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If `x=a (theta-sintheta), y=a (1+costheta), `find `(d^2 y)/(dx^2)` |
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Answer» `y = a+acostheta` `=>dy/(d theta) = - asin theta` `x = atheta - asintheta` `=>dy/(d theta) = a - acostheta = a( 1- costheta)` `:. dy/dx = (dy/(d theta))/(dx/(d theta)) = (-asin theta)/(a(1-costheta))` `=>dy/dx = -sintheta/(1-costheta)**(1+costheta)/(1+costheta)` `=>dy/dx = -(sintheta (1+costheta))/(1-cos^2theta)` `=>dy/dx = -(sintheta (1+costheta))/(sin^2theta)` `=>dy/dx = -cosec theta - cottheta ` Differentiating both sides w.r.t. `theta`, `=>d/(d theta) (dy/dx) = cosecthetacottheta + cosec^2theta` `=>d/(d theta) (dy/dx) = cosectheta(cottheta + cosectheta)` `=>(d^2y)/dx^2 = cosectheta(cottheta + cosectheta)**(d theta)/dx` `=>(d^2y)/dx^2 = (cosectheta(cottheta + cosectheta))/(a(1-costheta))` `=>(d^2y)/dx^2 = (1/sintheta((1+costheta)/sintheta))/(a(1-costheta))` `=>(d^2y)/dx^2 = (1+costheta)/(asin^2theta(1-costheta))` `=>(d^2y)/dx^2 =(1+costheta)/(asin^2theta(1-costheta))**(1-costheta)/(1-costheta)` `=>(d^2y)/dx^2 =sin^2theta/((asin^2theta)(1-costheta)^2)` `=>(d^2y)/dx^2 = 1/(a(1-costheta)^2)` As `y = a(1+costheta) => costheta = y/a-1` Putting this value in our equation, `=>(d^2y)/dx^2 = 1/(a(1-(y/a-1))^2)` `=>(d^2y)/dx^2 = a/(2a-y)^2` |
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