Some C4 questions Watch

loza3003
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Ive been going through some questions for C4 using the examination style questions at the back of the book and I have been getting stuck on a few of them and I do not know where I am going wrong. So I was wondering if someone could explain how to do them .

Firstly is question 4C. Which is a question on differential equations.

Hence or otherwise, obtain the solution of
(2x+1)(x-2) dy/dx= 10y

The answer from section 2 helps which asks to intergrate 1/ (2x+1)(x-2)
This is lnk((x-2)/(2x+1))^1/5

Next is question 5C. A population grows in such a way that the rate of change of the population P at time T in days is proportional to P.
The equation is: P=AK^t
C. Initially the population is 8 million and 7 days later it has grown to 8.5 million. Find the size of the population after a further 28 days.

6. Referred to an origion O the points A and B have position vectors i-5j-7k and 10i+10j+5k
Vector equation for the line passing through A and B is r=i-5j-7k+t(9i+15j+12k)
The position vector of point P is 4i-3k
C. Find the area of the triangle OAB
D. Find the ratio in which P divides the line AB.

7C. intergate 9tsintdt between pie/2 and pie/3 (-) intergral 12sin^2tcost st between pie/2 and pie/3.

Any help would be appreciated, would just like to finish the last steps of these off.
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Ch1pp0
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4C:
 (2x+1)(x-2) \frac{dy}{dx} = 10y

Divide by (2x+1)(x-2) and 10y. Multiply by dx.

 \frac{1}{10y}dy = \frac{1}{(2x+1)(x-2)} dx



\int \frac{1}{10y}dy = \int \frac{1}{(2x+1)(x-2)} dx

The hence or otherwise does the bit on the right. You should be able to do the bit on the left (hint involves lny)

5C: P=AK^t

Initially in C4 maths language is t=0. Therefore, initially, P=A, A=8 (measuring pop in millions to save time).

 8.5=8k^7

k^7 = 1.0625 (=8.5/8)

Solve to get k = (1.0625)^(1/7)
You now know A, K and t = 28. Find P.
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Ch1pp0
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For 6.C. It is a little bit out there but I would got with Heron's formula.
http://en.wikipedia.org/wiki/Heron's_formula

You see the length  OA = \sqrt{1^2 + (-5)^2 + (-7)^2} = \sqrt{75}

And the length  OB = \sqrt{10^2 + 10^2 + 5^2} = 15

And because you've probably got t=1 the length  AB = \sqrt{9^2 + 15^2 + 12^2} = \sqrt {450}

Or you could use the three lengths to find an angle (cosine rule) and use that angle and 2 other lengths to get  \text{AREA} = \frac{1}{2}absinC
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ProgressDesired
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7C. intergate 9tsintdt between pie/2 and pie/3 (-) intergral 12sin^2tcost st between pie/2 and pie/3.

To integrate 9tsint dt use integration by parts.

u = 9t
dv/dt = sin(t)

du/dt = 9
v = -cos(t)

So the integral is -9t*cos(t)-INT(-9cos(t))
So it's -9t*cos(t)+9*sin(t)

I think that's right, but make sure you do it yourself to make sure you understand the process.

To integrate the other one, suppose the integral is:

4(sin(t)^3)

What would the derivative be? Using the chain rule:

12*cos(t)*(sin(t)^2)

So that's it. Now just calculate the definite integral and you should be done.
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electriic_ink
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(Original post by Ch1pp0)
For 6.C. It is a little bit out there but I would got with Heron's formula.
http://en.wikipedia.org/wiki/Heron's_formula

You see the length  OA = \sqrt{1^2 + (-5)^2 + (-7)^2} = \sqrt{75}

And the length  OB = \sqrt{10^2 + 10^2 + 5^2} = 15

And because you've probably got t=1 the length  AB = \sqrt{9^2 + 15^2 + 12^2} = \sqrt {450}

Or you could use the three lengths to find an angle (cosine rule) and use that angle and 2 other lengths to get  \text{AREA} = \frac{1}{2}absinC
What you're probably supposed to do is use \mathbf{a} . \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos\theta to get the angle are then plug it into the area formula.
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Ch1pp0
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\displaystyle \int_\frac{\pi}{3}^\frac{\pi}{2} 9t\sin t dt - \int_\frac{\pi}{3}^\frac{\pi}{2} 12\sin^2 t\cos t dt

The part on the left is done by parts:
 u = 9t, du = 9, dv = \sin t, v=-\cos t

\int udv = uv - \int vdu



\displaystyle \int_\frac{\pi}{3}^\frac{\pi}{2} 9t\sin t dt = \left[-9t\cos t \right]_\frac{\pi}{3}^\frac{\pi}{2} - \int -9\cos t dt

 \displaystyle = \left[-9t\cos t \right + 9\sin t ]_\frac{\pi}{3}^\frac{\pi}{2}

The part on the right is quite easy:

 12\sin^2 t\cos t = \frac{d}{dt} 4(\sin t)^3

(Original post by ProgressDesired)
7C.
Sorry, when I started writing this wasn't posted. At least the answers match .
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loza3003
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Thanks to everyone who has helped =) Understand how to do them all now, except the vectors one.
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loza3003
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Sorry, but I also have a few more questions, these questions I would just like to know certain bits.
1. Subsitution: x/square root(x-2)
x=2+u -> Subsitution
Could I have help just up until the subbing in, as I cannot work out where i keep going wrong.

2. Differential equations.
dy/dx= 3y^2/ (2x+1)^3
y=2, x=1. Give answer in the form y= f(x)
Need help with all of it after rearranging it.

3. Using the subsitution u=1+2x
intergral 4x/(1+2x)^2

Then using this helps with differential equations. Given that y= pi/4 when x=0 solve the differential equation.
(1+2x)^2 dy/dx= x/sin^2y

Thank you all again =)
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Meriam_xx
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(Original post by Ch1pp0)
4C:
 (2x+1)(x-2) \frac{dy}{dx} = 10y

Divide by (2x+1)(x-2) and 10y. Multiply by dx.

 \frac{1}{10y}dy = \frac{1}{(2x+1)(x-2)} dx



\int \frac{1}{10y}dy = \int \frac{1}{(2x+1)(x-2)} dx

The hence or otherwise does the bit on the right. You should be able to do the bit on the left (hint involves lny)

5C: P=AK^t

Initially in C4 maths language is t=0. Therefore, initially, P=A, A=8 (measuring pop in millions to save time).

 8.5=8k^7

k^7 = 1.0625 (=8.5/8)

Solve to get k = (1.0625)^(1/7)
You now know A, K and t = 28. Find P.






PLEASE HELP ME SOLLVE THE DIFFERNTIAL EQUATION FOR PART 4C

Hence or otherwise, obtain the solution of
(2x+1)(x-2) dy/dx= 10y
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