2022-01-17

Justify that you have found the requested point by analyzing an appropriate derivative. $x=2\mathrm{sin}t$,
$y=\mathrm{cos}t$,
$0\le t\le \pi$
Rightmost point.

nick1337

Expert

Step 1 Since we need to find the lowest point of the curve, we need to find the minimum of the function $x=2\mathrm{sin}t$ where $0\le t\le \pi$ We know that function x has a relative maximum at point $t={t}_{0}$ if we have that $\frac{dx}{dt}\left({t}_{0}\right)=0$ $\frac{dx}{dt}>0$ for $t<{t}_{0}$, and $\frac{dx}{dt}<0$ for $t>{t}_{0}$. We also know that if function x has one relative maximum at point $t={t}_{0}$, then function x has the maximum at point $t={t}_{0}$ Using the previous reults, we have that $\frac{dx}{dt}=2\mathrm{cos}t⇒\frac{dx}{dt}\left(\frac{\pi }{2}\right)=0$ Because we know that $\mathrm{cos}\left(\frac{\pi }{2}\right)=0$. Since we know that $\mathrm{cos}t>0$ for $0\le t\le \frac{\pi }{2}$ and $\mathrm{cos}t<0$ for $\frac{\pi }{2}, then using the previous results, we can conclude that $\frac{dx}{dt}>0$ for $0\le t<\frac{\pi }{2}$ and $\frac{dx}{dt}<0$ for $\frac{\pi }{2}. According to the previous results, we can conclude that the function $x=2\mathrm{sin}t$, where $0\le t\le \pi$, has the maximum at point $t=\frac{\pi }{2}$. ince in this exercise we have that $y=\mathrm{cos}t$, where $0\le t\le \pi$, then we have that $x\left(\frac{\pi }{2}\right)=2\mathrm{sin}\left(\frac{\pi }{2}\right)=2×1=2$ $y\left(\frac{\pi }{2}\right)=\mathrm{cos}\left(\frac{\pi }{2}\right)=0$ Which means that the rightmost point of the curve is equal to