What angle does the tangent to the curve y(x)=sin^2(x/9)+16sin(x/9) at x=9\pi make with the x-axis?
if y' is slope, then slope is the tangent of that angle.
y'=2sin( )cos( ) + 16cos( )
at x/9, compute all that, then the angle is
arctan(y')
To find the angle that the tangent to the curve makes with the x-axis, we need to find the derivative of the curve at x = 9π. The tangent to the curve at a specific point has the same slope as the derivative at that point.
Step 1: Find the derivative of the curve y(x) = sin^2(x/9) + 16sin(x/9).
We can use the chain rule to calculate the derivative.
Let's denote u = x/9.
Differentiating sin^2(u), we get:
d/dx [sin^2(u)] = d/dx [sin^2(x/9)]
= 2sin(u)⋅cos(u)⋅(du/dx)
= 2sin(x/9)⋅cos(x/9)⋅(1/9)
= (2/9)sin(x/9)⋅cos(x/9)
Similarly, differentiating 16sin(u), we get:
d/dx [16sin(u)] = d/dx [16sin(x/9)]
= 16⋅cos(u)⋅(du/dx)
= 16⋅cos(x/9)⋅(1/9)
= (16/9)cos(x/9)
Thus, the derivative of y(x) is:
dy/dx = (2/9)sin(x/9)⋅cos(x/9) + (16/9)cos(x/9)
Step 2: Evaluate the derivative at x = 9π.
Substituting x = 9π in the derivative formula, we have:
dy/dx = (2/9)sin((9π)/9)⋅cos((9π)/9) + (16/9)cos((9π)/9)
= (2/9)sin(π)⋅cos(π) + (16/9)cos(π)
= (2/9)(0)⋅(-1) + (16/9)(-1)
= (-16/9)
Step 3: Find the slope of the tangent line at x = 9π.
The slope of the line is given by the value of the derivative at x = 9π.
In this case, the slope is (-16/9).
Step 4: Find the angle that the tangent line makes with the x-axis.
The tangent of the angle is equal to the slope of the line.
Therefore, the tangent of the angle is (-16/9).
To find the angle itself, we need to take the arctangent of (-16/9):
angle = arctan((-16/9))
Using a calculator, we can find the angle to be approximately -60.6 degrees.
So, the angle that the tangent to the curve y(x) = sin^2(x/9) + 16sin(x/9) at x = 9π makes with the x-axis is approximately -60.6 degrees.