Question

Separation of variables db/dt = b^1.1

Answer 1

Separation of variables db/dt = b^1.1

Answer given is b(t) = (10 / ( 6.31 -t )^10
can someone show me how to get that?

Answer 2

There must have been some more information to find the constant there...

Answer 3

\[\large \dfrac{db}{dt}=b^{1.1}\]These exponents might make things a little confusing. Let me know if you stuck on the integration.
Multiplying the dt to the other side, and dividing the b term to the other side gives us,\[\large \dfrac{db}{b^{1.1}}=dt \qquad \qquad \rightarrow \qquad b^{-1.1}db=dt\]Integrating gives us,\[\large \int\limits b^{-1.1}db=\int\limits dt \qquad \qquad \rightarrow \qquad \qquad \dfrac{b^{-0.1}}{-0.1}\quad =\quad t+c\]

Answer 4

Now we need to solve for b.
Let's start by converting our decimal to a fraction.
It's \(-\dfrac{1}{10}\). We're `dividing` by a fraction, so we'll rewrite it as multiplication.


\[\large \dfrac{b^{-0.1}}{\left(-\dfrac{1}{10}\right)} \qquad \rightarrow \qquad -10\;b^{-0.1}\]

Answer 5

Dividing each side by \(-10\) gives us,\[\large b^{-0.1}=-\dfrac{1}{10}(t+c)\]

Answer 6

\[\large b^{-1/10}=-\dfrac{1}{10}(t+c)\]We'll rewrite our exponent as a fraction so we can see what's going on.

Answer 7

cosntant can be expressed as k

Answer 8

or c, doesnt matter

Answer 9

We'll raise both sides to a power, we want to choose a power that will change the exponent on b to 1.

So let's raise both sides to the \(-10\)th power.\[\large \left(b^{-1/10}\right)^{-10}=\left(-\dfrac{1}{10}(t+c)\right)^{-10}\]

Answer 10

and thanks zepdrix!

Answer 11

the constant is b(0）= 100

Answer 12

Ah ok :) The initial condition.

Answer 13

Apply the negative exponent on the right gives us,
\[\large b=\dfrac{1}{\left(-\dfrac{1}{10}(t+c)\right)^{10}}\]

Answer 14

Again, we're dividing by a fraction, let's fix that.\[\large b=\dfrac{(-10)^{10}}{\left(t+c\right)^{10}}\]Since 10 is an EVEN power, the negative will disappear.\[\large b=\dfrac{(10)^{10}}{\left(t+c\right)^{10}}\]

Answer 15

Or I guess we should write it the way your answer key shows.\[\large b=\left(\dfrac{10}{t+c}\right)^{10}\]

Hmm yours has a -t in the bottom... I hope I didn't make a silly mistake somewhere.

Answer 16

Oh oh ok. We can't just absorb the negative into the power like I thought we could, we have to distribute it into the bottom part I suppose.\[\large b=\left(\dfrac{-10}{t+c}\right)^{10} \qquad \rightarrow \qquad b=\left(\dfrac{10}{c-t}\right)^{10}\]

Answer 17

it is negative when you move the negative to the right ,

Answer 18

Ok ok ok fine we'll do the initial conditions part now :) lol

Answer 19

haha thanks! :D

Answer 20

\[\large b(0)=100 \qquad \rightarrow \qquad 100=\left(\dfrac{10}{c-0}\right)^{10}\]

Answer 21

ah i didnt think of taking a power of 10 to get rid of the exponent, i did log and bring the exponenet down and turned into a mess lol

Answer 22

Hah :D

We're suppose to get 6.31 from this?? Hmm

Answer 23

Oh yah that I guess that is what I got when punching it into the calculator XD haha

I was just confused cause of all the 10's.. i figured we'd be getting a nice even number the way it looks.

Answer 24

Any trouble finding C?

Answer 25

yea, i know what to do now ,thanks hehe

Answer 26

yay c: