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 determine 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 take the derivative of c(t) with respect to t and set it equal to zero. Let's find the derivative first.
c'(t) = (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
Setting c'(t) equal to zero:
-3t^2 + 3 = 0
Simplifying the equation:
t^2 = 1
Taking the square root of both sides:
t = ± 1
Now, we have the critical points t = ± 1. To determine if these critical points correspond to the maximum, we need to check the second derivative.
Taking the second derivative of c(t):
c''(t) = [(-6t(t^2 + 1)^2) - (-3t^2 + 3)(2(t^2 + 1)(2t))]/(t^2 + 1)^4
= [(-6t^3 - 6t)(t^2 + 1) + (6t^3 - 6t)(2t)]/(t^2 + 1)^4
= [(-6t^5 - 6t + 6t^3 + 6t^3 - 6t^5 + 3t^2 - 3t^2 + 3)/(t^2 + 1)^4
= (12t^3)/(t^2 + 1)^4
Plugging in t = 1 and t = -1:
c''(1) = 12(1)^3/(1^2 + 1)^4 = 12/16 = 3/4
c''(-1) = 12(-1)^3/((-1)^2 + 1)^4 = -12/16 = -3/4
Since the second derivative at t = 1 is positive (3/4) and at t = -1 is negative (-3/4), we can conclude that t = 1 corresponds to the maximum concentration of the drug in the bloodstream.
To determine how long it takes for the drug to drop below 0.2, we need to solve the equation c(t) = 0.2 algebraically.
0.2 = 3t/(t^2 + 1)
Multiplying both sides by (t^2 +1) to remove the denominator:
0.2(t^2 + 1) = 3t
Expanding the equation:
0.2t^2 + 0.2 = 3t
Rearranging the terms:
0.2t^2 - 3t + 0.2 = 0
We can solve this quadratic equation for t. However, it seems to be too tedious to solve algebraically by factoring or using the quadratic formula. Instead, let's use numerical methods or a graphing calculator to approximate the solution.
To find the highest concentration of the drug reached in the bloodstream, we need to find the maximum value of the function c(x) = 3t / (t^2 + 1).
To find the maximum value, we can find the derivative of the function and set it equal to zero.
Let's find the derivative of c(x) with respect to t:
c'(t) = [3(t^2 + 1) - 3(2t)(t)] / (t^2 + 1)^2
= [3t^2 + 3 - 6t^2] / (t^2 + 1)^2
= [3 - 3t^2] / (t^2 + 1)^2
Now, set c'(t) = 0 and solve for t:
[3 - 3t^2] / (t^2 + 1)^2 = 0
3 - 3t^2 = 0
3t^2 = 3
t^2 = 1
t = ±1
Since we are looking for the highest concentration, we can disregard the negative solution. Therefore, t = 1.
Now, substitute t = 1 back into the original concentration function to find the maximum concentration:
c(1) = 3(1) / (1^2 + 1)
= 3/2
Hence, the highest concentration of the drug reached in the bloodstream is 3/2.
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To determine how long it takes for the drug to drop below 0.2, we need to set up the inequality:
c(x) < 0.2
Substitute the concentration function into the inequality and solve for t:
3t / (t^2 + 1) < 0.2
Multiply both sides by (t^2 + 1):
3t < 0.2(t^2 + 1)
Expand the right side:
3t < 0.2t^2 + 0.2
Move all terms to one side to form a quadratic inequality:
0.2t^2 - 3t + 0.2 > 0
To solve this inequality algebraically, we can first find the roots of the quadratic equation by setting it equal to zero:
0.2t^2 - 3t + 0.2 = 0
Next, we can factor or use the quadratic formula to find the roots:
The roots of the equation are t = (-b ± √(b^2 - 4ac)) / 2a
t = (3 ± √(3^2 - 4(0.2)(0.2))) / (2(0.2))
t = (3 ± √(9 - 0.16)) / 0.4
t = (3 ± √8.84) / 0.4
Now, we need to determine if the drug is below 0.2 before or after these roots. Evaluating the concentration function at these points, we can see:
c(0.4) ≈ 0.371 < 0.2
c(3.1) ≈ 0.217 < 0.2
Therefore, the drug drops below 0.2 between t = 0.4 and t = 3.1.
Hence, approximately, it takes around 0.4 to 3.1 units of time for the drug concentration to drop below 0.2.