[chemicalX]

Hi im doing the Jan 10 pastpaper for OCR Core 4 and ive gotten stuck on Q6,
ive gotten to the part where ive got dx/dt and dy/dt

dx/dt= 9-1/t and dy/dt= 3t² -3/t

then i flipped dx/dt to give dt/dx so i can cancel the dt's and get dy/dx
for dy/dx i got = 3t²/9 -3/9t -3t³ -3??

is that right for dy/dx and what shall i do know?

question is:

A curve has parametric equations
x = 9t − ln(9t), y = t3 − ln(t3).
Show that there is only one value of t for which
dy
dx = 3 and state that value.

thanks

[xkakesx]

I don't know how to do the method you're using, and ignore this if it confuses you. But I always do dy/dx = dy/dt divided by dx/dt

It's a lot simpler, and if you're on OCR you get credit for either method

[chemicalX]

sorry i dont really get that, thanks anyway

[chemicalX]

anyone? please help

[ghostwalker]

(Original post by chemicalX)
Hi im doing the Jan 10 pastpaper for OCR Core 4 and ive gotten stuck on Q6,
ive gotten to the part where ive got dx/dt and dy/dt

dx/dt= 9-1/t and dy/dt= 3t² -3/t

then i flipped dx/dt to give dt/dx so i can cancel the dt's and get dy/dx
for dy/dx i got = 3t²/9 -3/9t -3t³ -3??

is that right for dy/dx and what shall i do know?

thanks

Your value of dy/dx is incorrect.

From:





I'd start by placing each derivative over a common denominator (i.e. t), and then combine them as previously.

[Xarren]

You'd get more help if you posted the actual question.

I'd happily help you with it, but I'm not going to go looking for it, especially since I haven't done OCR.

[chemicalX]

(Original post by Xarren)
You'd get more help if you posted the actual question.

I'd happily help you with it, but I'm not going to go looking for it, especially since I haven't done OCR.

question is:

A curve has parametric equations
x = 9t − ln(9t), y = t3 − ln(t3).
Show that there is only one value of t for which
dy
dx = 3 and state that value.

thanks

[chemicalX]

(Original post by ghostwalker)
Your value of dy/dx is incorrect.

From:





I'd start by placing each derivative over a common denominator (i.e. t), and then combine them as previously.

sorry i dont get how to do that?

[chemicalX]

(Original post by ghostwalker)
Your value of dy/dx is incorrect.

From:





I'd start by placing each derivative over a common denominator (i.e. t), and then combine them as previously.

oh wait.. so for dx/dt = 9t-1 and dy/dt = 3t³ -3?

[ghostwalker]

(Original post by chemicalX)
sorry i dont get how to do that?

dx/dt= (9t-1)/t

dy/dt=(3t^3-3)/t


Now divide. Aside from the denominators cancelling, that's all you can really do with it.

Once you have dy/dx, set it equal to 3, and solve for t.

[tom28]

Can't you just make your equation for dy/dx equal to 3 and then solve to get t?

[chemicalX]

awesome thanks guys