SME
Grade 9 Science · Biology
Population growth and carrying capacity
Use this for “100 animals grow at a rate of 0.5 per day toward a carrying capacity of 1,000. How many are there after 6 days?”
Chart
Fastest growth at t ≈ 4.39, where it reaches half the limit, 500
Horizontal asymptotes N = 0 and N = 1,000
N(6) ≈ 691
This is an example. Type your own values to start a new problem.
N
=
K
÷
1
+
−
N₀
e
−
×
G
Growth now
worked out (not typed) · per day
Formulas
N
=
K
÷
(1
+
(
K
−
N₀
N₀
)
×
e
−
rt
)
691
=
1,000
÷
(1
+
(
1,000 − 100
100
)
×
e
−0.5 × 6
)
G
=
r
×
N
×
K
−
N
K
107
=
0.5
×
691
×
1,000 − 691
1,000
Assumptions
•
Growth G = rN(K − N) ÷ K is fastest at N = K ÷ 2, then slows as resources run short.
•
While N is small, food and space are plentiful and growth is nearly exponential.
•
The population levels off at the carrying capacity K, the most the habitat can support.
Step-by-step
WE KNOW
t = 6 days, K = 1,000, N₀ = 100, r = 0.5 per day
FIND
population (N), growth now (G)
Step 1 · Find population (N)
Use N = K ÷ (1 + ((K − N₀) ÷ N₀) × e⁻ʳᵗ)
Work
out
A
=
K
−
N₀
N₀
,
then
K
1 + A ÷
e
rt
:
N₀
at
t
=
0,
K
after
a
long
time.
N
=
1,000
÷
(1
+
(
1,000 − 100
100
)
×
e
−0.5 × 6
)
A
=
1,000 − 100
100
=
9
e
0.5 × 6
=
e
3
≈
20.0855
N
=
1,000
1 + 9 ÷ 20.0855
N
=
1,000
1.4481
N
=
691
(about
691
individuals)
Step 2 · Find growth now (G)
Use G = r × N × (K − N) ÷ K
The
exponential
rate
rN,
slowed
by
the
share
of
K
still
unused,
K
−
N
K
.
G
=
0.5
×
691
×
1,000 − 691
1,000
G
=
0.5
×
691
×
309
1,000
G
=
345.5
×
309
1,000
G
=
106,759.5
1,000
G
=
107
per
day
(about
107
individuals
a
day,
on
average)
Check
Put the numbers back in. Both sides should match.
691
≈
1,000
÷
(1
+
(
1,000 − 100
100
)
×
e
−0.5 × 6
)
✓
107
≈
0.5
×
691
×
1,000 − 691
1,000
✓
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