Calculus help

webby

New Member
I'm trying to help my cousin with calculus :confused:
Does anyone remember a way to solve the following:
2y^3 + 6y - 1=0 ?????
I know if it was a y-squared term you could plug it into the quadratic formula or whatever it is called, but it is a y-cubed term.

tia

Don't work it till your brain is numb, mine is numb enough for all of us.

Webby
 
calc

can't really remember, but i think we used to use the del Ferro formulas for cubics, but i don't remember what they were
 
What is the problem asking for? It appears that your taking Cal. 1?

What I did with the problem was derive to respect of Y, assuming that the problem ask to solve to simpliest term!

Therefore:
2y^3 + 6y - 1 = 0
= -2y^3 - 6y + 1 = 0
= -6y^2 - 6 = 0
= -6(y^2 - 1) = 0
= -6(y+1)*(y-1) = 0

Might be it! That's all I can think of.

Ted
 
I will have to double check the original equation, but we had to derive implicitly and then we were supposed to find the points that had horizontal tangent lines (points where the derivitave = 0). I will find out the original equation, but it was something similar to the following:
xy^2+x^3y+3x=6
That is not the equation, but it was that type of equation.
My cousin's high school Calculus teacher seems to be dropping the ball, as most of the class is doing very poorly. I'm trying to help keep her grade up long enough to get to the semester break when she can drop the class.

I will get back to you.

Webby
 
Okay, here's the original problem.

2y^3 + 6x^2y - 12x^2 + 6y = 1

Find an equation for each horizontal tangent line to the above curve.

And then find the point (x,y) where a line through the origin with a slope of -1 is tangent to the curve.

We seem to be getting nowhere fast and are looking for ideas.
 
Your original equation is confusing. Is this equivalent?

2*y^3 + 6*x^2*y - 12*x^2 + 6*y = 1 <--not too difficult

Or this?

2*y^3 + 6*x^(2*y) - 12*x^2 + 6*y = 1 <--more difficult
 
2y^3 + 6y - 1=0
has no solution? just by looking at it - is there any number you can put in y's place to make that equation true? cubic will retain the positive/negative aspect, so for example if y=-1, then you would have 2y^3 + 6y - 1 = 2(-1)^3 + 6(-1) -1 = -2 -6 -1 = -9 which is not equal to 0.

The second equation you put up does have a solution... I can't remember how to do it, but I would try to factor out a (y-1) and see what happens, since eyeballing it, y=1 will make x=7/6 i think. Good luck! If I can find my old calculus book and get the dust off, I'll post again...
 
You need to find dy/dx; by far the easiest way is using implicit differentiation. So:

2y^3+6x^2*y-12x^2+6y=1

Using implicit differentiation and the product rule on the 6x^2*y term, we have:

6y^2(dy/dx) +12xy+6x^2(dy/dx)-24x+6(dy/dx)=0

Solving for dy/dx: dy/dx= (4x-2xy)/(y^2+x^2+6)

Now set dy/dx=0 and solve to find your tangent points. For the 2nd part, find the equation of a line through the origin w/slope -1 (y=-x), set it equal to dy/dx, and solve.
 
I made it that far. By the way your dy/dx should end in a 1 instead of a 6.

I set it equal to zero and found horizontal tangent points at x=0 and y=2. I am not used to finding y values when doing that. Normally when it is not implicit you find just the x values and plug it in to the original equation to find the y values. If I plug 0 in for x, I end up with my first post on this topic: 2y^3 + 6y - 1 = 0

And I'm still stuck.

By the way, this problem came from an old AP Calculus exam.
 
Originally posted by webby
I made it that far. By the way your dy/dx should end in a 1 instead of a 6.

I set it equal to zero and found horizontal tangent points at x=0 and y=2. I am not used to finding y values when doing that. Normally when it is not implicit you find just the x values and plug it in to the original equation to find the y values. If I plug 0 in for x, I end up with my first post on this topic: 2y^3 + 6y - 1 = 0

And I'm still stuck.

By the way, this problem came from an old AP Calculus exam.

You are correct on the 1..I was scribbling in pen and missed canceling a 6. I looked at it a bit more; here are my thoughts, not guaranteed to be correct. :)

The roots for dy/dx=0 should be x=0 and y=1/2, I believe. Now, substituting these roots back into the original expression gives:

2y^3+6y-1=0

I believe the equation for the tangent line would then be its' derivative: 6y^2+6.

Substituting y=1/2 into the original expression gives x^2-1/4=0

The equation for its' tangent line would be 2x.

Now, on to part 2: we are looking for points on the original curve that satisfy two conditions: y=-x and dy/dx=-1 (in other words, there may be multiple points that satisfy the first condition but the curve isn't tangent to the line y=-x). Substituting y=-x into the original equation, we get:

8y^3-12y^2+6y-1=0.

I cheated and used a cubic root calculator at this point... :) All three roots are real and equal: y=1/2. Therefore x=-1/2. Checking this by plugging this value into dy/dx; it evaluates to -1, satisfying our second condition.
 
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