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Year 13 Maths Help Thread

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Original post by Zacken
Surely you know how to solve a basic quadratic?


ughfhg yeah - i wasnt thinking properly :redface:

u^2 + 2u - 1 = 0

completing square: (u +1)^2 - 1 - 1 = 0

(u +1)^2 = 2
u+1 = +- sqrt 2
u = (+- sqrt 2) -1

But how do i know its + sqrt 2?
Original post by kiiten
ughfhg yeah - i wasnt thinking properly :redface:

u^2 + 2u - 1 = 0

completing square: (u +1)^2 - 1 - 1 = 0

(u +1)^2 = 2
u+1 = +- sqrt 2
u = (+- sqrt 2) -1

But how do i know its + sqrt 2?


tan 22.5 is positive by remembering the graph of tan x. So you need to pick the positive value of u.
Original post by Zacken
tan 22.5 is positive by remembering the graph of tan x. So you need to pick the positive value of u.


How could i show that im picking the +ve answer?
Original post by kiiten
How could i show that im picking the +ve answer?


-sqrt(2) - 1 is negative.
+sqrt(2) - 1 is positive

i mean all you need to say is that sqrt(2) > 1
Original post by Zacken
-sqrt(2) - 1 is negative.
+sqrt(2) - 1 is positive

i mean all you need to say is that sqrt(2) > 1


Ok thanks :smile:
Could someone help me with understanding how to answer this question, please?

The question is written at the top and everything I have done so far (that is correct) is in black. After that is where it went wrong. The red is what I attempted to do but no doubt is wrong because I don't get the answer required, and the blue is the correct workings shown for the question, I don't understand the steps gone from the last black line to the first blue line. More specifically, why is the square no longer there? I think I understand everything else.

Attachment not found
(edited 7 years ago)
20170419_183636.jpg508.5KB
Original post by JaredzzC
Could someone help me with understanding how to answer this question, please?

The question is written at the top and everything I have done so far (that is correct) is in black. After that is where it went wrong. The red is what I attempted to do but no doubt is wrong because I don't get the answer required, and the blue is the correct workings shown for the question, I don't understand the steps gone from the last black line to the first blue line. More specifically, why is the square no longer there? I think I understand everything else.


You missed some negatives. If you use substitution u=1−xu=1-x then it should be obvious.
Currently stuck on question 6iii on ocr m1 june 2014: http://www.ocr.org.uk/Images/242450-question-paper-unit-4728-01-mechanics-1.pdf
I looked at mark scheme and really not sure why friction acts to oppose tension.
Original post by RDKGames
You missed some negatives. If you use substitution u=1−xu=1-x then it should be obvious.


Oh, ok. Yea, I am able to see it now. I thought it was perhaps a rule that I was unaware of or something but I was being stupid and didn't put the power as a negative when not as a fraction. Thanks for such a quick response and sorry for the rotated picture!
(edited 7 years ago)
Using the identity cos2θ = 1 – 2sin2 θ, find ∫sinθ dθ

i get 1/2θ -1/2sin2θ

but the answer is with a quater sin2θ

can someone please explain this to me
(edited 7 years ago)
Original post by Jassy16
Using the identity cos2θ = 1 – 2sin2 θ, find ∫sinθ dθ

i get 1/2θ -1/2sin2θ

but the answer is with a quater sin2θ

can someone please explain this to me


Do you mean find ∫sin⁡2θ dθ\int \sin^2 \theta \, \mathrm{d}\theta? If so, from the identity you get 1−cos⁡2θ=2sin⁡2θ⇒sin⁡2θ=12(1−cos⁡2θ)1 -\cos 2\theta = 2\sin^2 \theta \Rightarrow \sin^2 \theta = \frac{1}{2}(1- \cos 2\theta).

So ∫sin⁡2θ dθ=∫12−12cos⁡2θ dθ\displaystyle \int \sin^2 \theta \, \mathrm{d}\theta = \int \frac{1}{2} - \frac{1}{2}\cos 2\theta \, \mathrm{d}\theta.

Now I hope you know that ∫cos⁡2θ dθ=12sin⁡2θ\int \cos 2\theta \, \mathrm{d}\theta = \frac{1}{2}\sin 2\theta. So ∫12cos⁡2θ dθ=12(12sin⁡2θ)=14sin⁡2θ\int \frac{1}{2}\cos 2\theta \, \mathrm{d}\theta = \frac{1}{2}\left(\frac{1}{2}\sin 2\theta \right) = \frac{1}{4}\sin 2\theta.
Original post by Zacken
Do you mean find ∫sin⁡2θ dθ\int \sin^2 \theta \, \mathrm{d}\theta? If so, from the identity you get 1−cos⁡2θ=2sin⁡2θ⇒sin⁡2θ=12(1−cos⁡2θ)1 -\cos 2\theta = 2\sin^2 \theta \Rightarrow \sin^2 \theta = \frac{1}{2}(1- \cos 2\theta).

So ∫sin⁡2θ dθ=∫12−12cos⁡2θ dθ\displaystyle \int \sin^2 \theta \, \mathrm{d}\theta = \int \frac{1}{2} - \frac{1}{2}\cos 2\theta \, \mathrm{d}\theta.

Now I hope you know that ∫cos⁡2θ dθ=12sin⁡2θ\int \cos 2\theta \, \mathrm{d}\theta = \frac{1}{2}\sin 2\theta. So ∫12cos⁡2θ dθ=12(12sin⁡2θ)=14sin⁡2θ\int \frac{1}{2}\cos 2\theta \, \mathrm{d}\theta = \frac{1}{2}\left(\frac{1}{2}\sin 2\theta \right) = \frac{1}{4}\sin 2\theta.


yeah, thanks man appreciate
Original post by Jassy16
yeah, thanks man appreciate


No problem!
I've been having a hard time with rates of change and all that, was wondering if anyone had any tips/tricks to nail it 99% of the time.

I'll throw an example in here if its needed

Spoiler

Not sure what I have done wrong. The correct answer is (26/3)Pi cubic units.

Please forgive me for lacking the basic skills of being able to sketch a graph. My future looks bright.

20170425_134012.jpg
(edited 7 years ago)
volumes of revolution.jpgBy applying the formula you found the volume of the solid formed when R is rotated around y axis. I need to ask you: what kind of solid is formed when R and S together are rotated around y axis? Could you also have a second look at your workings?
For part d why is it the area of the triangle - part b

and not:
area of triangle - (part a + part b)?

Screenshot 2017-04-25 19.58.15.png
in part a, we are not talking about the area, but about the volume.
Original post by kiiten
For part d why is it the area of the triangle - part b

and not:
area of triangle - (part a + part b)?

Screenshot 2017-04-25 19.58.15.png


As above, the calculation you made in part (a) is the volume of a solid of revolution, and is hence not relevant to part (d)
Original post by _gcx
As above, the calculation you made in part (a) is the volume of a solid of revolution, and is hence not relevant to part (d)


Is that because part a gives the 3D area? But parts b and d are 2D?

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