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

math  difference, Friday, May 13, 2016 at 4:27am
xsin2y=ycos2x
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