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  1. K

    How does one get the form of a circle out of this equation?

    ofcause!! oh how foolish i am :) answer: 1/9 = (x+ 5/3)^2 + y^2 thanks guys
  2. K

    How does one get the form of a circle out of this equation?

    Homework Statement Sketch modulus((z+1)/(2z+3))=1 on the complex plane where z=x+iy Homework Equations The Attempt at a Solution I know it is a circle but i need help simplifying the equation into the form of a circle. i'm stuck at 0= 3x^2 + 3y^2 + 10x + 8 I usually...
  3. K

    How to count the number of occurences of an integer in excel?

    Homework Statement I have a thousand random numbers generated between -100 and 100 and want to count the number of occurences of each integer. I know i can use the COUNTIF function, but for so many numbers it takes way too long. Is there a way to create a loop in excel that can do this or...
  4. K

    Show that if v . v' = 0 (both vectors) then speed v is constant

    alright i understand the theory behind it. But assume we didn't know anything about how this question is related to kinetic energy, how would i begin to solve it? Since it was a question in my math class and you are not expected to know kinetic energy.
  5. K

    Show that if v . v' = 0 (both vectors) then speed v is constant

    ok, so if T' is zero, that means acceleration =0 so therefore the speed will be constant. right?
  6. K

    Show that if v . v' = 0 (both vectors) then speed v is constant

    dT/dt = mv.dv/dt when m =1 dT/dt= v.dv/dt i can simplify furthur, but why? v' is the derivative of v ( velocity )
  7. K

    Show that if v . v' = 0 (both vectors) then speed v is constant

    Homework Statement Show that for a particle moving with velocity v(t), if v . v'=0 then the speed v is constant. Homework Equations The Attempt at a Solution Let v = (v1,v2,...,vn) and let v'= (v'1,v'2,...,v'n) So, v . v'= v1v'1 + v2v'2 + ... + vnv'n = 0 This is the...
  8. K

    Finding x(t) from a time dependant force

    haha yeah stop doing my homework for me! Ok i figured it out properly F(t) = ma0e-bt m(dv/dt) = ma0e-bt ∫mdv = ∫ma0e-btdt (from v=0 to v=v and t=0 to t=t) mv = m(a0e-bt)/-b (dx/dt) = (a0e-bt)/-b ∫dx = ∫(a0e-bt)/-b dt (from x=0 to x=x and t=0 to t=t) x(t) = a0e-bt/b2
  9. K

    Finding x(t) from a time dependant force

    alright ill take your word for it!
  10. K

    Finding x(t) from a time dependant force

    wait..you can just integrate like that without the dt on the right hand side?
  11. K

    Finding x(t) from a time dependant force

    touche sir. That does not make sense. ahh i see now Question was F(t) = ma0e-bt now i know why it is a0 so in this case acceleration a = a0e-bt but i'm still stuck! How do i get to x(t)?
  12. K

    Finding x(t) from a time dependant force

    yeah... this is what i tried F(t) = m(dv/dt) e-bt F(t)ebtdt = mdv ∫F(t)ebtdt = ∫mdv (with bounds t=0 to t=t, and v=0 to v=v) ∫F(t)ebtdt = mv (only integrating right side yields) ∫F(t)ebtdt = m(dx/dt) [∫F(t)ebtdt]dt = dx (integrating both...
  13. K

    Finding x(t) from a time dependant force

    Homework Statement A particle of mass m is subject to a force F(t) = mae-bt The initial position and speed are zero. Find x(t) Homework Equations The Attempt at a Solution how would i begin to start this??
  14. K

    Is there a more rigorous way to prove this?

    lol what....so simple. So it's just r'r'' + r'r'' = 2r'r'' ?
  15. K

    Is there a more rigorous way to prove this?

    Sorry i'm still learning with the notation and everything. Awww ok. You mind telling me how to prove this please? thanks btw there were errors in the question which i have edited now
  16. K

    Is there a more rigorous way to prove this?

    Homework Statement Show that d(v^2)/dt = 2 . (d^2r/dt^2) . (dr/dt) HINT: v^2 = ||dr/dt||^2 = dr/dt . dr/dt Homework Equations The Attempt at a Solution I did it another method: d(v^2)/dt = d/dv(v^2) . dv/dt --------------chain rule = 2v . dv/dt since v=dr/dt and dv/dt = d^2r/dt^2...
  17. K

    Derivative of kinetic energy with respect to position help

    Thanks heaps HallsofIvy, this is very clear. :D
  18. K

    Derivative of kinetic energy with respect to position help

    oooooo right. Makes so much sense now. Thanks heaps :)
  19. K

    Derivative of kinetic energy with respect to position help

    Yeah i understand that part, but how exactly do i use the chain rule to get to this step? dT/dx = d/dx(1/2mv^2) to mv(dv/dx)
  20. K

    Derivative of kinetic energy with respect to position help

    Homework Statement Show dT/dx = ma Homework Equations T=1/2mv^2 F=ma The Attempt at a Solution dT/dx = d/dx(1/2mv^2) = mv.dv/dx <--------------i believe you use the chain rule. But can someone explain exactly how to get to this step? = mv. (dv/dt)...
  21. K

    Vector geometry - Intersection of lines

    alright. Now how do i solve it?
  22. K

    Vector geometry - Intersection of lines

    Homework Statement I have 2 parametric vector equations (of a line) r(t) = (2,-4,4) + t(1,-3,4) s(t) = (1,-1,0) + t(2,-1,1) how do i find the coordinates for which they intersect each other? The answers is (1,-1,0) Homework Equations x=a+λv, for some λ in ℝ (parametric vector...
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