harder 3u and mechanics (1 Viewer)

[totallybord]

hey,
induction:
2k+3>2* sqrt{(k+1)(k+2)}
can you just square this and move it to the other side to make it >0

x,y>0,
{(x+y)/2}^n <or=to {x^n+y^n}/2 for n>or = 1

prove by induction: for n>or = 1

d^n/dx^n of (x^nlogx) =n!(logx+1+1/2+1/3+...1/n)

inequalities:
prove that : (a+3b)(b+3c)(c+3a)>or= to 64abc

if x,y,z,a,b,c are real prove that
(x^2+y^2+z^2)(+a^2+b^2+c^2)>or= to (ax+by+cz)^2

prove that :
0.5integral1->0 (x^3*(1-x)^3 < integral 1->0 {[x^3*(1-x)^3] /(1+x)}<integral 1->0 x^3*(1-x)^3

mechanics:
how does a pendulum swing around in a complete circle? is that even possible?

this is a really simple q but i just dont get the answer:
a light inexstensible string is threaded through a asmall smooth ring B, and the ends of the string are attached to the fixed points A and C. the mass of the ring is m and the point A is at height h vertically above C. the ring moves in a horizontal circle with centre C and radius h and CB has constant angular velocity w. show that w^2 = {g(1+sqrt2)/h}
mainly i just dont get why the horizontal component would be T(1/sqrt2 +1) =mhw^2

and i think thats it for today
thank you!

 

[Affinity]

hey,
induction:
2k+3>2* sqrt{(k+1)(k+2)}
can you just square this and move it to the other side to make it >0

If they don't require you to use induction, for example: "By mathematical induction or otherwise"

x,y>0,
{(x+y)/2}^n or = 1 <=== don't quite get this

prove by induction: for n>or = 1

This question doesn't make sense

d^n/dx^n of (x^nlogx) =n!(logx+1+1/2+1/3+...1/n)

should be straightforward

inequalities:
prove that : (a+3b)(b+3c)(c+3a)>or= to 64abc


By the AM-GM inequality, you know that
[a + 3b] = [a + b + b + b] > 4*(ab^3)^(1/4)
similarly, [b+3c] > 4(bc^3)^(1/4) and [c+3a] > 4(ca^3)^(1/4)

multiply and you are done.


if x,y,z,a,b,c are real prove that
(x^2+y^2+z^2)(+a^2+b^2+c^2)>or= to (ax+by+cz)^2

It's a special case of the cauchy-schwarz inequality, you can prove this in many ways, for example, consider the quadratic in K
p(K) = (xK + a)^2 + (yK + b)^2 + (zK+c)^2
now P is a sum of squares so cannot be negative, hence it can have atmost 1 real root, so it's discriminant is no greater than 0.
Expanding p gives
p(K) = (x^2 + y^2 + z^2)K^2 +2(ax + by + cz)K + (a^2 + b^2 + c^2)

so [2(ax + by + cz)]^2 - 4(x^2 + y^2 + z^2)(a^2 + b^2 + c^2) <= 0

etc.




prove that :
0.5integral1->0 (x^3*(1-x)^3 < integral 1->0 {[x^3*(1-x)^3] /(1+x)}0 x^3*(1-x)^3

What's that last bit 0 x^3*(1-x)^3 after the braces }?


mechanics:
how does a pendulum swing around in a complete circle? is that even possible?

Hmm you can do it.. make one and try it

this is a really simple q but i just dont get the answer:
a light inexstensible string is threaded through a asmall smooth ring B, and the ends of the string are attached to the fixed points A and C. the mass of the ring is m and the point A is at height h vertically above C. the ring moves in a horizontal circle with centre C and radius h and CB has constant angular velocity w. show that w^2 = {g(1+sqrt2)/h}
mainly i just dont get why the horizontal component would be T(1/sqrt2 +1) =mhw^2

CB is a pure horizontal pull so it contributes T, AB is a 45 degree pull with same tension, so it contributes T/sqrt(2)// add them up. and that should equal to the centripetal force

 

[totallybord]

sorry affinity: didnt finish typing the questions
x,y>0,
{(x+y)/2}^n <= {x^n+y^n}/2 for n >= 1

 

[totallybord]

affinity:
i thought hte horizontal component would just be Tsin@=mhw^2 which wld jst be T/sqrt2?
the end of the other q is
0.5integral1->0 {x^3*(1-x)^3 }dx< integral 1->0 {[x^3*(1-x)^3] /(1+x)}dx < integral 1->0 {x^3*(1-x)^3}dx

 

[totallybord]

btw affinity:
can you please explain to me the answer for 2007 hsc q6 (iv) since i didnt understand terry lee's explanation
http://www.boardofstudies.nsw.edu.au/hsc_exams/exam-papers-2007/pdf_doc/maths-ex2-07.pdf

 

[Affinity]

btw affinity:
can you please explain to me the answer for 2007 hsc q6 (iv) since i didnt understand terry lee's explanation
http://www.boardofstudies.nsw.edu.au/hsc_exams/exam-papers-2007/pdf_doc/maths-ex2-07.pdf

part (a) iv) the limit is 4 since 0 < (n+2)/2^(n-1) < (4n+8)/[n(n-1)]
and the last bit tends to 0... so (n+2)/2^(n-1) tends to 0 and that makes the limiting sum 4.


(b) iv) thats air resistance proportional to the speed squared.

 

[Affinity]

i thought hte horizontal component would just be Tsin@=mhw^2 which wld jst be T/sqrt2?

there are effectively 2 strings pulling the ring, don't forget about the tension in BC.

the end of the other q is
0.5integral1->0 {x^3*(1-x)^3 }dx< integral 1->0 {[x^3*(1-x)^3] /(1+x)}dx < integral 1->0 {x^3*(1-x)^3}dx

they key is to notice that
1.) x^3*(1-x)^3 >= 0 on the interval (0,1)
2.) 0<0.5 < 1/(1+x) < 1 on the same interval

so you know that

0.5(x^3*(1-x)^3) < (x^3*(1-x)^3)/(1+x) < (x^3*(1-x)^3) when x is between 0 and 1.

so the inequalities for the integrals also hold


didnt finish typing the questions
x,y>0,
{(x+y)/2}^n <= {x^n+y^n}/2 for n >= 1


If you have to use induction, then
case n = 1 is obvious,

suppose {(x+y)/2}^k <= {x^k+y^k}/2


then
{(x+y)/2}^(k+1)
<= {x^k+y^k}/2 * (x+y)/2
<= {x^(k+1)+y^(k+1) + yx^k + xy^k}/4
[remark: compare with what we want, that is {x^(k+1)+y^(k+1)}/2 you can see that we would be done if we can prove yx^k + xy^k <= x^(k+1)+y^(k+1)]
to prove yx^k + xy^k <= x^(k+1)+y^(k+1) -----> if you move everything to the RHS and factorize, you will get x^k(x-y) + y^k(y-x) = (x^k - y^k)(x-y)
but that must be non negative since x >= y and x^k >= y^k are both true or both false at the same time..


so yx^k + xy^k <= x^(k+1)+y^(k+1)

{(x+y)/2}^(k+1)
<= {x^(k+1)+y^(k+1) + yx^k + xy^k}/4
<= 2{x^(k+1)+y^(k+1)}/4
<= {x^(k+1)+y^(k+1)}/2

 

[totallybord]

oh affinity,
for that cm question, <ACB is 90 so i thought because its 90 degrees there wldnt be a tension there...
so if there are 2 strings and the angle between them is 90, theres still a tension?