A man of height h is walking away from a stre | vedclass

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(A) Let $y$ be the distance of the man from the base of the street lamp and $x$ be the length of his shadow.Using similar triangles,the ratio of the h

Vedclass · Accepted Answer

$v/2$

Vedclass · Answer

(A) Let $y$ be the distance of the man from the base of the street lamp and $x$ be the length of his shadow.Using similar triangles,the ratio of the height of the lamp to the total distance from the base is equal to the ratio of the man's height to the length of his shadow:$\frac{3h}{x + y} = \frac{h}{x}$$3x = x + y$$2x = y$Differentiating both sides with respect to time $t$:$2 \frac{dx}{dt} = \frac{dy}{dt}$Given that the man is walking away at a constant speed $v$,we have $\frac{dy}{dt} = v$.Therefore,$\frac{dx}{dt} = \frac{v}{2}$.The rate of increase of the shadow length is independent of the distance from the lamp,so at $10h$,it is still $v/2$.

$A$ man of height $h$ is walking away from a street lamp with a constant speed $v$. The height of the street lamp is $3h$. The rate at which the length of the man's shadow is increasing when he is at a distance $10h$ from the base of the street lamp is:

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