Question

someone please help! :] http://puu.sh/7pFtG.png

Answer 1

do you know how to solve a separable differential equation?

Answer 2

yes I do!

Answer 3

ok, then solve it

Answer 4

ROFL! good luck

Answer 5

\[\Large\bf\sf \frac{dy}{dt}\quad=\quad \frac{64-y}{16}\]
So we need to find our temperature function, \(\Large\bf\sf y(t)\).
Come up with anything yet? :)

Answer 6

Oh they also gave us initial data,\[\Large\bf\sf y(0)=0\]At time t=0, the temperature was 0 degrees.
We'll use that later to solve for the constant of integration.

Answer 7

did you get A=64? :o

Answer 8

Mmm I dunno lemme try to run through it real quick +_+

Answer 9

this is what I get http://puu.sh/7pGoO.png

Answer 10

and you just plug in 0 and 0 for t and y? :o

Answer 11

To solve for A? Yes good.

Answer 12

is this the answer for t? http://puu.sh/7pGxR.png

Answer 13

Hmm hold up hold up.
I came up with a different general solution.
Lemme make sure I didn't screw anything up.

Answer 14

oh no ;[

Answer 15

\[\Large\bf\sf y(t)\quad=\quad 64+Ae^{-\frac{1}{16}t}\]

Answer 16

\[\Large\bf\sf \int\limits\frac{dy}{64-y}\quad=\quad \int\limits \frac{1}{16}\;dt\]Integrating gives,\[\Large\bf\sf -\ln|64-y|\quad=\quad \frac{1}{16}t+c\]Multiply both sides by -1,\[\Large\bf\sf \ln|64-y|\quad=\quad C-\frac{1}{16}t\]Exponentiate each side,\[\Large\bf\sf 64-y\quad=\quad Ke^{-\frac{1}{16}t}\]Again multiply -1, and then add 64 to each side,\[\Large\bf\sf y(t)\quad=\quad 64+Ae^{-\frac{1}{16}t}\]I changed the constant each time I absorbed something new into it :p
Hopefully that wasn't too confusing.

Answer 17

I thought it went like this? http://puu.sh/7pGWS.png

Answer 18

oooh

Answer 19

When you integrate you get an extra negative from the -y inside the denominator, yes?

Answer 20

yea!

Answer 21

so then

Answer 22

you get 16 ln .5?

Answer 23

I think ummm, -16ln.5
right? :o

Answer 24

I think A=-64

Answer 25

so when you solve for t, you get a positive number, because you can't have negative time, right x]

Answer 26

A=-64 so our temperature function is,
\[\Large\bf\sf y(t)\quad=\quad 64-64e^{-\frac{1}{16}t}\]

Answer 27

And then we need to solve,\[\Large\bf\sf 32\quad=\quad 64-64e^{-\frac{1}{16}t}\]For t, yes?

Answer 28

Subtracting 64 from each side,\[\Large\bf\sf -32\quad=\quad -64e^{-\frac{1}{16}t}\]Negatives go away,\[\Large\bf\sf 32\quad=\quad 64e^{-\frac{1}{16}t}\]Dividing by 64, then doing some log stuff gives us,\[\Large\bf\sf \ln \frac{1}{2}\quad=\quad -\frac{1}{16}t\]

Answer 29

Are we just disagreeing on the negative in the exponent?

Answer 30

yea :/

Answer 31

how is it negative?

Answer 32

Yah we can't have a negative time :) That's a good observation.
I'm trying to figure out what went wrong.... Hmm.
The exponent should be negative, even Wolfram is giving me the same solution.

Answer 33

Here is the function with and without the negative exponent:
https://www.desmos.com/calculator/m656rmw0rz

Answer 34

The blue line is the one that makes sense.
See how it starts at 0 degrees, then the line moves UP in temperature?
Eventually it will reach 32 degrees.

I must be goofing up the calculations somewhere in there :(

Answer 35

Oh derp... I'm so stupid -_-
Because ln(.5) = negative number.
So of course we need the negative in front of the 16.
To give us a positive t value.

Answer 36

If you look at the simplest case:
\[\frac{dy}{dt} = -y\]\[-\frac{dy}{y} = dt\]integrate each side\[-\ln y = t+C\]\[\ln y = -t-C\]\[e^{\ln y} = e^{-t-C}\]\[y = Ce^{-t}\]