Question

To test an​ individual's use of a certain​ mineral, a researcher injects small amount of a radioactive form of that mineral into the​ person's bloodstream. The mineral remaining in the bloodstream is measured each day for several days. Suppose the amount of the mineral remaining in the bloodstream​ (in milligrams per cubic​ centimeter) t days after the initial injection is approximated by

C(t) = 1/2(4t+2)^-1/2

Find the rate of change C with respect to time for each of the given number of days.

a. t=1

b. t=4

c. t=7

d. t=6.5

Answer 1

Solution: The given concentration function milligram 2 cct) percubic centimeter Ž. (4++2) Rate of Change of 'C' with respect to time ct with respect to 't for rate of change differentiating d dt dt C'(t) = d City 1.(4t+2332 Ź *(-4) (4t+95%** d (4t+2) » c'ét dt - & C'Ct) **( 4t+2522 x(411 +9 1 >> cit) (47+2992 -1 63/2 centim percubic the decrease in @ Rate of change of Cit) al t= 1 day C'(1) =-0.0600 milligran (4+1 +2) 3/2 Negative sign depicts day, concentration. : = 4 days 484934912 Z-0.0131 milligramper experday Rate of change of Cit) at t = 4 C'(4) = al 1832 C'(7) = 30² at t= 7 days Rate of change of cet) -1 (464+2932 C'CH) = -6.0850 x163 milligram percciperday d) Rate of change of act) at t = 615 days (4+65 +2) / c'(65) Ě -6.749 48103 milligramper acper day * Please give feedback! Thank you. C'16.5) - 1 -| 2032 Scanned with CamScanner