Polarization online hw help

**fogel1497** · October 27th 2008, 10:32 AM

Use the given equation to answer the following questions. (Enter your answers from smallest to largest.)
r2 = sin(2θ)
0 ≤ θ < π

(a) Find the points on the given curve where the tangent line is horizontal.

(b) Find the points on the given curve where the tangent line is vertical.

**Soroban** · October 27th 2008, 11:50 AM

Hello, fogel1497!

We are expected to know (or be able to derive) this formula:

. . $\frac{dy}{dx} \;=\;\frac{r\cos\theta + r'\sin\theta}{-r\sin\theta + r'\cos\theta}$ .[1]

Given: . $r^2 \:=\:\sin2\theta\quad 0 \leq \theta < 2\pi$

Differentiate implicitly: . $2r\!\cdot\!r' \:=\:2\cos2\theta \quad\Rightarrow\quad r' \:=\:\frac{\cos2\theta}{r}$ .[2]

Substitute [2] into [1]: . $\frac{dy}{dx} \;=\;\frac{r\cos\theta + \left(\frac{\cos2\theta}{r}\right)\sin\theta}{-r\sin\theta + \left(\frac{\cos2\theta}{r}\right)\cos\theta}$

Multiply top and bottom by $r\!:\;\;\frac{dy}{dx} \;=\;\frac{r^2\cos\theta + \sin\theta\cos2\theta}{-r^2\sin\theta + \cos\theta\cos2\theta}$

Since $r^2 = \sin2\theta$ we have: . $\frac{dy}{dx} \;=\;\frac{\sin2\theta\cos\theta + \sin\theta\cos2\theta}{-\sin2\theta\sin\theta + \cos\theta\cos2\theta}$ . $= \;\frac{\sin(2\theta + \theta)}{\cos(2\theta + \theta)}$

. . and we have: . $\frac{dy}{dx} \;=\;\frac{\sin3\theta}{\cos3\theta}$

(a) Find the points on the curve where the tangent is horizontal.

The tangent line is horizontal when $\frac{dy}{dx} \,=\,0$

So we have: . $\sin3\theta \:=\:0\quad\Rightarrow\quad 3\theta \:=\:0,\:\pi,\:2\pi,\:3\pi,\:4\pi,\:5\pi$

Horizontal tangents when: . $x\;=\;0,\:\frac{\pi}{3},\:\frac{2\pi}{3},\:\pi,\:\ frac{4\pi}{3},\:\frac{5\pi}{3}$

(b) Find the points on the curve where the tangent is vertical.

The tangents line is vertical when $\frac{dy}{dx}$ is undefined.

So we have: . $\cos3\theta \:=\:0 \quad\Rightarrow\quad 3\theta \;=\;\frac{\pi}{2},\:\frac{3\pi}{2},\:\frac{5\pi}{ 2},\:\frac{7\pi}{2},\:\frac{9\pi}{2},\:\frac{11\pi }{2}$

Vertical tangents when: . $x \;=\;\frac{\pi}{6},\:\frac{\pi}{2},\:\frac{5\pi}{6 },\:\frac{7\pi}{6},\:\frac{3\pi}{2},\:\frac{11\pi} {6}$

**fogel1497** · October 27th 2008, 05:56 PM

Thank you, now that you explained it i was able to complete the rest of my homework. i missed this day in class, this stuff is actually a lot easier then the trig sub and other stuff we were doing before. thanks a lot! +rep

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