Problem 136 - Calculus Volume 3

Question

Find the curvature of \(\mathbf{r}(t)=\langle 2 \sin t, 5 t, 2 \cos t\rangle\).

Step-by-Step Solution

Answer

The curvature \( \kappa(t) \) is \( \frac{10}{33 \sqrt{33}} \).

Step 1: Find the first derivative of the vector function

To find the curvature of the vector function \( \mathbf{r}(t) = \langle 2 \sin t, 5t, 2 \cos t \rangle \), we first need to compute its first derivative.\\( \mathbf{r}'(t) \) is found by differentiating each component of \( \mathbf{r}(t) \):\\[ \mathbf{r}'(t) = \left\langle \frac{d}{dt}(2 \sin t), \frac{d}{dt}(5t), \frac{d}{dt}(2 \cos t) \right\rangle = \langle 2 \cos t, 5, -2 \sin t \rangle. \]

Step 2: Find the second derivative of the vector function

Next, we find the second derivative \( \mathbf{r}''(t) \). This is done by differentiating \( \mathbf{r}'(t) \):\\[ \mathbf{r}''(t) = \left\langle \frac{d}{dt}(2 \cos t), \frac{d}{dt}(5), \frac{d}{dt}(-2 \sin t) \right\rangle = \langle -2 \sin t, 0, -2 \cos t \rangle. \]

Step 3: Calculate the cross product

To find the curvature, calculate the cross product of \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \):\\( \mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 \cos t & 5 & -2 \sin t \ -2 \sin t & 0 & -2 \cos t \end{vmatrix} \).\This results in \( \langle -10 \cos t, 8 \sin t \cos t - 8 \sin t \cos t, 10 \sin t \rangle \) = \( \langle -10 \cos t, 0, 10 \sin t \rangle \).

Step 4: Calculate the magnitudes

Find the magnitude of \( \mathbf{r}'(t) \) and the magnitude of the cross product:\\( ||\mathbf{r}'(t)|| = \sqrt{(2 \cos t)^2 + 5^2 + (-2 \sin t)^2} = \sqrt{4 + 25 + 4} = \sqrt{33}. \)\\( ||\mathbf{r}'(t) \times \mathbf{r}''(t)|| = \sqrt{(-10 \cos t)^2 + 0^2 + (10 \sin t)^2} = \sqrt{100 \cos^2 t + 100 \sin^2 t} = \sqrt{100} = 10. \)

Step 5: Apply the curvature formula

Finally, apply the formula to find the curvature \( \kappa \):\\( \kappa(t) = \frac{||\mathbf{r}'(t) \times \mathbf{r}''(t)||}{||\mathbf{r}'(t)||^3}. \)\Substitute in the values calculated previously: \( \kappa(t) = \frac{10}{(\sqrt{33})^3} = \frac{10}{33 \sqrt{33}}.\)

Key Concepts

Vector Calculus

Vector calculus is a powerful field in mathematics that extends calculus to vector fields. In contrast to ordinary calculus, which deals with single-variable functions, vector calculus focuses on functions that involve multiple variables and have vectors as outputs. A vector is an entity possessing both magnitude and direction, typically represented in two-dimensional or three-dimensional space.
Vector calculus is essential in physics and engineering, modeling physical quantities like force fields, fluid flow, and electromagnetic fields.
For this exercise, we deal with a vector function \( \mathbf{r}(t) = \langle 2 \sin t, 5t, 2 \cos t \rangle \), which describes a path in space using the parameter \( t \). By analyzing the derivatives of this vector function, we can find out how the curve behaves as \( t \) changes.
Understanding the behavior allows for the determination of various properties of the curve, such as its curvature.

Differentiation

Differentiation is a key technique in calculus used to find the rate of change of a function. In the context of vector calculus, we apply differentiation to each component of a vector function individually, just as we would to a regular function.
To begin solving the given exercise, we first differentiate the vector function \( \mathbf{r}(t) = \langle 2 \sin t, 5t, 2 \cos t \rangle \): The derivative of each component is computed separately:
\( \frac{d}{dt}(2 \sin t) = 2 \cos t \) \( \frac{d}{dt}(5t) = 5 \) \( \frac{d}{dt}(2 \cos t) = -2 \sin t \)
Thus, the first derivative \( \mathbf{r}'(t) \) becomes \( \langle 2 \cos t, 5, -2 \sin t \rangle \). To find the curvature, you also need the second derivative \( \mathbf{r}''(t) \), which requires differentiating \( \mathbf{r}'(t) \) again: \( \frac{d}{dt}(2 \cos t) = -2 \sin t \) \( \frac{d}{dt}(5) = 0 \) \( \frac{d}{dt}(-2 \sin t) = -2 \cos t \)
So, \( \mathbf{r}''(t) = \langle -2 \sin t, 0, -2 \cos t \rangle \).

Cross Product

The cross product is an operation used in vector calculus to find a vector that is perpendicular to two given vectors. It is especially useful in three-dimensional space for computing areas and solving problems involving rotation and torque.
For the solution of finding curvature, the cross product of the first and second derivatives, \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \), must be calculated. This is represented as: \( \mathbf{r}'(t) \times \mathbf{r}''(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \times \langle -2 \sin t, 0, -2 \cos t \rangle \)
The evaluation involves finding the determinant of a matrix formed by these vectors:
\( = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 \cos t & 5 & -2 \sin t \ -2 \sin t & 0 & -2 \cos t \end{vmatrix} \)
Upon calculation, the result is \( \langle -10 \cos t, 0, 10 \sin t \rangle \), a vector orthogonal to both \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \).
Each calculation step builds toward understanding how the path curves in space.

Magnitude

Magnitude refers to the length or size of a vector. Calculating the magnitude is crucial when determining vector attributes such as curvature.
The magnitude of a vector \( \mathbf{v} = \langle x, y, z \rangle \) is calculated using the formula: \( ||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2} \)
For the given exercise, we need the magnitude of each resulting vector during the process.
First, calculate the magnitude of \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \): \( ||\mathbf{r}'(t)|| = \sqrt{(2 \cos t)^2 + 5^2 + (-2 \sin t)^2} = \sqrt{4 + 25 + 4} = \sqrt{33} \)
Next, determine the magnitude of the cross product obtained earlier: \( ||\mathbf{r}'(t) \times \mathbf{r}''(t)|| = \sqrt{(-10 \cos t)^2 + 0^2 + (10 \sin t)^2} \) This simplifies using the identity \( \cos^2 t + \sin^2 t = 1 \), resulting in \( \sqrt{100} = 10 \).
Magnitudes help normalize vectors and compute ratios, essential in the curvature formula.

Curvature Formula

Curvature is a measure of how much a curve deviates from being straight. The curvature formula helps quantify this bendiness. In vector calculus, the curvature \( \kappa \) of a vector function \( \mathbf{r}(t) \) is calculated using the formula: \( \kappa(t) = \frac{|| \mathbf{r}'(t) \times \mathbf{r}''(t) ||}{|| \mathbf{r}'(t) ||^3} \)
This formula takes into account both the path and how quickly the direction of that path changes.
Let's apply the derived values from earlier: The magnitude of \( \mathbf{r}'(t) \) is \( \sqrt{33} \), and the magnitude of the cross product is \( 10 \).
Substituting into the formula gives: \( \kappa(t) = \frac{10}{(\sqrt{33})^3} \)
This expression represents how sharply the curve bends at any given \( t \). Understanding this is essential in fields like physics and engineering, where fluid dynamics or path optimization may depend on knowing the curvature.