differential equation problem help needed

[bubblegum5678]

Assume that the dynamics of caffeine absorption is given by
ct + 1 = 0.87ct + d,
where t is time in hours and d is the amount of caffeine taken every hour.


If the initial amount of caffeine is
c0 = 800
mg and no new caffeine is consumed, estimate the time needed for 60% of the caffeine to be eliminated from the body. (Round your answer to the nearest integer.)

 

[topsquark]

Assume that the dynamics of caffeine absorption is given by
ct + 1 = 0.87ct + d,
where t is time in hours and d is the amount of caffeine taken every hour.


If the initial amount of caffeine is
c0 = 800
mg and no new caffeine is consumed, estimate the time needed for 60% of the caffeine to be eliminated from the body. (Round your answer to the nearest integer.)

There's something wrong here. The equation says that d would have to be a function of t, not a constant. Please check your problem statement.

And this should be a differential equation...

-Dan

 

[HallsofIvy]

I'm going to guess that you do NOT mean "ct+ 1", 1 added to the number ct, but rather "\(\displaystyle c_{t+1}\)" and that this is NOT a "differential equation" but rather a "difference equation": \(\displaystyle c_{t+1}= 0.86c_t+ d\). But we are also told that d= 0 so this is just \(\displaystyle c_{t+1}= 0.86c_t\). That is, the amount of caffeine in the body at each hour is 0.86 times the amount of caffeine the previous hour. If \(\displaystyle c_0\) is the initial amount of caffeine in the body then the amount after one hour is \(\displaystyle c_1= 0.86c_0\), after 2 hours, \(\displaystyle c_2= 0.86c_1= 0.86(0.86c_0)= (0.86)^2c_0\), after 3 hours, \(\displaystyle c_3= 0.86c_2= (0.86)^3c_0\), etc. In general, after n hours, \(\displaystyle c_n= (0.86)^nc_0\).

60% of the caffeine will have been eliminated when 40% is left: You want to find n such that \(\displaystyle (0.86)^n= .40\). Take the logarithm of both sides to find n.