so I have the following parametric curve:

enter image description here

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:$\int [\sqrt{1+f'(x)^2}]$.

  1. 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.

  2. 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)

  • $\begingroup$ I don't understand ypur point for question "2". We have a parametrization $(x(t),y(y))$, what are the functions $f(x)$ and $g(x)$ you are referring to? $\endgroup$ – user May 28 '18 at 22:42
  • $\begingroup$ Well lets say we have to integrate the above parametric equation to find the AREA. The normal way would be taking f(x) - g(x) or g(x)-f(x) (who ever has the higher y value for the given limits) and integrate it at the limits. How would we go on about this when looking at parametric equations? Would we do x(t)-y(t) or rather y(t)-x(t)..... $\endgroup$ – user472288 May 28 '18 at 22:44
  • $\begingroup$ The area of what, pray tell? $\endgroup$ – amd May 28 '18 at 22:48
  • $\begingroup$ Lets say you graph the parametric equation and you want to find the area of say x=1 to x=2. How would we go on about that. I don't understand why the question seems so complicated. $\endgroup$ – user472288 May 28 '18 at 22:48

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

$$L=\int_0^{\pi/6} \sqrt{[x'(t)]^2+[y'(t)]^2}dt$$


  • $x(t)=\cos (2t)-\sin t \implies x'(t)=-2\sin (2t)-\cos t=-4\sin t\cos t-\cos t$

  • $y(t)=2\sin t \implies y'(t)=2\cos t$

To find the area under the curve $(x(t), y(t))$ we can use the following

$$A=\int_0^1 ydx=\int_0^{1} y(t)\,x'(t)\,dt$$

  • $\begingroup$ Hey again gimusi, thanks for the answer but as you can see in my post, I mentioned that I would really like some clarifications/answers and no hints... would appreciate some explanations, I already know the formula. $\endgroup$ – user472288 May 28 '18 at 22:35
  • $\begingroup$ @user472288 Your formula is wrong you don't have considered the square root. $\endgroup$ – user May 28 '18 at 22:36
  • $\begingroup$ Oh yeah I just forgot to add it as I was focused on the mathjax. $\endgroup$ – user472288 May 28 '18 at 22:36
  • $\begingroup$ Gimusi I hope I'm not bothering you, but I have to ask. You set 2sin(t)=0 and 2sin(t)=1 to find that it results in 0 and in pi/6. Usually we are looking for (x) limits, why are we evaluating y here? $\endgroup$ – user472288 May 28 '18 at 23:01
  • $\begingroup$ @user472288 Because in this case the expression for $y(t)$ is simpler, of course we have also to che the corresponding values fo $x(t)$. $\endgroup$ – user May 28 '18 at 23:08

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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