authenticricket
authenticricket
02-09-2017
Mathematics
contestada
can someone help with me with (bi)
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caylus
caylus
02-09-2017
Hello,
[tex] x=a*cos(t)==\ \textgreater \ \dfrac{dx}{dt} =-a*sin(t)\\ y=a*sin(t)==\ \textgreater \ \dfrac{dy}{dt}=a*cos(t)\\ \dfrac{dy}{dx} =- \dfrac{1}{tan(t)} \\\\ x= \dfrac{a}{2} ==\ \textgreater \ \\\\ cos(t)= \dfrac{1}{2} ==\ \textgreater \ t= \dfrac{\pi}{3} ==\ \textgreater \ y=a* \dfrac{\sqrt{3}}{2} \\\\ y'=- \dfrac{\sqrt{3}}{3} \\\\ \textrm{Equation of the tangent :}\\ y-a*\dfrac{\sqrt{3}}{2}=(x-\dfrac{a}{2} )*(- \dfrac{\sqrt{3}}{3})\\\\ y=- \dfrac{ \sqrt{3} }{3} *x+2*a*\dfrac{\sqrt{3}}{3}\\\\ \boxed{3y= \sqrt{3} *(2a-x)} [/tex]
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