Calculus Word Problem Related Rates 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.
 A: 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.
A: 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.
