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Chapter 14: Problem 7

Find the vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\); then sketch the graph of \(\mathbf{r}(t)\) in 2 -space and draw the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) $$\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle: \quad t_{0}=2$$

Short Answer

Expert verified
The tangent vector at \( t_0 = 2 \) is \( \langle 1, 4 \rangle \).

Step by step solution

01

Differentiate the Vector Function

The vector function given is \( \mathbf{r}(t) = \langle t, t^2 \rangle \). To find \( \mathbf{r}^{\prime}(t) \), we take the derivative of each component separately. The derivative of the first component \( t \) is 1, and the derivative of the second component \( t^2 \) is \( 2t \). Thus, \( \mathbf{r}^{\prime}(t) = \langle 1, 2t \rangle \).
02

Evaluate the Derivative at \( t_0 = 2 \)

To find \( \mathbf{r}^{\prime}(t_0) \) at \( t_0 = 2 \), we substitute \( t_0 = 2 \) into \( \mathbf{r}^{\prime}(t) = \langle 1, 2t \rangle \). This gives: \( \mathbf{r}^{\prime}(2) = \langle 1, 2 \cdot 2 \rangle = \langle 1, 4 \rangle \).
03

Sketch the Curve \( \mathbf{r}(t) \)

The function \( \mathbf{r}(t) = \langle t, t^2 \rangle \) describes a parabola opening upwards in the plane, as the second component, \( t^2 \), is a quadratic function of \( t \). To sketch this, plot points for various values of \( t \) and connect them smoothly to form the parabola. The points \( (0,0), (1,1), (2,4), (-1,1) \) and so on can be used.
04

Draw the Tangent Vector at \( t_0 = 2 \)

The tangent vector at \( t_0 = 2 \) is \( \langle 1, 4 \rangle \). At \( t = 2 \), the position vector is \( \mathbf{r}(2) = \langle 2, 4 \rangle \), which is the point on the parabola. From this point, draw a vector in the direction of \( \langle 1, 4 \rangle \). This tangent vector should point in the direction given by these components and is attached to the point \( (2, 4) \) on the parabola.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Differentiation
Vector differentiation is a method used in calculus to find the rate of change of a vector-valued function with respect to a variable. In simpler terms, it's about finding how the vector changes as the variable changes. This process is analogous to taking derivatives of scalar functions.
For the vector function \( \mathbf{r}(t) = \langle t, t^2 \rangle \):
  • To differentiate, handle each component separately, just like finding derivatives in basic calculus.
  • The first component \( t \) has a derivative of \( 1 \), because the derivative of \( t \) with respect to \( t \) is \( 1 \).
  • The second component \( t^2 \) has a derivative of \( 2t \), based on the power rule \( \frac{d}{dt}(t^n) = nt^{n-1} \).
The differentiated vector function is expressed as \( \mathbf{r}'(t) = \langle 1, 2t \rangle \). This tells us how the vector changes at any value of \( t \).
In exercises like these, vector differentiation helps in finding tangent vectors that describe the instantaneous direction of the curve described by \( \mathbf{r}(t) \).
Parametric Equations
Parametric equations involve expressing a set of quantities as explicit functions of independent parameters. In vector calculus, parametric equations often describe paths or trajectories in space.
For this example, the parametric equation \( \mathbf{r}(t) = \langle t, t^2 \rangle \) describes a path in 2-dimensional space:
  • As \( t \) changes, the vector \( \langle t, t^2 \rangle \) points to different positions along a path, which in this case, is a parabola.
  • The use of parameter \( t \) gives flexibility to calculate positions along the path by simply substituting different \( t \) values.
  • This approach is particularly useful in physics and engineering where time or another variable may dictate the path of motion.
To visualize, you can choose values for \( t \), like \( t = -1, 0, 1, 2 \), and find corresponding points such as \( (-1, 1), (0, 0), (1, 1), (2, 4) \).
This results in a curve which can be thought of as the tracing of a point moving through space, driven by changes in \( t \).
Tangent Vectors
Tangent vectors are a crucial concept in understanding the properties of curves in vector calculus. They represent the direction and rate of change of a curve at a particular point.
For example, if you have the vector \( \mathbf{r}(t) = \langle t, t^2 \rangle \) and its derivative \( \mathbf{r}'(t) \), the tangent vector at \( t_0 = 2 \) is:\(
\langle 1, 4 \rangle, \)
this tells you:
  • The curve direction is changing primarily upwards, as indicated by the larger second component \( 4 \).
  • The tangent vector is drawn starting at the point \( \langle 2, 4 \rangle \), which represents the position on the curve at \( t = 2 \).
  • The vector \( \langle 1, 4 \rangle \) shows the instantaneous direction of the curve at this point, tangent to the path.
In simple terms, tangent vectors are like arrows that touch the curve only at one point, heading in the direction the curve is going at that very spot.
This is informative in predicting the curve's immediate path and understanding its behavior at precise points.

