Problem 130 - Calculus Volume 3

Question

Find the curvature of the curve \(\mathbf{r}(t)=5 \cos t \mathbf{i}+4 \sin t \mathbf{j}\) at \(t=\pi / 3\).

Step-by-Step Solution

Answer

The curvature at \( t = \pi/3 \) is approximately 0.182.

Step 1: Differentiate the Curve

First, find the derivative of the curve vector function \( \mathbf{r}(t) = 5 \cos t \mathbf{i} + 4 \sin t \mathbf{j} \). Differentiate each component with respect to \( t \): - The derivative of \( 5 \cos t \) is \( -5 \sin t \).- The derivative of \( 4 \sin t \) is \( 4 \cos t \).Therefore, the derivative is \( \mathbf{r}'(t) = -5 \sin t \mathbf{i} + 4 \cos t \mathbf{j} \).

Step 2: Compute the Magnitude of r'(t)

Next, compute the magnitude \( ||\mathbf{r}'(t)|| \) using the components of the derivative: \[||\mathbf{r}'(t)|| = \sqrt{(-5 \sin t)^2 + (4 \cos t)^2} = \sqrt{25 \sin^2 t + 16 \cos^2 t}\]

Step 3: Differentiate r'(t) to Find r''(t)

Differentiate \( \mathbf{r}'(t) = -5 \sin t \mathbf{i} + 4 \cos t \mathbf{j} \) again to find \( \mathbf{r}''(t) \):- The derivative of \( -5 \sin t \) is \( -5 \cos t \).- The derivative of \( 4 \cos t \) is \( -4 \sin t \).Thus, \( \mathbf{r}''(t) = -5 \cos t \mathbf{i} - 4 \sin t \mathbf{j} \).

Step 4: Compute the Cross Product r'(t) x r''(t)

Find the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \). Since both vectors are in the plane, using the formula for scalar cross product magnitude yields:\[||\mathbf{r}'(t) \times \mathbf{r}''(t)|| = |-5 \sin t \cdot (-4 \sin t) - 4 \cos t \cdot (-5 \cos t)| = |20 \sin^2 t + 20 \cos^2 t| = 20\]

Step 5: Substitute t = π/3 into r'(t)

Evaluate \( ||\mathbf{r}'(t)|| \) at \( t = \pi / 3 \):\[||\mathbf{r}'(\pi/3)|| = \sqrt{25 (\sin(\pi/3))^2 + 16 (\cos(\pi/3))^2}\]Since \( \sin(\pi/3) = \sqrt{3}/2 \) and \( \cos(\pi/3) = 1/2 \), substituting these values gives:\[||\mathbf{r}'(\pi/3)|| = \sqrt{25 \left(\frac{3}{4}\right) + 16 \left(\frac{1}{4}\right)} = \sqrt{19.75}\]

Step 6: Calculate the Curvature

The formula for curvature \( \kappa \) is: \[\kappa = \frac{||\mathbf{r}'(t) \times \mathbf{r}''(t)||}{||\mathbf{r}'(t)||^3}\]Substituting \( t = \pi/3 \) values:\[\kappa = \frac{20}{(\sqrt{19.75})^3} \approx 0.182\]

Key Concepts

Derivative of Vector Functions

To understand the derivative of vector functions, it's important to start with the basics of differentiation. Differentiation allows us to find the rate at which values change, and when it's done on vector functions, it helps us observe how the direction and magnitude of a vector change over time. Each component of the vector function can be differentiated separately.

In the original exercise, we are given the curve \( \mathbf{r}(t) = 5 \cos t \mathbf{i} + 4 \sin t \mathbf{j} \). To find its derivative, \( \mathbf{r}'(t) \), we differentiate each component:

• The derivative of \( 5 \cos t \) is \( -5 \sin t \).
• The derivative of \( 4 \sin t \) is \( 4 \cos t \).

Therefore, the derivative vector function is \( \mathbf{r}'(t) = -5 \sin t \mathbf{i} + 4 \cos t \mathbf{j} \). Understanding this step is crucial as it's the foundational process for further computations in the curvature formula.

Cross Product

The cross product is a crucial concept in vector calculus, especially when computing areas and volumes. For two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cross product \( \mathbf{a} \times \mathbf{b} \) results in a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). This perpendicular nature makes cross products particularly useful for determining a vector's orientation in space.

In the context of curvature, we use another essential property of cross products: their magnitude. The magnitude of the cross product of the derivative functions \( ||\mathbf{r}'(t) \times \mathbf{r}''(t)|| \) measures how the direction of the curve changes. In the original solution, the scalar cross product's formula was applied, since the calculation simplifies in two-dimensional plane problems:

• \( ||\mathbf{r}'(t) \times \mathbf{r}''(t)|| = 20 \)

The value 20 indicates the intensity of the curve's directional change at that point, which is an essential part of calculating curvature.

Magnitude of Vectors

Magnitude helps measure the size or length of vectors. It is calculated using the components of the vector by applying the Pythagorean theorem in multi-dimensional space. The mathematical formula for the magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \) in two dimensions is \( ||\mathbf{v}|| = \sqrt{a^2 + b^2} \).

In our exercise, calculating the magnitude of \( \mathbf{r}'(t) \) was a key step. We transform the differentiated components into a magnitude:

• \( ||\mathbf{r}'(t)|| = \sqrt{(-5 \sin t)^2 + (4 \cos t)^2} \)

This equation provides the function needed to evaluate how the vector changes over time. Particularly for \( t = \pi/3 \), the result was \( \sqrt{19.75} \). This magnitude is essential for plugging into the curvature formula to understand the curve's behavior well.

Curvature Formula

Curvature \( \kappa \) is a scalar measure of how sharply a curve bends at a given point. The curvature formula for a vector-valued function \( \mathbf{r}(t) \) is \[ \kappa = \frac{||\mathbf{r}'(t) \times \mathbf{r}''(t)||}{||\mathbf{r}'(t)||^3} \]This formula combines the concepts of magnitude and cross products to describe the curve in motion.

In practice, you calculate the magnitudes of both the vector derivative and the cross product of derivatives to apply this formula. For the exercise at hand, we used:

• Cross product magnitude: 20
• Vector derivative magnitude at \( t = \pi/3 \): \( \sqrt{19.75} \)

Applying these, the curvature was found as approximately 0.182. This numerical value of curvature tells us about the curve's sharpness at the point \( t = \pi/3 \), serving as a powerful way to understand geometrical properties of curves in space.