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C3 - Parametric Equations - Help

20160221_192846-1.jpg
Here is the question, I don't know why but I don't understand this chapter, if anyone can solve this with working, it would be greatly appreciated.
Thanks
Reply 1
Avatar for KaylaB
KaylaB
20
Original post by BroodiestPear
20160221_192846-1.jpg
Here is the question, I don't know why but I don't understand this chapter, if anyone can solve this with working, it would be greatly appreciated.
Thanks


So first you would need to create the cartesian equation, so that's y=...
You can do this a numbers of ways but I usually do it by rearranging one equation to get t=... and substituting the value for t in to the other equation so you're dealing solely with x and ys. From there you can then rearrange to make y=... and then differentiate
for part a) you divide dy/dt by dx/dt
Original post by BroodiestPear
20160221_192846-1.jpg
Here is the question, I don't know why but I don't understand this chapter, if anyone can solve this with working, it would be greatly appreciated.
Thanks


Find dx/dt (and hence dt/dx) and dy/dt
Then dy/dx = dy/dt x dt/dx


Posted from TSR Mobile
dy/dt = 6 sec^2(t); dx/dt = 3(2cost)(-sint)=-6sintcost -> dt/dx = -1/6*sect*cosect
Thus dy/dx = 6sec^2(t) * (-1/6)*sect*cosect = -sec^3t*cosect, as required.
Reply 5
Avatar for KaylaB
KaylaB
20
Original post by the bear
for part a) you divide dy/dt by dx/dt


Original post by gdunne42
Find dx/dt (and hence dt/dx) and dy/dt
Then dy/dx = dy/dt x dt/dx


Posted from TSR Mobile

That's a much quicker and nicer way of doing it, I'd never been taught to do it like that :h:
Reply 6
Avatar for TeeEm
TeeEm
19
Original post by KaylaB
That's a much quicker and nicer way of doing it, I'd never been taught to do it like that :h:


the only sensible way ...
It's just using the fact that d/dx(f(g(x))=df(x)/dg(x)*dg(x)/dx
Reply 8
Avatar for Zacken
Zacken
22
Original post by constellarknight
...


Full solutions are against forum guidelines, by the way.
Original post by Zacken
Full solutions are against forum guidelines, by the way.


Alright, whatever.
Im trying to dx/dt but I get -3(sin2t)
can someone explain the steps ? Sorry:ashamed2:
Reply 11
Avatar for Zacken
Zacken
22
Original post by BroodiestPear
Im trying to dx/dt but I get -3(sin2t)
can someone explain the steps ? Sorry:ashamed2:


x=3cos⁡2t=3(cos⁡t)2\displaystyle x = 3\cos^2 t = 3 (\cos t)^2, so:

dxdt=3×2×ddt(cos⁡t)×(cos⁡t)2−1\displaystyle \frac{dx}{dt} = 3 \times 2 \times \frac{d}{dt}\left(\cos t\right) \times (\cos t)^{2-1}

Which is basically the same thing as yours. It's just that it's easier to get the show that using the form above.
(edited 8 years ago)
-3sin2t is an equivalent form of what I got, -6sintcost. It's just that it's easier to get to the required expression using -6sintcost.
Oh, I see it now :smile: Thanks for the help

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