Problem 130 - Calculus 3

Question

Find the curvature of the curve \(\mathbf{r}(t)=5 \cos t \mathbf{i}+4 \sin t \mathbf{j}\) at \(t=\pi / 3 .\) (Note: The graph is an ellipse.)

Step-by-Step Solution

Answer

The curvature at \( t=\pi/3 \) is \( \frac{80}{37\sqrt{37}} \).

Step 1: Compute the First Derivative

First, let's find the first derivative \( \mathbf{r}'(t) \). For the given vector function \( \mathbf{r}(t) = 5 \cos t \mathbf{i} + 4 \sin t \mathbf{j} \), calculate the derivatives with respect to \( t \). Using the derivatives of trigonometric functions, we get: \( \mathbf{r}'(t) = -5 \sin t \mathbf{i} + 4 \cos t \mathbf{j} \).

Step 2: Compute the Second Derivative

Now, find the second derivative \( \mathbf{r}''(t) \). Using the derivative of \( \mathbf{r}'(t) = -5 \sin t \mathbf{i} + 4 \cos t \mathbf{j} \), compute: \( \mathbf{r}''(t) = -5 \cos t \mathbf{i} - 4 \sin t \mathbf{j} \).

Step 3: Compute Magnitude of Derivatives

Next, calculate the magnitudes of the derivatives at \( t=\pi/3 \). Evaluate \( \mathbf{r}'(t) \) at \( t=\pi/3 \): \( \mathbf{r}'(\pi/3) = -5(\sqrt{3}/2) \mathbf{i} + 4(1/2) \mathbf{j} \). Its magnitude is \( \|\mathbf{r}'(\pi/3)\| = \sqrt{\left(-\frac{5\sqrt{3}}{2}\right)^2 + \left(2\right)^2} = \frac{\sqrt{37}}{2} \).

Step 4: Calculate Determinant for Curvature

Use the formula for curvature \( \kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} \). Compute the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \), which equals \(-5 \times (-4)\sin^2 t + 4 \times (-5)\cos^2 t \), yielding: \(20\sin t \cos t\). Specifically, at \( t=\pi/3 \), it is 10. The magnitude of this cross product is \(10 \).

Step 5: Compute the Curvature

Using the values found, substitute into \( \kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} \) with \( t=\pi/3 \): \( \kappa\left(\frac{\pi}{3}\right) = \frac{10}{(\frac{\sqrt{37}}{2})^3} = \frac{80}{37\sqrt{37}} \).

Key Concepts

Ellipse Calculus

Understanding the calculus of ellipses involves examining how they differ from circles. Unlike circles, ellipses have a major and minor axis due to their stretched nature. In parametric form, an ellipse can be represented using trigonometric functions. For example, the vector function \( \mathbf{r}(t) = 5 \cos t \mathbf{i} + 4 \sin t \mathbf{j} \) represents an ellipse with a semi-major axis of length 5 and a semi-minor axis of length 4.
To study the properties of such curves, we employ calculus tools, such as derivatives, to better understand their geometry and behavior at specific points.
The concept of curvature comes into play when we want to measure the bending of the curve at any point. For the given ellipse, this involves finding how the curve transitions across its path when moving from point to point.

Vector Functions Derivatives

Vector functions are functions of one or more variables that yield a vector as an output. The derivative of a vector function with respect to time, \( t \), denotes the rate at which the vector function changes. For our ellipse example, \( \mathbf{r}(t) = 5 \cos t \mathbf{i} + 4 \sin t \mathbf{j} \), we differentiate each component individually.
The first derivative, \( \mathbf{r}'(t) = -5 \sin t \mathbf{i} + 4 \cos t \mathbf{j} \), describes the velocity vector of the ellipse at any point \( t \). This indicates not only a change in position but also gives a direction tangential to the ellipse.
The second derivative, \( \mathbf{r}''(t) = -5 \cos t \mathbf{i} - 4 \sin t \mathbf{j} \), is the acceleration vector, showing how the velocity changes over time. Together, these derivatives are fundamental in calculating further dynamics, like measuring curvature or other changes in motion.

Trigonometric Derivatives

Trigonometric derivatives are crucial when dealing with problems involving circular or elliptical motion. The sine and cosine functions have well-known derivatives: \( \frac{d}{dt}[\sin t] = \cos t \) and \( \frac{d}{dt}[\cos t] = -\sin t \).
This knowledge is invaluable in calculating the first and second derivatives of any function that relies on trigonometric terms. In our example, the vector function \( \mathbf{r}(t) = 5 \cos t \mathbf{i} + 4 \sin t \mathbf{j} \) uses these principles directly to determine \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \).
Familiarity with these derivatives enables the computation of rates of change involving angles, which can describe oscillatory motion often seen in physics and engineering problems.

Cross Product Derivatives

The cross product is a vector operation that returns a vector perpendicular to two given vectors. In the context of derivatives, we use it to determine rotational characteristics and curvature. Given two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), their cross product is defined as \( \mathbf{a} \times \mathbf{b} \).
For a vector function, like our ellipse, finding the cross product of the first and second derivatives, \( \mathbf{r}'(t) \times \mathbf{r}''(t) \), helps us compute the curvature \( \kappa(t) \). This tells us about how sharply the curve bends at a specific point.
In the given problem, the calculated cross product results in \( 20\sin t \cos t \) and its magnitude at \( t = \pi/3 \) is crucial for the formula \( \kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} \). This theory extends to numerous fields, including physics, where it assists in understanding forces and trajectories.