fter a foreign substance is introduced into the blood, the rate at which antibodies are made is modeled by
f(t) = 27 t/8 t^2 + 3
where f(t) is in thousands of antibodies per minute, and t is measured in minutes with 0 ≤ t ≤ 60. Assuming that there are no antibodies present at time t = 0, find the total number of antibodies in the blood at the end of 60 minutes. Round your answer to two decimal places.
Number of Antibodies (in thousands) =
This makes no sense to me:
f(t) = 27 t/8 t^2 + 3
do you mean
f(t) = 27 t/(8 t^2 + 3 )
or what?
yeah...sorry
To find the total number of antibodies in the blood at the end of 60 minutes, we need to find the definite integral of the rate of antibody production function f(t) from 0 to 60.
First, let's simplify the function:
f(t) = 27t / (8t^2 + 3)
Now, let's find the definite integral:
∫[0,60] f(t) dt = ∫[0,60] 27t / (8t^2 + 3) dt
To solve this integral, we will use a substitution. Let's substitute u = 8t^2 + 3, then du = 16t dt:
∫[0,60] 27t / (8t^2 + 3) dt = (27/16) ∫[0,60] 1/u du
Now, we can solve the integral:
(27/16) ∫[0,60] 1/u du = (27/16) ln|u| |[0,60]
Substituting back u = 8t^2 + 3:
(27/16) ln|8t^2 + 3| |[0,60]
Plugging in the upper and lower limits:
(27/16) [ln|8(60^2) + 3| - ln|8(0^2) + 3|]
Simplifying:
(27/16) [ln|28803| - ln|3|]
Using a calculator, we find:
(27/16) [ln(28803) - ln(3)] ≈ 28.39
Therefore, the total number of antibodies in the blood at the end of 60 minutes is approximately 28.39 thousands of antibodies. Rounded to two decimal places, the answer is 28.39.
To find the total number of antibodies in the blood at the end of 60 minutes, we need to calculate the definite integral of the rate function f(t) over the interval [0, 60].
The rate function is given by:
f(t) = (27t) / (8t^2 + 3)
To calculate the integral, we'll start by finding the antiderivative of the rate function:
F(t) = ∫ f(t) dt
To find the antiderivative, we can use the power rule of integration. The power rule states that the integral of x^n dx is (1/(n+1))x^(n+1) + C, where C is the constant of integration.
Applying the power rule, we get:
F(t) = ∫ (27t) / (8t^2 + 3) dt
= (27/8) ∫ (t) / (t^2 + 3/8) dt
Now, we can use the substitution method. Let u = t^2 + 3/8. Then, du = 2t dt.
Rewriting the integral in terms of u:
F(t) = (27/8) ∫ (1/2) (1/u) du
= (27/16) ln|u| + C
Substituting back for u:
F(t) = (27/16) ln|t^2 + 3/8| + C
Since we know that there are no antibodies present at t = 0, we can determine the constant of integration C.
F(0) = (27/16) ln|(0^2 + 3/8)| + C
= (27/16) ln|(3/8)| + C
= (27/16) ln(3/8) + C
Now, we can find the total number of antibodies at the end of 60 minutes by evaluating F(60) - F(0):
Number of Antibodies (in thousands) = F(60) - F(0)
= (27/16) ln|(60^2 + 3/8)| - (27/16) ln(3/8)
Calculating this expression will give us the total number of antibodies in the blood after 60 minutes. Round the answer to two decimal places.