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)=-4f(t)(1+f(t))
If there is 5 grams of solid at time t=2 estimate the amount of solid 1 second later.

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solytexx2g3 solytexx2g3 · Answer 1 · 2019-12-08T21:04:10+01:00

Answer:

3 grams

Step-by-step explanation:

first of all

the weight in grams has been given to be f (t)

since there are 5 grams of solid in the beaker

then;

[tex]f(t)=5g[/tex]

but at a time t = 2

[tex]f(2)=5g[/tex] (t represents time)

the expression for the rate of change (in grams/minute)

[tex]f'(t)=-f(t)(1+f(t))[/tex]

subtituting the value of f(t) into the expression above

[tex]=-4f(2)(1+f(2))[/tex]

we already showed that f (2) = 5, substituting this into the rate of change expression

[tex]=-4(5)(1+5)\\=-20(1+5)\\=-20-100\\=-120grams/minute[/tex]

converting this into second, we will need to divide by 60 second

[tex]\frac{-120}{60} \\=-2g/s[/tex]

the weight is decreasing 2 g/s at a time 2 ( the minus sign shows decrease)

the amount of solid 1 second later will be

the amount of solid in the beaker subtract the amount of solid that dissolved

5 - 2 = 3 grams