Let f(t) be the weight (in grams) of a solid sitting in a beaker of water

Question

Let  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  can be determined from the weight using the forumula: 

If there is 4 grams of solid at time  estimate the amount of solid 1 second later.

Answer 1

If you plug in 2 to f'(x) you will get:

f'(2) = -5f(2)(3+f(2)) = -5*4(3+4) = -140 grams/minute

We need to convert this rate to grams per second which can be done by divide -140 by 60 to give you a rate of -2.333 grams/second. Now to get estimate the final answer you just do 4 - 2.333 because we're assuming the solid will lose about 2.333 grams every second. Therefore, one second later the solid will have about 1.667 grams.