Question

Hi I have question regarding to Problem set 2, Differentiation pdf, 1H-1b.

The solution says:

"by assumption, λ = − ln2/k"

Why assume the λ is negative ? I know "decay" probably means the quantity of something is going down. Is it because of that we use a negative sign here ?

thanks

Answer 1

Hi taking -ve is just 1 choice , an assumption. as u know quantity is becoming lesser. if you take it as +ve , it will not matter. I dont know what prob u are referring to - which would make it more clear to me

Answer 2

The question is to show that y1 is y1/2 after λ .
If λ is positive then we will get y1 = 2y1.
Just want to know why λ is negative. ( I know negative λ gives the right answer )

Answer 3

can u pl paste the complete problem, i dont have it

Answer 4

This is one of the Problem set 2 problems
you can find the question and answer at these locations:

go to section 1H, find question 1 b)

question is here:
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset1prb.pdf

answers are here:
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset1sol.pdf

Answer 5

hello mameko, in part (b) of the problem he used the law of growth instead of the law of decay for radioactive isotopes.

In nuclear physics λ must equal ln 2/k, where k is the decay constant and λ is the half life of the decaying mass.

But he used law of growth with a +ve sign in the equation y.e^kt (growth) instead of y.e^-kt (decay), and this is not proper for a decaying isotope.

At the same time he wanted you to assume a value for λ which will satisfy the condition of the problem part (b), the only solution to get around it is to assume a -ve value for λ.

If he had used law of decay instead, he could have solved the problem with λ = ln 2/k with no issue.

This also tells us a fact that a radioactive isotope will seize to lose its mass after its half life.

Answer 6

Thank you very much for your explanation. I understand it now.