Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

appleibeats · Mar 20, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

a man 1.5m is walking directly towards a street lamp 8 metres off the ground at a speed of 36km/h.

i) Find the rate at which the length of his shadow is decreasing

ii) Find the speed with which the tip of his shadow is moving along the path

davidgoes4wce · Mar 21, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
a man 1.5m is walking directly towards a street lamp 8 metres off the ground at a speed of 36km/h.

i) Find the rate at which the length of his shadow is decreasing

ii) Find the speed with which the tip of his shadow is moving along the path

do you happen to have the answer? Just want to check

appleibeats · Mar 26, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

sorry no.

Sien · Mar 26, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I think this is how you do it, I'm not too sure so don't quote me on this

All the numbers are postive btw

Sent from my D6503 using Tapatalk

drsabz101 · Mar 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

how do you do Q11 .c. Chapter 5

Drongoski · Mar 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For (a): tan @ = 1
.: eqn is: y - 3 = 1 (x - 0 )
i.e. y = x + 3 or x -y + 3 = 0 (as required in general form)

For (d): tan @ = -1
.: eqn is: y - [-2] = -1 (x - 0) or x + y + 2 = 0

PS
Sorry - you asked for (c) only.

drsabz101 · Mar 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I know how to do allof them execpt for c as the answer had surds in it, and so I got confused.

drsabz101 · Mar 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

dont worry, it was a stupid mistake, i copied the question out wrong

appleibeats · Apr 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-04-05 at 6.11.00 pm.png

Could someone help explain part b)

I got part a) out, the answers were v^2 = 2(4 - x^2), root6 cm/s , It starts at x = 2 , so on the first occasion it reaches x = 1, it must be moving backwards.

Now part b) answer : -2 less than or = x less than or = to 2 , 2root2 cm/s

appleibeats · Apr 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

View attachment 33069

Could someone help explain part b)

I got part a) out, the answers were v^2 = 2(4 - x^2), root6 cm/s , It starts at x = 2 , so on the first occasion it reaches x = 1, it must be moving backwards.

Now part b) answer : -2 less than or = x less than or = to 2 , 2root2 cm/s

davidgoes4wce · Apr 6, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33069

Could someone help explain part b)

I got part a) out, the answers were v^2 = 2(4 - x^2), root6 cm/s , It starts at x = 2 , so on the first occasion it reaches x = 1, it must be moving backwards.

Now part b) answer : -2 less than or = x less than or = to 2 , 2root2 cm/s

In reply to your question Ex3E Q 7a,b

Would be good if somebody could confirm.

davidgoes4wce · Apr 7, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Anyone have an idea on how to do Ex 4A Q26 b

My thinking is we have to show some geometric proof.

InteGrand · Apr 7, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Anyone have an idea on how to do Ex 4A Q26 b

My thinking is we have to show some geometric proof.

From isosceles triangle AED, ABC, etc., we get that their base angles are 36 deg, because they have an angle of 108 deg (the pentagon's interior angle) and the angle sum of a triangle is 180 deg.

So angle CAB = 36 deg, i.e. angle BAP = 36 deg.

Also, by adjacent angles, angle DAB = angle EAB – angle EAD = 108 – 36 = 72 deg.

Finally, by adjacent angles, angle DAP = angle DAB – angle CAB = 72 – 36 = 36 deg.

Drongoski · Apr 7, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Q26

(b)

angle AED = one of 5 equal internal angles = (5-2)* 180/5 = 108 deg

Since triangle EAD is isosceles, angle EAD = (180- 108)/2 = 36 deg

Similarly angle BAC = 36 deg

Now angle BAE = 108 deg

.: angle CAD = 108 - 36 - 36 = 36 deg

.: angle DAC = angle BAC

i.e. angle DAP = angle BAP

(c)

By symmetry, ADB is isosceles

.: angle DAB = angle DBA = 72 deg

.: angle ADB = 180 - 2 x 72 = 36 deg

.: angle ADB = angle BAP and angle DBA = angle ABP

.: triangles DAB and ABP are equiangular and .: similar

(d)

By symmetry DA = DB and .: triangle DAB is isosceles

.: since triangles DAB & ABP are similar, the latter triangle is also isosceles.

.: AP = AB = 1

Also triangle APD is isosceles (2 base angles of 36 deg)

.: PB = AP = 1

Now corresponding sides of similar triangles DAB and ABP are proportional

.: DA/AB = AB/BP ==> DA/1 = 1/x ==> DA = 1/x

(e)

? ? ? ?

(f)

Assuming results of (e), using cosine rule for triangle ADB:

AD² + BD² - 2 * AD * BD cos 36 = AB²

But BD = AD

.: AD² + AD² - 2 *AD*AD*cos 36 = 1²

We get: 2*AD² (1-cos 36) = 1

Whence: 1 - cos 36 = 2* sin² 18 = (1/2)(1/AD)²

.: sin 18 = (1/(2*AD)) = 1/(2 * 0.5 *(sqrt(5 + 1))) (taking +ve square root)

= (1/4)(sqrt(5) - 1) upon rationalising the denominator.

