Find an equation of a line tangent to y = 2sin x whose slope is a maximum value in the interval (0, 2π]
I believe the equation is y=2x-4pi. How is the b-value -4pi?
y' = 2cosx
y" = -2sinx
y' is max when y" = 0, at 0,π,2π
y'(π) = -2
y'(2π) = 2
So, the point (2π,0) has max slope.
y = 2(x-2π)
Note that x=0 is not in the given domain.
b is clearly not 4π
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D2sinx%2C+y%3D2x%2C+y%3D2x-4%CF%80
To find the equation of the line tangent to the curve y = 2sin(x) with the maximum slope in the given interval, we first need to determine the x-coordinate at which the maximum slope occurs.
To find this x-coordinate, we can differentiate the function y = 2sin(x) with respect to x using the chain rule. The derivative of y = 2sin(x) is dy/dx = 2cos(x). The maximum slope occurs where the derivative is equal to zero or undefined.
Setting dy/dx = 2cos(x) equal to zero and solving for x, we have:
2cos(x) = 0
This equation is satisfied when x = π/2 and x = 3π/2. However, we need to restrict our solution to the interval (0, 2π], so only x = 3π/2 is within this range.
Now, we have the x-coordinate where the maximum slope occurs, which is x = 3π/2. To find the corresponding y-coordinate, we substitute this x-value into the original equation:
y = 2sin(x)
y = 2sin(3π/2)
y = 2(-1)
y = -2
So the point where the line tangent to y = 2sin(x) with the maximum slope in the interval (0, 2π] is (3π/2, -2).
Now, we can use the point-slope form of the equation of a line to find the equation of the tangent line. The slope of the tangent line is the same as the slope of the original curve at the point (3π/2, -2), which is dy/dx evaluated at x = 3π/2.
dy/dx = 2cos(x)
dy/dx = 2cos(3π/2)
dy/dx = 2(0)
dy/dx = 0
So, the slope of the line tangent to y = 2sin(x) at x = 3π/2 is 0.
Using the point-slope form with the point (3π/2, -2) and a slope of 0, we have:
y - y1 = m(x - x1)
y - (-2) = 0(x - 3π/2)
y + 2 = 0
y = -2
Therefore, the equation of the line tangent to y = 2sin(x) with the maximum slope in the interval (0, 2π] is y = -2, which is a horizontal line. The b-value is 0, not -4π. It seems there was a calculation error in your previous answer.