O-Level Additional Math

[FrekiWang]

listener:

Q1) A particle P travels in a straight line so that its distance, s m, from a fixed point O is given by s = 2t + 18/(t+1), where t is the time in seconds measured from the start of the motion. Calculate

i) the initial acceleration of P.
ii) the velocity of P when it is next at starting point.
iii) the value of t when the particle is instantaneously at rest.
iv) the total distance travelled in the first 5 seconds.

Q2) The diagram shows parts of the curve y = 2e^(x/2) (not drawn to scale) and the lines x=1, x=k and x=4.
i) Find the total area of regions A, B and C giving your answer correct to 1 decimal place.
ii) Given that the area of Region A is equal to Region B, calculate the value of k giving your answer correct to 4 significant figures.

http://i56.tinypic.com/nya59z.jpg\">

Q1 (I assume there is no () for 2t+18)
i)
v= ds/dt = 2 - 18(t+1)^(-2), a = dv/dt = 36(t+1)^(-3)
when t=0, a = 36(1)^(-3)=36m/s2
ii)
for the position of the starting point, t=0, s = 2(0) + 18/(0+1)= 18
when it is at the starting point again, s=18
2t+18/(t+1)=18
2t(t+1)+18=18(t+1)
2t^2 -16t = 0
2t(t - = 0
t = 0(when it starts) or t = 8(next at starting point)
v = 2 - 18(8+1)^(-2) = 16/9 m/s
iii)
v=0 (at rest)
2-18(t+1)^(-2)=0
(t+1)^(-2)=1/9
(t+1)^2=9
t+1 = -3 or t+1 = 3
t = -4(reject) or t = 2
so t = 2
iv) it makes a turn at t = 2, so consider two segments 0 to 2, and 2 to 5.
0 to 2:
when t=0, s = 18
when t=2, s = 2(2)+18/(2+1)=10
distance travelled=|10-18|=8
0 to 5\"
when t=5, s = 2(5) + 18/(5+1)=13
distance travelled=|13-10|=3
So total distance travelled is 8+3=11m

[FrekiWang]

listener:

Q2) The diagram shows parts of the curve y = 2e^(x/2) (not drawn to scale) and the lines x=1, x=k and x=4.
i) Find the total area of regions A, B and C giving your answer correct to 1 decimal place.
ii) Given that the area of Region A is equal to Region B, calculate the value of k giving your answer correct to 4 significant figures.

http://i56.tinypic.com/nya59z.jpg\">

Q2.
i) when x=4, y=2e^2, total area of A,B,C and small unknown area = 4 x 2e^2=8e^2.
The small unknown area
=Integrate[2e^(x/2)]dx (from 0 to 1)
=4e^(x/2) (from 0 to 1)
=4e^0.5 - 4e^0
=4e^0.5 - 4
So total area of A+B+C
= 8e^2 - (4e^0.5 - 4)
= 56.52
= 56.5(1dp)
ii) Area of A
= Integrate[2e^(x/2)]dx (from 1 to k)
= 4e^(x/2) (from 1 to k)
= 4e^(k/2) - 4e^0.5
Area of B
= Integrate[2e^(x/2)]dx (from k to 4)
= 4e^(x/2) (from k to 4)
= 4e^(2) - 4e^(k/2)
Area of A = Area of B, we have
4e^(k/2) - 4e^0.5 = 4e^(2) - 4e^(k/2)
e^(k/2) - e^0.5 = e^(2) - e^(k/2)
2e^(k/2) = e^0.5 + e^2
e^(k/2) = 4.518889
k/2 = ln(4.518889)
k = 3.017(4sf)

[Xiao Hu]

Hi Freki,

Appreciate it if you could help to solve this Add Maths question. It's from the \"Double-Angle Formulae\" trigo topic.

If 270<x<360, simplify sqrt(2+sqrt(2+2cosx)).
Ans:2sin(x/4).

