Implicit differentiation

Daniel123 · 2008-04-23 05:51:27

Hello. Could someone please check my method for this question please? Particularly the part marked with the star (I'm confident that if this step is correct then the rest is).

"Find the coordinates of the point on the curve

where the curve is parallel to the y-axis"

I then reasoned that in order for the curve to be parallel to the y-axis, the denominator of the gradient function must be equal to 0 (is this legal???

). It then follows that y = x/2. Substiuting y = x/2 into the original equation for the curve gives:

By inspection, x = - 2 is a root. Dividing by (x+2) gives:

, the discriminant of which is -175. This therefore implies that x = 2 is the only (x) point on the curve where the curve is parallel to the y-axis. Substituting x = - 2 back into the original equation gives:

.

The final answer is then

.

Thanks!

Last edited by Daniel123 (2008-04-23 07:05:22)

JaneFairfax · 2008-04-23 06:18:18

Daniel123 wrote:

I then reasoned that in order for the curve to be parallel to the y-axis, the denominator of the gradient function must be equal to 0 (is this legal???
).

You can also differentiate x implicitly with respect to y and set

.

Last edited by JaneFairfax (2008-04-23 06:20:47)

Daniel123 · 2008-04-23 06:20:01

Ahh, that would be nicer. Should have thought of it myself. Thanks Jane.

JaneFairfax · 2008-04-23 06:24:02

No problem. And (−2,−1) looks correct: http://www.mathsisfun.com/graph/functio … =-6&ymax=7.

Daniel123 · 2008-04-23 07:04:45

Great, thanks! I'm looking forward to my C4 exam

Edit: Zooming in on the point (-2,-1) is interesting.

Last edited by Daniel123 (2008-04-23 07:06:43)

John E. Franklin · 2008-04-23 07:41:16

I'm baffled how Jane
got the graphing equation
y = (x +/- sqrt(4*x^3+x^2+28))/2

Daniel123 · 2008-04-23 08:01:34

Dragonshade · 2008-04-23 10:26:53

Or

(x+2)(x-a)(x-b)=0

2-b-a=1/4
2ab=7
ab-2b-2a=0

wow, only one solution

Last edited by Dragonshade (2008-04-23 10:52:24)

John E. Franklin · 2008-04-23 13:30:23

Thanks, I love the complete the square way to separate the xy !

Math Is Fun Forum

#1 2008-04-23 05:51:27

Implicit differentiation

#2 2008-04-23 06:18:18

Re: Implicit differentiation

#3 2008-04-23 06:20:01

Re: Implicit differentiation

#4 2008-04-23 06:24:02

Re: Implicit differentiation

#5 2008-04-23 07:04:45

Re: Implicit differentiation

#6 2008-04-23 07:41:16

Re: Implicit differentiation

#7 2008-04-23 08:01:34

Re: Implicit differentiation

#8 2008-04-23 10:26:53

Re: Implicit differentiation

#9 2008-04-23 13:30:23

Re: Implicit differentiation

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