The concentration C of a chemical in the bloodstream 't' hours after injection into muscle tissue is given by:
C=(3t^2+t)/(t^3+50) where t≥0
What is the rate of change over the interval [0,4]? What is the rate of change over the interval [4,8]? What is the horizontal asymptote of the function? Interpret the meaning of this data in the context of the problem.
To find the rate of change of a function, we need to differentiate it with respect to the independent variable. In this case, we need to find the derivative of the concentration function C(t) with respect to time (t).
1. Rate of change over the interval [0,4]:
To find the rate of change over this interval, we need to differentiate the concentration function C(t) with respect to t and evaluate it at t = 4.
Taking the derivative of C(t) using the quotient rule:
C'(t) = [(2t)(t^3+50) - (3t^2+t)(3t^2)] / (t^3+50)^2
Now, substituting t = 4 into the derivative:
C'(4) = [(2(4))(4^3+50) - (3(4)^2+4)(3(4)^2)] / (4^3+50)^2
Simplifying the expression gives the rate of change over [0,4].
2. Rate of change over the interval [4,8]:
We need to repeat the same process as above but evaluate the derivative at t = 8.
Taking the derivative of C(t):
C'(t) = [(2t)(t^3+50) - (3t^2+t)(3t^2)] / (t^3+50)^2
Substituting t = 8 into the derivative:
C'(8) = [(2(8))(8^3+50) - (3(8)^2+8)(3(8)^2)] / (8^3+50)^2
Simplifying the expression gives the rate of change over [4,8].
3. Horizontal asymptote:
To find the horizontal asymptote of the function, we take the limit of C(t) as t approaches infinity. If the limit exists, that value is the horizontal asymptote.
Taking the limit as t approaches infinity:
lim(t→∞)(C(t)) = lim(t→∞)((3t^2+t)/(t^3+50))
As t approaches infinity, the terms with lower degrees become negligible, and the limit can be evaluated as:
lim(t→∞)(C(t)) = lim(t→∞)((3t^2)/(t^3)) = 0
Therefore, the horizontal asymptote of the function C(t) is y = 0.
Interpretation of the data:
The rate of change over the interval [0,4] tells us how quickly the concentration of the chemical is changing during the first 4 hours after injection into muscle tissue. The rate of change over the interval [4,8] tells us how quickly the concentration is changing between 4 and 8 hours.
The horizontal asymptote at y = 0 means that as time approaches infinity, the concentration of the chemical approaches zero. In the context of the problem, this suggests that the chemical eventually gets completely metabolized and cleared from the bloodstream.
Since you want the rate of change over an interval, it is not "instantaneous" change, so we don't have to use Calculus.
C(0) = 0/50 = 0
C(4) = 52/114
rate of change for [0,4]
= (52/114 - 0)/(4-0) = 13/114
do the same for [4,8]
as t ---> ∞
C ---> 3 , the horizontal asymptote
It suggest that the concentration approaches and remains 3 units, which seems unreasonable