My answer is 2cos(x/4), different from txtbook's.
http://i56.tinypic.com/jrdxxs.jpg\">
Thanks in advance,
Xiao Hu

[FrekiWang]

Xiao Hu:

Hi Freki,

Appreciate it if you could help to solve this Add Maths question. It's from the \"Double-Angle Formulae\" trigo topic.

If 270<x<360, simplify sqrt(2+sqrt(2+2cosx)).
Ans:2sin(x/4).

My answer is 2cos(x/4), different from txtbook's.
http://i56.tinypic.com/jrdxxs.jpg\">
Thanks in advance,
Xiao Hu

sqrt[2+sqrt(2+2cosx)]
=sqrt{2+sqrt[2+2(2cos^2(x/2)-1)]}
=sqrt[2+sqrt(4cos^2(x/2))]
Note here, 135<x/2<180 (2nd), so cos(x/2) is negative.
=sqrt[2-2cos(x/2)]<--- I suppose your mistake is because you have +2cos(x/2) in this step, I will explain this at the end.
=sqrt[2-2(1-2(sin^2(x/4))]
=sqrt[4sin^2(x/4)]
Note here, 67.5<x/4<90 (1st), so sin(x/4) is positive.
=2sin(x/4)

So the common mistake here is that some students will assume sqrt(x^2)=x, which is not true.

In fact, sqrt(x^2)=|x|, which is equal to x if x is positive and is equal to -x if x is negative. (e.g. if x = -2, we have sqrt(x^2)=2 which is -x).

Therefore, whenever you need to pull a 'perfect square' from the square root, make sure you determine whether you are pulling the square of a positive or negative number. In this question, cos(x/2) is negative, that is why when you pull cos^(x/2) out of the square root, you have to add a negative sign in front.

Hope you could understand. Cheers

[Xiao Hu]

Hi FrekiWang,

Perfect explanation, you found my mistake. Yes, I undestand that. I think this is a very common pitfall.
Very happy to learn from you, it’s a great site to have you and others help us parents with maths questions that we can’t solve.

Thanks again!
Xiao Hu

[Xiao Hu]

Hi FrekiWang,


Need help on this question, please see the attached picture.

Ans in txtbook:2y=x+4

I just couldn't figure out how to get at least 1 more point on the chord in order to find the equation of the chord. Even with the centre and the radius of the circle given, and the mid-point given, I just couldn't figure it out.

http://i52.tinypic.com/2q0j7ty.jpg\">

(Can't add the picture, it kept prompting can't be more than 640 pixel wide. So I tried breaking it up into 3 lines, hope it's readable.)
Thanks in advance,
Xiao Hu

[FrekiWang]

I suppose you can understand my explanation below:


The line connecting the centre of circle and the midpoint of the chord, is perpendicular to the chord.

You have the coordinates of the centre and the midpoint, thus the gradient of the line. Then you can find the gradient of the chord.

Gradient of line + coordinates of one point will be enough for you to find the equation:D

[Xiao Hu]

Hi FrekiWang,

Oh mine! You are great!! When I first wrote my question, I was saying I just coundn’t find at least 1 more point or a gradient. But I have to rewrite bec was having trouble to load the image of the question. I left out the word gradient.

THanks so much for your prompt reply and kind help!!

You are good,
Xiao Hu

[FrekiWang]

Xiao Hu:

Hi FrekiWang,

Oh mine! You are great!! When I first wrote my question, I was saying I just coundn't find at least 1 more point or a gradient. But I have to rewrite bec was having trouble to load the image of the question. I left out the word gradient.

THanks so much for your prompt reply and kind help!!

You are good,
Xiao Hu

lol, you are welcome

[Xiao Hu]

Hi FrekiWang,


May I ask more about this question I posted.
What if the sign in the question is changed from + to -?

Quote part of your solution below:
=sqrt[2+sqrt(4cos^2(x/2))]
Note here, 135<x/2<180 (2nd), so cos(x/2) is negative.
=sqrt[2-2cos(x/2)]<--- I suppose your mistake is because you have +2cos(x/2) in this step, I will explain this at the end.

So we would have to change the sign to + when we take out the perfect square cos(x/2) out of the root?

Thanks in advance,
Xiao Hu