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# Find dy/dx in terms of t watch

1. x= t^2 -2t

y= t^4- 4t

The answer is 2 ( t^2+ t+1) How???
2. (Original post by APersonYo)
x= t^2 -2t

y= t^4- 4t

The answer is 2 ( t^2+ t+1) How???
What did you get when you tried it? Please post your working.
3. (Original post by notnek)
What did you get when you tried it? Please post your working.
x= t^2 -2t

dx/dt= 2t-2

therefore dt/dx= 1/ (2t-2)

y= t^4- 4t

dy/dt= 4t^3 -4

threfore dy/dx= 4t^3-4/ 2t-2

And my answer is wrong lol
4. (Original post by APersonYo)
x= t^2 -2t

dx/dt= 2t-2

therefore dt/dx= 1/ (2t-2)

y= t^4- 4t

dy/dt= 4t^3 -4

threfore dy/dx= 4t^3-4/ 2t-2

And my answer is wrong lol

Either factorise or if you can't do that then you could use long division to divide the denominator into the numerator.
5. (Original post by APersonYo)
x= t^2 -2t

dx/dt= 2t-2

therefore dt/dx= 1/ (2t-2)

y= t^4- 4t

dy/dt= 4t^3 -4

threfore dy/dx= 4t^3-4/ 2t-2

And my answer is wrong lol
not wrong, just unfinished:

4t^3-4 / 2t-2 = 4(t^3-1)/2(t-1)
= 2(t^3-1)/(t-1)
= 2(t-1)(t^2+t+1)/(t-1)
= 2(t^2+t+1)
6. (Original post by APersonYo)
x= t^2 -2t

dx/dt= 2t-2

therefore dt/dx= 1/ (2t-2)

y= t^4- 4t

dy/dt= 4t^3 -4

threfore dy/dx= 4t^3-4/ 2t-2

And my answer is wrong lol
7. (Original post by harrycompsci)
not wrong, just unfinished:

4t^3-4 / 2t-2 = 4(t^3-1)/2(t-1)
= 2(t^3-1)/(t-1)
= 2(t-1)(t^2+t+1)/(t-1)
= 2(t^2+t+1)
Thank you.

I also have another question:

x= cos t

y= t(4-sint)

I got dt/dx = 1/ -sin t

dy/dt= 4-tcos t

dy/dx= 4-tcos t / -sin t

The answer is dy/ dx= t cost t - 4 cosec t +1

How???
8. (Original post by APersonYo)
Thank you.

I also have another question:

x= cos t

y= t(4-sint)

I got dt/dx = 1/ -sin t

dy/dt= 4-tcos t

dy/dx= 4-tcos t / -sin t

The answer is dy/ dx= t cost t - 4 cosec t +1

How???
I think your dy/dt term is incorrect. try the product rule again
9. (Original post by ewanwest)
i think your dy/dt term is incorrect. Try the product rule again
thank you so much!!!

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Updated: January 25, 2017
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