Question

Medal!
Casey hit the bell in the school clock tower. Her pressure reader held nearby measured the sound intensity or loudness at 40 lb/in^2 after 4 s had elapsed and at a 4.7 lb/in^2 after 7s had elapsed. She remembers from her science class that sounds decays exponentially.

So the given points are (4,40) and (7,4.7)
-Find an exponential equation that models these data.
-How loud was the bell when it was struck at 0seconds?

Answer 1

man you really have a random sample of questions this evening, huh?

Answer 2

did you save them up all semester?

Answer 3

hahaha kinda.. we have a lot of different topics for hw

Answer 4

ok are you supposed to use something that looks like
\[A=P_0e^{rt}\] or can it just be \(A=a\times b^{\frac{t}{h}}\)?

Answer 5

lets go with the second one

Answer 6

I thought is was just the standard exponential equation y=ab^x

Answer 7

ok good, the second one

Answer 8

you have two points \((4,40) \) and \( (7,4.7)\)

Answer 9

first we can compute \(\frac{4.7}{40}\) as our \(b\)

Answer 10

or we can do it the way we did the previous one, which is kind of tedious, but probably what they want for this problem
do you have to hand this in?

Answer 11

yea this is hw that we have to turn in

Answer 12

ok then we do it the previous way

Answer 13

but quicker

Answer 14

\[7-4=3\] making \(b=\sqrt[3]{\frac{7.4}{40}}\) whatever that is

Answer 15

typo there sorry
\[b=\sqrt[3]{\frac{4.7}{40}}\]

Answer 16

i get \(b=.49\) rounded

Answer 17

then you can use
\[A=40\times 4.9^{x-4}\] as your model
damn this is weird

Answer 18

hahahaha XD

Answer 19

ok another typo, i must be getting tired
\[A=40\times .49^{x-4}\]

Answer 20

this is a really really dumb way to do this

Answer 21

so basically you reordered the y=ab^x equation to get b=to the square root and i agree it really is

Answer 22

no matter, that is what they want

Answer 23

honestly you explained way more than my teacher ever did in a whole semester!

Answer 24

careful, it is not the square root
since \(7-4=3\) it is the cubed root

Answer 25

the \(a\) in this case is what you get if \(x=4\) namely \(40\) and the \(b\) is the cubed root of the ratio \(\frac{4.7}{40}\)

Answer 26

so we have
\[A=40\times .49^{x-4}\] and the initial value, when \(x=0\) is
\[A(0)=40\times .49^{-4}\]

Answer 27

there is a much snappier way to do this
much snappier

Answer 28

but i don't want to confuse you, so lets leave it at that

Answer 29

ok.. :) thanks again so much!! so to find how loud the bell was you have to keep it in that form?

Answer 30

only an idiot starts counting at \(x=4\)
a normal person starts at \(x=0\) and if you want to know what have 4 seconds early you can compute using \(-4\)

Answer 31

but nvm you have the formula\[40\times .49^{x-4}\] which is what you are asked for

Answer 32

ok :) yea it is a little confusing in terms of that question that they ask from you lol

Answer 33

i can show you the other way to do this if you like
but i don't want to make it harder for you

Answer 34

sure! If you have time of course.. again thanks so much you have been really very helpful!

Answer 35

btw the answer to the last question is \[40\times .49^{-4}\]

Answer 36

i would do it like this
start counting at \(x=0\) not \(x=4\) then since \(7-4=3\) the model is
\[40\times \left(\frac{4.7}{40}\right)^{\frac{x}{3}}\]

Answer 37

if i want to know what it was 4 second earlier, compute
\[40\times \left(\frac{4.7}{40}\right)^{\frac{-4}{3}}\]

Answer 38

oh ok makes sense :)

Answer 39

http://www.wolframalpha.com/input/?i=40%28.49%29^%28-4%29
http://www.wolframalpha.com/input/?i=40%284.7%2F40%29^%28-4%2F3%29

you can check above that you get the same answer

Answer 40

Oh i see.. so basically this method is like the shortcut

Answer 41

almost in any case
i rounded the cube root, so it is not exact
first answer is more exact

Answer 42

kinda yeah
i used only the numbers given \[7-4=3\] which is why my exponent was \(\frac{x}{3}\) and the ratio \(\frac{4.7}{40}\) is what i used for \(b\) rather than \(\sqrt[3]{\frac{4.7}{40}}\)

Answer 43

but the other way works as well, and if you have to hand it in, use that way

Answer 44

I see :) thanks again!! This really helped a lot hope you enjoy the rest of your day ^.^

Answer 45

you too
i hope you are done with the random questions (for your sake)
good luck

Answer 46

haha yes all done :) now I just have to prep for a test next week! :)