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Mathematics: Post your doubts here!

Lama AN

Lama AN

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Al salam 3alaykom people <3
I have one question in M1 in session 2005 NOV . Question number 1 part (ii)
can anyone explain to me how is the working done for this question ? please i have a quiz tomorrow
 
Saad Mughal

Saad Mughal

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Lama AN said:
Al salam 3alaykom people <3
I have one question in M1 in session 2005 NOV . Question number 1 part (ii)
can anyone explain to me how is the working done for this question ? please i have a quiz tomorrow
Click to expand...

Walaikum Assalam.

Question 1:
(a) We know the velocities of the car at different points,
We need to find acceleration and distance,
Hence, 2as = v^2 - u^2 is to be used.

Solu:
(i) AB = d1, acceleration = a, initial velocity = 5 m/s, final velocity = 7 m/s
2as = v^2 - u^2
2a(d1) = (7)^2 - (5)^2
a*d1 = (49-25)/2
a*d1 = 12
a = 12/d1

(ii) BC = d2, acceleration = a, initial velocity = 7 m/s, final velocity = 8 m/s
2as = v^2 - u^2
2a(d2) = (8)^2 - (7)^2
a*d2 = (64-49)/2
a*d2 = 7.5
a = 7.5/d2

(b) a = 12/d1 ---> (i)
a = 7.5/d2 ---> (ii)
Solving simultaneously,
12/d1 = 7.5/d2
12*d2 = 7.5*d1
d1 = 12/7.5* d2
d1 = 1.6*d2

Note: You can also make d2 the subject of the equation.
 
Kumkum

Kumkum

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Hemdon said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_1.pdf
Question no: 3
The question is to find the length of CD.
From Mark Scheme:
2dsin30 + 2d√3sin60
= 2d.½+ 2d√3.√3/2 = 4d
Can someone explain how to find the length of ED?? I found CE as 3d
Click to expand...
Untitled.png
u see the rectangle BEDF, we can use that to find the length of ED.
we can find the length of BF (it is equal in length to ED)
so to find BF:
sin30 = BF/2d
BF = 2dsin30
= d (since sin 30 is half)
hence, ED = d
now u can find CD

hope i've helped :)
 
Hemdon

Hemdon

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Kumkum said:
View attachment 32061
u see the rectangle BEDF, we can use that to find the length of ED.
we can find the length of BF (it is equal in length to ED)
so to find BF:
sin30 = BF/2d
BF = 2dsin30
= d (since sin 30 is half)
hence, ED = d
now u can find CD

hope i've helped :)
Click to expand...
Thats alot of help....Thanks and may Allah bless u!!
 
Kumkum

Kumkum

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Hemdon said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_1.pdf
Question no. 2...
I tried to this question, but i dont get the sin^-1 part, how u do that??
Click to expand...


ok we are given that: x= sin^-1(2/5)
this simply means 'sin(x)=2/5' [ sin(x) = opposite/hypotenuse]

(i) (sin^2)x + (cos^2)x = 1
(cos^2)x = 1 - (sin^2)x [where (sin^2)x = (2/5)^2]
= 1 - (2/5)^2
= 21/25

(ii) (tan^2)x = (sin^2)x/(cos^2)x
=[(2/5)^2] / [(21/25)]
=4/21

hope i've answered ur question
 
Kumkum

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Dudu said:
Someone help me with this?

3) An oil pipeline under the sea is leaking oil and a circular patch of oil has formed on the surface of the
sea. At midday the radius of the patch of oil is 50 m and is increasing at a rate of 3 metres per hour.
Find the rate at which the area of the oil is increasing at midday.

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf
Click to expand...

i'll try my best to explain

here we are asked to find 'rate at which area is increasing' so that means we have to find 'dA/dt'
we are given 'the rate at which the radius is changing' which is 'dr/dt = 3'
to find 'dA/dt' we also need 'dA/dr'

to find 'dA/dr' we need to differentiate an equation of A(area) in terms of r(radius)
the only equation we can use is the area of a circle, which is "A=(pi)r^2"
so differentiating this we get:
dA/dr = 2(pi)r

we can find the value of dA/dr when r=50
===> dA/dr = 2(pi) * 50
= 100(pi)

so now we can combine the rates i.e dA/dt , dA/dr, and dr/dt to find what we want(dA/dt)
so,
dA/dt = dA/dr * dr/dt
= 100(pi) * 3
=300(pi)

hope i was clear enough :)
 
