Determining the change in radius of water flowing from a faucet - more general question on differentiation

Question

I've been outside of the academic world for several years now, and I'm forcing myself to go back through old textbooks and resources and work through the information in there. I can tell I'm losing my retention (and fast, too) and this is really bothering me. The boldface lower down is my specific question if you don't want to get hung up in the details of the problem itself. In it, I'm solving a problem by letting $\frac{\delta r}{\delta y}=\frac{\delta r}{\delta t} \div \frac{\delta y}{\delta t}$. I'm under the impression this is technically allowed, but I'd like a more rigorous solution to set my mind at ease.

I'm looking at problem 3.7 in The Mechanical Universe by Frautschi et al.:

A cylindrical stream of water flows at a speed $v_0$ out of a pipe. The amount of water flowing through a cross section of the stream is the same at all distances from the pipe; that is, if five liters (5 L) flow out of the pipe in 1 s, then 5 L also flow past a point 1 m below the end of the pipe. Since the falling water accelerates downward the stream must get narrower the farther it gets from the end of the pipe. Find an equation for the change $\frac{\delta r }{\delta y}$ in the radius r of the stream with respect to distance y below the end of the pipe.

I have worked out a tentative solution but it makes me a bit uneasy. Can someone look over it?

First, I considered the equation for the water flow for any radius and velocity, $\rho$ = $\pi r^2 v$. The initial flow would be $\rho_0 = \pi r_0^2 v_0$ where $r_0$ is the radius of the pipe and $v_0$ is the initial velocity. We know at any time in the future, the water will still have this constant flow $\rho_0$ but the radius and velocity will change with time. So $rho_0 = \pi r(t)^2 v(t) = \pi r_0^2 v_0$. Trivially, $v(t)=v_0 + 9.8t$, so we can solve for $r(t) = r_0 \sqrt{\frac{v_0}{v_0+9.8t}}$. Note $\frac{\delta r}{\delta t} = \frac{-4.9 \sqrt{v_0}}{(v_0+9.8t)^{3/2}}$.

Now, and this is what makes me nervous, I decide to "cheat" by treating differential operators like they're just letters and say $\frac{\delta r}{\delta y}=\frac{\delta r}{\delta t} \div \frac{\delta y}{\delta t}$. As $\frac{\delta y}{\delta t}=v(t)=v_0+9.8t$, this yields a result of $\frac{\delta{r}}{\delta y} = \frac{-4.9 \sqrt{v_0}}{(v_0 + 9.8t)^{5/2}}$. This answer seems pretty surprisingly nice, so I figured that was an alright indication of me doing something "allowed." My questions: Is this solution correct? If not, is it because of the "cheating" mentioned above? What would be a better way to move from $r(t)$ to $r(y)$? Even if it is correct, is there a more rigorous way to solve this so that I feel a little more assured?

Answer 1

If you are working through The Mechanical Universe I'd suggest having Tom Apostol's Calculus 1 on hand since Apostol was the math editor of The Mechanical Universe. The chain rule as found in Calculus 1 is this: Let f be the composition of two functions u and v, say $f=u\circ v$. Suppose that both derivatives v'(x) and u'(x) exist, where y=v(x). Then the derivative f'(x) also exists and is given by the formula $$f'(x) = u'(y)\cdot v'(x).$$ In other words, to compute the derivative of $u\circ v$ At x, we first compute the derivative of u at the point y, where $y=v(x),$ and multiply this by v'(x).

The chain rule is a product of two derivatives and while it is useful to think of them as fractions sometimes, they are not. That doesn't necessarily mean that your logic is wrong though. The questionable part is your assumption that $\frac{dt}{dy}=\frac{1}{\frac{dy}{dt}}$. Edit: I realized that I made a mistake while writing my example, and I haven't thought too much on another counterexample. My suspicion is growing that you can treat derivatives like you want to, but I haven't been able to come up with an argument as to why that should be the case. Even so, try what I suggested and you will arrive at the solution.

I'd use the distance formula $y=y_0 + v_0 t + \frac{1}{2}at^2$, setting the variables according to your coordinate system and solving for t, since t is quadratic. You'll find t as a function of y and from there the solution quickly follows.