Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-6, sketch the curve over the indicated domain for \(t\). Find \(\mathbf{v}, \mathbf{a}, \mathbf{T}\), and \(\kappa\) at the point where \(t=t_{1} .\) $$ \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} ; \quad 0 \leq t \leq 2 ; t_{1}=1 $$

Step-by-Step Solution

Answer

\(\mathbf{v}(1) = \mathbf{i} + 2\mathbf{j}\), \(\mathbf{a}(1) = 2\mathbf{j}\), \(\mathbf{T}(1) = \frac{1}{\sqrt{5}}(\mathbf{i} + 2\mathbf{j})\), \(\kappa(1) = \frac{4\sqrt{5}}{25}\)."

Step 1: Calculate Velocity Vector (\(\mathbf{v}\))

Given \(\mathbf{r}(t)=t\mathbf{i}+t^{2}\mathbf{j}\), find the velocity vector \(\mathbf{v}(t)\) by differentiating \(\mathbf{r}(t)\) with respect to \(t\): \[ \mathbf{v}(t) = \frac{d}{dt}(t\mathbf{i} + t^2\mathbf{j}) = \mathbf{i} + 2t\mathbf{j} \] Evaluate at \(t=1\): \[ \mathbf{v}(1) = \mathbf{i} + 2\times1\mathbf{j} = \mathbf{i} + 2\mathbf{j} \].

Step 2: Calculate Acceleration Vector (\(\mathbf{a}\))

The acceleration vector \(\mathbf{a}(t)\) is found by differentiating the velocity vector \(\mathbf{v}(t)\): \[ \mathbf{a}(t) = \frac{d}{dt}(\mathbf{i} + 2t\mathbf{j}) = 2\mathbf{j} \] Evaluate at \(t=1\): \[ \mathbf{a}(1) = 2\mathbf{j} \].

Step 3: Calculate Tangent Vector (\(\mathbf{T}\))

The tangent vector \(\mathbf{T}\) is the unit vector in the direction of \(\mathbf{v}(t)\). Compute \(||\mathbf{v}(1)||\): \[ ||\mathbf{v}(1)|| = \sqrt{1^2 + 2^2} = \sqrt{5} \] Therefore, \(\mathbf{T}(1) = \frac{1}{\sqrt{5}}(\mathbf{i} + 2\mathbf{j})\).

Step 4: Calculate Curvature (\(\kappa\))

The curvature \(\kappa(t)\) can be found using the formula: \[ \kappa(t) = \frac{||\mathbf{v}(t) \times \mathbf{a}(t)||}{||\mathbf{v}(t)||^3} \] Here, the cross product \(\mathbf{v}(t) \times \mathbf{a}(t)\) in two dimensions results in a scalar since it's only the z-component: \[ \mathbf{v}(1) \times \mathbf{a}(1) = (1)(0) - (2)(2) = -4 \] Thus, \[ \kappa(1) = \frac{| -4 |}{(\sqrt{5})^3} = \frac{4}{5\sqrt{5}} \] Rationalizing gives \[ \kappa(1) = \frac{4\sqrt{5}}{25} \].

Key Concepts

Velocity Vector

The velocity vector is essential in calculus to describe how an object's position changes over time. For the curve described by the vector function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \), the velocity vector \( \mathbf{v}(t) \) is obtained by differentiating \( \mathbf{r}(t) \) with respect to \( t \). This process gives us the rate at which the position changes.
To find the velocity vector, we calculate:

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

Evaluating the velocity vector at \( t = 1 \):

• \( \mathbf{v}(1) = \mathbf{i} + 2\mathbf{j} \)

In terms of components, this means at \( t=1 \), the object moves one unit in the x-direction and two units in the y-direction. Thus, understanding the velocity vector gives us insight into the instantaneous direction and speed of motion.

Acceleration Vector

Acceleration vector tells us how the velocity of an object is changing with time. For our function \( \mathbf{v}(t) = \mathbf{i} + 2t\mathbf{j} \), the acceleration vector \( \mathbf{a}(t) \) is found by differentiating \( \mathbf{v}(t) \).
Let's calculate it:

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

Evaluating at \( t = 1 \):

• \( \mathbf{a}(1) = 2\mathbf{j} \)

This result shows that there is no change in the x-direction, and a constant acceleration of 2 units in the y-direction. In simpler terms, the object's y-coordinate velocity increases uniformly, providing us a straightforward look at how its path is bending over time.

Tangent Vector

The tangent vector is a unit vector that points in the direction of the velocity vector. It tells us about the orientation of the path the object follows. To find the tangent vector \( \mathbf{T}(t) \), we first have to normalize the velocity vector by dividing it by its magnitude.
First, compute the magnitude of \( \mathbf{v}(1) \):

• \( ||\mathbf{v}(1)|| = \sqrt{1^2 + 2^2} = \sqrt{5} \)

Now, find the unit tangent vector:

• \( \mathbf{T}(1) = \frac{1}{\sqrt{5}}(\mathbf{i} + 2\mathbf{j}) \)

The tangent vector being a unit vector ensures it has a length of one, while its direction precisely shows the tangent to the curve at \( t=1 \). It helps us visualize the direction in which the curve is traveling at any specific point.

Curvature

Curvature is a measure of how sharply a curve bends at a particular point. A higher curvature means a tighter bend. To calculate the curvature \( \kappa(t) \) at \( t=1 \), we use:

• The formula, \( \kappa(t) = \frac{||\mathbf{v}(t) \times \mathbf{a}(t)||}{||\mathbf{v}(t)||^3} \)

Since we are in two dimensions, the cross product gives a scalar:

• \( \mathbf{v}(1) \times \mathbf{a}(1) = -4 \)

Now, we compute \( \kappa(1) \):

• \( \kappa(1) = \frac{4}{5\sqrt{5}} = \frac{4\sqrt{5}}{25} \)

Curvature provides insights into the local geometry of the path. It allows us to understand how the curve's shape changes, which has practical applications in fields like engineering and physics where motion and path design are crucial.