1a) We want an exponential equation starting at 8 and approaching 0, so y = 8·e^-λt should work.
To solve for λ, use the one other data point we know: when t = 10, y = 5... so 5 = 8·e^-10λ.
Isolate λ using a logarithm, and we find that λ = -0.1 ln(5/8) ≈ 0.047, so y = 8·e^-0.047t.

1b) To find the depth after 3 minutes, simply substitute 3 for t and calculate y:
y = 8·e^-0.047·3 ≈ 6.95 m.

1c) To find the time at which the depth will be 1 cm (0.01 m), set y = 0.01 m and solve for t:
0.01 = 8·e^-0.047t; again, using a logarithm, t = -ln(0.01/8) / 0.047 ≈ 142 minutes.

1d) The time constant τ is the time it takes for the exponent to become -1 (leaving 1/e of the original amount)
so set the exponent -0.047τ equal to -1. Thus the time constant is τ = 1/0.047 ≈ 21.3 minutes.
The half-life T_1/2 is ln(2) times the time constant, so T_1/2 ≈ 14.7 minutes.

1e) The graph should be an exponential decay curve starting at (0, 8) and approaching y=0 as an asymptote.
Some useful examples of points on the graph include:
(14.7, 4) after one half-life 1·T_1/2
(21.3, 2.9) after one time constant 1·τ
(29.5, 2) after two half-lives 2·T_1/2
(42.6, 1.1) after two time constants 2·τ
(44.2, 1) after three half-lives 3·T_1/2
(59.0, 0.5) after four half-lives 4·T_1/2
(63.8, 0.4) after three time constants 3·τ
Plot these points and connect with a smooth curve to get an impressively accurate graph.


2a) We want an exponential equation that approaches 54, so start with y = C·e^-λt + 54 to shift the asymptote.
We know that at time t=0 the depth is y=30, so substitute: 30 = C·e⁰ + 54 = C+54, so C is -24.
(This can be interpreted as the fact that the initial value is 24 below the asymptote!)
Thus y = -24·e^-λt + 54, and we can find λ using one more data point: (23, 53).
Substitute again: 53 = -24·e^-23λ + 54, and solve for λ using a logarithm.
Result: λ ≈ 0.138, so the function is y = -24·e^-0.138t + 54.

2b) As before, set t = 12 and calculate: y = -24·e^-0.138·12 + 54 ≈ 49.4 cm.

2c) As before, set y = 47 and solve: t ≈ 8.9 minutes.

2d) Don't be distracted by the nonzero asymptote: τ and T_1/2 depend only on the exponential itself.
The time constant is still just the t-value that makes the exponent -1: τ = 1/0.138 ≈ 7.25 minutes.
The half-life again is T_1/2 = τ·ln(2) ≈ 5 minutes.

2e) The graph should be an exponential decay curve starting at (0, 30) and approaching y=54 as an asymptote.
(Note: this still counts as decay even though it's an increasing function, because it's converging, not diverging.)
Important thing to keep in mind: since the asymptote is nonzero, "half-life" doesn't mean you're dropping
halfway to zero; it means you're going halfway towards the asymptote. For example, we start at y=30,
which is 24 below the asymptote. Half of that would be 12 below the asymptote, so y = 54-12 = 42.
I find it very helpful to think of it as ignoring the "zero" value and measuring everything from the asymptote instead.
Some useful examples of points on the graph include:
(5, 42) after one half-life 1·T_1/2
(7.2, 45.2) after one time constant 1·τ
(10, 48) after two half-lives 2·T_1/2
(14.5, 50.8) after two time constants 2·τ
(15.1, 51) after three half-lives 3·T_1/2
(20.1, 52.5) after four half-lives 4·T_1/2
(21.7, 52.8) after three time constants 3·τ
Plot points, connect smoothly, and you're good to go.