Author Topic: ANY DOUBTS HERE!!! (Read 34889 times)

Sue T · « **Reply #120 on:** May 07, 2010, 08:32:13 am »

thank u again nid404!

Freaked12 · « **Reply #121 on:** May 07, 2010, 09:06:03 pm »

October November 2008
Question 9
i did part 1...subtracting the area of rectangle from the area of curve
but part 2, about the volume etc
and part 3
Pleaseeee

Freaked12 · « **Reply #122 on:** May 07, 2010, 09:06:41 pm »

Quote from: Freaked12 on May 07, 2010, 09:06:03 pm

October November 2008
Question 9
i did part 1...subtracting the area of rectangle from the area of curve
but part 2, about the volume etc
and part 3
Pleaseeee

Its pure maths paper 1

astarmathsandphysics · « **Reply #123 on:** May 07, 2010, 09:49:29 pm »

Will do it tomorrow. Have been away for 2 days becos of a disk I had to recover.. Just going through all the probs I have to answer

Freaked12 · « **Reply #124 on:** May 07, 2010, 10:19:10 pm »

Quote from: astarmathsandphysics on May 07, 2010, 09:49:29 pm

Will do it tomorrow. Have been away for 2 days becos of a disk I had to recover.. Just going through all the probs I have to answer

alright then
but if your a tutor,you could have done it in minutes.
pLease

nid404 · « **Reply #125 on:** May 08, 2010, 07:41:23 am »

Quote from: Freaked12 on May 07, 2010, 09:06:03 pm

October November 2008
Question 9
i did part 1...subtracting the area of rectangle from the area of curve
but part 2, about the volume etc
and part 3
Pleaseeee

The volume obtained when the shaded region is rotated through
360 ° about the x-axis is given by
Volume of revolution of the curve , y = 2 - Volume of revolution of the curve $y=\sqrt{3x+1}$

Now Volume of revolution of the curve y = 2 is given by
Volume of revolution= $int^5_1\pi (2)^2dx$
$4\pi\int^1_0 dx$
= $4\pi[1-0]$ = $4\pi$

Volume of revolution of the curve= $y=\sqrt{3x+1}$ = $\int^b_a \pi (f(x))^2 dx$
therefore volume of revolution= $\int^1_0 \pi (3x+1) dx$
= $\int^1_0 3\pi xdx + \int^1_0 \pi dx$
= $3\pi \int^1_0 x dx + \pi \int^1_0 dx$
= $3\pi [(x^2)/2] ^1_0+\pi [(x)]^1_0$
= $3\pi [1/2-0] + \pi [1-0]$
= $1.5\pi + \pi=2.5\pi$

Volume of the shaded region= $4\pi - 2.5\pi$ = 1.5\pi

Next answer up in a while

nid404 · « **Reply #126 on:** May 08, 2010, 08:11:00 am »

lol....Now I finally know how to use the latex thingy...cool

ohk...check the pic to look how the tangents are

We next find the acute angle, between the two tangents.

The equation of the curve is $y= \sqrt (3x+1)$

i.e $y= (3x+1)^(1/2)$

differentiating with respect to x
$dy/dx= 1/2 (3x+1)^(-1/2) (3)$

so $dy/dx= 3/2 [1/ \sqrt(3x+1)]$

we know $dy/dx= tan\theta$

Gradient of tangent at P

$dy/dx= dy/dx| (0,1)= 3/2[ 1/\sqrt(3(0) +1)]=3/2$

If the angle made by the tangent with the +ve direction of x axis is $\theta 1$ then
$tan\theta1=1.5$
$\theta1=56.3099$

Gradient of tangent at Q is given by
$dy/dx|q= dy/dx|(1,2)= 3/2[ 1/\sqrt(3(1)+1)]=3/4$
If the angle made by the tangent with the +ve direction of x axis is $\theta 2$ then
$tan\theta 2=0.75$
$\theta 2= 36.8699$

Let the tangent to the curve at point P intersect the x axis at point S and the tangent at point Q intersect the x axis at the point T.
Let the two tangents intersect at the point R.
Consider $\delta RST$

In $\delta RST\angle S=180^o - \theta1, \angle T= \theta 2$

Now $\angle R + \angle S + \angle T =180^o$
so $\angle R= 180^o - \angle S - \angle T$
i.e $\angle R= 180^o - (180^o - \theta1)- \theta 2 = \theta 1 - \theta 2$
so angle between the tangents = $\theta 1- \theta 2$
== 56.3099° – 36.8699° = 19.44°

I feel so good to help....that satisfaction

Freaked12 · « **Reply #127 on:** May 08, 2010, 06:49:28 pm »

Quote from: nid404 on May 08, 2010, 07:41:23 am

The volume obtained when the shaded region is rotated through
360 ° about the x-axis is given by
Volume of revolution of the curve , y = 2 - Volume of revolution of the curve $y=\sqrt{3x+1}$

