Justify that you have found the requested point by analyzing an appropriate derivative. x=2sin⁡t, y=cos⁡t,...

Step 1 Since we need to find the lowest point of the curve, we need to find the minimum of the function $x=2\sin t$ where $0\le t\le \pi$ We know that function x has a relative maximum at point $t=t_0$ if we have that $\frac{dx}{dt}(t_0)=0$ $\frac{dx}{dt}>0$ 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 $\frac{dx}{dt}=2\cos t\Rightarrow \frac{dx}{dt}(\frac{\pi }{2})=0$ Because we know that $\cos (\frac{\pi }{2})=0$. Since we know that $\cos t>0$ for $0\le t\le \frac{\pi }{2}$ and $\cos t<0$ for $\frac{\pi }{2}<t<\pi$, 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 $\frac{\pi }{2}<t\le \pi$. According to the previous results, we can conclude that the function $x=2\sin t$, where $0\le t\le \pi$, has the maximum at point $t=\frac{\pi }{2}$. ince in this exercise we have that $y=\cos t$, where $0\le t\le \pi$, then we have that $x(\frac{\pi }{2})=2\sin (\frac{\pi }{2})=2\times 1=2$ $y(\frac{\pi }{2})=\cos (\frac{\pi }{2})=0$ Which means that the rightmost point of the curve is equal to $(2,0)$