C1 and C2 DOUBTS HERE!!!!!

[cooldude]

oh am so sorry
i meant question 9c

A=80x-2x^2-pix^2/2
dA/dx=80-4x-pix
d^2A/dx^2=-4-pi
and therefore since (-4-pi) this is negative, thus the value in part b is a maxima

[halosh92]

thx everyone
this question also:
Q3

[Saladin]

thx everyone
this question also:
Q3

2 hours.

[Saladin]

I think I got it now!

[Saladin]

Alrite here it comes.

First you need to find the angle of the sector of the circle. The angle is 60^o because the semi-circle is divided into three equal parts as two other quarter circles combine to for two separate circles. Its like this, you need to view the other two quartiles as triangles. Thus the sector of B and C is one of the three. And half a circle is 180^o. Thus dividing it by three will give you radians.

So, now you need to understand that the gap is formed when the two segments of the circle is taken away from sector.

So you simply do this.

Find the area of the sector!



Now the area of the section!



Now the area of the shaded region is the area of [Area of section - 2 X Area of Sector].

This will lead to the image attached.

[halosh92]

Alrite here it comes.

First you neejavascript:void(0);d to find the angle of the sector of the circle. The angle is 60^o because the semi-circle is divided into three equal parts as two other quarter circles combine to for two separate circles. Its like this, you need to view the other two quartiles as triangles. Thus the sector of B and C is one of the three. And half a circle is 180^o. Thus dividing it by three will give you radians.

So, now you need to understand that the gap is formed when the two segments of the circle is taken away from sector.

So you simply do this.

Find the area of the sector!



Now the area of the section!



Now the area of the shaded region is the area of [Area of section - 2 X Area of Sector].

This will lead to the image attached.

THX ALOT DUDE gr8

[halosh92]

Q5b
i got the cordinates but didnt get this one
(0, 1/sqrt2)

[saadmannan]

hi

i have doubt in Jan 2005 Q9a i am not able to solve the equation
the sign in the question that is given is positive while im getting negative
i checked the mark scheme if anything was wrong but my method was exactly same as in MS but the answer is different
i have posted the question below

[Saladin]

hi

i have doubt in Jan 2005 Q9a i am not able to solve the equation
the sign in the question that is given is positive while im getting negative
i checked the mark scheme if anything was wrong but my method was exactly same as in MS but the answer is different
i have posted the question below

In 2 hours.

[cooldude]

Q5b
i got the cordinates but didnt get this one
(0, 1/sqrt2)

y=cosx+cos(pi/4)
y=cosx+(1/root 2)
x=0
y=1+(1/root 2) (as cos0=1)
therefore intersection with the y-axis-->
(0,1+(1/root 2))

[cooldude]

hi

i have doubt in Jan 2005 Q9a i am not able to solve the equation
the sign in the question that is given is positive while im getting negative
i checked the mark scheme if anything was wrong but my method was exactly same as in MS but the answer is different
i have posted the question below

2x+2y+(pi*x)=80
2y=80-pi*x-2x
ar=2xy+((pi*(x^2))/2)
ar=x(80-pi*x-2x)+((pi*(x^2))/2)
ar=80x-(pi*(x^2))-2(x^2)+((pi*(x^2))/2)
ar=80x-((pi*(x^2))/2)-2(x^2)
ar=80x-(x^2)(pi/2+2)
hence proved

[halosh92]

y=cosx+cos(pi/4)
y=cosx+(1/root 2)
x=0
y=1+(1/root 2) (as cos0=1)
therefore intersection with the y-axis-->
(0,1+(1/root 2))

for the second step 
y=cosx+(1/root 2)
 am getting the 1/root 2  as 0.707 as decimals? how did u get it as fraction form?
otherwise i get the rest
thx

[cooldude]

for the second step 
y=cosx+(1/root 2)
 am getting the 1/root 2  as 0.707 as decimals? how did u get it as fraction form?
otherwise i get the rest
thx

yeah its the same thing 0.707 is the decimal form of 1/root 2, its actually a known fact the values of sin, cos, tan of 0,30,45,60,90 are all given in our cie book so i guess thats how i knw it, ill just give u a list of the values so u can learn them for future qs-->
sin0=0, sin30=1/2, sin45=1/root 2, sin60=(root 3)/2, sin 90=1, cos0=1, cos 30=(root 3)/2, cos 45=1/root 2, cos 60=1/2, cos 90=0, tan0=0, tan 30=1/root 3, tan45=1, tan 60=root 3, tan 90=1/0 (undefined)

[Sue T]

in jan 09 we had to use the formula for surface area of a cylinder 2 slove ques 10 a
the formula isnt in th data booklet so do we need 2 learn all those volume n area formulas 4 all th shapes?

[studz]

in jan 09 we had to use the formula for surface area of a cylinder 2 slove ques 10 a
the formula isnt in th data booklet so do we need 2 learn all those volume n area formulas 4 all th shapes?

yeah Sue we do..but not for all the shapes...m assuming only for cylinder sphere.. and some times for a cuboid too..but yeah thats it i guess..