Implict Differentation: 2*y^3+y^2-y^5=x^4-2*x^3+x^2

[SUNDRAGON06]

I'm having trouble with implicit differentiation.
We are told to calculate dy/dx and then state the coordinates I think I have dy/dx but I'm so confused my brain hurts.
The equation is

. . .2*y^3 + y^2 - y^5 = x^4 - 2*x^3 + x^2

Here is my work

. . .6y^2*(dy/dx) + 2y*(dy/dx) - 5y^4*(dy/dx) = 4x^3 - 6x^2 + 2x

. . .(dy/dx)(6y^2 + 2y - 5y^4) = 4x^3 - 6x^2 + 2x

. . .dy/dx = (4x^3 - 6x^2 + 2x) / (6y^2 + 2y - 5y^4)

. . .dy/dx = 2x(2x^2 - 3x + 1) / y(6y + 2 - 5y^3)

Then for the finding the coordinaes part I have set the numerator equal to zero and got three x values -- x = 0, x = 1/2, and x = 1 -- but I do not know how to get the Y values.

Could someone help me PLEASE.

 

[skeeter]

your derivative is fine ... are you supposed to find points on the curve where the curve has horizontal tangent lines?

if so, substitute each value of x that makes the derivative = 0 back into the the original equation of the curve ... then solve for y.

for instance ... for x = 0

2*y^3 + y^2 - y^5 = x^4 - 2*x^3 + x^2

2*y^3 + y^2 - y^5 = 0

y^2(2y + 1 - y^3) = 0

first factor ... y^2 = 0, y = 0

second factor ... y^3 - 2y - 1 = 0

one rational root is y = -1 ...

do some synthetic division to find the other roots ...

-1].........1...........0.........-2.........-1
.......................-1..........1..........1
-------------------------------------------------
............1..........-1.........-1..........0

the depressed polynomial is y^2 - y - 1

so ...

y^3 - 2y - 1 = (y + 1)(y^2 - y - 1) = 0

use the quadratic formula to find the othet two roots ...

y = [1 +/- sqrt(5)]/2

so, three points on the curve for x = 0 that have horizontal tangent lines ...

(0,-1) (0, [1 + sqrt(5)]/2), (0, [1 - sqrt(5)]/2)

now do the same for x = 1/2 and x = 1.

remember that the hardest part about calculus is the algebra.

 

[SUNDRAGON06]

I'm almost following but something isn't clicking. Where/how did you get the rational root?

 

[skeeter]

I'm almost following but something isn't clicking. Where/how did you get the rational root?

rational root theorem.

 

[SUNDRAGON06]

ok I did
2y+1-y^3=0
2y-y^3=-1
y(2-y^2)=-1
Y=-1
and
(2-y^2)=-1
-y^2=-3
y^2=3
y=+sqroot3,-sqroot3

and it looks like you are putting a 5 where my three is and you got a devide by two, so is my way compleatly wrong and I need to do it yours or what

 

[skeeter]

ok I did
2y+1-y^3=0
2y-y^3=-1
y(2-y^2)=-1
Y=-1
and
(2-y^2)=-1
-y^2=-3
y^2=3
y=+sqroot3,-sqroot3

sorry, but your method is completely wrong ... any polynomial of degree 2 or larger must be set equal to 0 to be solved.

I also recommend that you "google" the rational root theorem to research how it works.

In the meantime, I recommend you get out your calculator and calculate the zeros of 2y + 1 - y^3 = 0 by graphing.