Exponentials, logarithms and modelling · Exponenciais, logaritmos e modelagem
| English | Português |
|---|---|
| half-life/hɑːf laɪf/ | vida média |
Why does a decay model stay positive?
- A medicine concentration falls by the same percentage each hour. A constant subtraction would eventually predict a negative amount.
- This lesson studies half-life 半衰期: The time for a decaying quantity to fall to half its initial value.
Choose the mathematical structure
- For y=Ae^(kt), k is a proportional rate. Taking logs gives ln y=ln A+kt. Logarithms require positive arguments, and log(x+y) is not log x+log y.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines half-life? · Qual descrição define corretamente a meia-vida?
The time for a decaying quantity to fall to half its initial value. · O tempo para uma quantidade decrescente cair à metade de seu valor inicial.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
If y=80e^(-0.2t), y=40 gives e^(-0.2t)=1/2. Hence t=ln2/0.2≈3.466. For 3^x=20, x=ln20/ln3. Plotting ln y against t linearises this exponential model.
Exponentials, logarithms and modelling · Exponenciais, logaritmos e modelagem
For y=Ae^(kt), k is a proportional rate · Para y=Ae^(kt), k é uma taxa proporcional
Explain why the half-life depends on 0.2 and not on the initial amount 80. · Explique por que a meia-vida depende de 0.2 e não da quantidade inicial 80.
Solve 2^x=32. · Resolva 2^x=32.
32=2⁵, so x=5. · 32=2⁵, logo x=5.
Test a tempting shortcut
- A fitted exponential is a model, not a guarantee. Specify the time units and range of use. A negative k describes decay; a negative starting amount usually contradicts the context.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
For positive x and y, ln(x+y)=ln x+ln y. This claim is false. Explain which definition or assumption it violates.
Find ln(e³). · Encontre ln(e³).
The natural logarithm undoes exponentiation: ln(e³)=3. · O logaritmo natural inverte a exponenciação: ln(e³)=3.
For positive x and y, ln(x+y)=ln x+ln y. · Para x e y positivos, ln(x+y)=ln x+ln y.
A fitted exponential is a model, not a guarantee. Specify the time units and range of use. A negative k describes decay; a negative starting amount usually contradicts the context. · Um ajuste exponencial é um modelo, não uma garantia. Especifique as unidades de tempo e o intervalo de uso. Um k negativo descrece decaimento; um valor inicial negativo geralmente contradiz o contexto.
Interpret a new situation
- Compare actual observations with the model. Systematic departures may indicate changing conditions. In a report, explain what the rate and initial value mean, rather than giving a bare equation.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the half-life of 80e^(-0.2t), to three decimal places. · Encontre a meia-vida de 80e^(-0.2t), com três casas decimais.
Set e^(-0.2t)=1/2. Then t=ln(2)/0.2≈3.466 to three decimal places. · Igualdade e^(-0.2t)=1/2. Então t=ln(2)/0.2≈3.466 com três casas decimais.
Match each part of a complete solution to its purpose. · Associe cada parte de uma solução completa ao seu propósito.
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · Uma suposição justifica o modelo; uma verificação testa o resultado; a interpretação conecta-o à pergunta.
Use this in your course
- edexcel IAL mathematics; official unit P2. Other-unit enrichment is identified in the scope review; it is not an extra cash-in requirement.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The time for a decaying quantity to fall to half its initial value. Choose the relationship, show the method, check its assumptions and interpret the result.