The slope of the line normal to the graph of 4 sin x – 9 cos y = 9 at the point (x,0)?
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To find the slope of the line normal to the graph of the given equation at the point (x,0), we first need to rewrite the equation in terms of y. Let's rearrange the given equation:
4 sin(x) - 9 cos(y) = 9
Subtracting 4 sin(x) from both sides:
-9 cos(y) = 9 - 4 sin(x)
Dividing by -9:
cos(y) = (4/9)sin(x) - 1
Now, we can find the slope of the line normal to the graph by taking the derivative of both sides with respect to x. Differentiating both sides of the equation gives us:
-d/dx(cos(y)) = d/dx((4/9)sin(x) - 1)
The derivative of cos(y) with respect to x can be found using the chain rule:
-d(cos(y))/dx = (4/9)cos(x)
The derivative of sin(x) with respect to x is simply cos(x):
(4/9)cos(x) = (4/9)cos(x)
Now, at the point (x,0), the value of cos(y) will be the x-coordinate of the given point. Let's substitute this information into the equation above:
-d(cos(y))/dx = (4/9)cos(x)
-d(cos(0))/dx = (4/9)cos(x)
-d(1)/dx = (4/9)cos(x)
0 = (4/9)cos(x)
Therefore, the slope of the line normal to the graph at the point (x,0) is 0.
y = 4/9 sinx - 1
y' = 4/9 cosx
So, the slope of the tangent at x is 4/9 cosx
The slope of the normal there is -1/y' = -1/(4/9 cosx) = -9/4 secx