Posted by **deel** on Monday, August 13, 2012 at 12:45pm.

Please help me to find an equation for the tangent to the curve xsin2y=ycos2x at the point (phi/4, phi/2)

- math -
**Steve**, Monday, August 13, 2012 at 2:59pm
first, it's PI, not PHI

pi(π) = 3.14, the ratio of circumference to diameter

phi(φ) = 1.62 (1+√5)/2, the golden ratio

So, we use implicit differentiation to get

xsin2y = ycos2x

sin2y + 2xcos2y y' = cos2x y' - 2ysin2x

y' = -(sin2y + 2y sin2x)/(2x cos2y - cos2x)

at (π/4,π/2), that gives a slope of

y' = -(0 + π*(1))/(π/2*(-1) - 0)

= -π/(-π/2)

= 2

So, bow you want the line with slope=2 through (π/4,π/2):

y - π/2 = 2(x - π/4)

y = 2x

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