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| Math Gurus Help Solve This by badmus45: 4:31am On Jan 05, 2015 |
the parametric equations of a curve are x=cos@ ,y=1+sin2@.find dy/dx and d2y/dx2 at @=π/6.find also the equation of the curve as relationship between x and y. |
| Re: Math Gurus Help Solve This by Nobody: 8:54am On Jan 05, 2015 |
dx/d@ = -sin@ dy/d@ = 2cos2@ d@/dx = -1/sin@ dy/dx = dy/d@ x d@/dx = -2cos2@/sin@ 1 Like |
| Re: Math Gurus Help Solve This by Nobody: 9:02am On Jan 05, 2015 |
dy/dx can also be written as 2(1- 2x2)/√(1-x2) d2y/dx2 = {2.-4x.√(1-x2) + 2.(1-2x2).0.5.(1-x2)-1/2.2x}/(1-x2) = 2x(2x2-3)/(1-x2)3/2 y = 1+sin2@ =1+2cos@√(1-cos2@) =1+2x√(1-x2) |
| Re: Math Gurus Help Solve This by Nobody: 10:22am On Jan 05, 2015 |
Alternatively to get d2y/dx2 d2y/dx2 = d/dx{dy/d@ x d@/dx} = d@/dx x d/d@{dy/d@ x d@/dx} =-1/sin@ x d/d@{-2cos2@/sin@} = - {2cos2@cos@ + 4sin@sin2@}/sin3x = (4cos3@ - 6cos@)/sin3x =2x(2x2 - 3)/(1-x2)3/2 |
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