remember that the slope of a function is the first derivative of that function.
So for the slope to have a maximum, the derivative of slope must be zero
But the derivative of the slope is the second derivative of the given function, so ...
y = 2sinx
y ' = 2 cosx
y'' = -2sinx
-2sinx = 0
sinx = 0
x = 0, π, 2π, 3π, 4π etc , for each of those y = 0
points where the slopes are a max are (0,0), (2π,0), (4π,0) etc.
Subbing each of those into the slope of 2cosx gives
a slope of 2(1) or 2
(notice that the slope at the points (π,0), (3π,0) etc would give us a slope of -2, thus it would be a minimum slope at those points)
The first line would be
y = 2x + b, with (0,0) on it
0 = 0 + b
b = 0
first tangent is y = x
second tangent:
y = 2x + b, with (2π,0) on it
0 = 2(2π) + b
b = -4π
second tangent equation:
y = 2x - 4π
etc.
a) Find an equation of a line that is tangent to y=2sinx and whose slope is a maximum value.
b) Is there more than one solution? Explain.
2 answers
you mean y=2x for the first line...because I did something similar however I took the limit, since cos x can not have y values larger than 1 or -1, i took 2cosx=1 i get zero and using both methods:Y-Y1=m(X-X1) and y=mx+b i get y=2x. i checked it using a graphical calculator the same answer, y=2x
1 answer