# Another Linear Approx Problem

1. Mar 11, 2007

### Weave

1. The problem statement, all variables and given/known data
Let f(t) be the weight (in grams) of a solid sitting in a beaker of water. Suppose that the solid dissolves in such a way that the rate of change (in grams/minute) of the weight of the solid at any time t can be determined from the weight using the forumula:
$$f'(t) = -4 f(t) ( 5 + f(t) )$$

If there is 2 grams of solid at time t = 2 estimate the amount of solid 1 second later.

2. Relevant equations
$$f'(t) = -4 f(t) ( 5 + f(t) )$$

3. The attempt at a solution
So I get $$f'(2)=-56$$ Not sure what to do after that, do I use a differential?

2. Mar 11, 2007

### Gib Z

Do you know what linear approximations work?

Using the derivative of the function which gives the amount of solid, find the gradient of the tangent at the point you know the functions value.

You have the gradient, and a point this straight line passes though. Find the equation of this new line, and sub in the value of one second later.

3. Mar 11, 2007

### HallsofIvy

Staff Emeritus
Yes, the whole point is to use the differential.

If f '= H(f,t), then df= H(f,t)dt and, of course, the "next" value of f is f+ df.

In particular, if f '(2)= -56, then df= -56 dt. What is dt here?

(I don't think a linear approximation is going to be very accurate for this large a dt!)

4. Mar 11, 2007

### Weave

dt=1 second? so we still get -56? That doesn't seem right

5. Mar 11, 2007

### nrqed

They say that f' is in grams per minute. So The change in mass will be -56 grams/minute * 1/60 minute, so a bit less than one gram is lost. It is *still* strange to use a differential for such a large time interval, it would really make sense only for a delta t at least ten times smaller. Oh, well...

Patrick

6. Mar 11, 2007

### matt grime

That's a very subjective call. It might well be an extremely accurate approximation - depends on f, not what units of time you use.

7. Mar 12, 2007

### HallsofIvy

Staff Emeritus
True. When I said "I don't think a linear approximation is going to be very accurate for this large a dt!" I hadn't noticed that the problem gave df/dt in grams per minute (so t= 2 minutes but then asks for the value in one more second (so dt= 1/60, not 1). However, I had cleverly asked "what is dt here" so no one would notice!