Solving Tangent Line Problem: 2y^3 + (6x^2)y - 12x^2 + 6y = 1

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Discussion Overview

The discussion revolves around solving a problem related to the curve defined by the equation 2y^3 + (6x^2)y - 12x^2 + 6y = 1. Participants are exploring the conditions for horizontal tangent lines and finding a specific point on the curve where a line through the origin with a slope of -1 is tangent to the curve. The scope includes mathematical reasoning and problem-solving related to derivatives and tangent lines.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests finding horizontal tangent lines by setting the numerator of the derivative to zero, leading to the conclusion that y = 2.
  • Another participant confirms that the line with slope -1 through the origin is y = -x and relates it to the tangent condition at point P.
  • There is a discussion on how to use the derivative and the original curve equation to find the coordinates of point P, with some participants expressing confusion about the process.
  • Several participants inquire about the relationship between the derivative and the curve equation, seeking clarification on how to solve the equations simultaneously.

Areas of Agreement / Disagreement

Participants generally agree on the approach to find horizontal tangents and the tangent line through the origin, but there is no consensus on how to proceed with finding the coordinates of point P, as some express confusion about the method.

Contextual Notes

Some participants mention the need to solve two equations simultaneously, but the specific methods and steps to achieve this remain unclear, indicating potential gaps in understanding differentiation and its application in this context.

Who May Find This Useful

This discussion may be useful for students learning about derivatives, tangent lines, and solving equations in the context of calculus and analytical geometry.

akxt
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Hi, I was wondering if anyone could help.

Consider the curve given by: 2y^3 + (6x^2)y - 12x^2 + 6y = 1

I solved the derivative which came out to be (4x-2xy)/(x^2 + y^2 + 1

1. Write an equation of each horizontal tangent line to the curve.
2. The line through the origin with slope -1 is tangent ot the curve at point P. Find the x- and y- coordinates of point P.

for 1 ) do I find when the derivative is 0? so I set the top = 0 to get y = 2.

for 2) I am not even sure how to start this.. so I have m = -1 and point (0,0) .. now what:confused:
 
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y = -x then is the line that is tangent to the original curve 2y^3 + (6x^2)y - 12x^2 + 6y = 1. -1 also equals that tangent, the derivative you obtained. The two equations - the derivative and the original curve - should have a common point, which is the point P you are looking for.
 
akxt said:
Hi, I was wondering if anyone could help.

Consider the curve given by: 2y^3 + (6x^2)y - 12x^2 + 6y = 1

I solved the derivative which came out to be (4x-2xy)/(x^2 + y^2 + 1

1. Write an equation of each horizontal tangent line to the curve.
2. The line through the origin with slope -1 is tangent ot the curve at point P. Find the x- and y- coordinates of point P.

for 1 ) do I find when the derivative is 0? so I set the top = 0 to get y = 2.
Yes.

for 2) I am not even sure how to start this.. so I have m = -1 and point (0,0) .. now what:confused:
The line with m= -1, through (0,0) is, of course, y= -x. You now have two equations for the (x,y) coords of the point where that line is tangent to the curve: y'= (4x-2xy)/(x^2 + y^2 + 1)= -1 and the equation of the curve, 2y^3 + (6x^2)y - 12x^2 + 6y = 1. Knowing that y= -x should make that very easy. (If such a point exists.)
 
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can someone do out part 2 of this problem?
 
PF policy explicitly forbids doing homework for others. Just post any problems you encounter and we'll help.
 
i don't understand how to use both the equation and derivative to find the coordinate, can you explain how to relate the two?
 
surfsup887 said:
i don't understand how to use both the equation and derivative to find the coordinate, can you explain how to relate the two?

have you learned the topic differentiation?
 
surfsup887 said:
i don't understand how to use both the equation and derivative to find the coordinate, can you explain how to relate the two?
You have two equations to solve for x and y. Solve the equations simultaneously.
 
Do you mean set up the two equations equal to each other, then plug in -x for all y's then solve for x then plug that back into the equation to get the y?
 

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