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 \(\vec{r}(1)\). $$ \vec{r}(t)=\left\langle t^{2}+t, t^{2}-t\right\rangle $$

Short answer

\(\vec{r}^{\prime}(t) = \langle 2t+1, 2t-1 \rangle\) and the vector at \(t=1\) is \(\vec{r}^{\prime}(1)=\langle 3, 1 \rangle\) starting at \(\vec{r}(1)=\langle 2, 0 \rangle\).

Step-by-step solution

Understand the Problem

The function \( \vec{r}(t) = \langle t^2 + t, t^2 - t \rangle \) is given in vector form, and we need to find its derivative, which is \( \vec{r}^{\prime}(t) \). Additionally, we are required to sketch the vector function for some values of \( t \) and the derivative vector at \( t = 1 \) originating from \( \vec{r}(1) \).

Find the Derivative

To find the derivative \( \vec{r}^{\prime}(t) \), differentiate each component of \( \vec{r}(t) \) with respect to \( t \). The first component \( t^2 + t \) becomes \( 2t + 1 \), and the second component \( t^2 - t \) becomes \( 2t - 1 \). Therefore, \( \vec{r}^{\prime}(t) = \langle 2t + 1, 2t - 1 \rangle \).

Evaluate \(\vec{r}(1)\) and \(\vec{r}^{\prime}(1)\)

Substitute \( t = 1 \) into the original vector function: \( \vec{r}(1) = \langle 1^2 + 1, 1^2 - 1 \rangle = \langle 2, 0 \rangle \). Now substitute \( t = 1 \) into the derivative function: \( \vec{r}^{\prime}(1) = \langle 2*1 + 1, 2*1 - 1 \rangle = \langle 3, 1 \rangle \).

Sketch \(\vec{r}(t)\) and \(\vec{r}^{\prime}(1)\)

Draw the trajectory defined by \( \vec{r}(t) \), plotting key points for various \( t \) values. The path \( \vec{r}(t) \) follows the parametric curve \( x = t^2 + t \), \( y = t^2 - t \). Place the initial point of the vector \( \vec{r}^{\prime}(1) = \langle 3, 1 \rangle \) at \( \vec{r}(1) = \langle 2, 0 \rangle \). Draw the vector so that it starts at \( (2,0) \) and ends at the point \( (5,1) \).

Key concepts

Vector Differentiation

Vector differentiation is the process of finding the derivative of a vector-valued function. Just like differentiating scalar functions, we apply rules of differentiation to each component of the vector function separately. For a vector function \( \vec{r}(t) = \langle f(t), g(t) \rangle \), its derivative \( \vec{r}^{\prime}(t) \) is computed by differentiating each component function:

• The derivative of \( f(t) \) becomes \( f^{\prime}(t) \).
• The derivative of \( g(t) \) becomes \( g^{\prime}(t) \).

Consequently, \( \vec{r}^{\prime}(t) = \langle f^{\prime}(t), g^{\prime}(t) \rangle \).
This means you treat each component separately but keep them bound together in a vector form. In our exercise, differentiating the components \( t^2 + t \) and \( t^2 - t \) gave us the results \( 2t + 1 \) and \( 2t - 1 \) respectively, forming the derivative vector \( \vec{r}^{\prime}(t) = \langle 2t + 1, 2t - 1 \rangle \).
Understanding vector differentiation helps analyze changes and motion along a path defined by vectors.

Parametric Curves

When dealing with parametric curves, each point on the curve is the result of a pair of equations that define a continuous path. Unlike traditional \( (x,y) \) functions, where \( x \) is independent and \( y \) is dependent, parametric curves introduce a third parameter, commonly denoted \( t \), where both \( x \) and \( y \) depend on \( t \):

• \( x(t) = t^2 + t \)
• \( y(t) = t^2 - t \)

In parametric plots, the parameter \( t \) often represents time, defining a path over time.
In our case, as \( t \) varies, the vector \( \vec{r}(t) \) traces a trajectory on the plane. It can be visualized more dynamically than static curves since it encapsulates direction and positioning, reflecting movement and transformation effectively.
Parametric curves allow us to describe motion with ease when using vector calculus, offering great insights, especially in physics and engineering contexts.

Calculus Sketching

Sketching involves visualizing mathematical expressions and their properties. In calculus, it helps to see the behavior of functions across various intervals, determine patterns, identify critical points, and understand the impact of derivatives. When sketching vector functions like \( \vec{r}(t) \), it's all about showing how \( x \) and \( y \) values change over variations in \( t \).
The sketch visually represents the path traced out by the function. For the derivative \( \vec{r}^{\prime}(t) \), sketching shows the direction and rate of change at any point along the curve.

• Visualize \( \vec{r}(t) \) as a continuous line or curve.
• Plot specific points for chosen values of \( t \) to capture the essential shape.
• Draw the vector representing \( \vec{r}^{\prime}(1) \) starting from \( \vec{r}(1) \), highlighting its slope and direction.

Understanding these sketches aids in grasping how vectors change and inform about curvature, speed, and other dynamic properties. It lays a solid intuitive foundation for comprehending more complex movements and transformations.