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Chapter 2: Problem 20

Ssuppose that a particle following the given path \(\mathbf{c}(t)\) flies off on a tangent at \(t=t_{0} .\) Compute the position of the particle at the given time \(t_{1}.\) $$\mathbf{c}(t)=\left(e^{t}, e^{-t}, \cos t\right), \text { where } t_{0}=1, t_{1}=2$$

Short Answer

Expert verified
The position of the particle at \(t=2\) is approximately \((5.436, 0, -0.301)\).

Step by step solution

01

Differentiate the position vector

To find the direction of the tangent vector at time \(t_0\), we start by differentiating \(\mathbf{c}(t) = (e^t, e^{-t}, \cos t)\) with respect to \(t\). The derivative is \(\mathbf{c}'(t) = (e^t, -e^{-t}, -\sin t)\).
02

Evaluate the tangent vector at \(t_0\)

Substitute \(t_0 = 1\) into the derivative to get the tangent vector at \(t_0\). That is \(\mathbf{c}'(1) = (e^1, -e^{-1}, -\sin 1)\). Calculate the numerical values: \(\mathbf{c}'(1) \approx (2.718, -0.368, -0.841)\).
03

Find the position at \(t_0\)

Substitute \(t_0 = 1\) into the position function \(\mathbf{c}(t)\) to get the position vector at \(t_0\). Thus, \(\mathbf{c}(1) = (e^1, e^{-1}, \cos 1)\) which approximately equals \((2.718, 0.368, 0.54)\).
04

Compute the line equation from \(t_0\)

The line of motion has the form \(\mathbf{r}(t) = \mathbf{c}(t_0) + (t - t_0) \mathbf{c}'(t_0)\). Substitute our known values, \(\mathbf{r}(t) = (2.718, 0.368, 0.54) + (t-1)(2.718, -0.368, -0.841)\).
05

Evaluate the line equation at \(t_1\)

Substitute \(t_1 = 2\) into the line equation from Step 4.\(\mathbf{r}(2) =(2.718, 0.368, 0.54) + (2-1)(2.718, -0.368, -0.841) = (2.718 + 2.718, 0.368 - 0.368, 0.54 - 0.841)\).
06

Simplify to find the position

Simplify the expression: \(\mathbf{r}(2) = (5.436, 0, -0.301)\). This represents the position of the particle at \(t = 2\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus used to determine the rate at which a function is changing at any point. It serves as a powerful tool to find the slope or gradient of a curve described by a function. For a particle moving along a path defined by a position vector, differentiation is required to compute what is known as the tangent vector.
  • Differentiating the position vector gives you the rate of change in position, which is effectively the velocity of the particle.
  • This derivative is particularly important in understanding how the trajectory of the particle changes over time, allowing us to study its motion intricately.
  • In this exercise, we're differentiating the position vector \((e^t, e^{-t}, \cos t)\) to find the vector's derivative \((e^t, -e^{-t}, -\sin t)\).
By finding the derivative, we can find out the exact direction and speed in which the particle is moving at any given point in time.
Particle Path
The path of a particle is defined by its position vector, which indicates its location in a space x-y-z coordinate system at any point in time. This path is the trajectory that a particle follows as a function of time and can be expressed as a vector function like \((e^t, e^{-t}, \cos t)\).
  • This vector function is composed of three individual functions corresponding to the x, y, and z axes.
  • Each component describes how position on each axis changes with respect to time \(t\).
The particle path function allows us to predict where the particle will be at any given time \(t\), which is crucial for analyzing motion in physics and engineering.
Position Vector
The position vector is a key concept in vector calculus and physics. It specifies the exact position of a particle in space at a particular time. In this context, the position vector is noted as \(\mathbf{c}(t) = (e^t, e^{-t}, \cos t)\).
  • The position vector combines all three dimensions - x, y, and z - into a single vector function.
  • It is a function of time, meaning as time progresses, the position of the particle in space also changes.
The position vector not only tells the location but can also be differentiated to find the velocity of the particle, which is critical for dynamics studies. At \(t = 1\), for example, the position is \(2.718, 0.368, 0.54\), showcasing a snapshot of where the particle is at that moment.
Tangent Vector
The tangent vector represents the direction and speed of motion of a particle at any given instant along its path. In other words, it is the velocity vector obtained from differentiating the position vector.
  • At a given time \(t_0\), the tangent vector helps determine how the particle will move if it continues in a straight line from that point.
  • In our exercise, the tangent vector at \(t_0 =1\) is calculated as \((e^1, -e^{-1}, -\sin 1)\) which numerically evaluates to approximately \(2.718, -0.368, -0.841\).
This vector provides insight into the instantaneous direction and rate of movement of the particle, serving as a linear approximation of the particle's path at \(t_0\).
Line Equation
The line equation is derived to determine the future position of a particle, assuming it continues its path along the tangent vector from a given starting position. The line equation takes the form:
  • \(\mathbf{r}(t) = \mathbf{c}(t_0) + (t - t_0) \cdot \mathbf{c}'(t_0)\)
  • It originates at the particle's current position and is directed along the tangent vector, allowing us to track its subsequent motion.
In our problem, this leads to an equation that evaluates the line of motion at \(t_1 = 2\), giving us the new position as \(5.436, 0, -0.301\). Thus, the line equation predicts where the particle will be if it "flies off" in the tangent direction.

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Most popular questions from this chapter

Verify the first special case of the chain rule for the composition \(f \circ \mathbf{c}\) in each of the cases: (a) \(f(x, y)=x y, \mathbf{c}(t)=\left(e^{t}, \cos t\right)\) (b) \(f(x, y)=e^{x y}, \mathbf{c}(t)=\left(3 t^{2}, t^{3}\right)\) (c) \(f(x, y)=\left(x^{2}+y^{2}\right) \log \sqrt{x^{2}+y^{2}}, \mathbf{c}(t)=\) \(\left(e^{t}, e^{-t}\right)\) (d) \(f(x, y)=x \exp \left(x^{2}+y^{2}\right), \mathbf{c}(t)=(t,-t)\)

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