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Question Number 25868 by ajfour last updated on 16/Dec/17

Commented by ajfour last updated on 16/Dec/17

$F i n d t h e a r e a e n c l o s e d b y a$ $v a r i a b l e l i n e \frac{x}{t} + \frac{y}{l - t} = 1 a n d$ $t h e c o o r d i n a t e a x e s i n t h e$ $f i r s t q u a d r a n t;$ $(t h e l i n e s h i f t s a s t c h a n g e s$ $f r o m t = 0 t o t = l) .$
Commented by jota@ last updated on 16/Dec/17

$y_{m a x} f o r x = a$ $\frac{a}{t} + \frac{y}{l - t} = 1$ $y = (1 - a / t) (l - t) . . . (1)$ $y^{'} = (\frac{a}{t^{2}}) (l - t) - (1 - \frac{a}{t})$ $y^{'} = 0$ $a (l - t) - t (t - a) = 0$ $t^{2} - a l = 0 t = \sqrt{a l} . . . (2)$ $(2) i n t o (1)$ $y = (1 - \frac{a}{\sqrt{a l}}) (l - \sqrt{a l})$ $y = l - \sqrt{a l} - \sqrt{a l} + a b u t a = x$ $y = l - 2 \sqrt{x l} + x$ $A r e a = \int_{0}^{l} y d x = l^{2} - \frac{4 l^{2}}{3} + \frac{l^{2}}{2} = \frac{l^{2}}{6} .$
Commented by ajfour last updated on 09/Nov/24

$b e y o n d a l l p r a i s e! t h a n k s . (r e p o s t)$
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