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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 ( \frac{dP}{dt} = 3000e^{2t} ):

P(t) = \int 3000e^{2t} dt.

Using the integration rule for exponential functions, we have:

P(t) = 3000 \cdot \frac{e^{2t}}{2} + C = 1500 e^{2t} + C,

where $C$ is the constant of integration.

Step 2

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Answer

Given that when $t = 0$ , $P = 4000$ , we can substitute these values into the equation:

4000 = 1500e^{2 \cdot 0} + C

This simplifies to:

4000 = 1500 \cdot 1 + C,

which leads to:

C = 4000 - 1500 = 2500.$$

Step 3

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Answer

Thus, the expression for $P(t)$ is:

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

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