# After a foreign substance is introduced into the blood, the rate at which antibodies are made is...

## Question:

After a foreign substance is introduced into the blood, the rate at which antibodies are made is given by {eq}r(t) = \frac{t}{t^2 + 1} {/eq} thousands of antibodies per minute, where time, {eq}t {/eq}, is in minutes. Assuming there are no antibodies present at time {eq}t = 0 {/eq}, find the total quantity of antibodies in the blood at the end of 6 minutes.

## Defining a Function from its Rate of Change

Since the derivative of a quantity represents its rate of change, we can return from a rate of change to an original function by finding the opposite of a derivative. The opposite of a derivative is an antiderivative, so if we can find the antiderivative, we can find the original function.

The function given, r(t), represents the rate of change of the amount of antibodies in the bloodstream. To find the total amount of antibodies made at this rate, we need to find the antiderivative of this function. Since we have a rational expression, we need to consider substitution as a possible method. Let's define the following substitution.

{eq}u = t^2 + 1\\ du = 2t dt {/eq}

We can use this method to find the antiderivative of our rate of change, since we do indeed have a very similar expression to du in our function. We're only off by a coefficient, which we can introduce. This makes the rate of change have the following form, which we can then perform the substitution on.

{eq}r(t) = \frac{1}{2} \cdot \frac{2t}{t^2 + 1}\\ r(u) = \frac{1}{2} \cdot \frac{1}{u} {/eq}

The antiderivative of the reciprocal function is the natural logarithm, so we can find this antiderivative and then reverse the substitution.

{eq}R(u) = \frac{1}{2} \ln u + c\\ R(t) = \frac{1}{2} \ln (t^2 + 1) + c {/eq}

Since we have the initial value for this function, we can use it to find the value of this constant term.

{eq}R(0) = \frac{1}{2} \ln ((0)^2 + 1) + c = 0\\ \rightarrow c= 0 {/eq}

With this constant now defined, we have a definite function representing the amount of antibodies at any time. We can evaluate this at any time to find the amount of antibodies at that moment. For six minutes:

{eq}R(6) = \frac{1}{2} \ln ((6)^2 + 1) = 1.80545896 {/eq}

There are therefore approximately 1,805 antibodies after six minutes have passed.