Edexcel C3,C4 June 2013 Thread

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ahhh I was getting stuck at differentiating xlna, but lna is just a constant which is why x -> 1 or rather, x disappears

thanks very much
(is that biology? lol)

Reply 4841

Mallika

Original post by PythianLegume

Integration finds the total area, and area below the x-axis is considered 'negative area' therefore, if a graph has sections above and below the x-axis, you have to split it into integrals and take the modulus of the negative integral.

Thank you

I see how that would work for something like a sine graph (where it's easier to split the parts of the curve) but what about a graph where limits for both areas (above and below) are the same? How would we split it then?

Reply 4842

username878045

Original post by Fiyinad

Yh i get dat bit buh wah if cot3x came up in the exam? would it be 1/3ln[sin(3x)]?

Yeh.

Reply 4843

arcturus7

Original post by MedMed12

its all about the MBE, OBEs at the moment

Surely in a maths exam it's all about the ODE's? :wink:

How's everyone feeling for this tomorrow?

Reply 4844

username878045

Original post by Supes180

If you swap limits by substitution and you end up with limits that have a different size order, do you still go from high to low/top to bottom? For example

\int^5_3 \ dx

turns into

\int^2_4 \ du

Is that right? Or should it be

\int^4_2 \ du

Keep them the first way.

Reply 4845

Kreayshawn

Original post by Frankster

y = a^x

dy/dx = a^x lna

y = a^x

ln y = xlna

1/y . dy/dx = ln a

dy/dx = y ln a

(From y = a^x )

Therefore: dy/dx = a^x ln a

can someone explain to me why the x goes in the bolded line?

Reply 4846

username878045

Original post by Mallika

Thank you

Well, if the graph is symmetrical above and below the x-axis, you double the answer - all you have found is the area above the x-axis (assuming you didn't get an answer of zero).

Reply 4847

username878045

Original post by Kreayshawn

can someone explain to me why the x goes in the bolded line?

It's been differentiated.

Reply 4848

MedMed12

Original post by arcturus7

Surely in a maths exam it's all about the ODE's? :wink:

How's everyone feeling for this tomorrow?

haha good one!!
Good :smile:

I like c4

just so tired
wbu?

Reply 4849

bloomingblossoms

n an experiment a scientist considered the loss of mass of a collection of picked leaves. The
mass M grams of a single leaf was measured at times t days after the leaf was picked.
The scientist attempted to find a relationship between M and t. In a preliminary model she
assumed that the rate of loss of mass was proportional to the mass M grams of the leaf.
(a) Write down a differential equation for the rate of change of mass of the leaf, using this
model.
(2)
(b) Show, by differentiation, that M = 10(0.98)^t
satisfies this differential equation.

I am really stuck on part b, can anyone help?

I've got this far: dm/dt=-kv
For the diffrentiation part: dM/dt: 10x0.98^(t)xln(0.98)

Reply 4850

Econ1994

Someone PLEASE post their solution to question 7c from paper E!!! Emailed my teacher and they havent replied -_-

Reply 4851

LegendX

Original post by Tikara

ahhh I was getting stuck at differentiating xlna, but lna is just a constant which is why x -> 1 or rather, x disappears

thanks very much
(is that biology? lol)

Yeah lol it's biology, I'm running out of paper so using whatever I can find lol haha.

Reply 4852

ACBLISS

Is everyone ready for C4, i can't wait for it to be over :/ It's boring doing the same maths again and again and again and again.....:/

Reply 4853

Dayo

An integral a day keeps the doctor away...

Reply 4854

yojay121

Original post by Econ1994

Someone PLEASE post their solution to question 7c from paper E!!! Emailed my teacher and they havent replied -_-

post the question and ill try and help i dont know what paper E is tho

Reply 4855

Killer_94

Can someone please help me with part b?

Reply 4856

Mallika

Original post by PythianLegume

Well, if the graph is symmetrical above and below the x-axis, you double the answer - all you have found is the area above the x-axis (assuming you didn't get an answer of zero).

Just one more question :colondollar:

What if it wasn't symmetrical? Or would we never such a graph in C4? (There's a question like this in the mixed exercise for integration which is why I'm asking)
Thank you again

Reply 4857

101101

I haven't done C3 for a year so can anybody think of anybody i should read through thats in c3 that could potentially come up to tomorrow, or something to just remind myself about.