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Assorted questions
1)The cubic polynomial xΒ³ + ax + b is denoted by p(x). It is given that (x-1) is a factor of p(x), and that when p(x) is divided by (x+1) the remainder is -6. Find the values of the constants a and b.
Since 1 is a root of the p(x), so I get
1 + a + b = 0
a + 1 = -b
Using long division, my remainder is (a+1)x + b
So -bx + b = -6
But then how do we solve to get a and b?
2)
i)Expand (1-2x)^-1 in ascending powers of x, up to and including the term in xΒ³.
I've done this and the expansion is ( 1 + x + 3xΒ²/2 + 5xΒ³/2 +...) but I've problem in the next part.
ii)State the set values for which the expansion in part (i) valid.
3)
dh/dt = 0.01[3 - sqrt(h)]
Use substitution x = 3 - sqrt(h),show that the equation above becomes
(x-3)*(dh/dt) = 0.005x
I don't know if the question is less in detail because this is not the full question but this is the only part that I don't know.
Thank you!
EDIT: If the question 3 above is less in detail, you can see below for full question.
3)
A rectangular reservoir has a horizontal base of area 1000mΒ². At time t=0, it is empty and water begins to flow into it at a constant rate of 30mΒ³s^-1. At the same time, water begins to flow out at a rate proportional to sqrt(h), where h m is the depth of the water at time t s. When h=1, dh/dt=0.02.
i)Show that h satisfies the differential equation
dh/dt = 0.01[3 - sqrt(h)]
ii)It is given that, after the substitution x = 3 - sqrt(h), the equation in part (i) becomes
(x-3)*(dh/dt) = 0.005x
Yes, it is given that the equation becomes that after the substitution, but I hope to know how to get it with the substitution
Thank you/
Since 1 is a root of the p(x), so I get
1 + a + b = 0
a + 1 = -b
Using long division, my remainder is (a+1)x + b
So -bx + b = -6
But then how do we solve to get a and b?
2)
i)Expand (1-2x)^-1 in ascending powers of x, up to and including the term in xΒ³.
I've done this and the expansion is ( 1 + x + 3xΒ²/2 + 5xΒ³/2 +...) but I've problem in the next part.
ii)State the set values for which the expansion in part (i) valid.
3)
dh/dt = 0.01[3 - sqrt(h)]
Use substitution x = 3 - sqrt(h),show that the equation above becomes
(x-3)*(dh/dt) = 0.005x
I don't know if the question is less in detail because this is not the full question but this is the only part that I don't know.
Thank you!
EDIT: If the question 3 above is less in detail, you can see below for full question.
3)
A rectangular reservoir has a horizontal base of area 1000mΒ². At time t=0, it is empty and water begins to flow into it at a constant rate of 30mΒ³s^-1. At the same time, water begins to flow out at a rate proportional to sqrt(h), where h m is the depth of the water at time t s. When h=1, dh/dt=0.02.
i)Show that h satisfies the differential equation
dh/dt = 0.01[3 - sqrt(h)]
ii)It is given that, after the substitution x = 3 - sqrt(h), the equation in part (i) becomes
(x-3)*(dh/dt) = 0.005x
Yes, it is given that the equation becomes that after the substitution, but I hope to know how to get it with the substitution
Thank you/
Reply 1
Civ-217
Since 1 is a root of the p(x), so I get
1 + a + b = 0
a + 1 = -b
Correct - you also know that p(-1) = -6.
Civ-217
i)Expand (1-2x)^-1 in ascending powers of x, up to and including the term in xΒ³.
I've done this and the expansion is ( 1 + x + 3xΒ²/2 + 5xΒ³/2 +...) but I've problem in the next part.
ii)State the set values for which the expansion in part (i) valid.
The binomial expansion for (1-x)^n only converges for |x| < 1. (Except when n is an integer because it's a finite expansion.)
I'm not sure what you're asking in the third question.
