Respuesta :
Answer:
[tex](e^{\frac{3\pi}{4}},\frac{3\pi}{4}), (e^{\frac{7\pi}{4}},\frac{7\pi}{4})[/tex].
Step-by-step explanation:
REcall that given a parametric curve x(t),y(t), the tangent's line slope is given by
[tex] \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
To find where the tangent line is horizontal, we must find where the slope is 0. That is, to find the values of t for which [tex]\frac{dy}{dt} =0[/tex] and [tex]\frac{dx}{dt}\neq0[/tex] at the same time.
Consider the polar curve [tex]r=e^{\theta}[/tex]. If we use the polar coordinates, we have that [tex]x = r\cos(\theta) = e^{\theta}\cos(\theta), y = r\sin(\theta) = e^{\theta}\sin(\theta)[/tex], which gives us a parametric curve with parameter [tex]\theta[/tex]. So, let us use the above to find the desired points.
We have that
[tex]\frac{dy}{dt} = e^{\theta}(\sin(\theta)+\cos(\theta))[/tex]
[tex]\frac{dx}{dt} = e^{\theta}(\cos(\theta)-\sin(\theta))[/tex]
Recall that the function [tex]e^\theta[/tex] is never 0, so, for us to have the derivative of y to be 0, we must have that [tex]\sin(\theta) = -\cos(\theta)[/tex]. Note that if this happens, the derivative of x is different from 0. So, we must solve the following equation in the interval [tex]\theta \in [0,2\pi][/tex].
[tex] \sin(\theta) = -\cos(\theta)[/tex], which is equivalent to [tex] \tan(\theta) = -1[/tex]. Which gives us [tex]\theta=-\frac{\pi}{4}[/tex]. This solution is out of our desired interval, then , using the fact that tangent is a periodic function with period pi, we can find the solutions in the desired interval by adding multiples of pi. Thus, the desired solutions are
[tex]\theta_1 = \frac{-\pi}{4}+\pi = \frac{3\pi}{4}, \theta_2 = \frac{-\pi}{4}+2\pi = \frac{7\pi}{4}[/tex]
Note that [tex]e^{\frac{7\pi}{4}} = 244.15[/tex] and [tex]e^{\frac{3\pi}{4}} = 10.55[/tex],so both solutions are inside the restriction.
