# Related Rates / Calculating Rate of Change

1. May 1, 2012

### JacksonSolo

1. The problem statement, all variables and given/known data

A lake is polluted by water from a plant located on its shore. Ecologist determine that when the level of pollutant is x parts per million (ppm), there will be F fish in the lake. When there are 4,000 fish in the lake, the pollution is increasing at a rate of 1.4ppm/year. At what rate is the fish population changing at that time?

2. Relevant equations

F = $\frac{32,000}{3+\sqrt{x}}$

3. The attempt at a solution

Ok so if I understand correctly, Im trying to find the change in F over the change in time (one year) or:$\frac{dF}{dt}$. They give me $\frac{dx}{dt}$ = 1.4. But I dont know how to use that to find $\frac{dF}{dt}$. Any ideas?

2. May 1, 2012

### Staff: Mentor

Use the chain rule: dF/dt = dF/dx * dx/dt.

3. May 1, 2012

### JacksonSolo

I tried that, but isnt the derivative of F just equal to 1? I really cant understand this problem. Its on the practice test for our upcoming final and I have never even seen anything like it before. But I really would like to understand it in case something like it shows up on our final though.

Cause Ive used that form before for like price elasticity when $\frac{dx}{dp}$ * $\frac{p}{x}$. But i dont understand how it would work here.

And by the way, this is a related rates problem right? And not an exponential growth model?

4. May 1, 2012

### Staff: Mentor

No. Show me how you calculated dF/dx.

5. May 1, 2012

### Staff: Mentor

Yes, this is a related rates problem.

You are given the relationship between pollution level x and the number of fish F, and need to find the rate at which the fish population is changing, dF/dt.

6. May 1, 2012

### JacksonSolo

Oh I see, $\frac{df}{dx}$ = $\frac{-3x^{-3/2}}{2}$? So $\frac{df}{dx}$ * $\frac{dx}{dt}$ = $\frac{-3x^{-3/2}}{2}$ * 1.4?

7. May 1, 2012

### Staff: Mentor

Just how did you go from
F = 32000/(3 + √x)

to dF/dx = (-3/2)x-3/2?

Also, you are treating dx/dt as if it were a constant (1.4). It isn't. This is the rate of change of pollutant at a particular time, the time when there are 4,000 fish. You will also need to figure out what the value of x is at this same time.

8. May 1, 2012

### JacksonSolo

I derived it. 32000 & 3 are constants so they cancel out and you are left with 1/$\sqrt{x}$ with is 1/x$^{\frac{1}{2}}$ which is just x$^{\frac{-1}{2}}$ Then you add one to the exponent and multiply it in front of x which gives you (-3/2)x$^{-3/2}$.

Oh ok, so I rewrite the original equation in terms of x in order to find out how many pollutants are in the lake when there are 4000 fish.

That gives me (3+$\sqrt{x}$)*(4000) = 32000. Which simplifies to x = 25 ($\sqrt{x}$ = 5) Is that right?

9. May 1, 2012

### Staff: Mentor

You're kidding, right?

What you said is so incorrect, it's hard to know what to tell you.

What differentiation rules have you heard of?

BTW, you don't "derive" a function - you differentiate it to get the derivative.
At least you're on the right track, but x = 25 and √x = 5 are both incorrect.

x does not represent the number of pollutants in the lake - presumably there is one pollutant. x represents the parts per million of that pollutant.