We had this exercise in class, which went something like:
A balloon's height is described by the function $$ h(t)=\frac{1}{5} t \cdot (t-5)^2 \space (0 \leq t \leq 5) $$ $t$ denotes the passed time in hours, $h(t)$ the height in kilometers.
- At what time is the balloon's rising velocity at maximum? What is its value?
- When does the balloon drop the fastest? At what height does this happen?
We all know that $v = \frac{s}{t}$. $s$ is the traveled distance, so $s = h(t)$, at least until the balloon hits its maximum height. $v = \frac{s}{t}$ cannot describe velocity continuously, but if we let $s = h(t)$ and plug it into $v$'s equation, we got $\frac{h(t)}{t}$, from where we can get to $\lim_{\epsilon \to 0}\frac{h(t + \epsilon) - h(t)}{t}$, which is the very (or at least one) definition of differentiation. Hence, $h(t)$'s derivative $h'(t)$ should give me the the velocity $v$ at any time, so $v(t) = h'(t)$.
Long story short, I tried it but failed. I got
$$ h'(t) = \frac{3}{5}t^2 - 4t + 5 = v(t) $$
and to get the maximum velocity required in task 1, I need to set the derivative of $v(t)$ to $0$:
$$ v'(t) = \frac{6}{5}t - 4 = 0 \iff t = \frac{10}{3} $$
Plugging this into $v(t)$ now gives me:
$$ v(\frac{10}{3}) = \frac{3}{5}(\frac{10}{3})^2 - 4(\frac{10}{3}) + 5 = \frac{25}{3} $$
which is wrong. The solutions says that the highest rising velocity is at $t = 0$. (It should actually be solved graphically, but that seemed so unmathematical and vague to me I didn't think of it in the first place.)
I think it doesn't work because $v''(t) = \frac{6}{5} > 0$, so there is only a minimum and we can answer task 2, but not task 1.
However, why is this? I mean, it is right conceptually, isn't it? Why don't I get a maximum at all, there must be one, right?
When we set the derivative of a differentiable function equal to $0$ we are finding a so called "critical point". An absolute maximum for a differentiable function occurs either at a critical point (which you found by setting the derivative equal to $0$), OR it can occur at one of the boundary points, in this case $t=0$ or $t=5$. Therefore, by finding the critical points and the boundary points, plugging those into the velocity function, and finding whichever is largest, we can determine the maximum velocity. This is because the maximum MUST occur at one of these points. In your case, try the critical point you found ($t=\frac{10}{3}$), $t=0$, and $t=5$. Whichever gives the highest velocity is a maximum.
By the way, using graphs is not "unmathematical" at all, but I understand that you want to find it "analytically".