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A spotlight on the ground shines on a wall $12 m$ away. If a man $2m$ tall walks from the spotlight toward the wall at a speed of $1.6 m/sec$, how fast is the length of his shadow decreasing when he is $4m$ from the building.

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the height of the shadow is determined by drawing a line connecting the spotlight to the man's head, and then extending it to the wall. Notice that the shadow height will decrease down to $2m$ (it cannot be shorter than the man). Also, your second paragraph is not right. You are suggesting to subtract the height of the man from the volume of a cone. Assume everything is 2-dimensional. You are interested in the length of the shadow, not in any areas involved. – Alex R. Oct 23 '12 at 18:19
what should the formula be then? – dsta Oct 23 '12 at 18:19

Hint: if you draw a picture, there are similar triangles that let you find a relation between the distance between the man and the wall and the height of the shadow.

Added: Let us measure from the spotlight at $x=0$ with the wall at $x=12$. The height of the shadow is then $\frac {12}x \cdot 2$ meters from similar triangles.

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Can you please try to explain to me what the first step would be? I am really lost on this question. – dsta Oct 23 '12 at 18:43

First, throw variables at it:

$x$ = distance from man to wall,

$y$ = height of shadow on wall;

$12-x$ = distance from light to man.

The triangle formed by the light, the man's feet and the man's head is similar to the triangle formed by the light, the bottom of the wall, and where the light hits the wall. Using properties of similar triangles, we can say $${12-x \over 2} = {12 \over y}$$ $$12y-xy=24$$ Now, take the derivative of both sides: $$12dy-xdy - ydx=0$$ $$dy={ydx \over (12-x)}$$ $dx$, the change in $x$, is given as -1.6m/s(negative because the distance to the wall is decreasing) and $x$ is given as 4. Using similar triangles, we can calculate $y$ at that point to be 3. Substituting, we get $$dy = -.6m/s$$ The shadow is moving down the wall at 0.6 meters/sec.

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