The rate of decay of the mass of a particular substance is modelled by the differential equation $$\frac{dx}{dt} = -\frac{5}{2}x, \quad t > 0$$ where $x$ is the mass of the substance measured in grams and $t$ is the time measured in days. Given that $x = 60$ when $t = 0$, (a) solve the differential equation, giving $x$ in terms of $t$. You should show all steps in your working and give your answer in its simplest form. (b) Find the time taken for the mass of the substance to decay from 60 grams to 20 grams. Give your answer to the nearest minute.

A-Level Edexcel Maths Pure Integration: The rate of decay of the mass of

The rate of decay of the mass of a particular substance is modelled by the differential equation $$\frac{dx}{dt} = -\frac{5}{2}x, \quad t > 0$$ where $x$ is the mass of the substance measured in grams and $t$ is the time measured in days - Edexcel - A-Level Maths Pure - Question 6 - 2016 - Paper 4

Question 6

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The rate of decay of the mass of a particular substance is modelled by the differential equation $$\frac{dx}{dt} = -\frac{5}{2}x, \quad t > 0$$ where $x$ is the mass... show full transcript

Worked Solution & Example Answer:The rate of decay of the mass of a particular substance is modelled by the differential equation $$\frac{dx}{dt} = -\frac{5}{2}x, \quad t > 0$$ where $x$ is the mass of the substance measured in grams and $t$ is the time measured in days - Edexcel - A-Level Maths Pure - Question 6 - 2016 - Paper 4

Step 1

(a) solve the differential equation, giving x in terms of t. You should show all steps in your working and give your answer in its simplest form.

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Answer

To solve the differential equation, we start by separating the variables:

$\frac{dx}{x} = -\frac{5}{2} dt$

Next, we integrate both sides:

$\int \frac{dx}{x} = \int -\frac{5}{2} dt$

This gives us:

$\ln|x| = -\frac{5}{2}t + C$

Exponentiating both sides, we have:

$x = e^{-\frac{5}{2}t + C} = e^C e^{-\frac{5}{2}t}$

Letting $k = e^C$ , we then write:

$x = ke^{-\frac{5}{2}t}$

To find the value of $k$ , we use the initial condition $x = 60$ when $t = 0$ :

$60 = ke^{0} \Rightarrow k = 60$

Thus, the solution is:

$x = 60e^{-\frac{5}{2}t}$

This is $x$ in terms of $t$ . All steps have been shown.

Step 2

(b) Find the time taken for the mass of the substance to decay from 60 grams to 20 grams. Give your answer to the nearest minute.

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Answer

Using the equation derived in part (a), we set up the equation for the mass at two different times:

$20 = 60e^{-\frac{5}{2}t}$

Dividing both sides by 60, we have:

$\frac{1}{3} = e^{-\frac{5}{2}t}$

Taking the natural logarithm of both sides:

$\ln\left(\frac{1}{3}\right) = -\frac{5}{2}t$

This simplifies to:

$t = -\frac{2}{5} \ln\left(\frac{1}{3}\right)$

Calculating this value gives:

$t \approx -\frac{2}{5} \times (-1.0986) = 0.4394 \text{ days}$

To convert days into minutes:

$t \times 24 \times 60 = 0.4394 \times 1440 \approx 632.6 \text{ minutes}$

Rounding to the nearest minute, we find:

$t \approx 633 \text{ minutes}$

Thus, the time taken for the mass to decay from 60 grams to 20 grams is approximately 633 minutes.

The rate of decay of the mass of a particular substance is modelled by the differential equation $$\frac{dx}{dt} = -\frac{5}{2}x, \quad t > 0$$ where $x$ is the mass of the substance measured in grams and $t$ is the time measured in days - Edexcel - A-Level Maths Pure - Question 6 - 2016 - Paper 4

(a) solve the differential equation, giving x in terms of t. You should show all steps in your working and give your answer in its simplest form.

(b) Find the time taken for the mass of the substance to decay from 60 grams to 20 grams. Give your answer to the nearest minute.

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