Additional Mathematics ; post your doubts here!

[ThatReallyReallyWeirdDude]

oh ..at t=5, speed is 0m/s....so after dat we put values of t and found dat distance is decreasing...that is it is returning to the starting point?...is it not the way to think...correct me if i m wrong..

You could think of it that way, but you should know that whenever speed = 0 on particle questions that means the particle just turned around, and so while the distance from O would start decreasing the distance TRAVELLED would be increasing.
If you just found the distance at t=10s straight away you'd get 25m/s, which is wrong because it's not counting the distance between t=0 and t=5 and back

 

[hredoymohammad]

many many thx..for clearing out the conception

 

[omglaigcse]

if they ask you sketch/draw a graph function with a range, for example x>1, do you draw the whole graph, or only the part x>1??

 

[Chucky]

woah, I didn't think anyone here took Add math!
I've been practicing latest papers all WEEK. I'll definitely post for help!

 

[hredoymohammad]

confusion....suppose we have to find the area under a graph...sometimes that is broken into parts...for example in one part x= 2 to x=3 as limit ...and again x=4 to x=5 as a limit...when will we broke the part into two and when not?

 

[ThatReallyReallyWeirdDude]

if they ask you sketch/draw a graph function with a range, for example x>1, do you draw the whole graph, or only the part x>1??

Yeah, stick to the domain given. I've lost marks before not doing that
(i.e. draw only x>1)

 

[ThatReallyReallyWeirdDude]

confusion....suppose we have to find the area under a graph...sometimes that is broken into parts...for example in one part x= 2 to x=3 as limit ...and again x=4 to x=5 as a limit...when will we broke the part into two and when not?

I'm sorry I don't quite get what you mean.
I think what you're talking about is when the curve goes UNDER the X-Axis? In this case, you have to find the modulus of those areas under the X-Axis since it'll give you a negative value for the area.
For example if you have to find the area between x=2 and x=5, but between x=3 and x=5 the curve is under the x-axis, you'd have to find the area between x=2 and x=3, then find the area between x=3 and x=5. The area between x=3 and x=5 will come as a negative value so take the modulus (positive value) of it.

 

[hredoymohammad]

yes thanx ...i was talking about dat... so if a graph is below x axis and above x axis...we will divide them into two parts..and add them...n otherwise if the curve is above x axis...we dont need to divide the curve into two ? right

 

[hredoymohammad]

n brother could u tell me where to use rad and degree in very short.. most of the tym i mix dem up..and reach a wrong answer -_-

 

[ThatReallyReallyWeirdDude]

yes thanx ...i was talking about dat... so if a graph is below x axis and above x axis...we will divide them into two parts..and add them...n otherwise if the curve is above x axis...we dont need to divide the curve into two ? right

This should explain clearly:

​

n brother could u tell me where to use rad and degree in very short.. most of the tym i mix dem up..and reach a wrong answer -_-

Okay.
WHENEVER YOU HAVE A CALCULUS QUESTION USE RADIANS
Otherwise you can just tell from question:

If it is something along the lines of, solve sin/tan/cos(z) for 0<z<6, you can tell it's radians since the range of z is so small.
READ THE QUESTION CAREFULLY! Usually it SAYS it's in radians!
If range is big like 0<x<180 you can tell it's in degrees!​

​

 

[hredoymohammad]

Thanx a lot bhia btw r u giving IGCSE..or As Or a level ?
and pls spend 30sec reading this suppose a curve is only above x axis...but there is nothing below x axis.....
so now it has got only one part ...there area between x=a and x =b
so total area is A= integrate f x for limits a to b .
i m correct ?

 

[ThatReallyReallyWeirdDude]

Thanx a lot bhia btw r u giving IGCSE..or As Or a level ?
and pls spend 30sec reading this suppose a curve is only above x axis...but there is nothing below x axis.....
so now it has got only one part ...there area between x=a and x =b
so total area is A= integrate f x for limits a to b .
i m correct ?

Yes, you're right
I'm taking IGCSE

 

[hredoymohammad]

wow ...ur concepts are very clear ...i think u have kept in touch with ad maths...i have not practised it for many days..so some of the concepts i forgot :/ btw which subs u r giving dis time?

 

[ThatReallyReallyWeirdDude]

wow ...ur concepts are very clear ...i think u have kept in touch with ad maths...i have not practised it for many days..so some of the concepts i forgot :/ btw which subs u r giving dis time?

Actually I've been focusing mostly on other subjects!
Additional Maths, Chemistry, Physics, English First Language, Accounting, Mathematics, Arabic as a Foreign Language and ICT. You?

 

[hredoymohammad]

Ad maths, Chemistry, Physics, English First Lang, Accounting, Maths, Economics, Bengali are not u pakistani? i thot so

 

[ThatReallyReallyWeirdDude]

Ad maths, Chemistry, Physics, English First Lang, Accounting, Maths, Economics, Bengali are not u pakistani? i thot so

I'm not P:
Mm nearly same except I have ICT instead of Eco and Arabic instead of Bengali

 

[Ryan De Leeuw]

Help? This shit's drivin me nuts -_-

 

[ThatReallyReallyWeirdDude]

Help? This shit's drivin me nuts -_-

I remember that question!
Alright:
In the first 0.5 seconds, the spacecraft will go from 12j to 12j+20i+7.5j (0.5*it's velocity)
=20i+19.5j
Then the missile shoots
The position of spacecraft at t seconds (t seconds after missile shoots)
equals original position + distance moved in t seconds
=20i+19.5j+t(40i+15j)
=(20+40t)i+(19.5+15t)j
The position of missile at t seconds
=46i + t(ki+30j)
=(kt + 46)i + 30tj
Now if they collide, they should have an equal position vector at a time t
therefore
(20+40t)i+(19.5+15t)j=(kt+46)i+(30t)j
The j coefficients are equal
Therefore 19.5+15t=30t
15t=19.5
and t=1.3
we have to add the 0.5 seconds before the missile shot
so the time taken = 1.3+0.5=1.8 seconds.
next, k is easy
We know t
and using earliers "(20+40t)i+(19.5+15t)j=(kt+46)i+(30t)j"
20+40t=kt+46
t=1.3
20+52=1.3k+46
1.3k=26
k=20
Correct?

 

[kim anika]

0606/2/m/j/03
Q9i

 

[kim anika]

How to find term independent in binomial theorum ????