# Implicit Differentiation Problem

• Feb 23rd 2006, 02:04 PM
Vigo
Implicit Differentiation Problem
Consider the curve given by X^2+4y^2=7+3xy
a) show that dy/dx=3y-2x/8y-3x
b) show that there is a point P with x-cooridnate 3 at which the line tangent to the curve at P is horizontal. Find the y-cooridnate of P.
c)find the value of d^2y/dx^2 at the point P found in part (b). Does the curve have a local maximum, a local minimum, or neither at the point P? Justify your answer.

(a) is easy. All you do is find the derivative.

For (b), I got the point (3,2) by plugging 3 into the original equation and got 2.

For (c), the value I got was -2/7 by finding the second derivative and plugging (3,2) for the x's and y's. I need to know if this is right and if there are any max's or min's at this point. Thanks.
• Feb 23rd 2006, 05:13 PM
ThePerfectHacker
Use the second derivative test. Place that point into the second derivative and see whether it is positive or negative.

If the second derivative is positive then it is a minimum.
If the second derivative is negative then it is a maximum.
If the second derivative is zero the test is inconlusive.

Note, inconclusive means the test cannot be used. Not, that there is no minimum or maximum. If the second derivative is zero then you can use still use the first derivative test.
• Feb 23rd 2006, 07:53 PM
Vigo
Ok, so there is a local maximum at the point (3,2)?
• Feb 24th 2006, 01:20 AM
ticbol
Quote:

Originally Posted by Vigo
Consider the curve given by X^2+4y^2=7+3xy
a) show that dy/dx=3y-2x/8y-3x
b) show that there is a point P with x-cooridnate 3 at which the line tangent to the curve at P is horizontal. Find the y-cooridnate of P.
c)find the value of d^2y/dx^2 at the point P found in part (b). Does the curve have a local maximum, a local minimum, or neither at the point P? Justify your answer.

(a) is easy. All you do is find the derivative.

For (b), I got the point (3,2) by plugging 3 into the original equation and got 2.

For (c), the value I got was -2/7 by finding the second derivative and plugging (3,2) for the x's and y's. I need to know if this is right and if there are any max's or min's at this point. Thanks.

a).
x^2 +4y^2 = 7 +3xy -------(i)

Differentiate with respect to x,
2x +4[2y*y'] = 3[x*y' +y]
2x +8y*y' = 3x*y' +3y
(y')[8y -3x] = 3y -2x
y' = (3y -2x)/(8y -3x) -------shown.

-----------
b).
Point P has x=3. Plug that into (i) to get the y at point P,
3^2 +4y^2 = 7 +3*3*y
9 +4y^2 = 7 +9y
4y^2 -9y +2 = 0
(4y -1)(y -2) = 0
y = 1/4 or 2 ---------***
So P is either (3,1/4) or (3,2).

If P were (3,1/4), plug those into the y',
y' = (3*0.25 -2*3) / (8*0.25 -3*3) = (-5.25)/(-8) ----not zero, so not horizontal tangent line.

If P were (3,2),
y' = (3*2 -2*3)/(8*2 -3*3) = 0/7 = 0 ---zero slope, so horizontal tangent line if P were (3,2).

Therefore, per the problem, point P is (3,2). ----------answer.

---------------------------
c).
The second derivative of y at P(3,2).
y' = (3y -2x)/(8y -3x)
y'' = derivative of y' with respect to x,
so,
y'' = {(8y -3x)*[3y' -2] -(3y -2x)*[8y' -3]} / {(8y -3x)^2}
Since y'=0, x=3, and y=2, then,
y'' = {(8*2 -3*3)[0 -2] -(3*2 -2*3)[0 -3]} / {(8*2 -3*3)^2}
y'' = {(7)[-2] -(0)[-3]} / {(7)^2}
y'' = {-14}/{49}
y'' = -2/7 ----------***

[Hey, you got them all, from a) to c). Very good.]

The second derivative of y at P(3,2) is negative.
That means the concavity of the curve at P(3,2) is downward, so the y'=0 there is for a maximum point.
Therefore, P(3,2) is a maximum point.

Is it a local maximum point? To test that, try to find if there is a y-value more than y=2. If there is, or there are, then P(3,2) is only a local maximum point----at other x's, the y's maybe greater than 2.
If no y greater than 2 could be found, then P(3,2) is "global" or absolute maximum of the curve.
• Feb 24th 2006, 11:16 AM
Vigo
Alright thanks a lot for all of your help.