3U Maths Question Thread

[RuiAce]

Hey Rui,

Could you please explain your last few lines of working (after you equated the two expressions)

Shouldn't it be:
6x(x-3) = 10

There was a typo on the LHS in the line above it. Fixed

[anotherworld2b]

Thank you for your help rui ace
Could i please have some help with this question as well ? 

[jakesilove]

Thank you for your help rui ace
Could i please have some help with this question as well ?

Our curve is



So, our gradient of the tangent is



At the point x=1, the gradient is therefore 4+6=10. At the point x=-1, the gradient is therefore -4+6=2, just like you got.

Then, we figure out the coordinates. This will be (1, 3) and (-1, -1).









Great, so you got this much completely correct! Now, we're just looking for the intersection between these two lines, so we solve simultaneously.








Which are the coordinates of Q

[anotherworld2b]

I see now thank you for your help
I also wanted to ask how would you approach this question?

Our curve is



So, our gradient of the tangent is



At the point x=1, the gradient is therefore 4+6=10. At the point x=-1, the gradient is therefore -4+6=2, just like you got.

Then, we figure out the coordinates. This will be (1, 3) and (-1, -1).









Great, so you got this much completely correct! Now, we're just looking for the intersection between these two lines, so we solve simultaneously.








Which are the coordinates of Q

[jakesilove]

I see now thank you for your help
I also wanted to ask how would you approach this question?

Looks like you approached it correctly. We know that a food drop can feed 24 hikers for 6 days. We want the days to INCREASE from 6 to 10. This works by multiplying the days by 10/6. We expect the hikers to DECREASE, by the same ratio, as this is what the question stipulates. So, I'm going to DIVIDE 24 by 10/6. This gives me 14.4 hikers. Now, 15 hikers would be too much, so the answer is (I think) anywhere between 6 and 14 hikers

[Shadowxo]

For the earlier question, isn't b (or c) = -7 instead of 7? As 3 = 10+b
 Might change your answer a little

[jakesilove]

For the earlier question, isn't b (or c) = -7 instead of 7? As 3 = 10+b
 Might change your answer a little

Yep, you're absolutely right! Changes the final answer, but the method is the same

[Shadowxo]

Yep right method, just makes the answer a nice whole number

[Rathin]

∫sin^2(3x) dx

[RuiAce]

∫sin^2(3x) dx

[Rathin]

Isn't sin^2(x) = (1-cos(2x))/2?

EDIT: nvm you fixed it

[Caitlyn_]

Hey Jake,

 I struggle with both Circle geometry and Parametrics. I can't quite get my head around them. Do you have any tips or advice to help with coping with these two topics better?

[Rathin]

I got the answer as x/2 - 3sin(6x) +c .. but .. answer says x/2 (-sin(6x))/12 + c

[RuiAce]

Hey Jake,

 I struggle with both Circle geometry and Parametrics. I can't quite get my head around them. Do you have any tips or advice to help with coping with these two topics better?

As for geometry in any circumstance, though notably in particular with circle geometry, it all comes down to being able to "SEE" it. This is what takes a lot of practice.

To "see" it better, it helps to know what all the theorems look like. For example, the angles standing on same arc (angles in same segment) really just remind me of two triangles mirrored together. The angle in a semicircle is an opposite one. If by some coincidence there's a cyclic quadrilateral is there that's a giveaway, and the alternate segment theorem looks like a triangle with a tangent attached with it.

Once you are able to visualise it more clearly, focus on details. In a question, if they give you a tangent, you'll probably need to use a tangent theorem. Otherwise, stick to everything else and only that. Also, when focusing, don't look at the entire diagram at once; look at a few lines at a time. If you know what your theorems "look like", you can try to see their shapes better.


The first part of parametrics is simple - eliminate the parameter. Then your tangents and normals to the parabola should be a fairly standard process. Work through the proofs given by example in maths in focus and that's plenty. Finally, the section on locus problems tends to be the tougher one.

For locus problems, remember you want to eliminate the parameter. You need to be good at simultaneous equations enough so that you know how to throw away the useless t's, p's, q's and etc. (also theta but that's kinda rare in 3U). The difference is that unlike the first part of parametrics, sometimes these require skills. You need mastery over some standard cases, as well as proficiency at simultaneous equations.

Especially when you have to square things. Also, never forget to use what you know/proved earlier

I got the answer as x/2 - 3sin(6x) +c .. but .. answer says x/2 (-sin(6x))/12 + c

You integrated wrong.

[Rathin]

Oh, didn't learn that rule.
Similarity for ∫sin(x)=-1/a cos(ax) right?