આકૃતિમાં બતાવ્યા પ્રમાણે ખેંચી ન શકાય તેવી સ્ | vedclass

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Question

As $P$ and $Q$ fall down, the length l decreases at the rate of $U \,m/s.$

From the figure, ${l^2} = {b^2} + {y^2}$

Differentiating with respect

Vedclass · Accepted Answer

$\frac{U}{{\cos 	heta }}$

Vedclass · Answer

As $P$ and $Q$ fall down, the length l decreases at the rate of $U \,m/s.$

From the figure, ${l^2} = {b^2} + {y^2}$

Differentiating with respect to time

$2l 	imes \frac{{dl}}{{dt}} = 2b 	imes \frac{{db}}{{dt}} + 2y 	imes \frac{{dy}}{{dt}}$      $\left( {{m{As \,\,}}\frac{{db}}{{dt}} = 0,\frac{{dl}}{{dt}} = U} ight)$

$&rArr;$ $\frac{{dy}}{{dt}} = \left( {\frac{l}{y}} ight) 	imes \frac{{dl}}{{dt}}$ $ \Rightarrow \frac{{dy}}{{dt}} = \left( {\frac{1}{{\cos 	heta }}} ight) 	imes U = \frac{U}{{\cos 	heta }}$

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