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Most popular questions from this chapter

We assume that \(s\) is an arc length parameter for a smooth vector-valued function \(\mathbf{r}(s)\) in 3 -space and that \(d\) T/ds and \(d\) N/ds exist at each point on the curve. This implies that \(d\) B/ds exists as well, since \(\mathbf{B}=\mathbf{T} \times \mathbf{N} .\) The following derivatives, known as the Frenet-Serret formulas, are fundamental in the theory of curves in 3 -space: $$\begin{aligned} &d \mathbf{T} / d s=\kappa \mathbf{N} \quad \text { [Exercise } 61]\\\ &d \mathbf{N} / d s=-\kappa \mathbf{T}+\tau \mathbf{B} \quad \text { [Exercise } 63]\\\ &d \mathbf{B} / d s=-\tau \mathbf{N} \quad \text { [Exercise } 62(\mathrm{c})] \end{aligned}.$$ Use the first two Frenet-Serret formulas and the fact that \(\mathbf{r}^{\prime}(s)=\mathbf{T}\) if \(\mathbf{r}=\mathbf{r}(s)\) to show that $$\tau=\frac{\left[\mathbf{r}^{\prime}(s) \times \mathbf{r}^{\prime \prime}(s)\right] \cdot \mathbf{r}^{\prime \prime \prime}(s)}{\left\|\mathbf{r}^{\prime \prime}(s)\right\|^{2}} \text { and } \quad \mathbf{B}=\frac{\mathbf{r}^{\prime}(s) \times \mathbf{r}^{\prime \prime}(s)}{\left\|\mathbf{r}^{\prime \prime}(s)\right\|}$$

Solve the vector initial-value problem for \(y(t)\) by integrating and using the initial conditions to find the constants of integration. $$y^{\prime}(t)=t^{2} \mathbf{i}+2 t \mathbf{j}, \quad \mathbf{y}(0)=\mathbf{i}+\mathbf{j}$$

Suppose that the position function of a particle moving in 3-space is \(\mathbf{r}=3 \cos 2 t \mathbf{i}+\sin 2 t \mathbf{j}+4 t \mathbf{k}\) (a) Use a graphing utility to graph the speed of the particle versus time from \(t=0\) to \(t=\pi\) (b) Use the graph to estimate the maximum and minimum speeds of the particle. (c) Use the graph to estimate the time at which the maximum speed first occurs. (d) Find the exact values of the maximum and minimum speeds and the exact time at which the maximum speed first occurs.

Determine whether \(\mathbf{r}(t)\) is continuous at \(t=0 .\) Explain your reasoning. A. $$\mathbf{r}(t)=3 \sin t \mathbf{i}-2 t \mathbf{j}$$ B. $$\mathbf{r}(t)=t^{2} \mathbf{i}+\frac{1}{t} \mathbf{j}+t \mathbf{k}$$

Find the escape speed in \(\mathrm{km} / \mathrm{s}\) for a space probe in a circular orbit that is \(300 \mathrm{km}\) above the surface of the Earth.
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