Qualification > Queries

ANY DOUBTS HERE!!!

<< < (26/58) > >>

nid404:

--- 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

--- End quote ---

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

Now Volume of revolution of the curve y = 2 is given by
Volume of revolution=

= =

Volume of revolution of the curve==
therefore volume of revolution=
=
=
=
=
=

Volume of the shaded region= = 1.5\pi

Next answer up in a while

nid404:
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

  i.e

differentiating with respect to x


so

we know

Gradient of tangent at P



If the angle made by the tangent with the +ve direction of x axis is  then  



Gradient of tangent at Q is given by

If the angle made by the tangent with the +ve direction of x axis is   then




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

In

Now
so
i.e
so angle between the tangents =
                                           == 56.3099° – 36.8699° = 19.44°

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

Freaked12:

--- 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

Now Volume of revolution of the curve y = 2 is given by
Volume of revolution=

= =

Volume of revolution of the curve==
therefore volume of revolution=
=
=
=
=
=

Volume of the shaded region= = 1.5\pi

Next answer up in a while

--- End quote ---

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

nid404:
That's all I need. God's blessings. Pray for me :)

Good luck  ;)

halosh92:

--- 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

  i.e

differentiating with respect to x


so

we know

Gradient of tangent at P



If the angle made by the tangent with the +ve direction of x axis is  then  



Gradient of tangent at Q is given by

If the angle made by the tangent with the +ve direction of x axis is   then




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

In

Now
so
i.e
so angle between the tangents =
                                           == 56.3099° – 36.8699° = 19.44°

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

--- End quote ---

this is C2??
P.S nid ure awesome ;)

Navigation

[0] Message Index

[#] Next page

[*] Previous page