Carrying Capacity worksheet

1. Time to reach the target: 6.828 time units

1. Turn the logistic formula, N = K ÷ (1 + A·e^(−rt)), around so it gives the time t: t = ln[N(K − N₀) ÷ (N₀(K − N))] ÷ r

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{\mathrm{1,000}-52}{52}$$=18.23076923$

3. Work out the ratio inside the natural log (ln)

   $\frac{625\times \left(\mathrm{1,000}-52\right)}{52\times \left(\mathrm{1,000}-625\right)}$$=30.38461538$

4. Divide the natural log by the growth rate r to get the time

   $\frac{\ln \left(30.38461538\right)}{0.5}$$=6.827873$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(18.23076923\right)}{0.5}$$=5.806222$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.5\times \mathrm{1,000}}{4}$$=125$

2. Population at time t: 777.68

1. Use the logistic growth formula, which gives the population N at time t: N(t) = K ÷ (1 + A·e^(−rt))

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{975-45}{45}$$=20.66666667$

3. Work out the shrinking factor e^(−r·t), which gets smaller as time passes

   $e^{\mathrm{-4.4}}$$=0.01227734$

4. Put the numbers into the logistic formula to get the population

   $\frac{975}{1+20.66666667\times 0.01227734}$$=777.678356$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(20.66666667\right)}{0.4}$$=7.571305$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.4\times 975}{4}$$=97.5$

3. Time to reach the target: 10.025 time units

1. Turn the logistic formula, N = K ÷ (1 + A·e^(−rt)), around so it gives the time t: t = ln[N(K − N₀) ÷ (N₀(K − N))] ÷ r

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{\mathrm{1,200}-42}{42}$$=27.57142857$

3. Work out the ratio inside the natural log (ln)

   $\frac{800\times \left(\mathrm{1,200}-42\right)}{42\times \left(\mathrm{1,200}-800\right)}$$=55.14285714$

4. Divide the natural log by the growth rate r to get the time

   $\frac{\ln \left(55.14285714\right)}{0.4}$$=10.024818$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(27.57142857\right)}{0.4}$$=8.29195$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.4\times \mathrm{1,200}}{4}$$=120$

4. Time to reach the target: 11.062 time units

1. Turn the logistic formula, N = K ÷ (1 + A·e^(−rt)), around so it gives the time t: t = ln[N(K − N₀) ÷ (N₀(K − N))] ÷ r

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{\mathrm{1,000}-70}{70}$$=13.28571429$

3. Work out the ratio inside the natural log (ln)

   $\frac{950\times \left(\mathrm{1,000}-70\right)}{70\times \left(\mathrm{1,000}-950\right)}$$=252.42857143$

4. Divide the natural log by the growth rate r to get the time

   $\frac{\ln \left(252.42857143\right)}{0.5}$$=11.062257$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(13.28571429\right)}{0.5}$$=5.173379$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.5\times \mathrm{1,000}}{4}$$=125$

5. Time to reach the target: 12.931 time units

1. Turn the logistic formula, N = K ÷ (1 + A·e^(−rt)), around so it gives the time t: t = ln[N(K − N₀) ÷ (N₀(K − N))] ÷ r

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{\mathrm{1,000}-46}{46}$$=20.73913043$

3. Work out the ratio inside the natural log (ln)

   $\frac{700\times \left(\mathrm{1,000}-46\right)}{46\times \left(\mathrm{1,000}-700\right)}$$=48.39130435$

4. Divide the natural log by the growth rate r to get the time

   $\frac{\ln \left(48.39130435\right)}{0.3}$$=12.931067$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(20.73913043\right)}{0.3}$$=10.106741$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.3\times \mathrm{1,000}}{4}$$=75$

6. Time to reach the target: 10.581 time units

1. Turn the logistic formula, N = K ÷ (1 + A·e^(−rt)), around so it gives the time t: t = ln[N(K − N₀) ÷ (N₀(K − N))] ÷ r

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{\mathrm{1,000}-76}{76}$$=12.15789474$

3. Work out the ratio inside the natural log (ln)

   $\frac{850\times \left(\mathrm{1,000}-76\right)}{76\times \left(\mathrm{1,000}-850\right)}$$=68.89473684$

4. Divide the natural log by the growth rate r to get the time

   $\frac{\ln \left(68.89473684\right)}{0.4}$$=10.581449$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(12.15789474\right)}{0.4}$$=6.244947$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.4\times \mathrm{1,000}}{4}$$=100$

7. Time to reach the target: 8.204 time units

1. Turn the logistic formula, N = K ÷ (1 + A·e^(−rt)), around so it gives the time t: t = ln[N(K − N₀) ÷ (N₀(K − N))] ÷ r

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{\mathrm{1,125}-44}{44}$$=24.56818182$

3. Work out the ratio inside the natural log (ln)

   $\frac{800\times \left(\mathrm{1,125}-44\right)}{44\times \left(\mathrm{1,125}-800\right)}$$=60.47552448$

4. Divide the natural log by the growth rate r to get the time

   $\frac{\ln \left(60.47552448\right)}{0.5}$$=8.204477$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(24.56818182\right)}{0.5}$$=6.402904$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.5\times \mathrm{1,125}}{4}$$=140.625$

8. Population at time t: 810.49

1. Use the logistic growth formula, which gives the population N at time t: N(t) = K ÷ (1 + A·e^(−rt))

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{\mathrm{1,000}-34}{34}$$=28.41176471$

3. Work out the shrinking factor e^(−r·t), which gets smaller as time passes

   $e^{\mathrm{-4.8}}$$=0.00822975$

4. Put the numbers into the logistic formula to get the population

   $\frac{\mathrm{1,000}}{1+28.41176471\times 0.00822975}$$=810.48992$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(28.41176471\right)}{0.4}$$=8.367008$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.4\times \mathrm{1,000}}{4}$$=100$

9. Population at time t: 831.79

1. Use the logistic growth formula, which gives the population N at time t: N(t) = K ÷ (1 + A·e^(−rt))

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{950-52}{52}$$=17.26923077$

3. Work out the shrinking factor e^(−r·t), which gets smaller as time passes

   $e^{\mathrm{-4.8}}$$=0.00822975$

4. Put the numbers into the logistic formula to get the population

   $\frac{950}{1+17.26923077\times 0.00822975}$$=831.785482$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(17.26923077\right)}{0.4}$$=7.122316$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.4\times 950}{4}$$=95$

10. Population at time t: 729.19

1. Use the logistic growth formula, which gives the population N at time t: N(t) = K ÷ (1 + A·e^(−rt))

2. Work out the constant A from the starting point. It compares the room left to grow with the starting population: A = (K − N₀) ÷ N₀

   $\frac{\mathrm{1,000}-47}{47}$$=20.27659574$

3. Work out the shrinking factor e^(−r·t), which gets smaller as time passes

   $e^{\mathrm{-4}}$$=0.01831564$

4. Put the numbers into the logistic formula to get the population

   $\frac{\mathrm{1,000}}{1+20.27659574\times 0.01831564}$$=729.193127$

5. Find the inflection point. Growth is fastest when the population is half of K, which happens at t = ln(A) ÷ r

   $\frac{\ln \left(20.27659574\right)}{0.4}$$=7.523668$

6. The fastest the population can ever grow, reached at half of K: r × K ÷ 4

   $\frac{0.4\times \mathrm{1,000}}{4}$$=100$