Asked by Sara
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?
I believe the equation is y=2x-4pi. How is the b-value -4pi?
Answers
Answered by
Steve
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
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
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