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Water flows out of a tank through a hole in the bottom and, at time t minutes, the depth of water in the tank is x metres. At any instant, the rate at which the depth of the water is decreasing is proportional to the square root of the depth of water in the tank.

So the mark scheme is very vague in regard to the answer to this. I have come up with a solution to the following questions which makes perfect sense to me and is a sensible number yet... wrong.

(1) write down a differential equation which models this situation.

I got : dx/dt= kx^1/2 (which is correct)

(2) when t=0, x=2; when t=5, x=1. Find t when x=0.5.

I integrated with respect to t. Aftere all that I got t=4.5, it is meant to be 8.5 however.

Any help on the second part would be appreciated.

So the mark scheme is very vague in regard to the answer to this. I have come up with a solution to the following questions which makes perfect sense to me and is a sensible number yet... wrong.

(1) write down a differential equation which models this situation.

I got : dx/dt= kx^1/2 (which is correct)

(2) when t=0, x=2; when t=5, x=1. Find t when x=0.5.

I integrated with respect to t. Aftere all that I got t=4.5, it is meant to be 8.5 however.

Any help on the second part would be appreciated.

Yer, I actually had that Sorry

But the second part is the confusing part.

For me, and 6 mark worthy.

But the second part is the confusing part.

For me, and 6 mark worthy.

The problem being, I may just be being stupid, is that it doesnt work.

I thought of subbing in the values, however ive done it again slightly differently and still not got approximately 8.5.

I thought of subbing in the values, however ive done it again slightly differently and still not got approximately 8.5.

Ok it took me this long to work out that you are in fact right.

Things look good for the exam tomorrow -_-

Things look good for the exam tomorrow -_-

I just did this paper...It definitely works because I got it then but for some reason I can't get it now!

Original post by unknownalias

Water flows out of a tank through a hole in the bottom and, at time t minutes, the depth of water in the tank is x metres. At any instant, the rate at which the depth of the water is decreasing is proportional to the square root of the depth of water in the tank.

So the mark scheme is very vague in regard to the answer to this. I have come up with a solution to the following questions which makes perfect sense to me and is a sensible number yet... wrong.

(1) write down a differential equation which models this situation.

I got : dx/dt= kx^1/2 (which is correct)

(2) when t=0, x=2; when t=5, x=1. Find t when x=0.5.

I integrated with respect to t. Aftere all that I got t=4.5, it is meant to be 8.5 however.

Any help on the second part would be appreciated.

So the mark scheme is very vague in regard to the answer to this. I have come up with a solution to the following questions which makes perfect sense to me and is a sensible number yet... wrong.

(1) write down a differential equation which models this situation.

I got : dx/dt= kx^1/2 (which is correct)

(2) when t=0, x=2; when t=5, x=1. Find t when x=0.5.

I integrated with respect to t. Aftere all that I got t=4.5, it is meant to be 8.5 however.

Any help on the second part would be appreciated.

I know this is late but the mistake many people make is usually the part where you substitute t = 0 , x = 2 into your integrated equation

2(x)^0.5 = -kt + c

Let me show you what I mean. First we substitute the values given to solve for C The reason why we use t = 0 , x = 2 instead of t = 5 , x = 1 is because t = 0 cancels out ' k '

Substitute t = 0 , x = 2

2(2)^0.5 = -k(0) + c

simplify

2(2)^0.5 = c

Now here is the mistake people make. You may look at this and think that 2(2)^0.5 should be first multiplied by each other to get 4^0.5 which equals 2. However when you try and solve for k using c = 2 , t = 5 , x = 1 you will find that k will have no value as I will show.

2(x)^0.5 = -kt +2

Substitute t = 5 , x = 1

2(1)^0.5 = -k(5) +2

Simpify

2 = -5k + 2

Solve for k

2-2 = -5k

0 = -5k -----> Here we cant solve for k anymore.

Let's rewind back to finding c = 2(2)^0.5 By understanding the order of operation. Bidmas or whatever format you may have learned you will find that we cannot multiply first until we solve the bracket with the exponent first. If you put 2(2)^0.5 into a calculator you will find that you will not get a value of 2. But you will get an irrational number ( a number that cannot be represented as a fraction ).

So c = 2(2)^0.5 (NOT 4^0.5)

The rest of the question I expect is self explanatory

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