A street light is at the top of a 12 m pole. | vedclass

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(A) Let the man be at a distance $x$ from the pole at any time $t$,and let the height of his shadow on the wall be $y$. The total distance from the po

Vedclass · Accepted Answer

$3/4$

Vedclass · Answer

(A) Let the man be at a distance $x$ from the pole at any time $t$,and let the height of his shadow on the wall be $y$. The total distance from the pole to the wall is $12 \ m$. When the man is $8 \ m$ from the wall,his distance from the pole is $x = 12 - 8 = 4 \ m$. Given $dx/dt = 1/2 \ m/s$.Using similar triangles,the ratio of the height of the shadow to the height of the man relative to the pole is:$\frac{y}{2} = \frac{12}{x}$$y = \frac{24}{x}$Differentiating with respect to $t$:$\frac{dy}{dt} = -\frac{24}{x^2} \cdot \frac{dx}{dt}$At $x = 4$ and $dx/dt = 1/2$:$\frac{dy}{dt} = -\frac{24}{16} \cdot \frac{1}{2} = -\frac{3}{4} \ m/s$.The negative sign indicates that the length of the shadow is decreasing. Thus,the rate at which the shadow is decreasing is $3/4 \ m/s$.

$A$ street light is at the top of a $12 \ m$ pole. $A$ man $2 \ m$ tall walks away from the pole towards a wall $12 \ m$ away from the pole at a speed of $1/2 \ m/s$. The rate at which his shadow on the wall is decreasing when he is $8 \ m$ from the wall is:

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