Question

find the unit tangent vector \(\mathbf{T}(t)\) and the curvature \(\kappa(t)\) at the point where \(t=t_{1} .\) For calculating \(\kappa\), we suggest using Theorem \(A\), as in Example \(5 .\) $$ \mathbf{r}(t)=\frac{1}{3} t^{3} \mathbf{i}+\frac{1}{2} t^{2} \mathbf{j} ; t_{1}=1 $$

Short answer

The unit tangent vector is \( \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}} \mathbf{j} \) and curvature is \( \frac{1}{2\sqrt{2}} \).

Step-by-step solution

Find the derivative of the position vector

The position vector given is \( \mathbf{r}(t) = \frac{1}{3} t^3 \mathbf{i} + \frac{1}{2} t^2 \mathbf{j} \). First, we find the derivative of this vector to get the velocity vector \( \mathbf{v}(t) \). This is done by differentiating each component:\[ \mathbf{v}(t) = \frac{d}{dt} \left( \frac{1}{3} t^3 \right) \mathbf{i} + \frac{d}{dt} \left( \frac{1}{2} t^2 \right) \mathbf{j} = t^2 \mathbf{i} + t \mathbf{j}. \]

Evaluate the velocity vector at \(t_1\)

We substitute \(t_1 = 1\) into \(\mathbf{v}(t)\) to find the specific velocity vector at \(t_1\):\[ \mathbf{v}(1) = 1^2 \mathbf{i} + 1 \mathbf{j} = \mathbf{i} + \mathbf{j}. \]

Find the magnitude of the velocity vector

The magnitude of \( \mathbf{v}(1) = \mathbf{i} + \mathbf{j} \) is computed as follows:\[ \| \mathbf{v}(1) \| = \sqrt{1^2 + 1^2} = \sqrt{2}. \]

Compute the unit tangent vector \( \mathbf{T}(t_1) \)

The unit tangent vector is given by \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\| \mathbf{v}(t) \|} \). So, \( \mathbf{T}(1) \) is:\[ \mathbf{T}(1) = \frac{\mathbf{i} + \mathbf{j}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}} \mathbf{j}. \]

Find the second derivative of the position vector

To find the curvature, we need the second derivative of the position vector, \( \mathbf{a}(t) = \mathbf{r}''(t) \). So we differentiate \( \mathbf{v}(t) = t^2 \mathbf{i} + t \mathbf{j} \):\[ \mathbf{a}(t) = \frac{d}{dt} (t^2 \mathbf{i} + t \mathbf{j}) = 2t \mathbf{i} + \mathbf{j}. \]

Evaluate the acceleration vector at \(t_1\)

We substitute \(t_1 = 1\) into \(\mathbf{a}(t)\):\[ \mathbf{a}(1) = 2(1) \mathbf{i} + \mathbf{j} = 2 \mathbf{i} + \mathbf{j}. \]

Use Theorem A to compute curvature \( \kappa(t_1) \)

According to Theorem A, curvature \( \kappa(t) \) is given by:\[ \kappa(t) = \frac{\| \mathbf{v}(t) \times \mathbf{a}(t) \|}{\| \mathbf{v}(t) \|^3}. \]Let's calculate each component at \( t = 1 \):- **Cross Product:** \( \mathbf{v}(1) \times \mathbf{a}(1) = (\mathbf{i} + \mathbf{j}) \times (2 \mathbf{i} + \mathbf{j}) = \mathbf{k} \).- **Magnitude of Cross Product:** \( \| \mathbf{k} \| = 1. \)- **Curvature:** \[ \kappa(1) = \frac{1}{(\sqrt{2})^3} = \frac{1}{2\sqrt{2}}. \]

Key concepts

Curvature

Curvature is a measure of how sharply a curve bends at a given point. It provides crucial insights into the nature and geometry of the curve. When you encounter a curve, not all parts are equally bent. Curvature helps determine which sections are more curved than others, and it is often used across various fields of mathematics and physics to understand motion and path characteristics.

In our exercise, to find the curvature \( \kappa(t) \), we use the relation \[ \kappa(t) = \frac{\| \mathbf{v}(t) \times \mathbf{a}(t) \|}{\| \mathbf{v}(t) \|^3} \]. This involves both the velocity and the acceleration vectors.

Essentially, the curvature at a point gives information about the rate of change of the direction of the velocity vector. The steeper the curve, the higher the curvature value. Remember, the cross product of the velocity and acceleration vectors plays a significant role in giving us the numerator of the curvature formula.

Velocity Vector

The velocity vector tells us about the speed and direction of an object moving along a path. It is the first derivative of the position vector with respect to time. Think of it as describing how fast and in which direction the position is changing at any given instant.

For the given position vector \( \mathbf{r}(t) = \frac{1}{3} t^3 \mathbf{i} + \frac{1}{2} t^2 \mathbf{j} \), the velocity vector is obtained by differentiating each component:

• Derivative of \( \frac{1}{3} t^3 \) is \( t^2 \)
• Derivative of \( \frac{1}{2} t^2 \) is \( t \)

Thus, \( \mathbf{v}(t) = t^2 \mathbf{i} + t \mathbf{j} \).

The velocity vector at \( t_1 = 1 \) is \( \mathbf{i} + \mathbf{j} \. \) By understanding the velocity vector, you gain insight into the instantaneous motion at any point along the curve.

Acceleration Vector

The acceleration vector represents the rate of change of the velocity vector with respect to time. It's the second derivative of the position vector and gives us an idea of how quickly the object's speed or direction is changing.

We find acceleration by differentiating the velocity vector \( \mathbf{v}(t) = t^2 \mathbf{i} + t \mathbf{j} \):

• The derivative of \( t^2 \mathbf{i} \) is \( 2t \mathbf{i} \)
• The derivative of \( t \mathbf{j} \) is \( \mathbf{j} \)

This means \( \mathbf{a}(t) = 2t \mathbf{i} + \mathbf{j} \).

When evaluated at \( t_1 = 1 \), the acceleration vector becomes \( 2 \mathbf{i} + \mathbf{j} \. \) One of the key uses of the acceleration vector is in calculating curvature, as it helps understand how quickly a moving object is turning along a path.

Cross Product

The cross product is a mathematical operation that takes two vectors in three-dimensional space and produces another vector that is perpendicular to the plane containing the original vectors. This new vector's magnitude is related to the area of the parallelogram span by the two vectors. It's a crucial tool in physics and geometry.

In our context, to find the curvature \( \kappa(t) \), we compute the cross product of the velocity vector \( \mathbf{v}(1) = \mathbf{i} + \mathbf{j} \) and the acceleration vector \( \mathbf{a}(1) = 2 \mathbf{i} + \mathbf{j} \):

• The result is the vector \( \mathbf{k} \), a unit vector oriented orthogonally to the plane defined by \( \mathbf{i} \) and \( \mathbf{j} \).
• In simple terms, this cross product finds the direction and magnitude at which the two vectors diverge.

The magnitude of this cross product tells us about the extent of the bending of the curve at a specific point, forming a critical component of the curvature calculation.