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Chapter 4: Problem 3

Compute the velocity vector, the acceleration vector, the speed, and the equation of the tangent line. $$\mathbf{c}(t)=\left(e^{t}, \sin t, \cos t\right), \text { at } t=0$$

Short Answer

Expert verified
Velocity: (1, 1, 0), Acceleration: (1, 0, -1), Speed: \( \sqrt{2} \), Tangent Line: \( (1 + t, t, 1) \).

Step by step solution

01

Compute the Velocity Vector

The velocity vector, \( \mathbf{v}(t) \), is the derivative of the position vector \( \mathbf{c}(t) \). Differentiate each component of \( \mathbf{c}(t) = (e^t, \sin t, \cos t) \) with respect to \( t \). \[ \mathbf{v}(t) = \left( \frac{d}{dt}e^t, \frac{d}{dt}\sin t, \frac{d}{dt}\cos t \right) = (e^t, \cos t, -\sin t) \]Evaluate at \( t = 0: \mathbf{v}(0) = (e^0, \cos 0, -\sin 0) = (1, 1, 0) \).
02

Compute the Acceleration Vector

The acceleration vector, \( \mathbf{a}(t) \), is the derivative of the velocity vector. Differentiate each component of \( \mathbf{v}(t) = (e^t, \cos t, -\sin t) \) with respect to \( t \).\[ \mathbf{a}(t) = \left( \frac{d}{dt}e^t, \frac{d}{dt}\cos t, \frac{d}{dt}(-\sin t) \right) = (e^t, -\sin t, -\cos t) \]Evaluate at \( t = 0: \mathbf{a}(0) = (e^0, -\sin 0, -\cos 0) = (1, 0, -1) \).
03

Compute the Speed

Speed is the magnitude of the velocity vector. Calculate the magnitude of \( \mathbf{v}(t) = (e^t, \cos t, -\sin t) \).\[ \text{Speed} = \| \mathbf{v}(t) \| = \sqrt{(e^t)^2 + (\cos t)^2 + (-\sin t)^2} = \sqrt{e^{2t} + 1} \]Evaluate at \( t = 0: \text{Speed} = \sqrt{e^0 + 1} = \sqrt{2} \).
04

Equation of the Tangent Line

The equation of the tangent line can be determined using the formula: \[ \mathbf{T}(t) = \mathbf{c}(t_0) + t \cdot \mathbf{v}(t_0) \]Here, \( \mathbf{c}(0) = (e^0, \sin 0, \cos 0) = (1, 0, 1) \) and \( \mathbf{v}(0) = (1, 1, 0) \).Thus, the equation of the tangent line is:\[ \mathbf{T}(t) = (1, 0, 1) + t(1, 1, 0) = (1 + t, t, 1) \]

