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Mathematics: Post your doubts here!

sea_princess

sea_princess

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sea_princess

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reina81 said:
Q 10
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_3.pdf
Click to expand...
i - okay first notice that the question gave you the base area( m^2) and the rate of flow in m^3 so when making the formula you divide 30 by 1000 to get the rate of increase in height
so now you have dh/dt =0.03 -k square root h
put h =1 and dh/dt = 0.02
0.02 = o.o3 - k ( square root of 1 is 1)
k = 0.03 -0.02 = 0.01
so now the equation is
dh/dt =0.03 - 0.01 square root h
by taking 0.01 as a common factor
dh/dt= 0.01 (3- square root h)
ii- this part is integration , u have the equation
( x-3 ) dx/dt =0.005 x ( divide both sides by o.oo5 x and put the integration sign)
(x- 3)/ 0.005x dx= dt ( you multiply both sides by dt)
separate the left side
x/ 0.005 x - 3/0.005x dx = dt
200 -600/x dx = dt ( now u can integrate)
200x - 600 ln (x) = t +c
put x=3 , t= 0
600 - 600 ln(3) = c
thus t = 200x - 600 ln (x) - 600+ 600 ln (3)
t= 200x + 600 ln (3/x) - 600 ( u can take 200 as common factor )
t = 200 (x + 3 ln (3/x) -3)
 
sea_princess

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sea_princess said:
i - okay first notice that the question gave you the base area( m^2) and the rate of flow in m^3 so when making the formula you divide 30 by 1000 to get the rate of increase in height
so now you have dh/dt =0.03 -k square root h
put h =1 and dh/dt = 0.02
0.02 = o.o3 - k ( square root of 1 is 1)
k = 0.03 -0.02 = 0.01
so now the equation is
dh/dt =0.03 - 0.01 square root h
by taking 0.01 as a common factor
dh/dt= 0.01 (3- square root h)
ii- this part is integration , u have the equation
( x-3 ) dx/dt =0.005 x ( divide both sides by o.oo5 x and put the integration sign)
(x- 3)/ 0.005x dx= dt ( you multiply both sides by dt)
separate the left side
x/ 0.005 x - 3/0.005x dx = dt
200 -600/x dx = dt ( now u can integrate)
200x - 600 ln (x) = t +c
put x=3 , t= 0
600 - 600 ln(3) = c
thus t = 200x - 600 ln (x) - 600+ 600 ln (3)
t= 200x + 600 ln (3/x) - 600 ( u can take 200 as common factor )
t = 200 (x + 3 ln (3/x) -3)
Click to expand...

iii- here there's a trick , notice the question said the height of water is 4 m
so u go back to the equation x= 3- square root h and put h =4
then u get x = 3- 2 = 1
put this x in the equation obtained in part (ii)
t= 200 (1 + 3 ln(3/1) - 3)
t= 259.2 seconds
 
R

reina81

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sea_princess said:
iii- here there's a trick , notice the question said the height of water is 4 m
so u go back to the equation x= 3- square root h and put h =4
then u get x = 3- 2 = 1
put this x in the equation obtained in part (ii)
t= 200 (1 + 3 ln(3/1) - 3)
t= 259.2 seconds
Click to expand...


thank you so much!!! really helped :)
 
farhan143

farhan143

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This is so hard i cant understnd a thing...... can any 1 explain me this im in need of serious help!!!!!

doubt.jpg

Please help me solve this question .....
THIS IS 2011 OCT/NOVEMBER
STATISTICS PAPER 62
 
Uneeq

Uneeq

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How to solve this, 2-3x < |x-3|? The answer should be x > -1/2 only. This is from CIE Past Year Mathematics. Oct/Nov 09. Question 1. Variant 31
 
Esme

Esme

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you have to find the probabilities of the other 7 integers which will be 1-(1/10) / 7. this is because the proabilities of all the other integers are equal. now the formula for expectance is sum of (X x P) . so multiply the variable X (eg -2) with its probability (9/70) and then add them all.
 
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2pac

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DragonCub said:
The square root is not that terrible. It is just used to compensate certain squares.
Also to solve this question I think you need to master the trigonometric identities.
View attachment 7343
Click to expand...
Thank you so much.Really helped a lot.
Could you also please explain the second question of the same paper?
I did the first step,that is,simplifying it with logarithm but then substituting the values using the graph and then solving it led to many errors.Have seen a similar question in p2 and so I will bother you with some p2 questions as well.:)
 
DragonCub

DragonCub

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2pac said:
Thank you so much.Really helped a lot.
Could you also please explain the second question of the same paper?
I did the first step,that is,simplifying it with logarithm but then substituting the values using the graph and then solving it led to many errors.Have seen a similar question in p2 and so I will bother you with some p2 questions as well.:)
Click to expand...
y = A (x^n), natural log each side to get
ln y = ln A + n (ln x)
The gradient is n and y-intercept is ln A.
Just draw the line of best fit to find the two values. It should not be hard.
 
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