Flying Superhero The path through the air of a flying superhero has the shape of the graph: 
$$y=\frac{1}{100}\cos\left(\frac{x}{2}+3\right)+\frac{1}{4}$$
A sidekick on the ground measures the displacement of the superhero in the x-direction as a function of time:
$$x(t)=200t-100t^2 \text{ (in km)}$$
Determine the velocity $v(t)$ of the superhero in the vertical direction as a function of time.
So I know the height $y$ as a function of the horizontal location $x$. I know $x$ as a function of $t$. I am asked to find the VERTICAL speed. But first I have to figure out the vertical LOCATION as a function of $t$. Any suggestions on how to get out of this pickle?
 A: So, Paul's comment pretty much nails it, but I just wanted to add some illustrations, that somewhat qualitatively justifies it. So, below is the Superman flying according to his trajectory scheduled for today

Red circles are Sup's shadows, on the ground and on the vertical wall. Jimmy's measuring his horizontal speed, because he can run. Jimmy, unlike his pal can't fly, so no one can measure Clark's speed on that wall. But, Jimmy's smart, so he knows a little bit of math.
Jimmy's also a photographer, so he took Superman's pictures twice with the very close interval $\Delta t$ and overlaid both images into one and got the following

Jimmy knows that since $\Delta t$ is small, he can assume that Clark's trajectory was almost a straight line, and that line is slanted with the angle $y'(x)$, so $\Delta y = y' \Delta x$. By the time $\Delta t$ Clark moved the distance $\Delta x = v_x \Delta t$ in $x$ direction. So, as we already said, Jimmy's smart, so he did the math, which in this literal case was a substitution one into another
$$
\Delta y = y' v_x \Delta t
$$
Now, you don't need a to be a brain surgeon (just photographer would do) to find out that Superman's vertical speed (or should we say velocity?) is nothing but 
$$
v_y = \frac {\Delta y}{\Delta t} = y' v_x
$$
Luckily for Jimmy and us, we know both of those
$$
y' = -\frac 1{200} \sin \left( \frac x2 + 3\right ) = \frac 1{200} \sin \left( 50t^2 - 100t - 3\right ) \\
v_x = 200(1 - t^2)
$$
so Jimmy immediately faxed Lois Lane that
$$
v_y = \left( 1-t^2\right)\sin \left( 50t^2 - 100t - 3\right )
$$
PS: I hope everyone noticed that at some point Superman goes underground and continues doing that for uncertain amount of time leaving huge holes, so he totally deserves getting his butt kicked by Batman.
A: The above solution is wrong. Here's the solution
solution - Flying Superhero
Reyam Mustafa
