Grade 12 · Differential growth models · 377 of 600
Differential growth models · Initial amount a=2.5; Rate k=-0.5
Evaluate at t=2.4 time units. Find the amount, instantaneous rate, and multiplier relative to the initial amount; check the rate against the given differential equation. Givens: Initial amount a=2.5; Rate k=-0.5.
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
An ideal amount satisfies y′=(-0.5)y and y(0)=2.5, so y(t)=2.5e^((-0.5)t). Evaluate at t=2.4 time units. Find the amount, instantaneous rate, and multiplier relative to the initial amount; check the rate against the given differential equation.
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A starting point
Evaluate e^(kt), multiply by the initial amount, then multiply that amount by k to get its rate.
Work through the reasoning
Step 1
Identify the model and target
An ideal amount satisfies y′=(-0.5)y and y(0)=2.5, so y(t)=2.5e^((-0.5)t). Evaluate at t=2.4 time units. The governing relation is y′=ky; y(0)=a; y(t)=a e^(kt). Evaluate e^(kt), multiply by the initial amount, then multiply that amount by k to get its rate.
Step 2
Substitute and calculate
Substitution gives y=2.5 exp((-0.5)(2.4))=0.752986. The rate is k y=(-0.5)(0.752986)=-0.376493, rather than just the amount y.
Step 3
Check the mathematical meaning
Dividing y by 2.5 gives e^((-0.5)(2.4))=0.301194. The rate has negative sign because k=-0.5; an amount and its rate have different units. Time / input: 2.4; Current amount: 0.753; Instantaneous rate: -0.376; Growth multiplier: 0.301. Decimal values are rounded, so use unrounded intermediate values.
The answer
Substitution gives y=2.5 exp((-0.5)(2.4))=0.752986. The rate is k y=(-0.5)(0.752986)=-0.376493, rather than just the amount y. Dividing y by 2.5 gives e^((-0.5)(2.4))=0.301194. The rate has negative sign because k=-0.5; an amount and its rate have different units. Animation check: Time / input: 2.4; Current amount: 0.753; Instantaneous rate: -0.376; Growth multiplier: 0.301. 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.
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Watch the relationship
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The mathematical idea
A differential equation constrains an unknown function through its instantaneous rate. For proportional growth, the exponential solves both the rate relation and the initial condition. Checking only a curve's starting value is insufficient: the derivative must equal k times the amount at every time. An ideal amount satisfies y′=(-0.5)y and y(0)=2.5, so y(t)=2.5e^((-0.5)t). Evaluate at t=2.4 time units. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
y′=ky; y(0)=a; y(t)=a e^(kt)
03 · Reflect and transfer
Explain what changes and why.
Change the initial amount while holding the rate constant. Which readout stays unchanged, and why?
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.