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<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

How do I differentiate these parametric functions? FP1

y=2a1/2x1/2y = 2a^{1/2}x^{1/2}

y=c2xy = \frac{c^2}{x}
Reply 1
Avatar for .raiden.
.raiden.
10
For the first multiply by power of x then take 1 away from power

For second rewrite as this

y=c2x−1y=c^2x^{-1}

Then try differentiating.
Reply 2
Avatar for lizz-ie
lizz-ie
OP
18
Original post by raiden95
For the first multiply by power of x then take 1 away from power

For second rewrite as this

y=c2x−1y=c^2x^{-1}


y=−c2x−2y = -c^2x^{-2}

So this? y=−c2x2y = \frac{-c^2}{x^2}
Reply 3
Avatar for Lunch_Box
Lunch_Box
15
Differentiate normally leaving the constants a and c unchanged.
Reply 4
Avatar for Lunch_Box
Lunch_Box
15
Original post by lizz-ie
y=−c2x−2y = -c^2x^{-2}

So this? y=−c2x2y = \frac{-c^2}{x^2}


Yep
Reply 5
Avatar for .raiden.
.raiden.
10
Original post by lizz-ie
y=−c2x−2y = -c^2x^{-2}

So this? y=−c2x2y = \frac{-c^2}{x^2}


dydx=−c2x2\frac{dy}{dx} = \frac{-c^2}{x^2}

Yes, but don't forget the dy/dx :smile:
Reply 6
Avatar for lizz-ie
lizz-ie
OP
18
Original post by Lunch_Box
Differentiate normally leaving the constants a and c unchanged.


So y=a1/2x−1/2y = a^{1/2}x^{-1/2} ?
Reply 7
Avatar for .raiden.
.raiden.
10
Original post by SheldonWannabe
If you multiplied the whole thing by x and took away the power, you would be left with the same thing, just with yx on the left hand side. So that is wrong...
you simply move the power of x to the coefficient and minus one from the power of x, just differentiate like you would with any other equation.


That is exactly what I mean ? The OP clearly understood.

I didn't say multiply whole thing by x, i said multiply by power of x
(edited 11 years ago)
Reply 8
Avatar for lizz-ie
lizz-ie
OP
18
Original post by raiden95
dydx=−c2x2\frac{dy}{dx} = \frac{-c^2}{x^2}

Yes, but don't forget the dy/dx :smile:


Thank you so much! :smile:
Reply 9
Avatar for .raiden.
.raiden.
10
Original post by lizz-ie
So y=a1/2x−1/2y = a^{1/2}x^{-1/2} ?


Yes thats correct but with dy/dx
Original post by lizz-ie
So y=a1/2x−1/2y = a^{1/2}x^{-1/2} ?


Fast learner
Original post by raiden95
That is exactly what I mean ? The OP clearly understood.

I didn't say multiply whole thing by x, i said multiply by power of x

Haha oh damn yeah, I read that completely wrong... Now I feel stupid :tongue:
Reply 12
Avatar for Ateo
Ateo
9
Just a little note, these are not parametric equations. They are Cartesian equations of the parabola and the rectangular hyperbola respectively.
Reply 13
Avatar for Giveme45
Giveme45
Original post by Ateo
Just a little note, these are not parametric equations. They are Cartesian equations of the parabola and the rectangular hyperbola respectively.


I was also confused about that.
Reply 14
Avatar for davros
davros
16
Original post by Ateo
Just a little note, these are not parametric equations. They are Cartesian equations of the parabola and the rectangular hyperbola respectively.


If we're being really pedantic, the first equation gives the upper branch of the parabola because of the convention that we take positive square roots to define a function.

The alternative form y2=4axy^2 = 4ax gives both branches of the parabola.

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