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Let $P(t)$ be a function such that $ \frac{dP}{dt} = 3000e^{2t} $ - HSC - SSCE Mathematics Advanced - Question 13 - 2023 - Paper 1

Question 13

$Let-$P(t)$-be-a-function-such-that-$-\frac{dP}{dt}-=-3000e^{2t}-$-HSC-SSCE Mathematics Advanced-Question 13-2023-Paper 1.png$

Let $P(t)$ be a function such that $ \frac{dP}{dt} = 3000e^{2t} $. When $ t = 0, P = 4000 $. Find an expression for $ P(t) $.

Worked Solution & Example Answer:Let $P(t)$ be a function such that $ \frac{dP}{dt} = 3000e^{2t} $ - HSC - SSCE Mathematics Advanced - Question 13 - 2023 - Paper 1

Step 1

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Answer

To find ( P(t) ), we need to integrate the differential equation:

$\frac{dP}{dt} = 3000e^{2t}$

Integrating both sides with respect to ( t ):

$P(t) = \int 3000e^{2t} dt = 3000 \cdot \frac{1}{2}e^{2t} + C = 1500e^{2t} + C$

Step 2

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Answer

We use the initial condition ( P(0) = 4000 ) to find the constant ( C ):

$P(0) = 1500e^{0} + C = 4000 \implies 1500 + C = 4000$

Thus,

$C = 4000 - 1500 = 2500$

Step 3

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101 rated

Answer

Substituting back ( C ) into the equation for ( P(t) ):

$P(t) = 1500e^{2t} + 2500$

This is the final expression for ( P(t) ).

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