original curve: 2y^3+6(x^2)y-12x^2+6y=1
dy/dx=(4x-2xy)/(x^2+y^2+1)
a) write an equation of each horizontal tangent line to the curve
b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point P.
-when plugging y as 2 in original, it doesn't work. what did i do wrong?
I do not agree with your equation for dy/dx. You appear to have done the implicit differdntiation incorrectly
When you plugged y = 2 into the original equation, you mentioned that it did not work. The reason it didn't work is most likely due to an algebraic mistake. Let's go through the steps to find the mistake:
Starting with the original curve:
2y^3 + 6(x^2)y - 12x^2 + 6y = 1
To find the horizontal tangent lines, we need to find the points where the derivative of y with respect to x, dy/dx, is equal to 0. Taking the derivative of the original equation with respect to x, we get:
dy/dx = (4x - 2xy)/(x^2 + y^2 + 1)
Setting dy/dx equal to 0:
0 = (4x - 2xy)/(x^2 + y^2 + 1)
To find the points where this holds, we need to consider two cases: when the numerator is 0, and when the denominator is 0.
Case 1: Numerator = 0
4x - 2xy = 0
2x(2 - y) = 0
This gives us two possibilities:
1) 2x = 0 ⇒ x = 0
2) 2 - y = 0 ⇒ y = 2
So one possible point is (x, y) = (0, 2).
Case 2: Denominator = 0
x^2 + y^2 + 1 = 0
However, this case does not yield any real solutions, since the sum of squares is always non-negative. Therefore, we do not consider it.
Now let's move on to part b) of your question.
To find point P on the curve where the line through the origin with slope 0.1 is tangent, we need to set the slope of the curve equal to 0.1. In other words, we need to solve the equation:
dy/dx = 0.1
Substituting the expression for dy/dx from earlier:
(4x - 2xy)/(x^2 + y^2 + 1) = 0.1
Simplifying, we get:
4x - 2xy = 0.1(x^2 + y^2 + 1)
This is a nonlinear equation, so we need to solve it for x and y simultaneously. Unfortunately, it cannot be solved explicitly due to the presence of both x and y terms on both sides of the equation.
Therefore, we need to use numerical methods or approximation techniques to find the values of x and y at point P.