Now Volume of revolution of the curve y = 2 is given by
Volume of revolution= $int^5_1\pi (2)^2dx$
$4\pi\int^1_0 dx$
= $4\pi[1-0]$ = $4\pi$

Volume of revolution of the curve= $y=\sqrt{3x+1}$ = $\int^b_a \pi (f(x))^2 dx$
therefore volume of revolution= $\int^1_0 \pi (3x+1) dx$
= $\int^1_0 3\pi xdx + \int^1_0 \pi dx$
= $3\pi \int^1_0 x dx + \pi \int^1_0 dx$
= $3\pi [(x^2)/2] ^1_0+\pi [(x)]^1_0$
= $3\pi [1/2-0] + \pi [1-0]$
= $1.5\pi + \pi=2.5\pi$

Volume of the shaded region= $4\pi - 2.5\pi$ = 1.5\pi

Next answer up in a while

you just saved me 150 riyal i was about to pay to a tutor
God bless

nid404 · « **Reply #128 on:** May 08, 2010, 06:53:10 pm »

That's all I need. God's blessings. Pray for me

Good luck

halosh92 · « **Reply #129 on:** May 08, 2010, 09:14:13 pm »

Quote from: nid404 on May 08, 2010, 08:11:00 am

lol....Now I finally know how to use the latex thingy...cool

ohk...check the pic to look how the tangents are

We next find the acute angle, between the two tangents.

The equation of the curve is $y= \sqrt (3x+1)$

i.e $y= (3x+1)^(1/2)$

differentiating with respect to x
$dy/dx= 1/2 (3x+1)^(-1/2) (3)$

so $dy/dx= 3/2 [1/ \sqrt(3x+1)]$

we know $dy/dx= tan\theta$

Gradient of tangent at P

$dy/dx= dy/dx| (0,1)= 3/2[ 1/\sqrt(3(0) +1)]=3/2$

If the angle made by the tangent with the +ve direction of x axis is $\theta 1$ then
$tan\theta1=1.5$
$\theta1=56.3099$

Gradient of tangent at Q is given by
$dy/dx|q= dy/dx|(1,2)= 3/2[ 1/\sqrt(3(1)+1)]=3/4$
If the angle made by the tangent with the +ve direction of x axis is $\theta 2$ then
$tan\theta 2=0.75$
$\theta 2= 36.8699$

Let the tangent to the curve at point P intersect the x axis at point S and the tangent at point Q intersect the x axis at the point T.
Let the two tangents intersect at the point R.
Consider $\delta RST$

In $\delta RST\angle S=180^o - \theta1, \angle T= \theta 2$

Now $\angle R + \angle S + \angle T =180^o$
so $\angle R= 180^o - \angle S - \angle T$
i.e $\angle R= 180^o - (180^o - \theta1)- \theta 2 = \theta 1 - \theta 2$
so angle between the tangents = $\theta 1- \theta 2$
== 56.3099° – 36.8699° = 19.44°

I feel so good to help....that satisfaction

this is C2??
P.S nid ure awesome

nid404 · « **Reply #130 on:** May 09, 2010, 07:23:13 am »

This is P1

why thank you

ruby92 · « **Reply #131 on:** May 10, 2010, 11:58:42 am »

M/j 2006 paper 2 CIE chemistry
4 f (ii)
ally alcohol CH2=CHCH2OH
heated under reflux with acidified MnO4 -
according to the Ms the product is HO2CCO2H
what about the CH2OH part?

nid404 · « **Reply #132 on:** May 10, 2010, 01:57:35 pm »

that is what gets converted to CO2H

the extreme left Carbon forms CO2 and the middle one forms CO2H.

Gettin it

ny · « **Reply #133 on:** May 10, 2010, 03:51:29 pm »

hello.i have this vector question for P1.
actually it was my mock exam question.

the question is:
the points A and B are such that the unit vector in the direction AB(from A to B) is 1/3i + pj +2/3k , where p is a negative constant.

i) find the value of p

then, the position vectors of A and B, relative to an origin O, are 9i +2j +qk and 12i -4j -5k respectively.

ii) find the value of q

thats all.thank you soo much.please reply asap..

ruby92 · « **Reply #134 on:** May 10, 2010, 10:31:48 pm »

Posted by: nid404
Insert Quote
that is what gets converted to CO2H

the extreme left Carbon forms CO2 and the middle one forms CO2H.

Gettin it

sort of yes.but why does it form CO2?

News:

Author Topic: ANY DOUBTS HERE!!! (Read 34889 times)

Sue T

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Freaked12

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Freaked12

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astarmathsandphysics

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Freaked12

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nid404

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nid404

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Freaked12

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nid404

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halosh92

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nid404

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ruby92

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nid404

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ny

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ruby92

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