assume that during the first 3 minutes after a foreign substance is introduced into the blood,the rate R'(t) at which new antibodies are produced(in thousands of antibodies per minute) is given by R'(t)=t/(t^2+1) where t is in minutes.find the total quantity of new antibodies in the blood at the end of 3 minutes.
dR/dt = t/(t^2+1)
Assume R = 0 at t = 0
R = integral of t dt/(t^2 + 1)
= (1/2)ln(t^2 + 1)
At t = 3 minutes, R = (1/2) ln(10) = 1.15
Since R is measured in thousands, the answer is 1150.
i don't understand this part:dt/(t^2 + 1)
= (1/2)ln(t^2 + 1)
It's not dt/(t^2 + 1)
It is t dt/(t^2 + 1)
That is just (1/2) du/u if u=t^2+1. That's where the log comes from
To find the total quantity of new antibodies in the blood at the end of 3 minutes, we need to integrate the rate function R'(t) from 0 to 3 with respect to t. This will give us the total quantity of new antibodies produced during this time period.
The given rate function is: R'(t) = t / (t^2 + 1)
To integrate this function, follow the steps below:
Step 1: Write down the integral:
∫(0 to 3) t / (t^2 + 1) dt
Step 2: Simplify the integrand:
We can simplify the integrand by expressing it as a partial fraction. Let's write t / (t^2 + 1) as A / (t^2 + 1), where A is a constant:
t / (t^2 + 1) = A / (t^2 + 1)
Step 3: Find the value of A:
Multiply both sides of the equation by (t^2 + 1):
t = A
Comparing the coefficients on each side of the equation, we get:
A = 1
So, t / (t^2 + 1) can be expressed as 1 / (t^2 + 1).
Step 4: Rewrite the integral with the simplification:
∫(0 to 3) 1 / (t^2 + 1) dt
Step 5: Evaluate the integral:
To evaluate the integral, we can use the substitution method. Let's substitute u = t^2 + 1:
du = 2t dt
Rearranging the equation, we get:
dt = du / (2t)
Substituting these values into the integral, we have:
∫(0 to 3) 1 / u * (du / (2t))
Simplifying further:
∫(0 to 3) 1 / (2tu) du
Now, let's integrate with respect to u:
(1/2) ∫(0 to 3) 1 / u du
= (1/2) ln|u| |(0 to 3)
= (1/2) (ln|3^2 + 1| - ln|0^2 + 1|)
= (1/2) (ln 10 - ln 1)
= (1/2) ln 10
So, the total quantity of new antibodies in the blood at the end of 3 minutes is (1/2) ln 10.