Mathematics: Post your doubts here!

[RoOkaYya G]

wait! ill re do it.
*sigh*

 

[asadalam]

i saw jst ur post n corrected it there.
wait lets take it from beginning! ok?

OMG!!! im so sorry!! i misread. its x^2 in the equation i used x^3

nxt time plz put ur attachments upright -_______- thank u

Ok please just differentiate that one equation,it would be enough!=P

 

[RoOkaYya G]

Ok please just differentiate that one equation,it would be enough!=P

i got it *sigh*
PLEASE NEXT TIME UPRIGHT ATTACHMENT

 

[asadalam]

i got it *sigh*
PLEASE NEXT TIME UPRIGHT ATTACHMENT

Dont worry i got it!I did it no need

 

[RoOkaYya G]

dV/dx = dV/dt * dt/dx
PS: the 6 is mentioned to beautify the quest only
simplifying the equation :
6pi*x^2 - (pix^3) /3
differenciating it :
12pix-pix^2

when x= 2
dV/dx= 20 pi
replacing in chain rule :
20 pi =3* dt/dx
dt/dx =20pi/3

dx/dt = 3/20pi <---------ANSWER

 

[RoOkaYya G]

Dont worry i got it!I did it no need

thts better
see! u managed on ur own good!
good luck for ur exams

 

[asadalam]

dV/dx = dV/dt * dt/dx
PS: the 6 is mentioned to beautify the quest only
simplifying the equation :
6pi*x^2 - (pix^3) /3
differenciating it :
12pix-pix^2

when x= 2
dV/dx= 20 pi
replacing in chain rule :
20 pi =3* dt/dx
dt/dx =20pi/3

dx/dt = 3/20pi <---------ANSWER

Thanks a lot you're a legend!I'm sure you have an A* in store for your maths exam!

 

[Haya Ahmed]

40)
R = square root ( 1^2 + 1 ^2) = square root 2

alpha :
tan alpha = 1/1
alpha=tan inverse 1
= pi/4

Rcos (theta + alpha) = square root 2 (theta + pi/4) <----------ANSWER

cos theta + 1/ root 2 = sin theta
cos theta- sin theta = - 1 / root 2
Rcos (theta + alpha) = - 1 / root 2
square root 2 (theta + pi/4) = - 1 / root 2
solve for theta :
cos (theta+ pi/4) = - 0.5
(theta+ pi/4) = cos inverse 0.5
(theta+ pi/4) = pi/3
cos is negative so in quadrant S and T
in quadrant S (theta+ pi/4) = (pi -pi/3) = 2pi/3
in quadrant T (theta+ pi/4) = pi + pi/3 = 5 pi/3
theta = (2pi/3 - pi/4) and (5pi/3- pi/4)
=5pi/ 12 and 14pi/12
for acute angle, theta = 5pi/12 <---------------- ANSWER

44) ii) b)

solving part i) ull get it in R cos form as :
25sin(theta +73.7)....using this derive 25 cos (theta -73.7)

25 cos (theta-73.7) = -3
cos (theta-73.7) = -3/25
(theta-73.7) = cos inverse (3/25)
since cos is negative the solutions r found in S andT quadrant
in S, (theta-73.7) = 180 -83.1 = 96.9
in T,(theta-73.7) =180+ 83.1 = 263.1
(theta-73.7) =96.9 and 263.1

theta =(96.9+73.7) and (263.1 +73.7)
theta= 170.6 and 336.8
theta = 170.6 for 0< theta< 180 <-----------ANSWER

45)
equate the give equation of tan theta to 1/5
cross multiply

ull get x^2 +384 =5(8x)
x^2 + 384-40x =0
x^2 -40x +384 = 0

solving the quadratic :

(x-16)(x-24) =0
x= 16 and x=24 <------------ANSWER


ill post 46 in a while dinner

Thanks for your help !! I got it now !! RoOkaYya G

 

[RoOkaYya G]

Thanks a lot you're a legend!I'm sure you have an A* in store for your maths exam!

legend? y u say so? :O

was that sarcasm? :/

if u meant it then thanks loads! ^_^ In Sha Allah ^_^
i wish u the same u had an awesome As result keep it up! ^_^

 

[asadalam]

legend? y u say so? :O

was that sarcasm? :/

if u meant it then thanks loads! ^_^ In Sha Allah ^_^
i wish u the same u had an awesome As result keep it up! ^_^

lol,no sarcasm,i'm just really impressed,no one in my class got this one right so i meant it ^^

 

[RoOkaYya G]

Thanks for your help !! but can you explain 44 ii) b once more in detailed because I didn't get it !

