise starting from the positive \(x\)-axis, you increase the angle \(\theta\).
• As \(\theta\) increases, for the spiral \(r = 4\theta\), the radius expands proportionally.

Understanding this relationship facilitates solving problems involving arc lengths and areas bounded by curves, as it simplifies complex geometrical problems into more manageable mathematical expressions.

Calculus

Calculus, particularly differential calculus, plays a significant role in understanding and finding the arc length of curves defined in polar coordinates. Differentiation is used here to find the rate at which values change along the curve, allowing us to calculate the tangent at any point on the spiral.
The arc length formula is crucial in this exploration; for a smooth curve described parametrically by \((x(t), y(t))\), the arc length \(L\) is given by \(L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt\). This gives us a way to accumulate the 'curve distance' from one point to another along the spiral.

• The formula is adapted for polar coordinates; derivatives like \((\frac{dx}{d\theta}, \frac{dy}{d\theta})\) replace the standard Cartesian ones.
• Even if the integral becomes complex and unsolvable analytically, numerical methods help approximate the length.

Calculus thus provides the tools to convert and manage mathematical problems involving motion and change, which are inherent in any study of a spiral length.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with that of integration, establishing that they are inverse processes. This is essential when working with finding the length of a curve, as it allows us to derive functions describing growth or shrinkage in context.
In this problem, to find \(L'(\theta)\), the derivative of the length function \(L(\theta)\), we use the fundamental theorem. It tells us that \(L'(\theta)\) is equivalent to evaluating the integrand at the upper limit \(\theta\): \(L'(\theta) = \sqrt{16 + 8\theta^2 - 16\theta \sin(2\theta)}\)

• This derivative is positive in the domain we're exploring, indicating the spiral grows as \(\theta\) increases.
• The concavity or convexity, evaluated through the second derivative, \(L''(\theta)\), gives insights into the nature of the spiral's expansion.

Thus, the fundamental theorem not only provides a way to solve integrals but also gives a direct route to understanding how functions evolve over time. Understanding this core concept allows us to probe deeper into the behavior of spirals and many other complex phenomena.