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.

Question

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.

Answer 1

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.

Answer 2

i don't understand this part:dt/(t^2 + 1)

= (1/2)ln(t^2 + 1)

Answer 3

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

Answer 4

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.