To solve for λ, use the one other data point we know: when t = 10, y = 5... so 5 = 8·e

Isolate λ using a logarithm, and we find that λ = -0.1 ln(5/8) ≈ 0.047, so

1b) To find the depth after 3 minutes, simply substitute 3 for t and calculate y:

y = 8·e

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

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 ≈

The half-life T

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

(21.3, 2.9) after one time constant 1·τ

(29.5, 2) after two half-lives 2·T

(42.6, 1.1) after two time constants 2·τ

(44.2, 1) after three half-lives 3·T

(59.0, 0.5) after four half-lives 4·T

(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

We know that at time t=0 the depth is y=30, so substitute: 30 = C·e

(This can be interpreted as the fact that the initial value is 24 below the asymptote!)

Thus y = -24·e

Substitute again: 53 = -24·e

Result: λ ≈ 0.138, so the function is

2b) As before, set t = 12 and calculate: y = -24·e

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

2d) Don't be distracted by the nonzero asymptote: τ and T

The time constant is still just the t-value that makes the exponent -1: τ = 1/0.138 ≈

The half-life again is T

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

which is

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

(7.2, 45.2) after one time constant 1·τ

(10, 48) after two half-lives 2·T

(14.5, 50.8) after two time constants 2·τ

(15.1, 51) after three half-lives 3·T

(20.1, 52.5) after four half-lives 4·T

(21.7, 52.8) after three time constants 3·τ

Plot points, connect smoothly, and you're good to go.