You are given the polar curve r = e^θ
List all of the points (r,θ) where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of r and limit yourself to 1<_r<_1000 ( note the restriction on r) and 0≤θ<2????. If two or more points share the same value of r, list those starting with the smallest value of θ. If any blanks are unused, type an upper-case "N" in them.
Point 1: (r, θ) = ________
Point 2: (r, θ) = ________
Point 3: (r, θ) = ________

Respuesta :

The tangent to [tex]r(\theta)=e^\theta[/tex] has slope [tex]\frac{\mathrm dy}{\mathrm dx}[/tex], where

[tex]\begin{cases}x(\theta)=r(\theta)\cos\theta\\y(\theta)=r(\theta)=\sin\theta\end{cases}[/tex]

By the chain rule, we have

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}[/tex]

and by the product rule,

[tex]\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r(\theta)\sin\theta[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r(\theta)\cos\theta[/tex]

so that with [tex]\frac{\mathrm dr}{\mathrm d\theta}=e^\theta[/tex], we get

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}=-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}[/tex]

The tangent line is horizontal when the slope is 0; this happens for

[tex]-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}=0\implies\sin(2\theta)=-1\implies2\theta=-\dfrac\pi2+2n\pi\implies\theta=-\dfrac\pi4+n\pi[/tex]

where [tex]n[/tex] is any integer. In the interval [tex]0\le\theta\le2\pi[/tex], this happens for [tex]n=1,2[/tex], or

[tex]\theta=\dfrac{3\pi}4\text{ and }\theta=\dfrac{7\pi}4[/tex]

i.e at the points

[tex](r,\theta)=\left(e^{3\pi/4},\dfrac{3\pi}4\right)[/tex]

and

[tex](r,\theta)=\left(e^{7\pi/4},\dfrac{7\pi}4\right)[/tex]

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