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Edexcel C4 June 2016 - Unofficial Markscheme

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For the volume on 7b I got something like

V=\frac{1637}{5}\pi

I got same as you.
EDIT: nvm my bad

Reply 61

Kay Fearn

Hi guys i really need some help with this, i'm panicing so hard..

for question 4 i accidently put the x on the top instead of the bottom so differentiated and got -x^2 / 5 = t + c

then everything else i did was from this. So i think i got both of the questions wrong. My methodology was correct but i got the wrong equation because I made this silly error.

I ended up getting 640 as t and did 640x24x60 to get the minutes

How many marks would i lose for this? Would i get any marks carried forward?

for the very last vector question i used 1/2absinC to get the area but used 1/2(2route5)(2route5)(3/5) instead of using 2route2 so i got 6 instead of 12/5 for the area.

For the very last section I used P +/- linda(direction vector of l2) to work out the two points of B but worked out linda as 1 which is wrong.

How many marks in total would i lose for these errors? Any help would be very appreciated!

(edited 7 years ago)

Reply 62

username2130115

was the parametric x = 4 tant and y = 5root(3)sin2t

Reply 63

MrPooey

Anyone know how many marks would be lost for misreading the volume question as an area? I'm thinking all 4 lost.

Reply 64

Aruba852

I differentiated -cos (piy) wrong , didnt take the pi out. If my whole dy/dx is correct expect im missing the pi how much marks should i lose ... it was a 6 marks question i guess

Reply 65

NotNotBatman

Original post by Don Pedro K.

'Twas 544pi - 1024(pi)/5 = 1696(pi)/5

The 544 looks right, I can't remember the fraction though, what was the curve equation?

Reply 66

coolguy123456

Wasn't the area of the whole rectangle thing 512 pi?

Reply 67

Sapereaude54

Original post by edothero

Let E be a general point on the line

l_{2}

\mathbf{r} = \begin{pmatrix} 1-5\lambda \\ 5+4\lambda \\ 2+3\lambda \end{pmatrix}

Find PE:

\vec{PE} = \begin{pmatrix} 1-5\lambda \\ 5+4\lambda \\ 2+3\lambda \end{pmatrix} - \begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix}

\vec{PE} = \begin{pmatrix} -5\lambda \\ 4\lambda \\ 3\lambda \end{pmatrix}

Now we know that

\vec{AP} = \vec{PE}

Therefore the lengths are the same, as the length of

\vec{AP} = 2\sqrt{2}

You know that

|\vec{PE}| = \sqrt{(5\lambda)^{2} + (4\lambda)^{2} + (3\lambda)^{2}} = 2\sqrt{2}

\sqrt{50\lambda^{2}} = 2\sqrt{2}

50\lambda^{2} = 8

\lambda = \pm \dfrac{2}{5}

Then use this lambda with the general point equation

\mathbf{r} = \begin{pmatrix} 1-5\lambda \\ 5+4\lambda \\ 2+3\lambda \end{pmatrix}

Lol, lost all 5 marks on that question...shame
Also for the value of lambda that I obtained was 32, I know this is wrong know but everything else I got was correct.
Do you think the error will be carried forward to the next question.

Reply 68

Don Pedro K.

Original post by Aruba852

I differentiated -cos (piy) wrong , didnt take the pi out. If my whole dy/dx is correct expect im missing the pi how much marks should i lose ... it was a 6 marks question i guess

It was 5 marks for the dy/dx and 4 marks for the coordinates. I also got the dy/dx wrong but I reckon we will definitely get some method marks!

Reply 69

Don Pedro K.

Original post by NotNotBatman

The 544 looks right, I can't remember the fraction though, what was the curve equation?

Y = (2x-1)^(3/4) I believe?

Reply 70

gameofthrones1

http://www.thestudentroom.co.uk/showthread.php?t=4184271&p=66069183#post66069183 - Please vote!

Reply 71

Don Pedro K.

Original post by coolguy123456

Wasn't the area of the whole rectangle thing 512 pi?

We wanted volume not area :smile:

Reply 72

SirRaza97

OH GOD YES! After doing misreably in C3, I made up for it in c4!!!! 71-73/75. FEELSGOODMAN

Reply 73

JAllard

Any idea how many marks 6. b), c), d), & 8. a) were worth?

Reply 74

ImNervous

Hi guys
On Q3 I forgot the minus 2 and got an answer of 6/22+π instead of 8/22+π. This lead to my next part being wrong also. Bow many marks do you think I'll loose. All of them?

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Reply 75

Student403

Original post by edothero

Spoiler

Interesting.

I wrote because mod l2's direction = 5sqrt2 and we want lengths of +-2sqrt2, then we must multiply the line equation direction by factors of +-2/5