
V(t) = D′(t) =
A(t) = V′(t) = D″(t) =
1) Solve V(t)=0 to find where D(t) has a maximum or minimum.
t values: t₁ = t₂ =
2) Use A(t)=D″(t) to decide which is a max and which is a min.
D(t) has a minimum at t = and a maximum at t =
3) Inflection point for D(t): solve A(t)=0.
Inflection time: t =
| Interval (seconds) | 0–10 | 10 | 10–20 | 20 | 20–30 | 30 | 30–40 |
|---|---|---|---|---|---|---|---|
| D (ATP availability) | |||||||
| V = D′ | |||||||
| A = V′ = D″ |
| t | 0 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
|---|---|---|---|---|---|---|---|---|---|
| V(t) | |||||||||
| A(t) |
| t | 0 | 10 | 20 | 30 | 40 |
|---|---|---|---|---|---|
| D(t) |