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Chapter 13: Problem 3

Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$\mathbf{r}(t)=t \mathbf{i}+(2 / 3) t^{3 / 2} \mathbf{k}, \quad 0 \leq t \leq 8$$

Short Answer

Expert verified
The unit tangent vector is \( \frac{\mathbf{i} + t^{1/2} \mathbf{k}}{\sqrt{1 + t}} \), and the arc length is \( \frac{52}{3} \).

Step by step solution

01

Find the Velocity Vector

The velocity vector is the derivative of the position vector with respect to time. For \( \mathbf{r}(t) = t \mathbf{i} + \frac{2}{3} t^{3/2} \mathbf{k} \), differentiate each component: \( \mathbf{v}(t) = \frac{d}{dt}(t \mathbf{i}) + \frac{d}{dt}(\frac{2}{3} t^{3/2} \mathbf{k}) = \mathbf{i} + t^{1/2} \mathbf{k} \).
02

Calculate Speed

Speed is the magnitude of the velocity vector \( \mathbf{v}(t) = \mathbf{i} + t^{1/2} \mathbf{k} \). Calculate the magnitude: \[ |\mathbf{v}(t)| = \sqrt{1^2 + (t^{1/2})^2} = \sqrt{1 + t} \]
03

Determine the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is the velocity vector divided by its magnitude. Thus,\[ \mathbf{T}(t) = \frac{\mathbf{v}(t)}{|\mathbf{v}(t)|} = \frac{\mathbf{i} + t^{1/2} \mathbf{k}}{\sqrt{1 + t}} \]
04

Set Up the Integral for Arc Length

The arc length \( L \) from \( t = 0 \) to \( t = 8 \) is found by integrating the speed: \[ L = \int_0^8 |\mathbf{v}(t)| \, dt = \int_0^8 \sqrt{1 + t} \, dt \]
05

Solve the Integral

Calculate \( \int \sqrt{1 + t} \, dt \) using substitution. Let \( u = 1 + t \), so \( du = dt \). The integral becomes:\[ \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C \]Back-substitute \( u = 1 + t \):\[ \frac{2}{3} (1 + t)^{3/2} \]Evaluate from 0 to 8:\[ L = \left[ \frac{2}{3} (1 + t)^{3/2} \right]_0^8 = \frac{2}{3} [(1 + 8)^{3/2} - (1 + 0)^{3/2}] = \frac{2}{3} (9^{3/2} - 1^{3/2}) = \frac{2}{3} (27 - 1) = \frac{2}{3} \times 26 = \frac{52}{3} \]

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Key Concepts

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

Velocity Vector
To better understand the concept of a velocity vector, think of it as a mathematical tool that determines both the direction and the speed of an object at a given point in time. For the curve given by the position vector \( \mathbf{r}(t) = t \mathbf{i} + \frac{2}{3} t^{3/2} \mathbf{k} \), the velocity vector is obtained by differentiating each component of the position vector with respect to time.
  • The derivative of \( t \mathbf{i} \) with respect to \( t \) is \( \mathbf{i} \).
  • The derivative of \( \frac{2}{3} t^{3/2} \mathbf{k} \) with respect to \( t \) is \( t^{1/2} \mathbf{k} \).
By combining these components, the velocity vector is expressed as \( \mathbf{v}(t) = \mathbf{i} + t^{1/2} \mathbf{k} \). This indicates how the position of a point on the curve changes over time in terms of both direction and magnitude.
Arc Length
The arc length of a curve is a measure of the distance along the curve from one point to another. It can be envisioned as the length of a piece of string that perfectly follows the curve's path. To compute the arc length of the curve from \( t = 0 \) to \( t = 8 \), we can use integral calculus.
The formula for calculating arc length \( L \) is given by the integral of the speed over the specified interval:\[L = \int_{0}^{8} |\mathbf{v}(t)| \, dt\]Here, \( |\mathbf{v}(t)| \) represents the speed, which is the magnitude of the velocity vector. For this problem, it simplifies to the integral:\[L = \int_{0}^{8} \sqrt{1 + t} \, dt\]By evaluating this integral, the total arc length of the curve is found.
Integral Calculus
Integral calculus provides a method for calculating accumulations, like areas under curves or, in this case, arc lengths. To solve the integral for the arc length, we use a substitution technique to simplify the expression. The integral in this problem is:\[\int \sqrt{1 + t} \, dt\]Using substitution, let \( u = 1 + t \), then \( du = dt \). The integral becomes:\[\int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C\]Re-substituting back in terms of \( t \), we get:\[\frac{2}{3}(1+t)^{3/2}\]Evaluating this from \( t = 0 \) to \( t = 8 \) gives the final arc length.
Speed
Speed is the magnitude of the velocity vector and represents the rate at which an object is traveling along a path, regardless of direction. For the velocity vector \( \mathbf{v}(t) = \mathbf{i} + t^{1/2} \mathbf{k} \), the speed \( |\mathbf{v}(t)| \) is computed as follows:\[|\mathbf{v}(t)| = \sqrt{1^2 + (t^{1/2})^2} = \sqrt{1 + t} \]This expression quantifies how fast the position of the object described by the vector changes at time \( t \). Measuring speed is crucial as it gives insights into the dynamic nature of the object's motion over the path.

