The percentage of alcohol in a person's bloodstream t hr after drinking 8 fluid oz of whiskey is given by the following equation.
A(t) = 0.26te-0.3t , (0 t 12)
(a) What is the percentage of alcohol in a person's bloodstream after 1/2 hr? After 6 hr? (Round your answers to four decimal places.)
after 1/2 hr %
after 6 hr %
(b) How fast is the percentage of alcohol in a person's bloodstream changing after 1/2 hr? After 6 hr? (Round your answers to four decimal places.)
after 1/2 hr %/hr
after 6 hr %/hr
(a) To find the percentage of alcohol in a person's bloodstream after 1/2 hr, we need to substitute t = 1/2 into the equation A(t) = 0.26te^(-0.3t):
A(1/2) = 0.26 (1/2)e^(-0.3 * 1/2)
A(1/2) = 0.13e^(-0.15)
Using a calculator, we find that A(1/2) is approximately 0.1173, which corresponds to 11.73% (rounded to four decimal places).
To find the percentage of alcohol in a person's bloodstream after 6 hr, we substitute t = 6 into the equation:
A(6) = 0.26 (6)e^(-0.3 * 6)
A(6) = 1.56e^(-1.8)
Using a calculator, we find that A(6) is approximately 0.0454, which corresponds to 4.54% (rounded to four decimal places).
Therefore, after 1/2 hr, the percentage of alcohol in a person's bloodstream is 11.73%, and after 6 hr, it is 4.54%.
(b) To find how fast the percentage of alcohol in a person's bloodstream is changing after 1/2 hr, we need to find the derivative of the function A(t) with respect to t, and then substitute t = 1/2.
A'(t) = (0.26e^(-0.3t)) - (0.3te^(-0.3t))
Substituting t = 1/2 into the derivative:
A'(1/2) = (0.26e^(-0.3 * 1/2)) - (0.3 * 1/2 * e^(-0.3 * 1/2))
A'(1/2) = 0.26e^(-0.15) - 0.15e^(-0.15)
Using a calculator, we find that A'(1/2) is approximately -0.0087, which corresponds to -0.87%/hr (rounded to four decimal places).
To find how fast the percentage of alcohol in a person's bloodstream is changing after 6 hr, we substitute t = 6 into the derivative:
A'(6) = (0.26e^(-0.3 * 6)) - (0.3 * 6 * e^(-0.3 * 6))
A'(6) = 0.26e^(-1.8) - 1.08e^(-1.8)
Using a calculator, we find that A'(6) is approximately -0.0061, which corresponds to -0.61%/hr (rounded to four decimal places).
Therefore, after 1/2 hr, the percentage of alcohol in a person's bloodstream is changing at a rate of -0.87%/hr, and after 6 hr, it is changing at a rate of -0.61%/hr.
To find the percentage of alcohol in a person's bloodstream after a certain amount of time, we can substitute the given time values into the equation A(t) = 0.26te^(-0.3t).
(a)
To find the percentage of alcohol in the bloodstream after 1/2 hour, substitute t = 1/2 into A(t) equation:
A(1/2) = 0.26(1/2)e^(-0.3 * 1/2)
Simplifying the equation, we have:
A(1/2) = 0.13e^(-0.15)
Calculating the value of e^(-0.15), we find it to be approximately 0.86.
A(1/2) = 0.13 * 0.86
A(1/2) ≈ 0.1118
Therefore, the percentage of alcohol in the person's bloodstream after 1/2 hour is approximately 0.1118%.
To find the percentage of alcohol in the bloodstream after 6 hours, substitute t = 6 into A(t) equation:
A(6) = 0.26(6)e^(-0.3 * 6)
Simplifying the equation, we have:
A(6) = 1.56e^(-1.8)
Calculating the value of e^(-1.8), we find it to be approximately 0.1653.
A(6) = 1.56 * 0.1653
A(6) ≈ 0.2578
Therefore, the percentage of alcohol in the person's bloodstream after 6 hours is approximately 0.2578%.
(b)
To find how fast the percentage of alcohol in the bloodstream is changing after 1/2 hour, we need to find the derivative of A(t):
A'(t) = (0.26te^(-0.3t))'
Using the product rule, we differentiate the equation:
A'(t) = 0.26e^(-0.3t) + 0.26te^(-0.3t)(-0.3)
Simplifying the equation, we have:
A'(t) = 0.26e^(-0.3t) - 0.078te^(-0.3t)
To find A'(1/2), substitute t = 1/2 into the derivative equation:
A'(1/2) = 0.26e^(-0.3 * 1/2) - 0.078(1/2)e^(-0.3 * 1/2)
Simplifying the equation, we have:
A'(1/2) = 0.26e^(-0.15) - 0.039e^(-0.15)
To calculate the value of e^(-0.15), we find it to be approximately 0.86.
A'(1/2) = 0.26 * 0.86 - 0.039 * 0.86
A'(1/2) ≈ 0.2236 - 0.0335
A'(1/2) ≈ 0.1901
Therefore, the percentage of alcohol in the person's bloodstream is changing by approximately 0.1901% per hour after 1/2 hour.
To find how fast the percentage of alcohol in the bloodstream is changing after 6 hours, substitute t = 6 into the derivative equation:
A'(6) = 0.26e^(-0.3 * 6) - 0.078(6)e^(-0.3 * 6)
Simplifying the equation, we have:
A'(6) = 0.26e^(-1.8) - 0.468e^(-1.8)
To calculate the value of e^(-1.8), we find it to be approximately 0.1653.
A'(6) = 0.26 * 0.1653 - 0.468 * 0.1653
A'(6) ≈ 0.0429 - 0.0773
A'(6) ≈ -0.0344
Therefore, the percentage of alcohol in the person's bloodstream is changing by approximately -0.0344% per hour after 6 hours.
(a) just plug in t=1/2 and t=6
(b) A' = -0.078 e^-0.3t
plug in the values for t.