A drug in the bloodstream has a concentration of c(x) =3t/t^2 + 1
Approximate the highest concentration of the drug reached in the bloodstream
Determine how long it takes for the drug to drop below 0.2. Solve algaebraically
To approximate the highest concentration of the drug reached in the bloodstream, we need to find the maximum value of the concentration function c(x) = 3t / (t^2 + 1).
To find the maximum value, we need to find the critical points of the function. The critical points occur when the first derivative of the function is equal to zero.
So, let's find the first derivative of c(x).
c'(x) = [3(t^2 + 1) - 3t(2t)] / (t^2 + 1)^2
= (3t^2 + 3 - 6t^2) / (t^2 + 1)^2
= (-3t^2 + 3) / (t^2 + 1)^2
To find the critical points, we set c'(x) = 0 and solve for t:
-3t^2 + 3 = 0
Simplifying further, we get:
t^2 - 1 = 0
This can be factored as:
(t - 1)(t + 1) = 0
So, t = 1 or t = -1.
Now, let's find the second derivative of c(x) to determine the nature of these critical points.
c''(x) = [(-3t^2 + 3)'(t^2 + 1)^2 - (-3t^2 + 3)(2(t^2 + 1)(2t))] / (t^2 + 1)^4
= [(-6t)(t^2 + 1)^2 - (-3t^2 + 3)(2(t^2 + 1)(2t))] / (t^2 + 1)^4
= [-6t(t^2 + 1)^2 - 2t(-6t^3 - 6t + 6t^2 + 6)))] / (t^2 + 1)^4
= [-6t(t^2 + 1)^2 - 2t(-6t^3 + 6t^2 + 6))] / (t^2 + 1)^4
= [-6t(t^2 + 1)^2 + 2t(6t^3 - 6t^2 - 6))] / (t^2 + 1)^4
= [-6t(t^2 + 1)^2 + 2t(6t^3 - 6t^2 - 6))] / (t^2 + 1)^4
Plugging in t = 1 and t = -1 to c''(x), we find:
c''(1) = -12 / 4 = -3
c''(-1) = 12 / 4 = 3
Since c''(1) is negative, and c''(-1) is positive, the critical point t = 1 corresponds to a maximum value of the concentration function.
Therefore, the highest concentration of the drug reached in the bloodstream occurs when t = 1.
To determine how long it takes for the drug to drop below 0.2, we need to solve the equation c(t) = 0.2.
Given that c(t) = 3t / (t^2 + 1), we can set up the following equation:
3t / (t^2 + 1) = 0.2
Cross-multiplying, we get:
0.6t = t^2 + 1
Rearranging the equation:
t^2 - 0.6t + 1 = 0
This is a quadratic equation that can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c from the quadratic equation, we can solve for t.
a = 1
b = -0.6
c = 1
t = (-(-0.6) ± √((-0.6)^2 - 4(1)(1))) / (2(1))
= (0.6 ± √(0.36 - 4)) / 2
= (0.6 ± √(-3.64)) / 2
Since the square root of a negative number is not a real number, it means that the drug concentration does not drop below 0.2.
Therefore, algebraically, there is no real solution for the time it takes for the drug to drop below 0.2.