Question
Write the equation of lines tangent and normal to the following function at (0, π). To find derivative, use implicit differentiation.
x^2cos^y - siny = 0
x^2cos^y - siny = 0
Answers
Steve
I do not know what <u>cos^y</u> means. I'll just use cosy and you can fix it as needed.
x^2 cosy - siny = 0
2x cosy - x^2 siny y' - cosy y' = 0
y' = (2x cosy)/(x^2 siny + cosy)
so, at (0,π), y' = 0
Since the tangent is horontal, the normal is vertical, and the lines are
tangent: y=π
normal: x=0
x^2 cosy - siny = 0
2x cosy - x^2 siny y' - cosy y' = 0
y' = (2x cosy)/(x^2 siny + cosy)
so, at (0,π), y' = 0
Since the tangent is horontal, the normal is vertical, and the lines are
tangent: y=π
normal: x=0