If 2x^3 + 2xy - y^2 = 6, then approximate the equation of the tangent line at point x = 1.44
To find the equation of the tangent line to the curve at a specific point, we need two pieces of information: the slope of the tangent line and the point at which the tangent line intersects the curve.
First, let's find the derivative of the given equation to find the slope of the tangent line. The derivative represents the rate of change of a function at any given point and gives us the slope of the tangent line at that point.
Taking the derivative of the equation 2x^3 + 2xy - y^2 = 6 with respect to x, we get:
6x^2 + 2y + 2xy' - 2yy' = 0.
Now, substitute the x-value of the point of interest into the equation. In this case, x = 1.44. So, the equation becomes:
6(1.44)^2 + 2y + 2(1.44)y' - 2yy' = 0.
Next, we need to find the y-value at x = 1.44. Substitute x = 1.44 into the original equation 2x^3 + 2xy - y^2 = 6 and solve for y:
2(1.44)^3 + 2(1.44)y - y^2 = 6.
Now, we have two equations:
Equation 1: 6(1.44)^2 + 2y + 2(1.44)y' - 2yy' = 0 (the derivative equation)
Equation 2: 2(1.44)^3 + 2(1.44)y - y^2 = 6 (the equation of the curve)
These two equations can be solved simultaneously for the values of y and y'. Solve Equation 2 for y and substitute it into Equation 1. This will give you an equation in terms of y' only.
Solve this equation for y' using algebraic methods (such as isolating y' on one side of the equation).
Once you have the value of y', you can use the point-slope form of a linear equation to find the equation of the tangent line:
(y - y1) = m(x - x1),
where m is the slope of the tangent line and (x1, y1) is the point of tangency (in this case, x1 = 1.44).
Substitute the values of m and (x1, y1) into the equation to find the equation of the tangent line.