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1. Can you guys work these questions out? Please provide workings for both questions. I cant seem to get it right. Thank you.

2. (Original post by Jyashi)
Can you guys work these questions out? Please provide workings for both questions. I cant seem to get it right. Thank you.

Please show us what work you have done so far on both questions.
3. (Original post by rayquaza17)
Please show us what work you have done so far on both questions.
Mind letting me know whether my answer for the 2nd question is right?
Spoiler:
Show
X = 2M/3
4. (Original post by Dohaeris)
Mind letting me know whether my answer for the 2nd question is right?
Spoiler:
Show
X = 2M/3
Post working? I'm not sure if that looks right looking at the question.
Never mind.
5. (Original post by Dohaeris)
Mind letting me know whether my answer for the 2nd question is right?
Spoiler:
Show
X = 2M/3
That's right.
6. (Original post by rayquaza17)
That's right.
Thanks
7. Just came back from work so couldnt update post. Here is my working for Q6.

For Q7 i am getting to the equation of dX÷dt = kM

Is that a correct expression?

(Original post by rayquaza17)
Please show us what work you have done so far on both questions.
Attached Images

8. Can you tell me how you got to that answer please?

(Original post by Dohaeris)
Mind letting me know whether my answer for the 2nd question is right?
Spoiler:
Show
X = 2M/3
9. (Original post by Jyashi)
I don't mind helping you out. From what I can see, you used partial fractions, but there's no need for that (at least in the way I solved it.) I'll try and take it step-by-step.

First, they said M is the mass of the fuel that was burned, from this we know that M is a constant.

Next, we're told that the sum of the mass of waste produced and the mass of unburned fuel remaining is always constant, this'll come into play later on.

We're given that X is the mass of waste produced.

Then we're told that the rate of increase of X, dX/dt, is k times that the mass of fuel remaining.

The mass of fuel remaining will have to be M-X, where M is the original mass of unburned fuel, because we know that the sum of waste produced and unburned fuel remaining is constant. So if you take 'Y' as the unburned fuel remaining, then Y+X=M, so Y=M-X

From this, we can deduce the equation to be dX/dt = k*(M-X), and you can carry on the to solve the differential equation as you would any other, though you have to remember that M and k are constants.

Do you get it?
10. I kind of get it but still dont inderstand it deeply. Can you provide me a working of it in mathmatical form. I find that easier to understand then words.

(Original post by Dohaeris)
I don't mind helping you out. From what I can see, you used partial fractions, but there's no need for that (at least in the way I solved it.) I'll try and take it step-by-step.

First, they said M is the mass of the fuel that was burned, from this we know that M is a constant.

Next, we're told that the sum of the mass of waste produced and the mass of unburned fuel remaining is always constant, this'll come into play later on.

We're given that X is the mass of waste produced.

Then we're told that the rate of increase of X, dX/dt, is k times that the mass of fuel remaining.

The mass of fuel remaining will have to be M-X, where M is the original mass of unburned fuel, because we know that the sum of waste produced and unburned fuel remaining is constant. So if you take 'Y' as the unburned fuel remaining, then Y+X=M, so Y=M-X

From this, we can deduce the equation to be dX/dt = k*(M-X), and you can carry on the to solve the differential equation as you would any other, though you have to remember that M and k are constants.

Do you get it?
11. Hi,
Here's my worked solutions to question 7.
Hope this helps

Attachment 537719537721
Attached Images

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