Dudu

Dudu

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Kumkum said:
i'll try my best to explain

here we are asked to find 'rate at which area is increasing' so that means we have to find 'dA/dt'
we are given 'the rate at which the radius is changing' which is 'dr/dt = 3'
to find 'dA/dt' we also need 'dA/dr'

to find 'dA/dr' we need to differentiate an equation of A(area) in terms of r(radius)
the only equation we can use is the area of a circle, which is "A=(pi)r^2"
so differentiating this we get:
dA/dr = 2(pi)r

we can find the value of dA/dr when r=50
===> dA/dr = 2(pi) * 50
= 100(pi)

so now we can combine the rates i.e dA/dt , dA/dr, and dr/dt to find what we want(dA/dt)
so,
dA/dt = dA/dr * dr/dt
= 100(pi) * 3
=300(pi)

hope i was clear enough :)
Click to expand...


Try to explain to me? Ha. You pretty much nailed it!
Cheers :)
 
Hemdon

Hemdon

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Kumkum said:
ok we are given that: x= sin^-1(2/5)
this simply means 'sin(x)=2/5' [ sin(x) = opposite/hypotenuse]

(i) (sin^2)x + (cos^2)x = 1
(cos^2)x = 1 - (sin^2)x [where (sin^2)x = (2/5)^2]
= 1 - (2/5)^2
= 21/25

(ii) (tan^2)x = (sin^2)x/(cos^2)x
=[(2/5)^2] / [(21/25)]
=4/21

hope i've answered ur question
Click to expand...
LEMME GUESS....u topped in maths rite??? or r u a teacher?? :p
 
syed1995

syed1995

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Immadp said:
hey can you please help me with Oct/Nov 2008 paper 1 Q 9 iii ?
Click to expand...


find dy/dx of the curve.. and put the values of both P and Q's x co-ordinates in DY/DX equation to get the gradient of tangent at both P and Q...

then tan^-1(m1)-tan^-1(m2) to get the angle where m1 and m2 are both gradients.. the angle will always be positive if you're getting it negative just place the bigger angle on the left hand side.
 
kaushar

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i've got a prob with may/june 2013 paper 32 number 7
can i get some help plzz??
 
Kumkum

Kumkum

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kaushar said:
i've got a prob with may/june 2013 paper 32 number 7
can i get some help plzz??
Click to expand...

ok,
7(i) first expand 'cos(x+45)'
cos(x+45) = cosxcos45 - sinxsin45 [<---hope you know the addition formula for cos(A +B)]
=(√2 /2)cosx - (√2 /2)sinx

so now we have to express "cos(x + 45) - (√2)sinx " in the form "Rcos(x + a)" [cos(x + 45) - (√2)sinx = Rcos(x + a)]

first i'll simplify the LHS (the one in blue):
cos(x + 45) - (√2)sinx = (√2 /2)cosx - (√2 /2)sinx - (√2)sinx
= (√2 /2)cosx - (3√2 /2)sinx

now i'm expanding the RHS(the one in green):
Rcos(x + a) = Rcosxcosa - Rsinxsina

i'll equate the RHS and the LHS :
Rcosxcosa - Rsinxsina = (√2 /2)cosx - (3√2 /2)sinx <---i can re-write this as:
(Rcosa)cosx - (Rsina)sinx = (√2 /2)cosx - (3√2 /2)sinx

now to find R:
R = √[(√2 /2)^2 + (-√3 /2)^2]
= 2.236

now to find the angle 'a' :
(Rcosa)cosx - (Rsina)sinx = (√2 /2)cosx - (3√2 /2)sinx
from the equation above^, notice the coefficients of 'cosx' and 'sinx' respectively
so ===> Rcosa = √2 /2 and Rsina= 3√2 /2
to find the angle, we need to use 'tan'
hence; tana = [3√2 /2] / [√2 /2] because tan = sin/cos
angle 'a' = 71.57

so u'll have: 2.236cos(x + 71.57) nd u've expressed it in the form they want
(btw the working is not this long)

7(ii) now we solve: cos(x + 45) - (√2)sinx = 2 [the RHS is equivalent to the answer we found in (i)]
so,
2.236cos(x + 71.57) = 2
cos (x + 71.57) = 0.8945 [to make my working easier, i let 'x + 71.57" = y]
==> cosy = 0.8945
y = 26.56
cos is positive in the first and fourth quadrant so,
y = 360 - 26.56 = 333.4
nd i also added 360 to 26.56 because the range of 'x' changed i.e 26.56 < x + 26.56 < 386.56
so another value of y= 386.56
now we've got y as 26.56, 333.4 and 386.56
so frm these we can find our 'x' where "x = y - 71.57"
===> x = 261.8 and 315.0

hope i explained properly nd didn't confuse u evn more :)
 
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