Has anyone June 2006? Question 3b I know u use volume of revolution formula. I also get u use 1/2(1-cos2A) formla for sin^2(x/2) however I am not sure what happens to the x/2 when they write it out again it just disappears from the working out Not sure why that is ???
can someone help me with june 13 replacement paper. Q4b ? i understand what they done in the mark scheme, but why couldn't i have taken out (10,000)^1/3 as this would lead me to still getting x=0.81 and that is still valid ? as x is less than 8/9? i tried this way and multiplied (10,000)^1/3 at end to the expansion too but dont get the same answer, get something like 24... ? pls can someone help, this is driving me crazy! many thanks. Do you mind helping me out pls
x = 0.81? If you take out 1000^(1/3) from 7100^(1/3) you get 7.1^(1/3) and so set 8 - 9x = 7.1, I'm not quite sure what you did.
If you want to integrate cos^2 (y) you use the formula cos2y=2cos^2 (y) -1 which then becomes 1/2 + 1/2cos2y =cos^2 (y) and then you integrate that to get 1/2(x) + 1/4sin(2x) +c
I am aware that Ln |x| differentiates to 1/x, however when you have equations involving for example 48/x or 48/x^2 can I simply convert it to 48x^-1 and 48^x-2?
That would integrate to 48Ln |x| but then would the other integrate to -48^-1?
Starts from the regular formula for a cone (as proved in the video just fyi)
Basically, in this case the radius is 'h' because if you split the triangle up the middle you end up with two triangles, that each have two 45 degree angles. (which are isosceles) and one 90 degree angle. because there are isosceles, if the height in 'h' then the base is also 'h' (not the hypotenuse)
I wish i could attach a photo but my phone is fudged up
i took out (10,000)^1/3 so i would set 8-9x = 0.71 to get x=0.81 ? why cant you do it this way ?
Sorry, must have misread I think that it might because these expansions are more accurate with less terms for smaller values of x When you have something like 0.81 which isn't too far from the upper limit 8/9, by just using the first few terms of the series you don't get a very good estimate; whereas if you use a smaller x, after a few terms the terms quickly become very small so your estimate isn't too far from the actual value.
Sorry, must have misread I think that it might because these expansions are more accurate with less terms for smaller values of x When you have something like 0.81 which isn't too far from the upper limit 8/9, by just using the first few terms of the series you don't get a very good estimate; whereas if you use a smaller x, after a few terms the terms quickly become very small so your estimate isn't too far from the actual value.