Find the equation of the normal to the curve f(x) = 3cos(2x) at the point x = 0.
In class, we learnt to find the derivate of the function, which would represent the slope.
After I did that, I found the slope to be 0.. but that is not what my graphing calculator tells me.. So, I'm not too sure where I went wrong :S
To find the equation of the normal to the curve at a given point, we need both the slope of the tangent line and the coordinates of the point.
Let's start by finding the derivative of the function f(x) = 3cos(2x). The derivative gives us the slope of the curve at any point on the graph. In this case, we apply the chain rule:
f'(x) = -6sin(2x).
Now, let's find the slope of the tangent line at x = 0 by substituting x = 0 into the derivative:
f'(0) = -6sin(2(0)) = -6sin(0) = 0.
So, your calculation of the slope as 0 is correct.
To find the equation of the normal, we need the slope of the normal line, which is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent is 0, the negative reciprocal is undefined or infinite.
However, since we are looking for the equation of the normal at x = 0, we can proceed with some additional steps.
At x = 0, the function f(x) = 3cos(2x) evaluates to f(0) = 3cos(0) = 3.
So, we know the point on the curve is (0, 3).
The equation of a line passing through a point (x1, y1) with an undefined slope is of the form x = x1. In this case, since we are looking for the equation of the normal to the curve, it is a vertical line.
Therefore, the equation of the normal to the curve f(x) = 3cos(2x) at the point x = 0 is x = 0.
I hope this explanation helps! Let me know if you have any further questions.