# Calculus

A chemical reaction proceeds in such a way that after the first second, the amount of a
certain chemical involved in the reaction changes at a rate that’s inversely proportional to the product of the mass of the chemical present (in grams) and the time elapsed since the reaction began (in seconds).
A. The mass m = m(t) of this chemical is modeled with what differential equation for time t ≥ 1 second?
B. The solution to the differential equation modeling the mass m of the chemical at time t
seconds is m(t) = 2k lnt + C for t ≥ 1, where k and C are undetermined constants. Show how this equation is derived from your answer in part A.
C. Suppose that the amount of this chemical involved in the reaction is 40 grams at time t = 1 second and 30 grams at time t = 10 seconds. Find an explicit equation for the mass m of the chemical as a function of t, for t ≥ 1. Your equation should not involve any unknown constants or any calculator numbers.
D. According to your equation for m(t) in part C, at what time does the mass of the chemical involved in the reaction become zero? (You may use your calculator here.)

Could you also explain your answer? I would really appreciate it so that I could understand it. Thanks!

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1. I don't like the solution proposed in part B.
We are told that

the amount of a certain chemical involved in the reaction changes at a rate that’s inversely proportional to the product of the mass of the chemical present (in grams) and the time elapsed

That means that
dm/dt = k/(mt)
m dm = k/t dt
1/2 m^2 = k lnt + c
m^2 = 2k lnt + c

Anyway, once you have the equation, pluggin in the numbers for C and D should be no problem. C'mon back if you get stuck, and show how far you got.

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oobleck

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