Find the equation of the line normal to the curve at (0,0). (Normal lines are perpendicular to the tangent lines)
2y + sinx = xcoxy
I found the derivative to be (cosy - cosx)/ (2+ xsiny)
then the slope tangent is 0...then the slope perpendicular is undefined...so then is the normal equation..x=0?
If dy/dx = as you stated, yes, the slope is undefined. x=0
To find the equation of the line normal to the given curve at the point (0,0), you are on the right track, but your conclusion is not correct.
First, let's find the derivative of the given equation. The equation is:
2y + sin(x) = xcos(xy)
To find the derivative, differentiate both sides of the equation with respect to x. Using the product rule, we get:
2(dy/dx) + cos(x) - (sin(xy))(1 + xy(dy/dx)) = cos(xy)
Simplifying the equation, we get:
2(dy/dx) - xy(sin(xy))(dy/dx) = cos(xy) - cos(x)
Now, let's find the slope of the tangent line at (0,0), which is the derivative evaluated at x = 0 and y = 0.
Plugging in x = 0 and y = 0 into the equation, we get:
2(dy/dx) = cos(0) - cos(0)
2(dy/dx) = 1 - 1
2(dy/dx) = 0
(dy/dx) = 0/2 = 0
Therefore, the slope of the tangent line at (0,0) is 0.
To find the slope of the normal line, we can use the fact that the product of the slopes of two perpendicular lines is -1. Therefore, the slope of the normal line is the negative reciprocal of the slope of the tangent line.
Since the slope of the tangent line is 0 (undefined), the slope of the normal line is also 0.
Now, you want to find the equation of the line. Since the slope is 0, the line is a horizontal line. However, we need to find the y-intercept (the value of y when x = 0).
To find the y-intercept, plug in x = 0 into the original equation:
2y + sin(0) = 0*cos(0*y)
2y + 0 = 0
2y = 0
y = 0
The y-intercept is 0.
Therefore, the equation of the line normal to the curve at (0,0) is y = 0.