Question

Two calculus questions!:-)

Answer 1

Typing them

Answer 2

Let I[0] be the intesnity of sound at threshold of human hearing. The decibel level of a sound with intesnity I is given by\[dB=10\log_{10}(\frac{ I }{ I_{0} })\].
If the decibel ratings of two sounds differ by 2, how do the intesnities of the sound compare? Round to the nearest 10th

Answer 3

And I'll post the next one after this one is solved:-)

Answer 4

@eashy

Answer 5

\[\Large\bf dB_1=10\log\left(\frac{ I_1 }{ I_{o} }\right)\]\[\Large\bf dB_2=10\log\left(\frac{ I_2 }{ I_{o} }\right)\]\[\Large\bf dB_2-dB_1=2=10\log\left(\frac{ I_2 }{ I_{o} }\right)-10\log\left(\frac{I_1}{I_o}\right)\]Take a look at that a sec, does it make sense what I've done so far?

Answer 6

Makes sense

Answer 7

`how do the intesnities of the sound compare?`
Hmm so I think what they're asking for is.. hmm
I think they want us to find:\[\Large\bf \frac{I_2}{I_1}\]That will tell us the ratio of the intensities, or how they compare to one another.

Answer 8

So we'll have to use some rule of logs and do some fancy stuff to figure this out.

Answer 9

So we have this equation:\[\Large\bf 2\quad=\quad 10\log\left(\frac{ I_2 }{ I_{o} }\right)-10\log\left(\frac{I_1}{I_o}\right)\]Let's start by dividing each side by 10,\[\Large\bf \frac{2}{10}\quad=\quad \log\left(\frac{ I_2 }{ I_{o} }\right)-\log\left(\frac{I_1}{I_o}\right)\]

Answer 10

A rule of logs:\[\Large\bf\color{#DD4747}{\log a-\log b\quad=\quad \log\left(\frac{a}{b}\right)}\]We can use this rule of logs to simplify the right side of our equation.

Answer 11

Looks a little ugly since we have fractions on fractions :) lol

Answer 12

\[\Large\bf \frac{2}{10}\quad=\quad \log\left[\frac{\left(\dfrac{I_2}{I_o}\right)}{\left(\dfrac{I_1}{I_o}\right)}\right]\]

Answer 13

hmm? :o

Answer 14

Nevermind :O Just keep working haha

Answer 15

So when we divide fractions, there is another way we can rewrite that, yes? :o

Answer 16

\[\Large\bf \frac{a/b}{c/d}\quad=\quad \frac{a}{b}\cdot\frac{d}{c}\]We can flip the bottom fraction and write it as multiplication, sound familiar?

Answer 17

Hmm right

Answer 18

Since that is how you divide fractions :P

Answer 19

So we've got something like this:
\[\Large\bf \frac{2}{10}\quad=\quad \log\left[\left(\dfrac{I_2}{I_o}\right)\cdot\left(\dfrac{I_o}{I_1}\right)\right]\]Anddddd, looks like we have a nice cancellation at this point.

Answer 20

I got 1.6 for my answer.

Answer 21

Wait, so cancel out the I[0] and 2/10=log(I[2]*I[1])

Answer 22

And you are incredibly speedy with the equation typing

Answer 23

\[\Large\bf \frac{2}{10}\quad=\quad \log\left[\left(\dfrac{I_2}{\cancel{I_o}}\right)\cdot\left(\dfrac{\cancel{I_o}}{I_1}\right)\right]\]Gives us:\[\Large\bf \frac{2}{10}\quad=\quad \log\left(\dfrac{I_2}{I_1}\right)\]

Answer 24

XD

Answer 25

O sorry I actually meant / not *

Answer 26

1.6? Mmmmmm yah that sounds right!
I think we did everything correctly there, hopefully :)

Answer 27

Jesus, long problem! I think I have the answer to this one, so don't need to show work :P
The temp of a hot liquid is 100 and the room temp is 70, the liquid cools to 93.9 in 6 minutes. What is the temp after 12 minutes?

Answer 28

So when sounds differ by 2 decibels, the intensity of the larger sound is 1.6 times as .. intense as the other sound :o

Answer 29

My answer: 88

Answer 30

k sec :x checking

Answer 31

Posted that kind of question a few times and nobody knew :((

Answer 32

Hmmm does require us to use Newton's Law of Cooling or something similar? :o
Maybe it's simpler than that and I'm just not remembering heh

Answer 33

One last question since I actually skipped it.. http://gyazo.com/25c7718e7d1ce3a2b03edce6b9a5490a

Answer 34

Look familiar? :o
\[\Large\bf \frac{dT}{dt}\quad=\quad k(T_a-T)\]

Answer 35

Not at all!

Answer 36

Mmm darn :( I'm not sure how to solve the temperature problem without using calculus stuff.

Oh I guess I can at least check your answer of 88 though.

Answer 37

Mmmm yah I think you got it correctly!
I came up with 89 degrees.

Answer 38

89?! O no! That was an option!

Answer 39

D:

Answer 40

I'll take your word :-)

Answer 41

That's funny, those numbers are so close together XD I didn't think they'd give you both of those options lol

Answer 42

Options: 88, 89, 87, 90

Answer 43

http://gyazo.com/25c7718e7d1ce3a2b03edce6b9a5490a
So for that one, part C, I don't exactly understand how'd you get it to equal 75?

Answer 44

I came up with 89.040.. I thinkkk I did it correctly.
I tried to avoid rounding when possible.
But you never no :p

Answer 45

Well let's hold off on part C for a moment :P
Your earlier parts look a little bonkers.. we should fix that that up first.

Answer 46

Can you accurately figure out what the `initial value` of the stock would be?
V(0) = ?

I know it's not one of the questions, but it should give you a good idea why your numbers are wayyyyyy off :D

Answer 47

1785.. right?:(

Answer 48

\[\Large\bf V(t)\quad=\quad 51(1-e^{-1.2t})+35\]Plugging in 0 for t,\[\Large\bf V(0)\quad=\quad 51(1-e^{-1.2\cdot0})+35\]Which simplifies the exponential to:\[\Large\bf=\quad 51(1-1)+35\]So what is the initial value of the stock? :x

Answer 49

I SEE WHAT IM DOING WRONG

Answer 50

IVE BEEN PUTTING 35 IN THE BRACKET

Answer 51

Do you? XD oh good

Answer 52

So it's 1.3!

Answer 53

the initial value? :o

Answer 54

Oh no, part C

Answer 55

Grr stop worrying about part C +_+
You really need to fix part A first.

Answer 56

V(1)=70.64
V(12)=86
V'(t)=61.2e^(-1.2t)
V(t)=75=1.3
lim-->infinity=86

Answer 57

Mmmm ok good! They all look correct now! :)

Answer 58

I think last question

Answer 59

differentiate
x(6.7)^x

Answer 60

Nevermind I got it! :-)

Answer 61

\[\Large\bf \left[x\;6.7^x\right]'\quad=\quad \color{#DD4747 }{(x)'}\;6.7^x+x\color{#DD4747 }{(6.7^x)'}\]

Answer 62

Got it? K cool :x

Answer 63

(6.7^x+ln(6.7)*x*(6.7)^x

Answer 64

Thanks for all of your help!

Answer 65

\c:/ np