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Chapter 10: Problem 4

Find the unit tangent vector to the curve at the specified value of the parameter. $$ \mathbf{r}(t)=6 \cos t \mathbf{i}+2 \sin t \mathbf{j}, \quad t=\frac{\pi}{3} $$

Short Answer

Expert verified
The unit tangent vector to the curve \( \mathbf{r}(t)=6 \cos t \mathbf{i}+2 \sin t \mathbf{j} \) at \( t=\frac{\pi}{3} \) is \( \mathbf{T}\left(\frac{\pi}{3}\right) = -\frac{3}{2} \mathbf{i}+ \frac{1}{2\sqrt{3}} \mathbf{j} \).

Step by step solution

01

Calculate the Derivative of the curve

To find the tangent vector, the first step is to compute the derivative of the curve. The derivative gives us a vector that points in the direction of the curve's motion at any point, which is the tangent vector. The derivative of \( \mathbf{r}(t) \) is:\[ \mathbf{r}'(t) = -6 \sin t \mathbf{i}+2 \cos t \mathbf{j}\]
02

Evaluate the Derivative at the given point

Now we substitute \( t=\frac{\pi}{3} \) into the derivative:\[ \mathbf{r}'\left(\frac{\pi}{3}\right) = -6 \sin \left(\frac{\pi}{3}\right) \mathbf{i}+2 \cos \left(\frac{\pi}{3}\right) \mathbf{j} = -6 \times \frac{\sqrt{3}}{2} \mathbf{i}+2 \times \frac{1}{2} \mathbf{j} = -3\sqrt{3} \mathbf{i}+ \mathbf{j}\]
03

Normalize the vector obtained

To make this tangent vector a unit vector, we normalize it by dividing it by its magnitude. The magnitude of vector \( \mathbf{r}'\left(\frac{\pi}{3}\right) \) is:\[ ||\mathbf{r}'\left(\frac{\pi}{3}\right)|| = \sqrt{(-3\sqrt{3})^2 + 1^2} = 2\sqrt{3}\]So the unit tangent vector \( \mathbf{T}(t) \) at \( t=\frac{\pi}{3} \) is:\[ \mathbf{T}\left(\frac{\pi}{3}\right) = \frac{\mathbf{r}'\left(\frac{\pi}{3}\right)}{||\mathbf{r}'\left(\frac{\pi}{3}\right)||} = \frac{-3\sqrt{3} \mathbf{i}+ \mathbf{j}}{2\sqrt{3}} = -\frac{3}{2} \mathbf{i}+ \frac{1}{2\sqrt{3}} \mathbf{j}\]

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Most popular questions from this chapter

Consider the motion of a point (or particle) on the circumference of a rolling circle. As the circle rolls, it generates the cycloid \(\mathbf{r}(t)=b(\omega t-\sin \omega t) \mathbf{i}+b(1-\cos \omega t) \mathbf{j}\) where \(\omega\) is the constant angular velocity of the circle and \(b\) is the radius of the circle. Find the maximum speed of a point on the circumference of an automobile tire of radius 1 foot when the automobile is traveling at 55 miles per hour. Compare this speed with the speed of the automobile.

Prove the property. In each case, assume that \(\mathbf{r}, \mathbf{u},\) and \(\mathbf{v}\) are differentiable vector-valued functions of \(t,\) \(f\) is a differentiable real-valued function of \(t,\) and \(c\) is a scalar. $$ \begin{array}{l} D_{t}\\{\mathbf{r}(t) \cdot[\mathbf{u}(t) \times \mathbf{v}(t)]\\}=\mathbf{r}^{\prime}(t) \cdot[\mathbf{u}(t) \times \mathbf{v}(t)]+ \\ \mathbf{r}(t) \cdot\left[\mathbf{u}^{\prime}(t) \times \mathbf{v}(t)\right]+\mathbf{r}(t) \cdot\left[\mathbf{u}(t) \times \mathbf{v}^{\prime}(t)\right] \end{array} $$

Use the properties of the derivative to find the following. (a) \(\mathbf{r}^{\prime}(t)\) (b) \(\mathbf{r}^{\prime \prime}(t)\) (c) \(D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)]\) (d) \(D_{t}[3 \mathbf{r}(t)-\mathbf{u}(t)]\) (e) \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)]\) (f) \(D_{t}[\|\mathbf{r}(t)\|], \quad t>0\) $$ \begin{array}{l} \mathbf{r}(t)=t \mathbf{i}+2 \sin t \mathbf{j}+2 \cos t \mathbf{k} \\ \mathbf{u}(t)=\frac{1}{t} \mathbf{i}+2 \sin t \mathbf{j}+2 \cos t \mathbf{k} \end{array} $$

The position vector \(r\) describes the path of an object moving in the \(x y\) -plane. Sketch a graph of the path and sketch the velocity and acceleration vectors at the given point. $$ \mathbf{r}(t)=3 \cos t \mathbf{i}+2 \sin t \mathbf{j},(3,0) $$

In Exercises \(49-52,\) evaluate the definite integral. $$ \int_{0}^{1}(8 t \mathbf{i}+t \mathbf{j}-\mathbf{k}) d t $$
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