# Another Linear Approx Problem

## Homework Statement

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.

## Homework Equations

$$f'(t) = -4 f(t) ( 5 + f(t) )$$

## The Attempt at a Solution

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

## Answers and Replies

Gib Z
Homework Helper
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.

HallsofIvy
Homework Helper
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!)

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

nrqed
Homework Helper
Gold Member
dt=1 second? so we still get -56? That doesn't seem right

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

matt grime
Homework Helper
it would really make sense only for a delta t at least ten times smaller. Oh, well...

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

HallsofIvy