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Key Concepts

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

Velocity Vector
The velocity vector is a crucial concept in vector calculus as it describes how a point moves through space over time. To find the velocity vector of a position vector \( \mathbf{c}(t) \), such as \( \mathbf{c}(t)=\left(e^{t}, \sin t, \cos t\right) \), we take its derivative with respect to \( t \). This involves differentiating each component:
  • The derivative of \( e^t \) is \( e^t \)
  • The derivative of \( \sin t \) is \( \cos t \)
  • The derivative of \( \cos t \) is \( -\sin t \)
So, the velocity vector is \( \mathbf{v}(t) = (e^t, \cos t, -\sin t) \). Evaluating at \( t=0 \) gives \( \mathbf{v}(0) = (1, 1, 0) \). This vector indicates the direction and speed of the motion at that specific time.
Acceleration Vector
Acceleration is the rate of change of velocity, so to find the acceleration vector, we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Like with the velocity calculation, we find the derivative for each component:
  • The derivative of \( e^t \) remains \( e^t \)
  • The derivative of \( \cos t \) becomes \( -\sin t \)
  • The derivative of \(-\sin t\) is \( -\cos t \)
As a result, the acceleration vector is \( \mathbf{a}(t) = (e^t, -\sin t, -\cos t) \). At \( t=0 \), it evaluates to \( \mathbf{a}(0) = (1, 0, -1) \). This vector describes how the velocity is changing at any given moment.
Tangent Line Equation
The tangent line to a curve at a point gives us a linear approximation of the curve at that point. To compute the tangent line equation, we use the position vector \( \mathbf{c}(t) \) at the given point, and the velocity vector at the same point, \( \mathbf{v}(t_0) \). For \( \mathbf{c}(0) = (1, 0, 1) \) and \( \mathbf{v}(0) = (1, 1, 0) \), the formula is:\[ \mathbf{T}(t) = \mathbf{c}(0) + t \cdot \mathbf{v}(0) = (1, 0, 1) + t(1, 1, 0) \]Which simplifies to:\[ (1 + t, t, 1)\]This equation outlines a straight line that touches the curve exactly at \( t=0 \) and follows the instantaneous direction given by the velocity vector.
Speed Calculation
Speed is derived from the magnitude of the velocity vector and provides a scalar measure of instantaneous velocity, essentially telling us how fast something is moving regardless of direction. To find this, we calculate the magnitude of \( \mathbf{v}(t) = (e^t, \cos t, -\sin t) \):\[ \text{Speed} = \| \mathbf{v}(t) \| = \sqrt{(e^t)^2 + (\cos t)^2 + (-\sin t)^2}\]Evaluating this expression at \( t=0 \), we simplify:\[ \text{Speed} = \sqrt{(1)^2 + (1)^2 + (0)^2} = \sqrt{2}\]Thus, the speed at \( t=0 \) is \( \sqrt{2} \), describing how fast the point moves along the curve at that instant.

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Most popular questions from this chapter

Develop some of the classic differential geometry of curves. Let \(c:[a, b] \rightarrow \mathbb{R}^{3}\) be an infinitely differentiable path (derivatives of all orders exist). Assume \(\mathbf{c}^{\prime}(t) \neq \mathbf{0}\) for any \(t .\) The vector \(\mathbf{c}^{\prime}(t) /\left\|\mathbf{c}^{\prime}(t)\right\|=\mathbf{T}(t)\) is tangent to \(\mathbf{c}\) at \(\mathbf{c}(t),\) and, because \(\|\mathbf{T}(t)\|=1, \mathbf{T}\) is called the unit tangent to \(c\) (a) Show that \(\mathbf{T}^{\prime}(t) \cdot \mathbf{T}(t)=0 .\) [HINT: Differentiate \(\mathbf{T}(t) \cdot \mathbf{T}(t)=1 .]\) (b) Write down a formula for \(\mathbf{T}^{\prime}(t)\) in terms of \(\mathbf{c}\)

If \(r(t)=6 t i+3 t^{2} j+t^{3} k,\) what force acts on a particle of mass \(m\) moving along \(\mathbf{r}\) at \(t=0 ?\)

(a) Let \(g(x, y, z)=x^{3}+5 y z+z^{2}\) and let \(h(u)\) be a function of one variable such that \(h^{\prime}(1)=1 / 2 .\) Let \(f=h \circ g .\) Starting at \((1,0,0),\) in what directions is \(f\) changing at \(50 \%\) of its maximum rate? (b) For \(g(x, y, z)=x^{3}+5 y z+z^{2},\) calculate \(\mathbf{F}=\nabla g\) the gradient of \(g,\) and verify directly that \(\nabla \times \mathbf{F}=0\) at each point \((x, y, z)\)

Find the arc length of \(\mathbf{c}(t)=t \mathbf{i}+(\log t) \mathbf{j}+2 \sqrt{2 t} \mathbf{k}\) for \(1 \leq t \leq 2\)

Sketch the given vector field or a small multiple of it. $$\mathbf{F}(x, y)=\left(\frac{y}{\sqrt{x^{2}+y^{2}}}, \frac{x}{\sqrt{x^{2}+y^{2}}}\right)$$
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