Additional Mathematics ; post your doubts here!

[kim anika]

ok i gt term independent......bt i need de answer

 

[somaya]

can u rite da stepz 4 question no 5 (ii)in maj/june2011/12

 

[Ryan De Leeuw]

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?

Thanks dude
Whatabout this one sir?
A curve has equation y = e–x (Acos 2x + Bsin 2x). At the point (0, 4) on the curve, the gradient of
the tangent is 6.
(i) Find the value of A. [1]
(ii) Show that B = 5. [5]
(iii) Find the value of x, where 0 < x < π2
radians, for which y has a stationary value. [5]

 

[Ashay]

I have a doubt in question 9 from O/N 2010 variant 21

 

[Ryan De Leeuw]

ok i gt term independent......bt i need de answer

Whats the question?

 

[ThatReallyReallyWeirdDude]

Thanks dude
Whatabout this one sir?
A curve has equation y = e–x (Acos 2x + Bsin 2x). At the point (0, 4) on the curve, the gradient of
the tangent is 6.
(i) Find the value of A. [1]
(ii) Show that B = 5. [5]
(iii) Find the value of x, where 0 < x < π2
radians, for which y has a stationary value. [5]

Is that (e-x)(Acos2x+Bsin2x)?
Because if it's y=e-x(Acos2x+Bsin2x) then when x =0, y=e, not 4 :|

 

[ThatReallyReallyWeirdDude]

I have a doubt in question 9 from O/N 2010 variant 21

​

 

[kim anika]

look @ my prevoius question.......

 

[kim anika]

0606/2/m/j/03
Q9i

 

[Ashay]

Thank you so much!!

 

[kim anika]

a=e^lnb
a=b
is this formula right?????

 

[ThatReallyReallyWeirdDude]

a=e^lnb
a=b
is this formula right?????

lnb gives a value x
such that e^x =b.
So yeah, e^lnb=b.

 

[kim anika]

y isnt nebody answering my questions????

 

[kim anika]

lnb gives a value x
such that e^x =b.
So yeah, e^lnb=b.

thank you so much!!!!!
0606/2/m/j/03
Q9i

 

[ThatReallyReallyWeirdDude]

0606/2/m/j/03
Q9i

I dont think questions like this are in our syllabus anymore but nevertheless:

 

[kim anika]

I dont think questions like this are in our syllabus anymore but nevertheless:
View attachment 11603[/quote
thnk u......gud 2 knw dese question myt nt cum
nehw can u suggest wat shud i revise nw my xm starts after 4.5 hrs

 

[ThatReallyReallyWeirdDude]

Hmm
Simple
Formulas for trigonometric substituions, sin/cosine theorem etc. come in the exam so you should know the different integration/differentiation rules
Small things like, remember how to find unit vectors, binomial theorem/permutations/combinations formula...

 

[somaya]

Hmm
Simple
Formulas for trigonometric substituions, sin/cosine theorem etc. come in the exam so you should know the different integration/differentiation rules
Small things like, remember how to find unit vectors, binomial theorem/permutations/combinations formula...

can any1 clear my doubt

 

[silverstar]

Sorry for late reply, had my physics exam
Alright then, the particle comes at rest at t=5, so the summary of it's movement is as shown:
View attachment 10807
So the total distance travelled is the distance travelled between t=0 and t=5 (i.e. s5-s0) plus the distance travelled between t=10 and t=5 (s10-s5)
{v=1.4t-0.3t2+c
when t=0, v=0.5
v=1.4(0)-0.3(0)+c=0.5
therefore c=0.5
and v=1.4t-0.3t2+0.5} Though you woulda calculated this in (i)
So as for distance, s= integration of v
=0.7t2-0.1t3+0.5t + c
Since s=0 when t=0, c=0.
So s=0.7t2-0.1t3+0.5t
Now all you have to do is find |s5-s0| + |s10-s5|
=|7.5-0|+|-25-7.5|
=7.5+32.5=40 meters
The alternative way to understand it (instead of s5-s0 + s10-s5) is to think that first it gets to t=5, which is distance of 7.5m, then it goes back to 0 (another distance of 7.5m) and to t=10 (a distance of 25m FROM O)
So the total distance will be 2x7.5 + 25=40 meters.
Hope I helped!

hi cud someone plz get some notes on add maths ???