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C4 - Differentiating Parametric Equations

The question is:

Find dydx \frac{dy}{dx} for the following parametric equations:

x=12t−1x= \frac{1}{2t-1}y=t22t−1y= \frac{t^2}{2t-1}


First I rearranged them as

x=(2t−1)−1x=(2t-1)^{-1} and y=t2(2t−1)−1y=t^2(2t-1)^{-1}

and then I differentiated them each using the chain and product rules to get

dxdt=−2(2t−1)−2 \frac{dx}{dt}=-2(2t-1)^{-2} and dydt=−2t2(2t−1)−2+2t(2t−1)−1 \frac{dy}{dt}=-2t^2(2t-1)^{-2}+2t(2t-1)^{-1}

Then I divided the second equation by the first and ended up with a messy equation which I couldn't seem to simplify to anything near the correct answer, which is t(1−t)t(1-t) according to the book.

Can anyone point out where I went wrong? I'm not sure if I went about differentiating them the wrong way or if I just didn't simplify the final answer right.
Try differentiating with the quotient rule :smile:

Edit: No indices confusion
(edited 11 years ago)
Reply 2
Avatar for msmith2512
msmith2512
1
Your method is right and your derivative looks right.

I would leave them as:

dx/dt= -2/(2t-1)^2 and put dy/dt in the same form i.e. a fraction over (2t-1)^2 and then do the division.
It would be much easier to divide and simplify if you were to express dydt\frac{dy}{dt} as a single fraction
Right, I simplified the differentiated y equation into one fraction:

dydt=−2t(2t−1)2+2t(2t−1) \frac{dy}{dt}= \frac{-2t}{(2t-1)^2}+ \frac{2t}{(2t-1)}

=−2t(2t−1)2+4t2(2t−1)2= \frac{-2t}{(2t-1)^2}+ \frac{4t^2}{(2t-1)^2}

=2t2(2t−1)2= \frac{2t^2}{(2t-1)^2}

then I divided that by the x equation to get

2t2(2t−1)2×(2t−1)2−2 \frac{2t^2}{(2t-1)^2} \times \frac{(2t-1)^2}{-2}

=2t2−2= \frac{2t^2}{-2}

=−t2=-t^2

That's still not correct though according to the book. I've been looking for any silly mistakes I've made but I can't see any. Any more advice?
Original post by HandmadeTurnip
Right, I simplified the differentiated y equation into one fraction:

dydt=−2t(2t−1)2+2t(2t−1) \frac{dy}{dt}= \frac{-2t}{(2t-1)^2}+ \frac{2t}{(2t-1)}

=−2t(2t−1)2+4t2(2t−1)2= \frac{-2t}{(2t-1)^2}+ \frac{4t^2}{(2t-1)^2}

=2t2(2t−1)2= \frac{2t^2}{(2t-1)^2}

then I divided that by the x equation to get

2t2(2t−1)2×(2t−1)2−2 \frac{2t^2}{(2t-1)^2} \times \frac{(2t-1)^2}{-2}

=2t2−2= \frac{2t^2}{-2}

=−t2=-t^2

That's still not correct though according to the book. I've been looking for any silly mistakes I've made but I can't see any. Any more advice?


2nd line doesn't follow on from the first. And the 3rd doesn't follow on from your current second.

2t(2t−1)≠4t2(2t−1)2\dfrac{2t}{(2t-1)} \neq \dfrac{4t^2}{(2t-1)^2}
Reply 6
Avatar for leo99756
leo99756
4
Original post by HandmadeTurnip
Right, I simplified the differentiated y equation into one fraction:

dydt=−2t(2t−1)2+2t(2t−1) \frac{dy}{dt}= \frac{-2t}{(2t-1)^2}+ \frac{2t}{(2t-1)}

=−2t(2t−1)2+4t2(2t−1)2= \frac{-2t}{(2t-1)^2}+ \frac{4t^2}{(2t-1)^2}

=2t2(2t−1)2= \frac{2t^2}{(2t-1)^2}





I think you mistook -2t for -2t^2 here.



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Original post by hassi94
2nd line doesn't follow on from the first. And the 3rd doesn't follow on from your current second.

2t(2t−1)≠4t2(2t−1)2\dfrac{2t}{(2t-1)} \neq \dfrac{4t^2}{(2t-1)^2}


Ah, thank you. changing that to 2t(2t−1)(2t−1)2\frac{2t(2t-1)}{(2t-1)^2} gives me the right answer.

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