Undergraduate · Logistic differential equations · 624 of 650
Logistic differential equations · Capacity K=4; Growth rate r=0.8
Evaluate at t=3 time units. Find the amount, instantaneous growth rate, and equilibrium capacity. Givens: Capacity K=4; Growth rate r=0.8.
Try the question, use a hint when you need one, and compare your reasoning with the worked solution. The animation starts with this chapter’s values; changing its controls explores a new case.
01 · Make a prediction
Your practice question
The logistic model is y′=0.8y(1−y/4) with y(0)=1/2 and capacity K=4. Evaluate at t=3 time units. Find the amount, instantaneous growth rate, and equilibrium capacity.
Use this scratch space or work on paper. Your notes stay on this page and clear when you leave. Answers are for self-checking; they are not automatically graded.
A starting point
Use y=K/[1+(2K−1)e^(−rt)] and substitute that amount into the differential equation for the rate.
Work through the reasoning
Step 1
Identify the model and target
The logistic model is y′=0.8y(1−y/4) with y(0)=1/2 and capacity K=4. Evaluate at t=3 time units. The governing relation is y′=ry(1−y/K); y(0)=1/2. Use y=K/[1+(2K−1)e^(−rt)] and substitute that amount into the differential equation for the rate.
Step 2
Substitute and calculate
Substituting t=3 gives y=4/[1+(2(4)−1)exp(−0.8(3))]=2.446445. The rate r y(1−y/K)=(0.8)(2.446445)(1−2.446445/4)=0.760137.
Step 3
Check the mathematical meaning
The value 2.446445 lies between the initial 0.5 and K=4; the rate 0.760137 is positive. At finite t the exact value remains below K, even when rounding makes the gap appear small. Time: 3; Model amount: 2.446; Instantaneous growth: 0.76; Capacity: 4. Decimal values are rounded, so use unrounded intermediate values.
The answer
Substituting t=3 gives y=4/[1+(2(4)−1)exp(−0.8(3))]=2.446445. The rate r y(1−y/K)=(0.8)(2.446445)(1−2.446445/4)=0.760137. The value 2.446445 lies between the initial 0.5 and K=4; the rate 0.760137 is positive. At finite t the exact value remains below K, even when rounding makes the gap appear small. Animation check: Time: 3; Model amount: 2.446; Instantaneous growth: 0.76; Capacity: 4. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
Use Play, Pause, and the timeline to inspect the construction. Reset restores the question’s original settings. The displayed assumptions describe where this model applies.
Watch the relationship
PausedHD animation studio
From experiment to screen.
Present a crisp Canvas scene, save a full-HD image, or capture your model as a silent video.
The mathematical idea
Logistic growth reduces its proportional rate as the amount approaches a carrying capacity. The equilibrium K is not reached in finite time from the selected smaller initial value. The exact solution separates a model's asymptotic prediction from a finite animation endpoint. The logistic model is y′=0.8y(1−y/4) with y(0)=1/2 and capacity K=4. Evaluate at t=3 time units. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
y′=ry(1−y/K); y(0)=1/2
03 · Reflect and transfer
Explain what changes and why.
Why does a small growth rate near capacity indicate saturation rather than a suddenly negative population?
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.