Hi, thank you. I now know how to do the first question. This is the actual question that I don't even know how to do a little bit.
1)A rectangular water tank has a horizontal square base of side 1 metre. Water is being pumped into the tank at a constant rate of 400 cmΒ³/s. Water is also flowing out at any time t seconds is proportional to the square root of the depth, h cm, of water in the tank at that time. When t=0, the depth of water in the tank is 81 cm and the rate at which the water is flowing out is 900cmΒ³/s.
i)Explain how the information given above leads to the differential equation
dh/dt = 0.04 - 0.01*sqrt(h)
ii)Show that the solution of the differential equation in part (i) is given by
t =
Use the substitution x = sqrt(h) - 4 to find the time for the depth of water in the tank to decrease from 81 cm to 64 cm.
Thank you.
I've no clue for the first part, but using the d.e. from first part, I am able to obtain the result required for the second part but I've problem in the next part.
About the substitution, do we have to use the substitution to form a new d.e.and then use it to answer the question? But then again, I've no clue how we could do it.
Thank you.
1)A rectangular water tank has a horizontal square base of side 1 metre. Water is being pumped into the tank at a constant rate of 400 cmΒ³/s. Water is also flowing out at any time t seconds is proportional to the square root of the depth, h cm, of water in the tank at that time. When t=0, the depth of water in the tank is 81 cm and the rate at which the water is flowing out is 900cmΒ³/s.
i)Explain how the information given above leads to the differential equation
dh/dt = 0.04 - 0.01*sqrt(h)
ii)Show that the solution of the differential equation in part (i) is given by
t =
Unparseable latex formula:
[100 / 4-sqrt(h)] dh\int_
Use the substitution x = sqrt(h) - 4 to find the time for the depth of water in the tank to decrease from 81 cm to 64 cm.
Thank you.
I've no clue for the first part, but using the d.e. from first part, I am able to obtain the result required for the second part but I've problem in the next part.
About the substitution, do we have to use the substitution to form a new d.e.and then use it to answer the question? But then again, I've no clue how we could do it.
Thank you.
Reply 3
Civ-217
Hi, thank you. I now know how to do the first question. This is the actual question that I don't even know how to do a little bit.
1)A rectangular water tank has a horizontal square base of side 1 metre. Water is being pumped into the tank at a constant rate of 400 cmΒ³/s. Water is also flowing out at any time t seconds is proportional to the square root of the depth, h cm, of water in the tank at that time. When t=0, the depth of water in the tank is 81 cm and the rate at which the water is flowing out is 900cmΒ³/s.
i)Explain how the information given above leads to the differential equation
dh/dt = 0.04 - 0.01*sqrt(h)
1)A rectangular water tank has a horizontal square base of side 1 metre. Water is being pumped into the tank at a constant rate of 400 cmΒ³/s. Water is also flowing out at any time t seconds is proportional to the square root of the depth, h cm, of water in the tank at that time. When t=0, the depth of water in the tank is 81 cm and the rate at which the water is flowing out is 900cmΒ³/s.
i)Explain how the information given above leads to the differential equation
dh/dt = 0.04 - 0.01*sqrt(h)
If you just consider the water flowing in, it's obvious that dh/dt = 400 cm^3/s = 0.04 m^3/s. And since metres is the unit we're working in, dh/dt = 0.04. However, there is also water flowing out, at a rate proportional to sqrt h. So dh/dt = 0.04 - k sqrt h. Substitute dh/dt = -900 cm^3/s (remember to convert to m^3/s though), h = 81cm (again, change to m), and you should get k = 0.01.
Civ-217
ii)Show that the solution of the differential equation in part (i) is given by
t =
t =
Unparseable latex formula:
[100 / 4-sqrt(h)] dh\int_
Integrate the differential equation and you should get this.
