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Chapter 10: Problem 52

How is the unit tangent vector related to the orientation of a curve? Explain.

Short Answer

Expert verified
The unit tangent vector is directly related to the orientation of a curve. The orientation of the curve means the direction in which the curve is being followed. The unit tangent vector, obtained from the derivative of a curve function, points in the direction of the curve's path at any given point. Therefore, the orientation of the curve at any point is in the same direction as the unit tangent vector at that point.

Step by step solution

01

Defining the Unit Tangent Vector

The unit tangent vector describes the direction of the curve at any given point. It is obtained by dividing the derivative of the curve function, also known as the tangent vector, by its magnitude (length). Given a curve function \(y = f(x)\), the unit tangent vector \(T\) at a point \(P(x, y)\) is calculated as \(T = \frac{f'(x)}{||f'(x)||}\) where \(f'(x)\) is the derivative of the curve function and \(||f'(x)||\) is the magnitude of \(f'(x)\).
02

Understanding the Orientation of a Curve

The orientation of a curve refers to the direction in which the curve is traversed or followed. For example, the curve could be oriented from left to right, or from bottom to top.
03

Relating the Unit Tangent Vector and Curve Orientation

The unit tangent vector gives the direction of the curve at any given point, i.e., it points in the direction where the curve is heading next. This means that the orientation of the curve at a given point is the same as the direction of the unit tangent vector at that point. In other words, the way the unit tangent vector is pointing shows the path the curve is following at that point. Therefore, the orientation of the curve is related to the unit tangent vector.

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Most popular questions from this chapter

Use the given acceleration function to find the velocity and position vectors. Then find the position at time \(t=2\) $$ \begin{array}{l} \mathbf{a}(t)=2 \mathbf{i}+3 \mathbf{k} \\ \mathbf{v}(0)=4 \mathbf{j}, \quad \mathbf{r}(0)=\mathbf{0} \end{array} $$

The position vector \(r\) describes the path of an object moving in space. Find the velocity, speed, and acceleration of the object. $$ \mathbf{r}(t)=\langle 4 t, 3 \cos t, 3 \sin t\rangle $$

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. $$ D_{t}[f(t) \mathbf{r}(t)]=f(t) \mathbf{r}^{\prime}(t)+f^{\prime}(t) \mathbf{r}(t) $$

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 velocity and acceleration vectors of the particle. Use the results to determine the times at which the speed of the particle will be (a) zero and (b) maximized.

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. $$ \begin{array}{l} D_{t}\\{\mathbf{r}(t) \cdot[\mathbf{u}(t) \times \mathbf{v}(t)]\\}=\mathbf{r}^{\prime}(t) \cdot[\mathbf{u}(t) \times \mathbf{v}(t)]+ \\ \mathbf{r}(t) \cdot\left[\mathbf{u}^{\prime}(t) \times \mathbf{v}(t)\right]+\mathbf{r}(t) \cdot\left[\mathbf{u}(t) \times \mathbf{v}^{\prime}(t)\right] \end{array} $$
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