Question

A 0.7 ml dose of a drug is injected into a patient steadily for 0.4 secs.At the nd of this time, the quantity Q, of the drug in the body starts to decay exponentially at a continuous rate of 0.65 percent per second. Using formulas, express Q as acontinuous function of time,t , in secs
then Q(t)=?

Answer 1

dQ/dt = - k Q

k = 0.0065/s

Q(t) = Q(0) exp(-k t)

If you have not studied calculus, this would be too hard.

Answer 2

I know... but what is Q(0)? Is it 7/4

Answer 3

Mary:

The problem starts out by stating that the drug is injected at the rate of 0.7 ml per second for 0.4 second. thus, the initial amount of the drug in the patient's body is

0.7 ml 0.4 sec
------ * ------- = 0.28 ml = Q(0)
sec 1

Answer 4

Thank you so much

Answer 5

My great pleasure! MM

Answer 6

but it does not say 0.7 ml per second

Answer 7

True. I had to make an assumption here: that the 0.7 represents the RATE at which the drug is injected into the patient's body, and 0.4 represents the period of time over which the drug is being injected.

the only alternative that I can think of off hand is that a total of 0.7 ml of the drug is injected into the patient's body, period, meaning that the 0.4 sec is irrelevant. i prefer my original interpretation.

Answer 8

sure ... that was the confusion because if we don't take the o.4 into consideration we are getting the answer wrong

Answer 9

"dose" means the total quantity. rate of injection is almost irrelevant, except to indicate this time factor is not very small compared with the decay time of interest. Q should be 0.7. I think the 0.4s factor may be a delay that you want to include in your calculation, starting Q(0) at t-0.4s? They state it starts to decay after the injection, suggesting you need not consider small decay during it.

Answer 10

ok.. Thank you!!!! :)

Answer 11

o thank you so much ....this might be correct

Answer 12

The question is a bit ambiguous. they may want to see

Q(t) = 1.75 t (0≤t≤0.4 sec)
Q(t)= 0.7 exp(-0.0065(t-0.4)) ( 0.4< t)

(forgot the minus sign in the exponential decay)

Answer 13

yea :)

Answer 14

after looking at this problem,
I would say the second equation should be

1-0.0065 = 0.9935 as the base

Q(t) = 0.7 (0.9935)^(t-0.4) for 0.4 sec < t

Answer 15

ok thanks

Answer 16

if you want to use the base e, we could write this as

e^(ln(0.9935) * (t-0.4) )

e^ln(0.9935) is e^(-0.0065212...)
(the exponent is pretty close to 0.0065,but not exactly 0.0065)

Answer 17

ok thanks :)

Answer 18

counting time from completion of injection,
Q(t) = 0.7e^-0.0065t