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Find the radius of curvature at (1,1) to the curve y2-x = x3. |
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Answer» y2-x = x3 at (1, 1) (2 - x) log y = 3 log x-----(i) (by taking log both sides) (2-x)1/y (dy/dx) - log y = 3/x-----(ii) (by differentiating (i) w.r.t x) (2 - x)1/y(dy/dx) = 3/x + (3logx)/(2 - x) \(\frac{dy}{dx}=y\left(\frac{3}{x(2-x)}+\frac{3logx}{(2-x)^2}\right)\) \(\frac{dy}{dx}\) at (1, 1) = 1\((\frac{3}{1\times1}+\frac{3log 1}{1^2})\) = 3 \(\frac{(2-x)}y\frac{d^2y}{dx^2}-\frac{(2-x)}{y^2}(\frac{dy}{dx})^2\) - \(\frac1y\frac{dy}{dx}-\frac1y\frac{dy}{dx}=\frac{-3}{x^2}\)......(iii) (By differentiating equation (ii) w.r.t. x) Put x = 1, y = 1, \(\frac{dy}{dx}=3\) in equation (iii), we get ⇒ \(\frac{2-1}1\frac{d^2y}{dx^2}-\frac{(2-1)}1\times9-3-3=-3\) ⇒ \(\frac{d^2y}{dx^2}=12\) radius of curvature at (1, 1) to given curve is p = \(\cfrac{(1+(\frac{d^2y}{dx^2})^2)^{3/2}}{\frac{d^2y}{dx^2}}\) = \(\frac{(1+3^2)^{3/2}}{12}=\frac{(10)^{3/2}}{12}\) |
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