Let Y=Y(X) be a curve lying in the first quadrant such that the area enclosed by the line …

Question

Let $Y=Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $$Y-y=Y^{\prime}(x)(X-x)$$ and the co-ordinate axes, where $(x, y)$

Accepted Answer

$A=\frac{1}{2}(\frac{-y}{{Y}^{'}(x)}+x)(y-xY/x)=\frac{-{y}^{2}}{2{Y}^{'}(x)}+1$

$\Rightarrow (-y+x{Y}^{'}(x))(y-x{Y}^{'}(x))=-{y}^{2}+2{Y}^{'}(x)$

$\begin{gathered} -{y}^{2}+xy{Y}^{'}(x)+xy{Y}^{'}(x) -{x}^{2}{[{Y}^{'}(x)]}^{2} \ = -{y}^{2}+2{Y}^{'}(x) $

$ 2xy-{x}^{2}{Y}^{'}(x)=2 \ \frac{dy}{dx}=\frac{2xy-2}{{x}^{2}} \ \frac{dy}{dx}-\frac{2}{x}y=\frac{-2}{{x}^{2}} \ ={e}^{-2\ln x}=\frac{1}{{x}^{2}} \ y\cdot \frac{1}{{x}^{2}}=\frac{2}{3}{x}^{-3}+c \ x=1,y=1 \ 1=\frac{2}{3}+c\Rightarrow c=\frac{1}{3} \ Y=\frac{2}{3}\cdot \frac{1}{X}+\frac{1}{3}{X}^{2} \ \Rightarrow \, 12Y(2)=\frac{5}{3}	imes 12=20 \end{gathered}$

Solutions · 1