Civ-217
Use the substitution x = sqrt(h) - 4 to find the time for the depth of water in the tank to decrease from 81 cm to 64 cm.
x = sqrt h - 4
h = (x+4)^2
dh = 2(x+4) dx
t =
Unparseable latex formula:
[100 / 4-sqrt h] dh\int_
=
Unparseable latex formula:
100/x * 2(x+4) dx\int_
Does this help?
generalebriety
If you just consider the water flowing in, it's obvious that dh/dt = 400 cm^3/s = 0.04 m^3/s. And since metres is the unit we're working in, dh/dt = 0.04. However, there is also water flowing out, at a rate proportional to sqrt h. So dh/dt = 0.04 - k sqrt h. Substitute dh/dt = -900 cm^3/s (remember to convert to m^3/s though), h = 81cm (again, change to m), and you should get k = 0.01.
Integrate the differential equation and you should get this.
x = sqrt h - 4
h = (x+4)^2
dh = 2(x+4) dx
t =
=
Does this help?
Integrate the differential equation and you should get this.
x = sqrt h - 4
h = (x+4)^2
dh = 2(x+4) dx
t =
Unparseable latex formula:
[100 / 4-sqrt h] dh\int_
=
Unparseable latex formula:
100/x * 2(x+4) dx\int_
Does this help?
Thank you. But should
t =
Unparseable latex formula:
[100 / 4-sqrt h] dh\int_
be
t =
Unparseable latex formula:
[100 / -(4-sqrt h)] dh\int_
t =
Unparseable latex formula:
[100 / -x] dh\int_
Should there be a negative? I proceed from there obtain answer t= +1.85s or -1.85s, I don't know the correct answer but does this answer make sense?
And for the first part, you said that
dh/dt = 0.04 - k sqrt h. --(denoted by (1))
Substituting dh/dt = -900 cm^3/s into (1) but I didn't get 0.01. Did I substitute it wrong?
Sorry but how 400 cm^3/s = 0.04 m^3/s? Because I thought 1cmΒ³=10^-6 mΒ³ so 400 cm^3/s = 0.0004 m^3/s?
Thank you.
Reply 5
Civ-217
t =
Unparseable latex formula:
[100 / 4-sqrt h] dh\int_
be
t =
Unparseable latex formula:
[100 / -(4-sqrt h)] dh\int_
t =
Unparseable latex formula:
[100 / -x] dh\int_
Should there be a negative?
Oops, yes, sorry.
Civ-217
dh/dt = 0.04 - k sqrt h. --(denoted by (1))
Substituting dh/dt = -900 cm^3/s into (1) but I didn't get 0.01. Did I substitute it wrong?
Argh...
Sorry... my fault. Misread the question and I seem to have confused you by seemingly making up rubbish as I went along.
I'll try again.Civ-217
1)A rectangular water tank has a horizontal square base of side 1 metre. Water is being pumped into the tank at a constant rate of 400 cmΒ³/s. Water is also flowing out at any time t seconds is proportional to the square root of the depth, h cm, of water in the tank at that time. When t=0, the depth of water in the tank is 81 cm and the rate at which the water is flowing out is 900cmΒ³/s.
i)Explain how the information given above leads to the differential equation
dh/dt = 0.04 - 0.01*sqrt(h)
i)Explain how the information given above leads to the differential equation
dh/dt = 0.04 - 0.01*sqrt(h)
The bit I misread here was dh/dt - h is the depth. As the base area is 10000cm^2 and the water is being pumped in at 400cm^3/s, this means h is increasing by 400/10000 cm/s = 0.04 cm/s. Water is also being drained at a rate of k sqrt h.
dh/dt = 0.04 - k sqrt h.
When t = 0, dh/dt = -900/10000 = -0.09 cm/s, h = 81. Again, you should get 0.01. Try it from there.
Sorry about that. Seem to have had one of my dumb moments.
Reply 7
Civ-217
Thank you! I will look at that later because I'm going to bed. For the last part I get t = -1.85s, does the negative sign matter?
Afraid so, you can't have negative time. I would check your answers but I'm really busy right now, sorry.
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