calculus question

Question

Suppose you start with one liter of vinegar and repeatedly remove 0.05L, replace with water, mix, and repeat.

a) Find a formula for the concentration after n steps.

b)After how many steps does the mixture contain less than 12% vinegar?

Answer 1

Think of it in this way: initially, your solution (referred to as x_0) is 100% vinegar. After step 1, your solution is 95% of that original solution:

x₀ = 1

x₁= 0.95(1)

The second time, x₂, you now have 95% of your solution from x₁:

x₂ = 0.95(x₁)=0.95(0.95(1)) = 0.95² (1)= 0.95²

The third time you remove solution you have:

x₃ = 0.95(x₂) = 0.95 (0.95 (0.95 (1))) = 0.95³ (1) = 0.95³

At this point, we can begin generalizing. Each time you remove solution, you have 95% of whatever the solution previously was. So the nth time you remove solution, you have:

x_n = 0.95 (x_n-1) = 0.95ⁿ

Now for part b, we want to know when x_n< 0.12. So we need to solve:

0.12 < 0.95ⁿ

The easiest way to do this (though not the fastest) is to just start calculating 0.95, 0.95², 0.95³... until you get something less than 0.12. That may drive us up the wall though, so another way of doing this is:

0.12 = 0.95ⁿ

ln(0.12) = ln(0.95ⁿ)

ln(0.12) = (n)ln(0.95)

ln(0.12)/ln(0.95) = n

n = 41.33

So at step 41.33, the solution is exactly 12% vinegar. Thus at step 42, the solution is less than 12% vinegar.

Hope this helps!

Answer 2

Hi,

start with one liter

Second, 1- 0.05 = 95/100 (1- 0.05)^0

Third, 95/100 * 0.05 decreases of vinegar, so remain 95/100 - 95/100 * 0.05 = 95/100 ( 1 - 0.05)

Forth, 95/100 ( 1 - 0.05) - 95/100 ( 1 - 0.05)* 0.05 = 95/100 ( 1- 0.05)^2

Fifth, 95/100 ( 1- 0.05)^2 - [95/100 ( 1- 0.05)^2] * 0.05 = 95/100 ( 1- 0.05)^3

nth time = 95/100 ( 1- 0.05)^(n-2)

95/100 ( 1- 0.05)^(n-2) < 0.12 both sides times 100/95------> ( 1- 0.05)^(n-2) < 0.12* 100/95 = 12/95

( 1- 0.05)^(n-2) < 12/95 take a Ln of both sides

(n-2) Ln( 1- 0.05) < Ln12/95 both sides divided by Ln( 1- 0.05)

n-2 < 40.33 ------> n < 42.33

now you can prove, if you choose n= 42 and by calculator find 95/100 ( 1- 0.05)^(42-2)= 0.122086

and if you choose n= 43 ======> 95/100 ( 1- 0.05)^(43-2) = 0.115982 that it is less than 12%

After how many steps does the mixture contain less than 12% vinegar? after 42 or 43th time the container will be include less than 12% vinegar

I hope it is useful,

Minoo

Answer 3

a_n = 1(0.75)^n-1

0.12 > 0.75^n-1