Thread: Maths question lul

Why is 0.25sin^4(0.5pie) - 0.25sin^4(0) = 0.25 :S

It's easier with fractions. Note that sin(pi/2) = 1, and sin(0) = 0.
Rewrite it as:

= (1/4) [sin(pi/2)]^4 - (1/4) [sin(0)]^4
= (1/4) (1^4) - (1/4) (0^4)
= (1/4)(1) - (1/4)(0)
= 1/4 - 0
= 1/4
= 0.25

Did you really need a thread for this!?

Yeah... because if you input the values for sin that's what you get, lul.

I'm stuck on a maths question as well. Might as well just post it here to get an answer.

Find the general solution of the differential equation: tan(x)(dy/dx) +y = e^(2x)*tanx

I found that the integrating factor is e^(1/tanx) = e^(lnsinx) = sinx

Multiplying every term by sinx:
sinx (dy/dx) + y (sinx/tanx) = e^(2x)*sinx

But now when I do integration by parts it just goes in a loop..

Check out the homework guideline sticky...

If you've studied DEs I don't see why you shouldn't be able to do that. What's your specific confusion?

Originally Posted by Xei

If you've studied DEs I don't see why you shouldn't be able to do that. What's your specific confusion?

Sorry I put up the wrong question. I actually managed to do the previous one (just a standard 2nd order homogeneous DE). It is in fact this trigonometric function (above) that I'm stuck on right now; reasons for confusion stated.

Flaw in my math knowledge soz, didnt realise that sin^4 is the same as the 0.5pie^4 if that makes sense.

You'll be trying to integrate 1/tanx, yesh?

Write down the definition of tan. You should be able to integrate by substitution.

sinx (dy/dx) + y (sinx/tanx) = e^(2x)*sinx

=>I started by using some basic trig sinx/tanx = sinx/(sinx/cosx) = cosx

=>So now the equation looks like: sinx y'' + ycosx = e^(2x)*sinx

I then tried to do integration by parts by letting u = e^(2x) , u'=2e^(2x) , v = -cosx , v' = sinx

Was this the wrong step to take?

Yeah, to use that method you first have to get it into the form y' + fy = g; in other words, divide the whole thing by sinx.

Oh I see. Thanks man

Out of curiosity, what subjects did you take before university? Further maths I presume?

Yeah, and phys, chem, bio. They're not very demanding. The only exam I had to spend a long time training for was STEP, although the syllabus is just maths and further maths.

Wow that's impressive. I wouldn't be able to cope with that many subjects.
And how is it that you don't find further maths demanding?!

Meh I found it to just be learning set methods. I don't think there's much conceptually hard stuff... they just give you a list of different things you have to be able to do, and the set of instructions you follow to do them. There's no thinking involved, just memorisation of a list of things. Science A-Levels are exactly the same, it gets boring once you get used to it.

Check out some of these which have the same syllabus as A-Level but the opposite emphasis:

STEP Test Preparation | University Admissions Tests from Cambridge Assessment

I had to pass II and III.

That stuff looks intimidating... But I suppose it ought to be if its the admissions test to one of the best universities in the world.

Now I know who to ask when I'm stuck on a question

A mechanics problem:

A particle is projected from a point P on an inclined plane, up the line of greatest slope through P, with initial speed V. The angle of the plane to the horizontal is θ.

(i) If the plane is smooth, and the particle travels for a time 2Vcosθ/g before coming instantaneously to rest, show that θ=pi/4

(ii) If the same plane is given a roughened surface, with a coefficient of friction 0.5, find the distance travelled before the particle comes to rest.

---
I managed to do (i) by considering the components of gravity on an inclined plane, then using v=u+at. V, g terms cancel and you get θ=pi/4.

Not sure how to do part (ii)...

Help is appreciated.

Originally Posted by ThePieMan

(ii) If the same plane is given a roughened surface, with a coefficient of friction 0.5, find the distance travelled before the particle comes to rest.

Not sure if this approach is correct at all, but I'll just have a go at it.
---

F=mu x R as Xei had previously stated.

If you consider the particle the second it comes to rest, the reaction force = the component of gravity perpendicular to the slope upwards.
In other words R=gcosθ.

Now consider all the forces acting on the particle during its motion: -mgsinθ - mF = ma. (since the decelerating component is -gsinθ parallel to the slope downwards, the accelerating component upwards is +gsinθ. F is friction, which is equal to mu x R)

m terms cancel, leaving: a = gsinθ - (mu x R)
substituting θ=pi/4: a = gsin(pi/4) - (0.5 x gcos(pi/4)) = (5root2)/2

Now use the equation of motion to find distance travelled.

The key with pretty much all of those mechanics modules is just to draw a diagram... it's basically the same deal in this question, except you have a frictional force. I can't really remember how those things work, but isn't it just that a frictional force of mu times the reaction force of the surface opposes the motion?

In this case it's pretty much the same question as before, except you have to work out the reaction force (which is done in pretty much the same way as you found the decelerating component of the weight), and add that to the deceleration. Then it's just equations of motion again.

I'm guessing you'd have to find the mass, right? Then maybe use the work-energy theorem?

Originally Posted by Wayfaerer

I'm guessing you'd have to find the mass, right? Then maybe use the work-energy theorem?

Yeah, I apologise. This question can be answered by considering initial KE, and final PE = mgh = work done.

Originally Posted by ThePieMan

Yeah, I apologise. This question can be answered by considering initial KE, and final PE = mgh = work done.

For a fun but rewarding experience, try it using the Lagrangian or Hamiltonian.

Hmm not quite - I think the correct approach is to actually consider the components of gravity (gsinθ, gcosθ etc), but I have no idea how to deduce the reaction force from that..

I just figured the coefficient of friction wouldn't be of use without knowledge of the normal force on the particle.

Meh I'll try to solve it at some later date. I'll be sure to post detailed solutions to satiate your curiosity.