The slope of the line normal to the graph of 4 sin x – 9 cos y = 9 at the point (x,0)?
How do you do this?
3 answers
Help me??
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
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
alright thennnn
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