Additional Math Help HERE ONLY...!

[cooldude]

am quite a week student in add math...can sum1 plz xplain question 12 second one on velocity for me...its may/june 09 p1...


I still have accouting,chems litt n econs n add math left so i wont b able to do alot of pp for add maths...can any1 plz recommend which yrs i shuld do..like gud hard papers...atleast 3 yrs plz apart 4rm 09 Thanks

i) 0=kcos4t
cos4t=0
90=4t
t=22.5 sec
ii) dv/dt=a=-4ksin4t
iii) 12=4k as sin(4*(3/8))=-1
k=3
iv) u just have to sketch the graph of 3cos4t
v) integrate the expression for velocity-->
x=3sin4t/4+k
however since at the origin the displacement is 0, therefore k=0
now substitute t=pi/24
x=0.375 m

[Ghost Of Highbury]

am quite a week student in add math...can sum1 plz xplain question 12 second one on velocity for me...its may/june 09 p1...


I still have accouting,chems litt n econs n add math left so i wont b able to do alot of pp for add maths...can any1 plz recommend which yrs i shuld do..like gud hard papers...atleast 3 yrs plz apart 4rm 09 Thanks

i) substitute 0 for velocity.. 0 = cos 4t

  4t = y

   cos y = 0

     y = cos^-1 0 = 90

here its 90 degrees, but v gotta take it in radians, so Pi radians = 180 degree

4t = 0.5pi radians
t = pi/8

t = pi/4

ii) differentiate v = k cos 4t

     k = a
     cos 4t = b

     d/dx(a) = 0

     d/dx(b) = -4sin4t

     dv/dx = a(db) + b(da) = -4ksin4t + 0 = -4k sin 4t

iii)  
    k = 3

iv) make a table and plot the points.

v) integration of 3cos4t with limits pi/24 and 0

                  

  

[syedz123]

Thanks guys 

[J.Darren]

here its 90 degrees, but v gotta take it in radians, so Pi radians = 180 degree

4t = 180

t = pi/4

Cosine equals to 0 at 90 degreees (first instantaneously at rest) / 0.5 pi radian, 90 / 4 = 22.5 degrees / 0.125 pi radian

[Ghost Of Highbury]

Cosine equals to 0 at 90 degreees (first instantaneously at rest) / 0.5 pi radian, 90 / 4 = 22.5 degrees / 0.125 pi radian

my bad.edited. thanks

[syedz123]

i) substitute 0 for velocity.. 0 = cos 4t

  4t = y

   cos y = 0

     y = cos^-1 0 = 90

here its 90 degrees, but v gotta take it in radians, so Pi radians = 180 degree

4t = 0.5pi radians
t = pi/8

t = pi/4

ii) differentiate v = k cos 4t

     k = a
     cos 4t = b

     d/dx(a) = 0

     d/dx(b) = -4sin4t

     dv/dx = a(db) + b(da) = -4ksin4t + 0 = -4k sin 4t

iii)  
    k = 3

iv) make a table and plot the points.

v) integration of 3cos4t with limits pi/24 and 0

                  

  

sorri A@di bt i still dnt get da second one...it says find da position vector of P at 12:00...so do we jst leave da answer as -4k sin 4t? dont we hv to use da '12:00' in this and da answer shuld b in a vector f0rm?? c0z its findin da position vector...sorri i told u am a weak add math student

[Ghost Of Highbury]

Dude the question u r referring to is question 9...the one with the quote is different answer which is question 12...

[syedz123]

Dude the question u r referring to is question 9...the one with the quote is different answer which is question 12...

omgness!! LOL wts happning to me...sorri man lol Thanks agen 

[syedz123]

can someone plz help me wiv this
Find the coefficient of x4 in the expansion of
(i) (1-x/4)(1+2x)^6

and integration 4rm q7 (ii) may/june 09 p1

plz explain step by step

[cooldude]

can someone plz help me wiv this
Find the coefficient of x4 in the expansion of
(i) (1-x/4)(1+2x)^6

k the first q-->
first expand (1+2x)^6--> 1+12x+60x^2+160x^3+240x^4........
now we need the coefficient of x^4--> (1-(x/4))(1+12x+60x^2+160x^3+240x^4........)
now look at the above line ive typed, look for powers of x which multiply to give x^4, we see that 1 and 240x^4 multiply to give x^4 i.e. 240x^4, we also see that the product of -x/4 and 160x^3 is -40x^4 which again has x^4, so now we have 2 terms which have x^4, -40x^4 and 240x^4, so when we add them we get 200x^4, and the coefficient of this is 200, the answer required, the way i told u of finding terms which multiply to give x^4 is much easier than expanding the whole thing, u can do that for this q, as it will give u a clear idea of what ive done, and ive only expanded (1+2x)^6 uptil x^4 as that is what we need and not the whole expansion

[Ghost Of Highbury]

(1-x/4) ( 12x+60x^2+160x^3+240x^4.....)

(....160x^3 + 240x^4)(1-x/4)

160x^3 - (160x^4/4) + 240x^4 = ....240x^4 - 40x^4.... = 200x^4

hence, 200

[syedz123]

Thanks agen guys 

[cooldude]

now the second q-->
in the first part we found that the derivate is 3xe^3x
the simplest explanation i can give is that the integral of xe^3x will be 1/3 of the expression given to differentiate in part i is cuz the derivate we found in the second part is 3 times the integrand in part ii, thus since differentiation is the opposite of integration we divide it by 3 as the derivate we got in part i is 3 times the integrand in part ii, the other way i knw to find the integral is integration by parts but im not sure whether u add maths guys knw that as ive never done add maths myself so i have no idea what the syll is  , neway it wont make a diff since the q says hence, so just try and understand what i said or wait until sum1 can give a better explanation

[J.Darren]

There you go.

[jellybeans]

HIIII, um, i need help with Q12ii) pleaseeee
Thanks

i knw you integrate the velocity to get the displacement but what values do you sub in?  
thanks again. XD