look. the equations given above...there was 7sin theta...
so its Rsin is calculated as i did there
but the 2nd part (44ii)...the question is 7cos theta...
so u need Rcos form here!!! n not Rsin!!!

so the R remains the same n the alpha also..
so u replace in Rcos formula to get it

 

[RoOkaYya G]

lol,no sarcasm,i'm just really impressed,no one in my class got this one right so i meant it ^^

hmm ok
actually i read it wrong at first thts y i got the wrong answer whch was close enough
its a direct question no need to panic

whnever theres rate of change....write down a chain rule at first. note down all the infos given. then as per ur chain rule. find the missing parts n replace n solve

 

[asadalam]

hmm ok
actually i read it wrong at first thts y i got the wrong answer whch was close enough
its a direct question no need to panic

whnever theres rate of change....write down a chain rule at first. note down all the infos given. then as per ur chain rule. find the missing parts n replace n solve

Yeah our sir said the same,but actually we were all confused by the 6 and tried to insert it somehow in our answers so thats why we were wrong :/
In CIEs do they also give useless information?Cuz in o lvls all the info we had was to be used in some way,so thats why we thought we had to include 6.

 

[RoOkaYya G]

Yeah our sir said the same,but actually we were all confused by the 6 and tried to insert it somehow in our answers so thats why we were wrong :/
In CIEs do they also give useless information?Cuz in o lvls all the info we had was to be used in some way,so thats why we thought we had to include 6.

yeah they do CIE likes to screw up with students
but u dont worry about it. u need to be focussed n know the steps. dont bother about additional information

 

[RoOkaYya G]

yeah they do CIE likes to screw up with students
but u dont worry about it. u need to be focussed n know the steps. dont bother about additional information

i guess the question u gave...had other parts too...where the 6 shuldve been used.
but while making tht sheet of quest. they cut it off.
this is possible

 

[Haya Ahmed]

40)
R = square root ( 1^2 + 1 ^2) = square root 2

alpha :
tan alpha = 1/1
alpha=tan inverse 1
= pi/4

Rcos (theta + alpha) = square root 2 (theta + pi/4) <----------ANSWER

cos theta + 1/ root 2 = sin theta
cos theta- sin theta = - 1 / root 2
Rcos (theta + alpha) = - 1 / root 2
square root 2 (theta + pi/4) = - 1 / root 2
solve for theta :
cos (theta+ pi/4) = - 0.5
(theta+ pi/4) = cos inverse 0.5
(theta+ pi/4) = pi/3
cos is negative so in quadrant S and T
in quadrant S (theta+ pi/4) = (pi -pi/3) = 2pi/3
in quadrant T (theta+ pi/4) = pi + pi/3 = 5 pi/3
theta = (2pi/3 - pi/4) and (5pi/3- pi/4)
=5pi/ 12 and 14pi/12
for acute angle, theta = 5pi/12 <---------------- ANSWER

44) ii) b)

solving part i) ull get it in R cos form as :
25sin(theta +73.7)....using this derive 25 cos (theta -73.7)

25 cos (theta-73.7) = -3
cos (theta-73.7) = -3/25
(theta-73.7) = cos inverse (3/25)
since cos is negative the solutions r found in S andT quadrant
in S, (theta-73.7) = 180 -83.1 = 96.9
in T,(theta-73.7) =180+ 83.1 = 263.1
(theta-73.7) =96.9 and 263.1

theta =(96.9+73.7) and (263.1 +73.7)
theta= 170.6 and 336.8
theta = 170.6 for 0< theta< 180 <-----------ANSWER

45)
equate the give equation of tan theta to 1/5
cross multiply

ull get x^2 +384 =5(8x)
x^2 + 384-40x =0
x^2 -40x +384 = 0

solving the quadratic :

(x-16)(x-24) =0
x= 16 and x=24 <------------ANSWER


ill post 46 in a while dinner

well in question Q45 how did .. you get the equation 8x/x^2 +384 ... ?!

 

[RoOkaYya G]

well in question Q45 how did .. you get the equation 8x/x^2 +384 ... ?!

i tld u there...we cross mutiply. the equation given in form of tan...and the 1/5

 

[Haya Ahmed]

i tld u there...we cross mutiply. the equation given in form of tan...and the 1/5

well I want to know where did we actually use the formula tan(A-B) and what is A and what is B ?

 

[RoOkaYya G]

well I want to know where did we actually use the formula tan(A-B) and what is A and what is B ?

we dint use compound angle here dr

 

[RoOkaYya G]

well I want to know where did we actually use the formula tan(A-B) and what is A and what is B ?

the whole question was a nice story. but it was there to scare u

u jst need to equate the equation give AS IT IS!!! to 1/5 (no need to apply compound angle formula or whtever....)