# P3 - Parametric Integration ... ?

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16 years ago
#1
I think I've forgotten something here, but is there such a thing as Parametric Integration ?

It doesn't stick out in my notes, and i'm probably missing something glaringly obvious, but i'm stuck here ...

y = 5Cos(t)
x = 4Sin(t)

find area under curve between 0 & pi/4

Cheers, Rob
0
16 years ago
#2
I'll give it a go mate:

y = 5Cos(t)
x = 4Sin(t)

Find area under curve between 0 & pi/4
------------------------------------------------------------
ok, area is integral[y dx] between 0 and pi/4
=integral[5Cos(t) dx]

We can't integrate t with respect to x, so we do the following, and differentiate x:

x=4sin(t)

dx/dt=4Cos(t)

dx=4Cos(t) dt

If we substitute this back into the integral we have

area=integral[5Cos(t)*4Cos(t) dt]=int[20Cos^2(t) dt]

Now I don't know whether or not to change the limits, as I am not entirely sure if you have stated the limits for x or t. I'll assume t as that looks easier (lazy me).

rearranging the integral, area =int[10 + 10cos(2t) dt]
which equals 10t + 5sin(2t) + C

assuming your limits were for t we have

5/2(pi)+5sin(pi) - (0+5sin(o))

=5/2(pi)

I hope [assuming of course, those limits were for t]
Im also not so confident if I plugged in the differential of x correctly. Perhaps someone can help?

Chris
0
16 years ago
#3
Yep i think there is such thing as parametric integration in P3. I agree with Chris too. Got 5(pi)/2 Good luck.
0
16 years ago
#4
Thanks a lot .. immediately your post sent bells a chiming in my head, and also the thought of my tutors face as I asked yet another one of "those questions" that I really, *really* should know.

y = f(t);
x = g(t);

Int[y]dx = Int[f(t) dx/dt] dt

looks kinda familiar ish, well, thats what I seem to get :]

Thanks

Rob
0
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