Question

Calculus Help
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Answer 1

@ganeshie8

Answer 2

\[\frac{ dP }{ dt } = kP \] right

Answer 3

right

Answer 4

If you're interested, we can actually re arrange this equation and then intregrate both sides to get us an ice equation.

\[\frac{ dP }{ kP } = dt \]

\[\int\limits \frac{ dP }{ kP } = \int\limits dt \]

Answer 5

Works for me.

Answer 6

So now we just integrate,

\[\int\limits dt = t \]

Answer 7

Okay, 1/k = t right?

Answer 8

We can just move out the constant and then integrate. are you following?

\[\int\limits \frac{ dP }{ kP } = k * \int\limits \frac{ 1 }{ P }*dP\]

so actually we have an identity here

\[\int\limits (\frac{ 1 }{ x }) = \ln|x|+C\]

Sp now we can put everything together and get a nice equation

\[k*\int\limits_{p_{0}}^{p} \frac{ 1 }{ p }*dp = k*\ln(p)-\ln(p_{0}) = k*\ln\frac{ p }{ p_{0} }\]

\[k*\ln(\frac{ P }{ P_{0} }) = t\]

Answer 9

so now we're told that P is the population and t is the time and k is a positive constant.

so If P0 = A what is the time for the population to triple its initial value

now that we've built our equation do you have an idea of how to solve this?

Answer 10

@arianna1453 so far do you follow?

Answer 11

Yes, im following.

Answer 12

So if our initial population is A then P must be 3 times A because its triple.

\[P_{0} = A; P; 3A \]

Answer 13

ln(3)/k ?

Answer 14

Right,

Answer 15

let me check my integration

Answer 16

Okay, I think its ln(3)/k, answer B then.

Answer 17

\[\ln(\frac{ P }{ p_{0} }) = kt \]

Answer 18

so solving for t we get this

\[\frac{ Ln\frac{ P }{ P_{0}} }{ k } = t \]

Answer 19

i forgot to move the constant over to where it needed to be but you can easily see that we plug in and P/P0 is 3

Answer 20

@arianna1453 do you understand how I did the integration in step #1

Answer 21

Yes, I do. Thank you!!!