# 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...

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 formula: f'(t)=-4f(t)(3+f(t))

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

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To start, we know that since t = 2, and f(t) represents the amount of solid at time t, we can say:

`f(2) = 1 gram`

Using your equation given, we can plug this in for all values where f(t) is used, like a variable.

`f'(t) = -4f(t)(3+f(t))`

`f'(2)=-4f(2)(3+f(2))`

`f'(2)=-4(1)(3+1)`

`f'(2)=-16 (grams)/(minute)`

Next we need to divide this by 60 to change if from grams per minute to grams per second.

`(16(g/m))/(60 seconds)= 0.267 g/(seconds)`

With this number, we can now see that after one second approximately 0.267 grams of solid will dissolve. So after 1 second, if we begin with just 1 gram of solid remaining at 2 seconds, we will have 1 gram minus 0.267 grams of solid with 0.733 grams remaining as our answer! Hope that helped!

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