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Mathematics: Post your doubts here!

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dV/dx = dV/dt * dt/dx
PS: the 6 is mentioned to beautify the quest only :p
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 :cool:
 
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dV/dx = dV/dt * dt/dx
PS: the 6 is mentioned to beautify the quest only :p
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 :cool:
Thanks a lot you're a legend!I'm sure you have an A* in store for your maths exam! :D
 
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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
 
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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 :)
 
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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 :p
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 :)
 
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hmm ok :)
actually i read it wrong at first thts y i got the wrong answer whch was close enough :p
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.
 
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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 :p CIE likes to screw up with students :p
but u dont worry about it. u need to be focussed n know the steps. dont bother about additional information :)
 
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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 ... ?!
 
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