so I have the following parametric curve:
Now I'm trying to figure out how I should tackle this, I know the formula for finding lengths is $\int[\sqrt{g'(x)^2+f'(x)^2}]$ or when its a single function:$\int [\sqrt{1+f'(x)^2}]$.
Now, to solve this I need to find the limits first, and I'm really confused as to how to do so. When I try to picture the graph, as stated, at point A it's going to be at $(1,0)$ and B at $(1,0)$. I thought I should do the same as if this were a normal function, and just equal x=1 and y=0 and see where it gets me, but it just confused me further.
Not related to length, but lets say the question would ask for total area, even after we determine the correct integration limits, well normally, to know which function I should integrate first (as in $\int g(x)-f(x)$ or $\int f(x)-g(x)$ I would usually look at the graph and see from there which function has the higher $y$ value, how should I go on about parametric functions?
Would love some clarifications. Thanks in advance.
(Please no hints on this one, would really like some explanation)
$\endgroup$ 42 Answers
$\begingroup$To find the limits let observe that $y(t)=2\sin t$ and thus since in $A$ we have $y=0$ and in $B$ we have $y=1$ it is easy to find that
for $t=0 \implies x(0)=1, y(0)=0$
for $t=\pi/6 \implies x(\pi/6)=0, y(\pi/6)=1$
then the set up for the length is as follow
$$L=\int_0^{\pi/6} \sqrt{[x'(t)]^2+[y'(t)]^2}dt$$
with
$x(t)=\cos (2t)-\sin t \implies x'(t)=-2\sin (2t)-\cos t=-4\sin t\cos t-\cos t$
$y(t)=2\sin t \implies y'(t)=2\cos t$
To find the area under the curve $(x(t), y(t))$ we can use the following
$$A=\int_0^1 ydx=\int_0^{1} y(t)\,x'(t)\,dt$$
$\endgroup$ 5 $\begingroup$I prefer for the integral the Cartesian form of the curve which corresponds to an arc of parabola $$y=\frac{\sqrt{9-8x}}{2}-\frac 12$$ so we have $$\int_0^1\sqrt{\frac{25-8x}{9-8x}}dx=\left[-\frac{\sqrt{25-8x}\sqrt{9-8x}}{8}-\log\left|\frac{\sqrt{9-8x}+\sqrt{25-8x}}{4}\right|\right]_0^1$$
$$\int_0^1\sqrt{\frac{25-8x}{9-8x}}dx\approx 2.2509$$
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