C4 Exponential Decay- rep on offer :)

Liquid if pouring into a container at a constant rate of 20cm^3/s and is leaking out at a rate proportional to the volume of the liquid already in the container.

dV/dt = 20 -kV

the container is initially empty. by solving the differential equation, show that

V = A + Be^-kt

I've tried referring back to the textbook and still can't work this out =S.

I get this far:

(after rearranging and integrating the differential equation)

-1/k ln(20-kV) = t + C

and then i used V = 0 and t = 0 to find c

c= -1/k ln 20

then i get totally stuck and don't know where to go from here. someone please point me in the right direction.

oh

and don't forget

about the rep

:biggrin:

Reply 1

Clarity Incognito

Tommy Jay

Liquid if pouring into a container at a constant rate of 20cm^3/s and is leaking out at a rate proportional to the volume of the liquid already in the container.

Multiply up by -k, then how would you get rid of the log on the LHS?

Reply 2

nuodai

You need to sub in your value of C (in terms of k) and then rearrange for V. Remember to use the laws of logs.

Reply 3

Tommy Jay

nuodai

You need to sub in your value of C (in terms of k) and then rearrange for V. Remember to use the laws of logs.

okay subbing in the value of c i get:

-1/k ln(20-kV) = t - 1/k ln20

t= 1/k ln20 - 1/k ln(20-kV)

and then i think using rule of logs you go

t= 1/k ln(20/20-kV)

then would you times both sides by k

so tk = ln(20/20-kV)

and then e both sides

so e^kt = 20/20-kV

and then do i just keep going until i get V on its own?
also according to the answers it should be e^-kt not e^kt =S

Reply 4

Clarity Incognito

Tommy Jay

okay subbing in the value of c i get:

-1/k ln(20-kV) = t - 1/k ln20

Why did you rearrange to have t by itself, just multiply your first equation in this post by -k and get rid of logs and then rearrange to get V by itself.

Reply 5

Tommy Jay

Clarity Incognito

Why did you rearrange to have t by itself, just multiply your first equation in this post by -k and get rid of logs and then rearrange to get V by itself.

okay timesing both sides by -k i get:

ln(20-kV) = -kt + ln20

e both sides

20-kV = e^-kt + 20

kV= 20 - e^-kt - 20

kV= -e^-kt

V = -e^-kt/k

which doesn't fit in with what's written in the answers =S

ahhhh i hate C4!!!!!!

Reply 6

Clarity Incognito

Tommy Jay

Don't worry, this really is the most common error with this type of question.

When you get rid of logs. You have to put the whole RHS in the power of e.

e^{-kt + ln20}

You have now got to simplify that using a law of indices.

Reply 7

Tommy Jay

Clarity Incognito

Don't worry, this really is the most common error with this type of question.

When you get rid of logs. You have to put the whole RHS in the power of e.

e^{-kt + ln20}

You have now got to simplify that using a law of indices.

i thought i did that =S

the RHS: e^-kt + 20

eln20 becomes just 20 doesnt it?

Reply 8

Clarity Incognito

Tommy Jay

i thought i did that =S

the RHS: e^-kt + 20

eln20 becomes just 20 doesnt it?

Yes and no. eln20 becomes 20.

e^{-kt + ln20}

does not become

e^{-kt} + 20

Apply the indice law

a^b \times a^c = a^{b+c}

backwards.

Reply 9

Tommy Jay

Clarity Incognito

Yes and no. eln20 becomes 20.

e^{-kt + ln20}

does not become

e^{-kt} + 20

Apply the indice law

a^b \times a^c = a^{b+c}

backwards.

but surely it means that you're still left with a ln :frown:

20 - kV = e^{-kt+ln20}

where do i go from here? to get rid of the e i would have to ln it again. and then back to square one =S

Reply 10

Clarity Incognito

Tommy Jay

but surely it means that you're still left with a ln :frown:

20 - kV = e^{-kt+ln20}

where do i go from here? to get rid of the e i would have to ln it again. and then back to square one =S

No no, do not ln it back, just simplify the RHS.

a^{b+c} = a^b \times a^c

e^{-kt+ln20} =...

What's a, b and c in the latter?
Can you see where to go?

Reply 11

Tommy Jay

Clarity Incognito

No no, do not ln it back, just simplify the RHS.

a^{b+c} = a^b \times a^c

e^{-kt+ln20} =...

Can you see where to go?

yesss thankyou :smile:

so 20 -kV = e^-kt x e^ln20

20 - kV = e^-kt x 20

kV = 20 - 20e^-kt

V = 20/k - (20/k)e^-kt

AND IT MATCHES WHATS WRITTEN IN THE ANSWERS. WOOOOO :woo:

thankyou for your help :smile:

. the rep is on the way :smile:

Reply 12

Clarity Incognito

Tommy Jay

yesss thankyou :smile:

so 20 -kV = e^-kt x e^ln20

20 - kV = e^-kt x 20

kV = 20 - 20e^-kt

V = 20/k - (20/k)e^-kt

AND IT MATCHES WHATS WRITTEN IN THE ANSWERS. WOOOOO :woo: