Question

Sketch the plane curve represented by the vector-valued function, and sketch the vectors \(\mathbf{r}\left(t_{0}\right)\) and \(\mathbf{r}^{\prime}\left(t_{0}\right)\) for the given value of \(t_{0} .\) Position the vectors such that the initial point of \(\mathbf{r}\left(t_{0}\right)\) is at the origin and the initial point of \(\mathbf{r}^{\prime}\left(t_{0}\right)\) is at the terminal point of \(\mathbf{r}\left(t_{0}\right) .\) What is the relationship between \(\mathbf{r}^{\prime}\left(t_{0}\right)\) and the curve? $$ \mathbf{r}(t)=(1+t) \mathbf{i}+t^{3} \mathbf{j}, \quad t_{0}=1 $$

Short answer

The vector \(\mathbf{r}(t_{0})\) is at the point (2,1) on the curve and the derivative \(\mathbf{r}'(t_{0})\) is tangent to the curve at that point, pointing in the curve's direction of progress and reflecting the speed of movement at that point.

Step-by-step solution

Express given vectors

Given vector-valued function is \(\mathbf{r}(t)=(1+t)\mathbf{i}+t^{3}\mathbf{j}\) and \(t_{0} = 1\). The vector at \(t_{0} =1\) is \(\mathbf{r}(t_{0})=(1+1)\mathbf{i}+1^{3}\mathbf{j}=2\mathbf{i}+\mathbf{j}\).

Find the derivative of the vector-valued function

The derivative \(\mathbf{r}'(t)\) of the vector-valued function \(\mathbf{r}(t)\) is found by taking the derivatives of its i-component and j-component separately. This gives \(\mathbf{r}'(t)= \mathbf{i} + 3t^2\mathbf{j}\). Given \(t_0=1\), evaluating \(\mathbf{r}'(t_{0})\) at \(t_0=1\), we obtain \(\mathbf{r}'(t_{0})= \mathbf{i} + 3(1)^2\mathbf{j} = \mathbf{i} + 3\mathbf{j}\).

Sketch the vectors

To sketch the vectors, start by sketching the curve dictated by the given vector-valued function \(\mathbf{r}(t)\) for a suitable range of t-values. Then sketch the vector \(\mathbf{r}(t_{0})\) from the origin to the point (2,1) on the curve. Lastly, sketch the vector \(\mathbf{r}'(t_{0})\) starting from the terminal point of \(\mathbf{r}(t_{0})\) towards the direction of the curve at that point.

Discuss the relationship

The vector \(\mathbf{r}'(t_{0})\) represents the tangent to the curve at point \(\mathbf{r}(t_{0})\) and shows the direction in which the curve is heading at that point. The direction of \(\mathbf{r}'(t_{0})\) reflects the direction of motion of a particle moving along the curve at a given time \(t_{0}\). The magnitude of \(\mathbf{r}'(t_{0})\) indicates the speed of movement at that point.

Key concepts

Tangent Vector

Understanding the concept of a tangent vector is crucial in studying the motion and geometry of curves. A tangent vector at a point on a curve represents the direction and rate of change of the curve at that specific point. Think of a curve as the path traced by a moving particle; a tangent vector would illustrate the velocity of the particle at a particular moment in time.

In the context of our exercise, the vector \(\mathbf{r}'(t_0)\) is the tangent vector at the point \(t_0\). It not only points in the direction the curve is heading but also indicates the speed of movement along the curve. When \(t_0=1\), we calculate the tangent vector as \(\mathbf{r}'(t_0)=\mathbf{i} + 3\mathbf{j}\), visually representing the instantaneous direction and speed of the curve at that point. This concept is particularly helpful in fields such as physics and engineering, where understanding motion and change is essential.

Vector Calculus

Vector calculus is an extension of calculus to vector fields. This branch of mathematics involves differential and integral calculus over vector fields, enabling us to handle functions that have vectors as outputs. Vector calculus is pivotal in physics and engineering as it delves into concepts like velocity, force, and field flux.

In our exercise, we use vector calculus to derive the tangent vector of a curve. Starting from \(\mathbf{r}(t)=(1+t)\mathbf{i}+t^{3}\mathbf{j}\), we apply the concepts of differentiation to find \(\mathbf{r}'(t)\), which is the derivative of the vector function. This derivative, evaluated at \(t_0\), gives us the tangent vector that is essential for analyzing the curve's behavior at a given point. Vector calculus thus provides the tools to explore the dynamic properties of vector-valued functions such as velocity and acceleration.

Derivative of Vector Function

The derivative of a vector function is analogous to the derivative of a scalar function, but it involves vectors instead of scalar quantities. When we differentiate a vector function, we're looking at how each component of the vector changes with respect to the independent variable, often time. In simple terms, it measures the rate at which the vector function is changing at any given point, providing a vector output that describes changes in both magnitude and direction.

In the given problem, differentiation is applied component-wise to the vector function \(\mathbf{r}(t)\) to yield \(\mathbf{r}'(t)\). This process involves taking the derivatives of the i-component and j-component separately. For \(\mathbf{r}(t)=(1+t)\mathbf{i}+t^{3}\mathbf{j}\), we find the derivative as \(\mathbf{r}'(t) = \mathbf{i} + 3t^2\mathbf{j}\). The resulting derivative \(\mathbf{r}'(t)\) gives us the velocity vector of the curve, which is essential for understanding its motion and direction at any point in time. Through this, the relationship between the function and its derivative expresses not just the instantaneous speed, but also the path's changing direction—integral knowledge for predicting the curve's future position.