Problem 151 - Calculus 3

Question

A particle moves along the plane curve \(\mathrm{C}\) described by \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} .\) Solve the following problems. Describe the curvature as \(t\) increases from \(t=0\) to \(t=2\)

Step-by-Step Solution

Answer

The curvature decreases as \( t \) increases from 0 to 2.

Step 1: Calculate the First Derivative

To find the curvature, we begin by computing the first derivative of the vector function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \). The derivative is \( \mathbf{r}'(t) = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j}) = \mathbf{i} + 2t \mathbf{j} \).

Step 2: Calculate the Second Derivative

Next, find the second derivative of \( \mathbf{r}(t) \). The derivative of \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \) is \( \mathbf{r}''(t) = \frac{d}{dt}(\mathbf{i} + 2t \mathbf{j}) = 2 \mathbf{j} \).

Step 3: Calculate the Magnitude of Derivatives

Find the magnitudes of \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \). We have \( \|\mathbf{r}'(t)\| = \sqrt{1 + (2t)^2} = \sqrt{1 + 4t^2} \) and \( \|\mathbf{r}''(t)\| = \|2 \mathbf{j} \| = 2 \).

Step 4: Find Curvature Formula

Use the curvature formula \( \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \) to find the curvature. Since the curves are planar, the cross product has only the k-component and \( \mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2t & 0 \ 0 & 2 & 0 \end{vmatrix} = \begin{vmatrix} 2t & 0 \ 2 & 0 \end{vmatrix} \mathbf{k} = (0 - 4t) \mathbf{k} = -4t \mathbf{k} \). The magnitude is \( \| -4t \mathbf{k} \| = 4t \).

Step 5: Compute the Curvature

Substitute into the curvature formula: \( \kappa(t) = \frac{4t}{(1 + 4t^2)^{3/2}} \). Calculate the curvature as \( t \) increases from \( t=0 \) to \( t=2 \). Evaluate \( \kappa(0) = \frac{4 \times 0}{(1+0)^{3/2}} = 0 \) and \( \kappa(2) = \frac{4 \times 2}{(1 + 4\times4)^{3/2}} = \frac{8}{(17)^{3/2}} \).

Step 6: Examine the Result

The curvature \( \kappa(t) \) decreases as \( t \) goes from 0 to 2, starting from \( \kappa(0) = 0 \) and reaching some positive value \( \kappa(2) = \frac{8}{(17)^{3/2}} \). This indicates the curve becomes less curved as \( t \) increases.

Key Concepts

Derivative of Vector Functions

Understanding the derivative of vector functions plays a crucial role in analyzing the motion and characteristics of a curve. In our particle problem, we have a vector function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \). Computing the derivative of this function, denoted as \( \mathbf{r}'(t) \), provides us with the velocity vector of the particle as it moves along the path.

The derivative is taken component-wise, meaning we derive each component of the vector separately. For the function \( \mathbf{r}(t) \), the derivative is calculated as:

• \( \frac{d}{dt}(t \mathbf{i}) = \mathbf{i} \)
• \( \frac{d}{dt}(t^2 \mathbf{j}) = 2t \mathbf{j} \)

Hence, the velocity vector becomes \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \). This new vector describes the rate of change of the position vector \( \mathbf{r}(t) \) with respect to time \( t \), providing insight into how the particle travels along the curve.

Magnitude of a Vector

The magnitude of a vector, often referred to as the length or norm of the vector, is a critical concept in understanding the properties of vector functions. The magnitude is essentially the distance of the vector from the origin when it is placed in a coordinate system.

To find the magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), use the formula:

• \( \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \)

For our velocity vector \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \), the magnitude is computed as:

• \( \|\mathbf{r}'(t)\| = \sqrt{1 + (2t)^2} = \sqrt{1 + 4t^2} \)

This value represents the speed of the particle at any point on the curve, providing a scalar quantity that quantifies its velocity.

Cross Product in Vectors

The cross product of two vectors is a fundamental operation in vector calculus that results in a new vector perpendicular to both original vectors. It is especially useful in determining the curvature of a plane curve.

In the case of planar curves like ours, the cross product helps us calculate the numerator of the curvature formula. For vectors \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) and \( \mathbf{v} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \), the cross product is a vector \( \mathbf{u} \times \mathbf{v} \) given by:

• \( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a & b & c \ x & y & z \end{vmatrix} \)

For our problem, combining \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \), the cross product results in the vector \( -4t \mathbf{k} \), emphasizing that the primary influence on curvature comes from changes in the trajectory's direction.

Parametrized Curves

Parametrized curves are a way of representing paths in a coordinate system. They use a parameter, typically time \( t \), to define the position on the curve at every instance of \( t \). In our exercise, \( \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} \) defines the position of a particle over time.

Each component in the vector corresponds to a coordinate with respect to the parameter \( t \).

• \( t \mathbf{i} \) describes the horizontal or x-position as a linear function of \( t \).
• \( t^2 \mathbf{j} \) represents the vertical or y-position, illustrating quadratic growth over time.

This method allows for a flexible description of the motion, making it easier to compute derivatives and analyze properties such as speed, velocity, acceleration, and curvature as the particle moves through the plane.