The Student Room Group

AQA A2 Mathematics MPC3 Core 3 - Wednesday 15th June 2016 [Official Thread]

Scroll to see replies

Original post by Zacken
Post your working then.


I tried using chain rule this time;

dy/dx = 4 + 4tan^2(4x)
the 4 differentiates to 0 so i did chain rule of 4tan^2(4x);
=4(tan^2(4x))
=4[ (tan4x)^2 ]

u= tan4x y = u^2
du/dx = 4 + 4tan^2(4x) (as worked out in part a) dy/du = 2u (2tan4x)

dy/dx = 4[ du/dx x dy/du]
= 4[ (4 + 4tan^2(4x) ) x (2tan4x) ]
= 4[ 8tan(4x) + 8tan^3(4x) ]
= 32tan(4x) + 32tan^3(4x)
= 32tan(4x) [ 1 + tan^2(4x) ]
y = tan4x so;
= 32y(1 + y^2)
Original post by Abbas.13
I tried using chain rule this time;

dy/dx = 4 + 4tan^2(4x)
the 4 differentiates to 0 so i did chain rule of 4tan^2(4x);
=4(tan^2(4x))
=4[ (tan4x)^2 ]

u= tan4x y = u^2
du/dx = 4 + 4tan^2(4x) (as worked out in part a) dy/du = 2u (2tan4x)

dy/dx = 4[ du/dx x dy/du]
= 4[ (4 + 4tan^2(4x) ) x (2tan4x) ]
= 4[ 8tan(4x) + 8tan^3(4x) ]
= 32tan(4x) + 32tan^3(4x)
= 32tan(4x) [ 1 + tan^2(4x) ]
y = tan4x so;
= 32y(1 + y^2)


Yeah, that's correct, what's the problem?
Original post by Zacken
Yeah, that's correct, what's the problem?


I didn't get what you meant by use the rule tan^2(x) + 1 = sec^2(x)
Original post by Abbas.13
I didn't get what you meant by use the rule tan^2(x) + 1 = sec^2(x)


...uhm, you used it several times in your answer.
Original post by Zacken
...uhm, you used it several times in your answer.


I did? I don't recall making use of it??
Reply 125
June 2013 made me lose the will to live. There's like 20 marks based on sketching/interpreting graphs. Give me 20 marks of integration instead :frown:
Original post by Fudge2
June 2013 made me lose the will to live. There's like 20 marks based on sketching/interpreting graphs. Give me 20 marks of integration instead :frown:

Yeah I bloody hated it
Original post by Fudge2
June 2013 made me lose the will to live. There's like 20 marks based on sketching/interpreting graphs. Give me 20 marks of integration instead :frown:


Evaluate 1/41/21x21x2dx \displaystyle \int_{1/4}^{1/2} \frac{1}{x^2\sqrt{ 1-x^2}} dx .
May not be 20 marks. But it may be quite a few.
(edited 7 years ago)
Reply 128
Original post by Ano123
Evaluate 1/41/21x21x2dx \displaystyle \int_{1/4}^{1/2} \frac{1}{x^2\sqrt{ 1-x^2}} dx .
May not be 20 marks. But it may be quite a few.

image.jpg

How'd I do? :smile: cool question
Reply 129
Original post by Fudge2
image.jpg

How'd I do? :smile: cool question


I got the same as you (2.14093) 2.14 to 3 s.f:smile:
Original post by Fudge2
image.jpg

How'd I do? :smile: cool question


Original post by xs4
I got the same as you (2.14093) 2.14 to 3 s.f:smile:


Very good, but exact values are preferable - but I suppose it doesn't matter.
I used u=1/x sub. Transforms it into a very easy integral. The exact value is 15 -3.
(edited 7 years ago)
Hi,

I've literally just read the last page of this thread so I don't know if this question is considered difficult/easy so I might be posting inappropriately! And I've just had to look how to use LaTex...

I don't know if these types of questions are still on spec, but I tried to make it difficult by combining quite a few rules and identities.

Solve:

Unparseable latex formula:

\int\frac{4+x}\sqrt{(4-2x^{2})}



Giving your answer in the simplest rationalised form.

I have worked out the solution and will post it tomorrow!
(edited 7 years ago)
Reply 132
Original post by Ano123
Very good, but exact values are preferable - but I suppose it doesn't matter.
I used u=1/x sub. Transforms it into a very easy integral. The exact value is 15 -3.


Yeah I tried to get an exact value but the lower limit of arcsin(0.25) isn't nice so my calculator wouldn't give me nice values :tongue: u=1/x...I'd never spot that haha
(edited 7 years ago)
Original post by Fudge2
Yeah I tried to get an exact value but the lower limit of arcsin(0.25) isn't nice so my calculator wouldn't give me nice values :tongue: u=1/x...I'd never spot that haha


You could have still got an exact answer using your sub instead of mine. You just need to know what cot(arcsinx) is. :smile:
Original post by Suits101
Hi,

I've literally just read the last page of this thread so I don't know if this question is considered difficult/easy so I might be posting inappropriately! And I've just had to look how to use LaTex...

I don't know if these types of questions are still on spec, but I tried to make it difficult by combining quite a few rules and identities.

Solve:

Unparseable latex formula:

\int\frac{4+x}\sqrt{(4-2x^{2})}



Giving your answer in the simplest rationalised form.

I have worked out the solution and will post it tomorrow!

Integral answer.jpg
√(1+sin2x) dx
This one is tricky.
[latex doesn't seem to be working properly.]
(edited 7 years ago)
Original post by Suits101
Hi,

I've literally just read the last page of this thread so I don't know if this question is considered difficult/easy so I might be posting inappropriately! And I've just had to look how to use LaTex...

I don't know if these types of questions are still on spec, but I tried to make it difficult by combining quite a few rules and identities.

Solve:

Unparseable latex formula:

\int\frac{4+x}\sqrt{(4-2x^{2})}



Giving your answer in the simplest rationalised form.

I have worked out the solution and will post it tomorrow!

This would not come up on C3, only fp2
Original post by Qcomber
This would not come up on C3, only fp2


Hi.

In the AQA textbook these questions are in the Core 3 section, especially since it's just a u substitution mixed with arcsin formula which the chapter includes.

I haven't seen any come up, but just incase I posted it.
Original post by Suits101
Hi.

In the AQA textbook these questions are in the Core 3 section, especially since it's just a u substitution mixed with arcsin formula which the chapter includes.

I haven't seen any come up, but just incase I posted it.


If it does come up, they shouldn't have because knowledge of inverse trigonometrical integration is not on the spec for c3. Whereas the fp2 spec clearly states that you need to know that it even exists in the formula book.
Original post by Ano123
Integral answer.jpg


yep thats what i got. I'm guessing you're doing fp2

Quick Reply

Latest

Trending

Trending