Chemical spill word problem in calculus

Question

When a chemical from a factory is spilled into a lake, the function $$c(t)=\frac{70t}{5t^2+6}$$ is used to model the concentration (in grams/ liter) of the chemical in a nearby stream, where $t$ is the time (in days) since the spill. Determine how fast the concentration is decreasing when it reaches $3$ grams per liter.

Answer 1

$c(t) = \frac {70t}{5t^2 + 6}$

So you need to do two things:

1) When does $c(t) = 3$. Solve for $c(t) = 3 \implies t =???$.

I expect you to be able to do that on your own.

$\frac {70t}{5t^2 + 6} = 3$

There will be two possible answers.

2) Calculate $c'(t)$. You should use the quotient rule for that. I expect you to be able to do that on your own.

$f(t) = 70t; g(t) = 5t^2 +6$ so $c(t) = \frac {f(t)}{g(t)}$ and $c'(t) = \frac {f'(t)g(t) - f(t)g'(t)}{g^2(t)}$.

Evaluat $c'(t)$ for the two values of $t$ above.

One will be positive and one will be negative. The concentration will be increase at that time with the positive derivative and decreasing at the time with the negative derivative.

At that time $t$ we will have the concetration is $c(t) = 3 \frac gl$ and that time the rate of change will be $c'(t)$ grams per liter per day.

Ask if you have any further question.