Find an equation of a line tangent to y = 2sin x whose slope is a maximum value in the interval (0, 2π]
i found the derivative to be y'=2cosx but i don't know what to do next
well, where is cosx a maximum? At x=0.
So, at x=0, the slope of 2sinx = 2.
2sin(0) = 0, so, at (0,0) the slope is 2.
The line is y=2x.
By now you should know that sin(x) has its maximum slope where it crosses the x-axis. That's what we have calculated here.
how is the slope of 2sinx=2?
because f'(0)=2cos(0)=2?
also, how is y=2x? did you use y=mx+b? which numbers did you substitute?
he said at (0,0) the slope is 2
That is a line through the origin with slope = 2
y = 2 x + 0
To find the equation of a line tangent to the curve y = 2sin(x) with a maximum slope in the interval (0, 2π], we need to find the point(s) where the maximum slope occurs.
1. Start by finding the derivative of the function y = 2sin(x) with respect to x. The derivative gives us the slope of the function at any point.
dy/dx = 2cos(x)
2. Now, to find the maximum slope, we need to find the critical points of the derivative function. These are the points where the derivative is equal to zero or undefined.
Set dy/dx = 0:
2cos(x) = 0
Solving for x, we get:
cos(x) = 0
In the interval (0, 2π], the values of x where cos(x) = 0 are x = π/2 and x = 3π/2.
3. To find the corresponding y-values for these critical points, substitute x = π/2 and x = 3π/2 into the original function y = 2sin(x).
For x = π/2, y = 2sin(π/2) = 2.
For x = 3π/2, y = 2sin(3π/2) = -2.
So, the two points where the maximum slope occurs are (π/2, 2) and (3π/2, -2).
4. Now, we can find the equation of the tangent line using the point-slope form of a line.
Let's consider the point (π/2, 2):
Slope of the tangent line = slope of the function at that point = dy/dx evaluated at x = π/2 = 2cos(π/2) = 0.
Using the point-slope form with the point (π/2, 2) and the slope 0:
y - 2 = 0(x - π/2)
Simplifying, we get:
y = 2
So, the equation of the tangent line to y = 2sin(x) with a maximum slope in the interval (0, 2π] is y = 2.