Question

Find \(\vec{r}^{\prime}(t) .\) Sketch \(\vec{r}(t)\) and \(\vec{r}^{\prime}(1),\) with the initial point of \(\vec{r}^{\prime}(1)\) at \(\vec{r}(1)\). $$ \vec{r}(t)=\left\langle t^{2}+1, t^{3}-t\right\rangle $$

Short answer

\( \vec{r}^{\prime}(t) = \langle 2t, 3t^{2}-1 \rangle \), and \( \vec{r}^{\prime}(1) = \langle 2, 2 \rangle \).

Step-by-step solution

Differentiate Each Component

To find \( \vec{r}^{\prime}(t) \), we need to differentiate each component of \( \vec{r}(t) \). The vector \( \vec{r}(t) \) is given by \( \langle t^{2}+1, t^{3}-t \rangle \). Thus, we need to differentiate \( t^2+1 \) and \( t^3-t \) with respect to \( t \).

Differentiate the First Component

Differentiate the first component \( t^{2}+1 \). The derivative is \( \frac{d}{dt}(t^{2}+1) = 2t \).

Differentiate the Second Component

Differentiate the second component \( t^{3}-t \). The derivative is \( \frac{d}{dt}(t^{3}-t) = 3t^{2}-1 \).

Combine the Components

Now combine the derivatives of the components to find \( \vec{r}^{\prime}(t) \). Thus, \( \vec{r}^{\prime}(t) = \langle 2t, 3t^{2}-1 \rangle \).

Find \( \vec{r}^{\prime}(1) \)

To find \( \vec{r}^{\prime}(1) \), substitute \( t = 1 \) into \( \vec{r}^{\prime}(t) = \langle 2t, 3t^{2}-1 \rangle \). This yields \( \vec{r}^{\prime}(1) = \langle 2 \cdot 1, 3 \cdot 1^{2} - 1 \rangle = \langle 2, 2 \rangle \).

Evaluate \( \vec{r}(1) \)

Find the position vector \( \vec{r}(1) \) by substituting \( t = 1 \) into the original vector \( \vec{r}(t) = \langle t^2+1, t^3-t \rangle \). This gives \( \vec{r}(1) = \langle 1^2 + 1, 1^3 - 1 \rangle = \langle 2, 0 \rangle \).

Sketch the Vectors

To sketch \( \vec{r}(t) \) and \( \vec{r}^{\prime}(1) \), plot the point \( (2, 0) \) for \( \vec{r}(1) \) on a coordinate plane. Then, plot the vector \( \vec{r}^{\prime}(1) = \langle 2, 2 \rangle \) starting from \( (2, 0) \), which moves 2 units right and 2 units up.

Key concepts

Differentiation

Differentiation is a crucial concept in calculus, particularly vector calculus, where it involves finding the rate at which a quantity changes. When we differentiate a function, we calculate its derivative, which helps determine how the function value changes with respect to its variable.
In the given problem, the function is a vector function, where each component needs to be differentiated separately. This involves:

• Identifying each component of the vector function.
• Applying basic differentiation rules to find the derivative of each component. For example, the power rule where the derivative of \(t^n\) is \(n \cdot t^{n-1}\).

For the vector \( \vec{r}(t) = \langle t^2 + 1, t^3 - t \rangle \), differentiation is applied to each component to find \( \vec{r}'(t) \). This process will give us the rate of change of both components of the vector function as \(t\) changes.

Position Vector

A position vector describes the position of a point in space relative to the origin (0,0) in a coordinate system. In our exercise, \( \vec{r}(t) = \langle t^2 + 1, t^3 - t \rangle \) represents a position vector at time \( t \). Each value of \( t \) provides a specific location or point in the plane, mapped from the given function components.
When \( t = 1 \), substituting this into the position vector function, results in \( \vec{r}(1) = \langle 2, 0 \rangle \). This point \((2, 0)\) is where the position vector points at that specific time on the coordinate plane.

• Position vectors trace the trajectory or path of a moving point.
• They provide a snapshot of the point's location at any time \( t \).

Understanding position vectors helps visualize how a point moves as time progresses, making them fundamental in studying motion and dynamics in vector calculus.

Derivative of a Vector Function

The derivative of a vector function reflects how the vector changes as its parameter, often time \(t\), changes. In vector calculus, calculating the derivative of a vector function differs from differentiating scalar functions because each component vector function must be differentiated independently.
For our function \( \vec{r}(t) = \langle t^2+1, t^3-t \rangle \), after differentiating each component, we achieve \( \vec{r}'(t) = \langle 2t, 3t^2-1 \rangle \). This derivative defines how the direction and magnitude of the vector \( \vec{r}(t) \) change with \( t \).

• At \( t = 1 \), the derivative vector \( \vec{r}'(1) = \langle 2, 2 \rangle \) indicates the vector's direction and speed at that point in time.
• The derivative vector helps visualize motion, akin to velocity, where each component denotes a rate of change along the axis.

By sketching, we place \( \vec{r}'(1) \) starting from \( \vec{r}(1) \), illustrating in the plane how the vector progresses, ensuring a comprehensive understanding of direction and alteration over time.