What is the slope of the curve x = cos(y) at the point where y=-pi/3
To find the slope of the curve x = cos(y) at the point where y = -π/3, we need to take the derivative of x with respect to y and evaluate it at that specific point.
To start, let's rewrite the equation x = cos(y) in terms of y as y = arccos(x). Taking the derivative of both sides with respect to y gives us:
1 = -sin(y) * (dy/dx)
Now, we need to find dy/dx. To do this, we can rearrange the equation to isolate dy/dx:
dy/dx = 1 / (-sin(y))
Substituting y = -π/3, we have:
dy/dx = 1 / (-sin(-π/3))
Next, we need to simplify the expression. Remember that sin(-θ) = -sin(θ):
dy/dx = 1 / (-sin(π/3))
Knowing that sin(π/3) = √3/2, we can simplify further:
dy/dx = 1 / (-√3/2)
To rationalize the denominator, we multiply the numerator and denominator by √3:
dy/dx = 1 / (-√3/2) * (√3/√3)
Simplifying this expression gives us:
dy/dx = -2/√3
Therefore, the slope of the curve x = cos(y) at the point where y = -π/3 is -2/√3.