QED

davidgoes4wce · Apr 8, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Drongoski said:
Q26

(b)

angle AED = one of 5 equal internal angles = (5-2)* 180/5 = 108 deg

Since triangle EAD is isosceles, angle EAD = (180- 108)/2 = 36 deg

Similarly angle BAC = 36 deg

Now angle BAE = 108 deg

.: angle CAD = 108 - 36 - 36 = 36 deg

.: angle DAC = angle BAC

i.e. angle DAP = angle BAP

(c)

By symmetry, ADB is isosceles

.: angle DAB = angle DBA = 72 deg

.: angle ADB = 180 - 2 x 72 = 36 deg

.: angle ADB = angle BAP and angle DBA = angle ABP

.: triangles DAB and ABP are equiangular and .: similar

(d)

By symmetry DA = DB and .: triangle DAB is isosceles

.: since triangles DAB & ABP are similar, the latter triangle is also isosceles.

.: AP = AB = 1

Also triangle APD is isosceles (2 base angles of 36 deg)

.: PB = AP = 1

Now corresponding sides of similar triangles DAB and ABP are proportional

.: DA/AB = AB/BP ==> DA/1 = 1/x ==> DA = 1/x

(e)

? ? ? ?

(f)

Assuming results of (e), using cosine rule for triangle ADB:

AD² + BD² - 2 * AD * BD cos 36 = AB²

But BD = AD

.: AD² + AD² - 2 *AD*AD*cos 36 = 1²

We get: 2*AD² (1-cos 36) = 1

Whence: 1 - cos 36 = 2* sin² 18 = (1/2)(1/AD)²

.: sin 18 = (1/(2*AD)) = 1/(2 * 0.5 *(sqrt(5 + 1))) (taking +ve square root)

= (1/4)(sqrt(5) - 1) upon rationalising the denominator.

QED

I was having a bit of trouble with part (e) of that question as well

My thinking is AP=1, DP=1, angle APD=108 degrees, AD=1/x , by using the cosine rule trying to solve for length AD.

My only problem at the moment using the cosine rule I can get the answer , but only as a decimal. I'm having trouble:

$AD^{2}=DP^{2}+AP^{2}-2 \times AP \times DP \times cos \ \theta$

$AD^{2}=1^{2}+1^{2}-2 \times 1 \times 1 \times cos \ 108^\circ$

InteGrand · Apr 8, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
I was having a bit of trouble with part (e) of that question as well

My thinking is AP=1, DP=1, angle APD=108 degrees, AD=1/x , by using the cosine rule trying to solve for length AD.

My only problem at the moment using the cosine rule I can get the answer , but only as a decimal. I'm having trouble:

$AD^{2}=DP^{2}+AP^{2}-2 \times AP \times DP \times cos \ \theta$

$AD^{2}=1^{2}+1^{2}-2 \times 1 \times 1 \times cos \ 108^\circ$

$\noindent For (e), note $PC = AC -AP = \frac{1}{x} -1$, because from the previous part, $AP=1$, $AC = DA$ (by symmetry), and $DA = \frac{1}{x}$. But note that triangle $BPC$ is isosceles$^\dagger$ with $PC = BC = x$. So from what we showed earlier, we obtain $PC = x = \frac{1}{x}-1$. Now, letting $y=\frac{1}{x}$ (clearly $x\neq 0$ as it is a length), this is $\frac{1}{y} = y-1$, whence $y^2 -y -1=0$. This is the famous defining quadratic equation of the golden ratio, so $y= \varphi =\frac{1+\sqrt{5}}{2}$, as $y$ is the positive root. The claim follows since $AD =\frac{1}{x}=y$. Q.E.D.$

$\noindent $^\dagger$ \small Because its base angles $\angle C$ and $\angle B$ are equal by symmetry.$

appleibeats · Apr 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-04-09 at 11.11.06 am.png

I have done part a).

Part b) is where I am stuck.

Answer to b) : x = 1/2 inverse tan (2t), x --> pi/4 and v = 1/(1 + 4t^2)

Also unsure how to show part c) acceleration result.

Instead I get accel = -4sin2xcos2x

Not sure how to make that into -4(cos2x)^3 (sin2x)

porcupinetree · Apr 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33075

I have done part a).

Part b) is where I am stuck.

Answer to b) : x = 1/2 inverse tan (2t), x --> pi/4 and v = 1/(1 + 4t^2)

Also unsure how to show part c) acceleration result.

Instead I get accel = -4sin2xcos2x

Not sure how to make that into -4(cos2x)^3 (sin2x)

appleibeats · Apr 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-04-09 at 2.05.14 pm.png

Could someone help me with the first part of part b).

Not sure how to explain why the time T for the particle to travel to x = 1 is T = Integral .....

I though the e^... should be in the denominator and x^2 in the numerator, but it isn't. Why is it switched around??

leehuan · Apr 9, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

v=x^2.e^(-x^2)

dx/dt = x^2.e^(-x^2)

dt/dx = e^(+x^2)/x^2

T = \int_{1/2}^{1} e^(x^2)/x^2 dx

Check to see if you remembered to reciprocate.

Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

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