Mathematics: Post your doubts here!

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abbey789

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freezingfires said:
Here it says the rate of increase of the surface area of the sphere is proportional to its volume.
so dA/dt=kV
dA/dt=4pi r^2 differentiating this gives us 8pir dr/dt
now equate
8pirdr/dt=k(4/3pi r^3)
put values of dr/dt and r given in the question to get k=12/25
Now put this value of k back in 8pirdr/dt=k(4/3pi r^3) to get dr/dt=0.08r^2
Now apply differential equation solve it and you will get
-1/r=0.08t+k
put t=o and r=5 to get k=-1/5
Now solve it further to get in terms of r and t
your final equation will be r=5/(1-0.4t)
for the last part equate the denominator to 0 you will get t=2.5 which means that if denominator is 0 equation is not valid so values of t must be grater than 0 but less than 2.5.HOPE YOU GOT IT!! PLEASE PRAY FOR ME!! :)
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THANK YOU SO MUCH! Yes I will! :)
I didn't get this part dA/dt=4pi r^2 differentiating this gives us 8pir dr/dt
 
VinnCh

VinnCh

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LMGD33 said:
Hello guys, I'm solving http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_33.pdf and I had this beautifully posted solutions to all the paper, i', just wondering in Question 5 (ii) How did they get that formula? Why did they divide by 70 and multiply by 100? Thank you so much!!
View attachment 26637

Credit to Anika Raisa
Click to expand...

After I went through the question. The initial mass (t=0) of substance B, which is also y is 70. The question asks for percentage mass remaining. Since the mass remaining as t increase to infinite is 70/e, the percentage mass remaining should be (remaining/initial) x 100 , which is also [(70/e)/70]x100.
I hope you understand it. =)
 
VinnCh

VinnCh

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LindaKim123 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_31.pdf

question 9 ii) please!!
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Since 1+√2i and 1-√2i are the two imaginary roots, which means x=1+√2i and x=1-√2i . Equate them, ( x - [1+√2i] ) ( x - [1-√2i] ) and expand them. Use the resulting equation ( x^2 -2x + 3 ) to do long division with P(x). You'll get a quadratic equation in return, then just use the usual way to find the roots. =)
 
littlecloud11

littlecloud11

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LindaKim123 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_31.pdf

question 9 ii) please!!
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This is how I do it.
The two complex roots are 1+√2i and 1-√2i
If you kept it in the form of factors it would be [x-(1+√2i)] and [x-(1-√2i)]. These are the two factors of p(x)
If you multiply these two factors you get a quadratic equation.
(x-1-√2i) * (x-1+√2i)
= x^2 -x +√2xi -x +1 -√2i -√2ix +√2i -(√2i)^2
=x^2 -2x +1 +2
=x^2 -2x +3

Now use long division to divide p(x) by this quadratic factor to get the other two factors.
Hope this helps!
 
Gordious

Gordious

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LMGD33

LMGD33

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VinnCh said:
After I went through the question. The initial mass (t=0) of substance B, which is also y is 70. The question asks for percentage mass remaining. Since the mass remaining as t increase to infinite is 70/e, the percentage mass remaining should be (remaining/initial) x 100 , which is also [(70/e)/70]x100.
I hope you understand it. =)
Click to expand...
Is it something like this? Then i tend to infinity and obtain P = 100/e

IMG019.jpg
THANK YOU SO MUCH!! Remembered in my prayers!
 
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