You are Here: Home >< Maths

# The Proof is Trivial! Watch

1. (Original post by Star-girl)
I think you, bananarama and I (once I've corrected it) should all post our solutions as they seem to be different.

See above.

No worries.
I would, but I still have a stray minus Care to assist?

Spoiler:
Show
2. Guys, can't you do the question by considering ie. by considering the discriminant of a quadratic in . In fact I tried and I got something very similar but a factor of off.
3. (Original post by bananarama2)
I would, but I still have a stray minus Care to assist?

Spoiler:
Show
In line 6 you have inexplicably multiplied by -1!

4. OK guys, here was how I did the question before posting it:

Spoiler:
Show

Let the distance travelled be and the mass of the rope be . Then:

Gravitational force =

Frictional force =

Putting this together, we get a second order differential equation:

Solving this differential equation yields the complimentary function:

, where

We can guess our particular integral as some constant:

Which gives us the general solution:

We can then differentiate this to get:

Next, let’s apply boundary conditions. Firstly, :

Secondly, :

We can then solve these simultaneously to get:

We want conditions on whether the rope falls off the table. If it does not, then there must be a solution to the equation:

We can see very easily that B is always going to be negative. As such, for solutions to exist A must be negative:

5. (Original post by DJMayes)
OK guys, here was how I did the question before posting it:

Spoiler:
Show

Let the distance travelled be and the mass of the rope be . Then:

Gravitational force =

Frictional force =

Putting this together, we get a second order differential equation:

Solving this differential equation yields the complimentary function:

, where

We can guess our particular integral as some constant:

Which gives us the general solution:

We can then differentiate this to get:

Next, let’s apply boundary conditions. Firstly, :

Secondly, :

We can then solve these simultaneously to get:

We want conditions on whether the rope falls off the table. If it does not, then there must be a solution to the equation:

We can see very easily that B is always going to be negative. As such, for solutions to exist A must be negative:

You've got less than and I have equals... uh oh...
6. (Original post by Star-girl)
You've got less than and I have equals... uh oh...
The question has to by nature be an inequality, as there will be more than one value for v which is sufficiently slow to prevent it from falling.
7. (Original post by DJMayes)
x
8. Problem 55*

The problem is concerned with constructing lines of length root , where is not a perfect square. Suppose you only have a ruler which can only measure integer lengths and assume you can draw perfect right-angled triangles. Show that unless is of the form , then you only ever need one right-angled triangle to construct a line of length root , e.g. To construct , you create a triangle, to construct you create a triangle.
9. (Original post by DJMayes)
The question has to by nature be an inequality, as there will be more than one value for v which is sufficiently slow to prevent it from falling.
Ah yes, I suppose. So equality denotes the maximum value of v^2, then?
10. (Original post by und)
Could you elaborate? As far as I can see it isn't...
11. (Original post by bananarama2)
I would, but I still have a stray minus Care to assist?

Spoiler:
Show
I will indeed, I just need to first finish all the modifications on the first part of my solution.
12. When I come to the student room Mr M always turns up, like :

13. Solution 50

Using as the initial velocity.

Since the function of is continuous, if for some , then at that instant the frictional force must be greater than the weight, thus the rope will remain stationery. It remains to find a range of values of for which will become .

Creating and solving the differential equation, we obtain , so if is to be for some positive , then there must exist a positive root , so a necessary and sufficient condition (assuming all the constant terms are positive) is as required.

If the rope has mass , then the impulse applied to accelerate the rope to a speed is given by the change of momentum . Hence , giving the maximum impulse as .
14. (Original post by Star-girl)
Ah yes, I suppose. So equality denotes the maximum value of v^2, then?
No, I don't think so. If you look at my DE equations solution, equality implies that B = 0, which means that there aren't solutions to x' = 0 as it implies A =0, which is impossible.
15. (Original post by DJMayes)
Could you elaborate? As far as I can see it isn't...
Sorry, I meant why is it positive? When you plug into the DE with a positive complimentary function, the constant term has the wrong sign.
16. (Original post by DJMayes)
No, I don't think so. If you look at my DE equations solution, equality implies that B = 0, which means that there aren't solutions to x' = 0 as it implies A =0, which is impossible.
Hmmm...
17. (Original post by und)
Sorry, I meant why is it positive? When you plug into the DE with a positive complimentary function, the constant term has the wrong sign.
No it doesn't, because the co-efficient of x is negative.
18. (Original post by bananarama2)
When I come to the student room Mr M always turns up, like :

Right - I'll have a look now at your solution. Could you check over mine (I've modified it now) as well as it seems to be some discrepancies with DJ's...
19. (Original post by DJMayes)
No it doesn't, because the co-efficient of x is negative.
Oh wait nvm...
20. (Original post by bananarama2)
When I come to the student room Mr M always turns up, like :

Why run when you can teleport?

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 11, 2017
Today on TSR

...in schools

### I think I'm transgender AMA

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.