Question
A lamp of negligible height is placed on the ground $${l_1}\,m$$ away from a wall. A man $${l_2}\,m$$ tall is walking at a speed of $$\frac{{{l_1}}}{{10}}\,m/s$$ from the lamp to the nearest point on the wall. When he is midway between the lamp and the wall, the rate of change in the length of this shadow on the wall is :
A.
$$ - \frac{{5{l_2}}}{2}\,m/s$$
B.
$$ - \frac{{2{l_2}}}{5}\,m/s$$
C.
$$ - \frac{{{l_2}}}{2}\,m/s$$
D.
$$ - \frac{{{l_2}}}{5}\,m/s$$
Answer :
$$ - \frac{{2{l_2}}}{5}\,m/s$$
Solution :
Let $$BP = x.$$
From similar triangle property, we get
$$\eqalign{ & \frac{{AO}}{{{l_1}}} = \frac{{{l_2}}}{x} \cr & {\text{or }}AO = \frac{{{l_1}{l_2}}}{x} \cr & {\text{or }}\frac{{d\left( {AO} \right)}}{{dt}} = \frac{{ - {l_1}{l_2}}}{{{x^2}}}\frac{{dx}}{{dt}} \cr & {\text{When, }}x = \frac{{{l_1}}}{2},\,\frac{{d\left( {AO} \right)}}{{dt}} = - \frac{{2{l_2}}}{5}\,m/s. \cr} $$
Let $$BP = x.$$
From similar triangle property, we get
$$\eqalign{ & \frac{{AO}}{{{l_1}}} = \frac{{{l_2}}}{x} \cr & {\text{or }}AO = \frac{{{l_1}{l_2}}}{x} \cr & {\text{or }}\frac{{d\left( {AO} \right)}}{{dt}} = \frac{{ - {l_1}{l_2}}}{{{x^2}}}\frac{{dx}}{{dt}} \cr & {\text{When, }}x = \frac{{{l_1}}}{2},\,\frac{{d\left( {AO} \right)}}{{dt}} = - \frac{{2{l_2}}}{5}\,m/s. \cr} $$