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The equation of the curve passing through the point `(1,pi/4)` and having a slope of tangent at any point (x,y) as `y/x - cos^2(y/x)` is |
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Answer» Here, slope is given as `y/x-cos^2(y/x)` `:. dy/dx = y/x-cos^2(y/x)->(1)` Let, `y/x = v => y = vx => dy/dx = v+x(dv)/dx` so, (1) becomes, `=>v+x(dv)/dx = v- cos^2v` `=>(dv)/cos^2v = -dx/x` Integrating both sides, `int (dv)/cos^2v = - int dx/x` `=>int sec^2v (dv) = - int dx/x` `=>tanv = -ln x+c` `=>tan(y/x) = -ln x+c` As, the curve is passing through `(1,pi/4)`, it will satisfy the above equation. `:. tan (pi/4) = -ln (1) +c` `=> c = 1` So, the rfequired equation becomes, `tan(y/x) = -ln x+1` `=> tan(y/x) - 1= -ln x` `=> x = e^(1-tan(y/x))`, which is the required equation. |
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