Question

@whpalmer4 ok! i think i only have this one left. lol
we did a similar one but i just have such a hard time with the formula.
atmosphere pressure decreases by 1/2 every 3.6 miles gained in altitude. pressure is 14.7 psi at sea level to determine 9 miles above sea level.

Answer 1

so tentatively C is initial amount, x is alt in miles, k is half life.
so... A=C(1/2)^x/k

Answer 2

Have to be careful about how you write these things...that formula, it does not mean what you think means (with apologies to the Princess Bride)...

\[A = \frac{C(\frac{1}{2})^x}{k}\]That's the proper interpretation of what you wrote. Exponentiation happens before division, so you need to wrap \(x/k\) in parentheses to force that to be evaluated before use as an exponent.

C(x) = C_0 (1/2)^(x/k) would be acceptable, though I would personally write
C(x) = C_0 (2)^(-x/k) and ditch the fraction.

Answer 3

A=14.7(1/2)^3.6/(9) is that right? so... 14.7(2)^(-3.6/9) is that correct?

Answer 4

Uh, you've got the exponent inverted. Remember, every 3.6 miles, we multiply by 1/2, right? That means we have a sequence of multipliers: 1, 1/2, 1/4, 1/8, 1/6

\[2^0 = 1\]\[2^{-1} = 1/2\]\[2^{-2} = 1/4\]\[2^{-3} = 1/8\]etc.

But we want that to happen every 3.6 "clicks" instead of every "click" so we divide by 3.6. When the value of t, or x, or whatever gets to 3.6, -t/3.6 = -1, and that gets us our first division by 2. When the value gets to 7.2, -t/3.6 = -2 and we have a division by 4. Does that make sense?

Answer 5

So here we have \(k=3.6\), and our formula is\[C(x) = C_0(2)^{-x/3.6}\]so at 9 miles, we have \[C(9) = 14.7(2)^{-9/3.6}\]\[C(9) = 14.7*2^{-2.5} \approx 2.5986\]

Answer 6

Here's the ubiquitous graph...I've put on grid lines showing the halving of the pressure every 3.6 miles, plus a pair that shows the final answer for 9 miles.