So I have the following problem:
Find the the circumference of a melon, if the boundary line of the melon is a double cycloid and following equation describe the boundary line of the melon.
{x(t)=5 (t-sin(t)), y(t)=5(1-cos(t))} and
{x(t)=5 (t-sin(t)), y(t)=-5(1-cos(t))}
I have to give the following:
1) integrand for the one side of the melon: $$\ g(t)= $$ 2)indefinite integral for the half of the melon's circumference: $$\ C(t)= ... + C $$ 3) the circumference of the melon: $$\ C=$$
I don't know how to find the upper and lower bound. So it would be very helpful, if somebody could give me the equations to draw a graph in geogebra or in wolframalpha.
What I have found out: 1) integrand for the one side of the melon: $$\ g(t)=\sqrt{50-50cos(t)} $$
3) the circumference of the melon: it should be $$\ C=80$$, but if I take t from $ 0$ to $ 2\pi $ $\ g(t)=\sqrt{50-50cos(t)} $ the answer comes 0. What is wrong?
This might help, so I am going to put it here. Calculate the length of a cycloid curve with a radius of 5 inches. The cycloid parametric equations are $x(t)=r(t-sin(t))$ and $y(t)=r(1-cos(t))$. $$f(t)=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt=\sqrt{r^{2}(t-sin(t))^{2}+r^{2}(1-cos(t))^{2}}dt$$ $$=r\cdot\sqrt{2-2cos(t)}dt$$ Normally, I would stop at this point and integrate, but this can be beneficially simplified using a trig identity. $cos(\theta)=1-2sin^{2}(\theta/2)$ Factoring the radical gets $\sqrt{2(1-cos(t)).}$ Applying the identity gets $\sqrt{2(1-(1-2sin^{2}(t/2))}=\sqrt{4sin^{2}(t/2)}=2sin(\frac{t}{2})$. $$f(t)=r\cdot2sin(\frac{t}{2})dt$$ This last eqn is a great deal easier to integrate. $$\int_{0}^{2\pi}r\cdot2sin(\frac{t}{2})dt=-r\cdot4cos(\frac{t}{2})\Big|_{0}^{2\pi}=8r$$