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Most popular questions from this chapter

In Section \(13.4,\) you found \(\mathbf{T}, \mathbf{N},\) and \(\kappa .\) Now, in the following, find \(\mathbf{B}\) and \(\tau\) for these space curves. $$\mathbf{r}(t)=(3 \sin t) \mathbf{i}+(3 \cos t) \mathbf{j}+4 t \mathbf{k}$$

You will use a CAS to explore the osculating circle at a point \(P\) on a plane curve where \(\kappa \neq 0 .\) Use a CAS to perform the following steps: a. Plot the plane curve given in parametric or function form over the specified interval to see what it looks like.b. Calculate the curvature \(\kappa\) of the curve at the given value \(t_{0}\) using the appropriate formula from Exercise 5 or \(6 .\) Use the parametrization \(x=t\) and \(y=f(t)\) if the curve is given as a function \(y=f(x).\)c. Find the unit normal vector \(\mathbf{N}\) at \(t_{0} .\) Notice that the signs of the components of \(\mathbf{N}\) depend on whether the unit tangent vector \(\mathbf{T}\) is turning clockwise or counterclockwise at \(\left.t=t_{0} . \text { (See Exercise } 7 .\right)\) d. If \(\mathbf{C}=a \mathbf{i}+b \mathbf{j}\) is the vector from the origin to the center \((a, b)\) of the osculating circle, find the center \(\mathbf{C}\) from the vector equation $$\mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right),$$ The point \(P\left(x_{0}, y_{0}\right)\) on the curve is given by the position vector \(\mathbf{r}\left(t_{0}\right)\) e. Plot implicitly the equation \((x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}\) of the osculating circle. Then plot the curve and osculating circle together. You may need to experiment with the size of the viewing window, but be sure the axes are equally scaled. $$y=x(1-x)^{2 / 5}, \quad-1 \leq x \leq 2, \quad x_{0}=1 / 2.$$

Suppose that \(\mathbf{r}_{1}(t)=f_{1}(t) \mathbf{i}+f_{2}(t) \mathbf{j}+f_{3}(t) \mathbf{k}, \mathbf{r}_{2}(t)=g_{1}(t) \mathbf{i}+g_{2}(t) \mathbf{j}+g_{3}(t) \mathbf{k}\) \(\lim _{t \rightarrow t_{0}} \mathbf{r}_{1}(t)=\mathbf{A},\) and \(\lim _{t \rightarrow t_{6}} \mathbf{r}_{2}(t)=\mathbf{B} .\) Use the determinant formula for cross products and the Limit Product Rule for scalar functions to show that $$\lim _{t \rightarrow t_{0}}\left(\mathbf{r}_{1}(t) \times \mathbf{r}_{2}(t)\right)=\mathbf{A} \times \mathbf{B}$$

a. Show that if \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are differentiable vector functions of \(t,\) then $$\frac{d}{d t}(\mathbf{u} \cdot \mathbf{v} \times \mathbf{w})=\frac{d \mathbf{u}}{d t} \cdot \mathbf{v} \times \mathbf{w}+\mathbf{u} \cdot \frac{d \mathbf{v}}{d t} \times \mathbf{w}+\mathbf{u} \cdot \mathbf{v} \times \frac{d \mathbf{w}}{d t}$$ b. Show that $$\frac{d}{d t}\left(\mathbf{r} \cdot \frac{d \mathbf{r}}{d t} \times \frac{d^{2} \mathbf{r}}{d t^{2}}\right)=\mathbf{r} \cdot\left(\frac{d \mathbf{r}}{d t} \times \frac{d^{3} \mathbf{r}}{d t^{3}}\right)$$

Find the arc length parameter along the curve from the point where \(t=0\) by evaluating the integral $$s=\int_{0}^{t}|\mathbf{v}(\tau)| d \tau$$ from Equation (3). Then find the length of the indicated portion of the curve. $$\mathbf{r}(t)=\left(e^{t} \cos t\right) \mathbf{i}+\left(e^{t} \sin t\right) \mathbf{j}+e^{t} \mathbf{k}, \quad-\ln 4 \leq t \leq 0$$
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