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Chapter 12: Problem 23

Find the unit tangent vector for the following parameterized curves. $$\mathbf{r}(t)=\langle 8, \cos 2 t, 2 \sin 2 t\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Short Answer

Expert verified
Answer: \( \mathbf{T}(t) = \left\langle 0, \frac{-2\sin 2t}{\sqrt{4\sin^2 2t + 16\cos^2 2t}}, \frac{4\cos 2t}{\sqrt{4\sin^2 2t + 16\cos^2 2t}} \right\rangle \)

Step by step solution

01

Find the derivative of the parameterized curve

First, we need to find the derivative of the parameterized curve, which represents the tangent vector. To find the derivative of the curve \(\mathbf{r}(t)=\langle 8, \cos 2 t, 2 \sin 2 t\rangle\), we need to take the derivative with respect to \(t\) for each component: $$\mathbf{r}'(t) = \left\langle \frac{d}{dt}(8), \frac{d}{dt}(\cos 2t), \frac{d}{dt}(2\sin 2t) \right\rangle$$
02

Calculate the derivatives

Now let's calculate the derivatives for each component of the curve: $$\mathbf{r}'(t) = \left\langle 0, -2\sin 2t, 4\cos 2t \right\rangle$$
03

Find the magnitude of the tangent vector

Next, we need to find the magnitude of the tangent vector \(\mathbf{r}'(t)\) in order to normalize it. The magnitude of a vector \(\mathbf{v} = \langle a, b, c \rangle\) is given by: $$\| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2}$$ Applying this to our tangent vector \(\mathbf{r}'(t)\): $$\| \mathbf{r}'(t) \| = \sqrt{(0)^2 + (-2\sin 2t)^2 + (4\cos 2t)^2}$$
04

Simplify the magnitude expression

Let's simplify the magnitude expression: $$\| \mathbf{r}'(t) \| = \sqrt{4\sin^2 2t + 16\cos^2 2t}$$
05

Find the unit tangent vector

Finally, we can find the unit tangent vector by dividing the tangent vector by its magnitude: $$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \left\langle \frac{0}{\sqrt{4\sin^2 2t + 16\cos^2 2t}}, \frac{-2\sin 2t}{\sqrt{4\sin^2 2t + 16\cos^2 2t}}, \frac{4\cos 2t}{\sqrt{4\sin^2 2t + 16\cos^2 2t}} \right\rangle$$ The unit tangent vector of the parameterized curve \(\mathbf{r}(t)=\langle 8, \cos 2 t, 2 \sin 2 t\rangle\) is: $$\mathbf{T}(t) = \left\langle 0, \frac{-2\sin 2t}{\sqrt{4\sin^2 2t + 16\cos^2 2t}}, \frac{4\cos 2t}{\sqrt{4\sin^2 2t + 16\cos^2 2t}} \right\rangle$$

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Most popular questions from this chapter

Given a fixed vector \(\mathbf{v},\) there is an infinite set of vectors \(\mathbf{u}\) with the same value of proj\(_{\mathbf{v}} \mathbf{u}\). Let \(\mathbf{v}=\langle 0,0,1\rangle .\) Give a description of all position vectors \(\mathbf{u}\) such that \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}=\operatorname{proj}_{\mathbf{v}}\langle 1,2,3\rangle\).

Consider the curve \(\mathbf{r}(t)=(\cos t, \sin t, c \sin t),\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. It can be shown that the curve lies in a plane. Prove that the curve is an ellipse in that plane.

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w}\)

Suppose water flows in a thin sheet over the \(x y\) -plane with a uniform velocity given by the vector \(\mathbf{v}=\langle 1,2\rangle ;\) this means that at all points of the plane, the velocity of the water has components \(1 \mathrm{m} / \mathrm{s}\) in the \(x\) -direction and \(2 \mathrm{m} / \mathrm{s}\) in the \(y\) -direction (see figure). Let \(C\) be an imaginary unit circle (that does not interfere with the flow). a. Show that at the point \((x, y)\) on the circle \(C\) the outwardpointing unit vector normal to \(C\) is \(\mathbf{n}=\langle x, y\rangle\) b. Show that at the point \((\cos \theta, \sin \theta)\) on the circle \(C\) the outward-pointing unit vector normal to \(C\) is also $$ \mathbf{n}=\langle\cos \theta, \sin \theta\rangle $$ c. Find all points on \(C\) at which the velocity is normal to \(C\). d. Find all points on \(C\) at which the velocity is tangential to \(C\). e. At each point on \(C\) find the component of \(v\) normal to \(C\) Express the answer as a function of \((x, y)\) and as a function of \(\theta\) f. What is the net flow through the circle? That is, does water accumulate inside the circle?

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Assuming the curve lies in a plane, show that it is a circle centered at the origin with radius \(R\) provided \(a^{2}+c^{2}+e^{2}=b^{2}+d^{2}+f^{2}=R^{2}\) and \(a b+c d+e f=0\).
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