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 find the highest concentration of the drug reached in the bloodstream, we can take the derivative of the concentration function and find where it reaches a maximum.
The concentration function is given as c(x) = 3t/(t^2 + 1).
Taking the derivative with respect to t, we get:
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
To find the maximum, we need to find where c'(t) = 0.
Setting -3t^2 + 3 = 0 and solving for t, we get:
-3t^2 + 3 = 0
t^2 - 1 = 0
(t + 1)(t - 1) = 0
This gives us two possible values for t: t = -1, t = 1.
Now we need to test these values to see if they correspond to a maximum or a minimum.
For t = -1:
c'(-1) = (-3(-1)^2 + 3)/( (-1)^2 + 1)^2
= (-3 + 3)/2^2
= 0
For t = 1:
c'(1) = (-3(1)^2 + 3)/( (1)^2 + 1)^2
= (3 + 3)/2^2
= 1
Since c'(1) > 0 and c'(-1) = 0, the maximum concentration occurs at t = 1.
To determine how long it takes for the drug to drop below 0.2, we can set the concentration function equal to 0.2 and solve for t:
0.2 = 3t/(t^2 + 1)
Multiply both sides by (t^2 + 1):
0.2(t^2 + 1) = 3t
Rearrange the equation:
0.2t^2 + 0.2 = 3t
0.2t^2 - 3t + 0.2 = 0
Multiply both sides by 10 to get rid of the decimal:
2t^2 - 30t + 2 = 0
Now, we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 2, b = -30, and c = 2.
t = (-(-30) ± √((-30)^2 - 4(2)(2)))/(2(2))
t = (30 ± √(900 - 16))/4
t = (30 ± √884)/4
Simplifying further:
t ≈ (30 ± 29.748)/4
t ≈ 14.937 or t ≈ -0.237
Since time cannot be negative, the drug drops below 0.2 after approximately 14.937 units of time.
To find 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 do this, we can first find the derivative of c(x) with respect to t, which will give us an expression for the rate of change of concentration with respect to time.
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
To find the maximum concentration, we need to find the critical points of c(t), which occur when c'(t) = 0. We can set the numerator equal to 0 and solve for t:
-3t^2 + 3 = 0
t^2 = 1
t = ±1
Now we need to evaluate the concentration function at these critical points to find the maximum value.
c(1) = 3(1) / (1^2 + 1) = 3/2
c(-1) = 3(-1) / ((-1)^2 + 1) = -3/2
Since c(1) is positive and greater than c(-1), the maximum concentration reached in the bloodstream is 3/2.
Now let's determine how long it takes for the drug to drop below 0.2. To do this algebraically, we can set the concentration function equal to 0.2 and solve for t:
0.2 = 3t / (t^2 + 1)
0.2(t^2 + 1) = 3t
0.2t^2 + 0.2 = 3t
0.2t^2 - 3t + 0.2 = 0
Now we can solve this quadratic equation using the quadratic formula or factoring to find the values of t where the concentration drops below 0.2.