[CIE] All Pure Mathematics (P1, P2 and P3) doubts here !

[doubtigetastar]

Help! :

A quantity u varies with respect to t so that du/dt = a+bt where a & b are constants.
Given that it has a maximum value of 5.5 when t=1 and that its rate of change when t=2 is -3 , find u in terms of t.

[Vin]

CIE November 2010 Variant 2
Q7 iii)

I got the answer using a very very long method, does anyone have a shorter method since its 1 mark only?

Don't you complete the square for f(x) and substitute in hg? I think that's the smallest way. :/

THANK YOU!

Sorry there was a typo. (corrected- bold)  please check again?

Help! :

A quantity u varies with respect to t so that du/dt = a+bt where a & b are constants.
Given that it has a maximum value of 5.5 when t=1 and that its rate of change when t=2 is -3 , find u in terms of t.

Do you have the answer? Cant be sure of mine actually :/

[SkyPilotage]

Help! :

A quantity u varies with respect to t so that du/dt = a+bt where a & b are constants.
Given that it has a maximum value of 5.5 when t=1 and that its rate of change when t=2 is -3 , find u in terms of t.

-du/dt=-3 when t=2 soo :                          a+2b=-3
-maximum value of u is 5.5 so du/dt=0  :       a+b=0
So solve the eqns you get   b=-3 and a=3
Then integrate du/dt to get u= 3t-1.5t^2+c  --->subsitute (1,5.5) to get c=4 -----> so u=3t-1.5t^2+4
Hope it helps...

[bulono]

M/J 2004 question 8 i....
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s04_qp_1.pdf
in the answer scheme the perimeter of the semi circle is written as (pie)r...dont understand

[SkyPilotage]

M/J 2004 question 8 i....
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s04_qp_1.pdf
in the answer scheme the perimeter of the semi circle is written as (pie)r...dont understand

Circumference of circle is 2pi(r) since its half a circle the perimeter of semicircle is pi(r)..

[bulono]

i no but in the mark scheme it is written as (pi)r
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s04_ms.pdf

[Vin]

because its half the circle?

2(pi) r /2

= (pi) r

[bulono]

ohhhhhhhhhhhh...lol...thankxxx

[doubtigetastar]

-du/dt=-3 when t=2 soo :                          a+2b=-3
-maximum value of u is 5.5 so du/dt=0  :       a+b=0
So solve the eqns you get   b=-3 and a=3
Then integrate du/dt to get u= 3t-1.5t^2+c  --->subsitute (1,5.5) to get c=4 -----> so u=3t-1.5t^2+4
Hope it helps...

Thanks 

[iluvme]

Don't you complete the square for f(x) and substitute in hg? I think that's the smallest way. :/

This was what I did,

hg(x) = f(x)
h(x) = x²+bx+c

h(x-2)= (x-2)² + b(x-2) +c  = x²-4x+7
            x²-4x+4+bx-2b+c= x²-4x+7

b=0 c=3
h(x) = x²+3

[Arthur Bon Zavi]

the marked one is done! but i need help with the second one.. cant really understand how do i integrate it :/

No mistake in it. It is TOTALLY correct.

[SkyPilotage]

If 2 vectors are parallel and have same magnitude, do they have the same vector eqn?

[iluvme]

Yup, and the angle between them will be zero.

[Tohru Kyo Sohma]

i have a doubt here.............pls pls pls help me because i have my exam tomorrow......its P11 oct/nov 2009 qns 2!!i found the point on the curve but i cant find the points of the line x+2y= ? ( ? that pi, just incase its not clear)
PLEASE PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!this is very important for me...!!!!
and i have a few more doubts while doing the papers....how do we find whe to put <,> signs in function question...i usually get the value right but the signs are wrong!!!any tips!!

[SkyPilotage]

i have a doubt here.............pls pls pls help me because i have my exam tomorrow......its P11 oct/nov 2009 qns 2!!i found the point on the curve but i cant find the points of the line x+2y= ? ( ? that pi, just incase its not clear)
PLEASE PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!this is very important for me...!!!!
and i have a few more doubts while doing the papers....how do we find whe to put <,> signs in function question...i usually get the value right but the signs are wrong!!!any tips!!

-If you want to plot the point of the line just substitute x with any value from 0 to pi. ex  0.25pi+2y=pi--->0.75pi=2y find y and u have a point....
Do the same for another value of x and u get another point then join them to draw the line...
-Regarding the signs you put them on a number line and put the values on the number...outside the range of values is the same sign as a(coefficiient of x^2)....
for example if you have : x^2-x<0 so u put in the number line x=0 and x=1 soo outside this range is +ve since x^2 is +ve and inside the range is -ve. Since the question asks for <0 (-ve) it is 0<x<1. I hope that helps.