Mathematics: Post your doubts here!

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i prefer the s1 and m1 combo even fot those who are poor at stats (like me :cry: ). it will help commerce and humanitarian science student as they will get a idea of mechanics and also do stats which is directly related. for science welll we have a full fledge course for stats in bachelers . for medical i gues stats might help and mechanical is purely physics
 
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9ii put x=0 in eq of tangent and ssee if value of y coordinate is above 1.5 so nearer to B otherwise O
 
Mayedah

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A star said:
from solving first part you are able to get to sinx=0.5 . now if you compare both equations the only diff is n so we can say that
30/10=n hence n= 3 now for greatest solution find all solutions up to 3 x 360. then divide the highest with 3 you will get the answer
Click to expand...
Thankyouu soo much :D i got it now
 
externityxzx

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can anybody please explain how you put work done, work done against resistance, PE gain/loss, KE gain/loss in an equation? im lost :((( thank you!!
 
externityxzx

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iKhaled said:
sin 3x + 2cos 3x = 0

divid the equation by cos x and u will get tan 3x +2 = 0
tan 3x = -2
3x = -64.3 and 116.5
x= 158.9 and 38.9 but i have no idea how the mark scheme got 98.9 as one of x values sorry :/
Click to expand...
you can substitute 3x to something else like y (anything) so 3x = y
so you have tan y =-2, you have to change the range 0<x<180 to 0<y(which is 3x)<540
then you have y= -63.4 + 180 (keep adding 180)
y2= 116.6 and you put it back into y =3x which gives you 38.9
y3= 296.6 => 98.9
y4=476.6 => 158.9 :)
 
Rutzaba

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applepie1996 said:
ok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=o
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2

x= pi/4

so the limits be pi/4 and 0

by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx

now substituting

(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24

No need to thank me :p
because i didn't solve it :p
Rutzaba solved it for me :D
Click to expand...
Silent Hunter
 
iKhaled

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externityxzx said:
you can substitute 3x to something else like y (anything) so 3x = y
so you have tan y =-2, you have to change the range 0<x<180 to 0<y(which is 3x)<540
then you have y= -63.4 + 180 (keep adding 180)
y2= 116.6 and you put it back into y =3x which gives you 38.9
y3= 296.6 => 98.9
y4=476.6 => 158.9 :)
Click to expand...
oh that shows how to get it, thank you :)
 
abcde

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I'd really appreciate if someone can direct me to some vectors notes that cover all or most of the A'level p3 sy. I need them for someone else. :)
 
iKhaled

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