Question:
1) Suppose you have a street light at a height 10 meters. You drop a rock vertically
so that it hits the ground at a distance 5 meters from the street light. The shadow of the
rock moves along the ground. Find the speed of the shadow of the rock at the moment that
the rock is 3 meters above the ground and moving with the speed of 5 m/s.
2) You say goodbye to your friend at the intersection of two perpendicular roads.
At time t = 0 you drive off North at a (constant) speed v and your friend drives West at
a (constant) speed w. How fast is the distance between you and your friend increasing at
time t?
What should start with these two?
Best Answer
1) Let $x$ be the distance of the shadow from the spot of landing and $h$ be the height of the stone. From similar triangles,
$$\frac xh = \frac{x+5}{10}, \implies x=\frac{5h}{10-h}$$
Take the time derivatives,
$$x' = \frac{50}{(10-h)^2}h'$$
Given that $h=3m$ and $h'=-5m/s$, the speed of the shadow is $x'=-\frac{250}{49}m/s$.
2) Let $d$ be the distance. From the right triangle,
$$d(t)=\sqrt{(vt)^2+(wt)^2}$$
Take the time derivatives to get
$$d'=\frac{v^2+w^2}{\sqrt{v^2+w^2}}=\sqrt{v^2+w^2}$$