Question

Calc question:
An isotope of neptunium (Np-240) has a half-life of 65 minutes. If the decay of Np-240 is modeled by the differential equation dy/dt= -ky, where t is measured in minutes, what is the decay constant k? (My teacher didn't want to spend class time on this stuff because she says it is Algebra 2)

Answer 1

Any help would be much appreciated

Answer 2

\(\frac{dy}{dx} = -kdt\\ ln|y|=-kt+c\\ y=e^ce^{-kt} \\ y=Qe^{-kt}, Q=e^c\\ y=Qe^{-kt}\\ \frac{1}{2}Q=Qe^{-k(65)}\)

then solve for k you'll get approx. 0.0106.

Answer 3

Whoa, its dy/dt at the beginning. not dy/dx

Answer 4

Where did all of the e's and Q's come from

Answer 5

http://www.slader.com/textbook/9780132014083-calculus-graphical-numerical-algebraic-3rd-edition/357/exercises/22/

Answer 6

This made sense to me, but I didn't know where the 2 came from or why the k became positive

Answer 7

Q is the constant for an exponential function.. i just used it as a variable. But you don't need it because it will cancel out at the end. e and q came from the general exponential function equation since you are dealing with half lives.

not sure how to type in latex so, if you understand the process, you'll know the missing symbols..
first line should be:
\[\int\limits \frac{ dy }{ y }=\int\limits -kdt\] i used the eqn button (not sure if it will show properly)

Answer 8

yes

Answer 9

So why does the k become positive

Answer 10

and where did the 2 in the ln come from?

Answer 11

1/2 not 2,, as I mentioned before, you're calculating 1/2 life.
I assume you're familiar with the general formula for half life right?
that formula is based from exponential function, which I used and derived it to calculate the half life

last part:
\(\frac{-1}{65} ln \frac{1}{2} = k\)
evaluate and you'll get k

Answer 12

Yeah. I'm a bit familiar with it

Answer 13

So they just divided both sides by -1?

Answer 14

And used the rules of logs?

Answer 15

Ah I see now

Answer 16

That's really sneaky :3

Answer 17

yep

Answer 18

Ty for the help

Answer 19

no problem

Answer 20

@Jadzia
question, i have often wondered about raising a lower to the absolute value
e^ln|y| = y , but shouldnt it be |y| ?

Answer 21

yes it should be equal to |y|