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)
So, bow you want the line with slope=2 through (π/4,π/2):
y - π/2 = 2(x - π/4)
y = 2x
Answer this Question
- Physics - The electric flux through each of the six sides of a rectangular box ...
- Calculus - Evaluate, in spherical coordinates, the triple integral of f(rho,...
- Precalculs - I have no idea how to do these type of problems. -------Problem...
- maths - If sin(theta)=3/5 , tan(phi)= 1/2 and 90กใ< theta <180กใ<phi<...
- Algebra - Let \phi(n) be the Euler Phi Function. If 1 \leq n \leq 1000, what is ...
- Trigonometry - Slopes stuff - Point A is the terminal point of angle phi and ...
- precalculus PLEASE HELP!! - Rewrite 4 sin(x) -1 cos(x) as A sin(x + phi) A= phi...
- Maths trigonometry - using basic trigonometry derive the following equations, ...
- Math - What is the limit of cosx/log(x-phi/2+1) if lim x is to phi/2
- Math - Prove that if gcd(m,n)=1 then m^phi(n)+n^phi(m)=congruent 1 (mod mn)