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[CALCULUS III—CHAIN RULE] Find dw/dt by the chain rule where w = xy cos(z) and x = t, y = t^2, and z = arccos(t).

  • one year ago
  • one year ago

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    \[w=xy\ cos(z)\]\[w=t(t^2)\ cos(arccos(t))\]\[w=t^3t\]\[w=t^4\]\[dw/dt=4t^3\] Is that all there is to it or am I mistaken?

    • one year ago
  2. amistre64
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    by the chain rule eh \[w=xy\ cos(z)\] \[w(t)=x(t)y(t)\ cos(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))+x(t)y(t)\ cos'(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))-x(t)y(t)z'(t)\ sin(z(t))\]

    • one year ago
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    So did I goof?

    • one year ago
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    Or would all that simplify to 4t^3

    • one year ago
  5. amistre64
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    we can ignore the (t) parts to clean it up \[w'=x'y\ cos(z)+xy'\ cos(z)-xyz'\ sin(z)\] \[w'=cos(z)(x'y+xy')-xyz'\ sin(z)\] \[w'=cos(z)(x'y+xy')-xyz'\ sin(z)\] x = t, y = t^2, and z = arccos(t) x = 1, y = 2t, and z = \(\frac{1}{\sqrt{1+...}}\) \[w'=cos(cos^{-1}z)(y+2t^2)-t^3z'\ sin(cos^{-1}z)\]

    • one year ago
  6. amistre64
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    \[w'=z(t^2+2t^2)-t^3z'\ sin(cos^{-1}z)\] \[w'=3t^2z-t^3z'\ sin(cos^{-1}z)\] is that looking right so far?

    • one year ago
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    You forgot your x' and y' and z' on your derivatives but it is ok because I recognize what you did.

    • one year ago
  8. amistre64
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    it does simplify out to 4t^3

    • one year ago
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    \[w′=cos(z)(x′y+xy′)−xyz′ sin(z)\] x'y = t^2 xy' = 2t x'y + xy' = 2t + t^2 = t(2+t) cos(z) = cos(arccos t) = t t( t(2+t) ) - xyz' sin(z)

    • one year ago
  10. amistre64
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    provided t not equal -1 or 1 due to that z'

    • one year ago
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    I am not quite seeing the simplification to 4t^3

    • one year ago
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    I understand the first part where you showed the formal chain rule of w(t) = x(t) y(t) cos(z(t)) The simplification of it to 4t^3 is still eluding me.

    • one year ago
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    ps thank you for your assistance so far

    • one year ago
  14. amistre64
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    i got lost, let me go back to the w' chains :) and sort it back out

    • one year ago
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    ok

    • one year ago
  16. amistre64
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    \[w'=x'y\cos(z)+xy'\cos(z)+xy~\cos'(z(t))\] \[w'=x'y\cos(z)+xy'\cos(z)-xyz'~\sin(z)\] \[w'=x'y\cos(z)+xy'\cos(z)-xyz'~\sin(z)\] \[x = t,~ y = t^2, ~z = cos^{-1}(t)\]\[x' = 1, ~y' = 2t, ~z' =(1-t^2)^{(-1/2)}\] \[w'=t^2t+t~2t~t-tt^2(1-t^2)^{(-1/2)}~\sin(\cos^{-1}(t))\] \[w'=t^3+2t^3-t^3(1-t^2)^{(-1/2)}~\sin(\cos^{-1}(t))\] \[w'=t^3(3-(1-t^2)^{(-1/2)}~\sin(\cos^{-1}(t))\] the key here is in definine sin(cos^-1 (t)) better

    • one year ago
  17. amistre64
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    |dw:1361983895329:dw|

    • one year ago
  18. amistre64
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    we still have to include the caveat of t not equal to -1 or 1

    • one year ago
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    sin(arccos(t)) = sqrt( 1 - t^2 )

    • one year ago
  20. amistre64
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    so that reduces to w' = (3-1) t^3 which means i prolly missed a sign in there someplace along the way :)

    • one year ago
  21. amistre64
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    z' = -(1-t^2)^1/2

    • one year ago
  22. amistre64
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    i forgot the - part :)

    • one year ago
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    w' = (3-1)t^3 w' = 2t^3 ?? :(

    • one year ago
  24. amistre64
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    check out z', its spose to be negative ... not positive

    • one year ago
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    Ok. I will ponder all of this and do some work on paper to verify everything. Thank you again for your time.

    • one year ago
  26. amistre64
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    youre welcome

    • one year ago
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    @amistre64 Here is the derivative of x(t) y(t) cos z(t) http://www.wolframalpha.com/input/?i=derivative+of+x%28t%29y%28t%29+cos%28z%28t%29%29+

    • one year ago
  28. amistre64
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    yes, but z' is negative so that changes the sign .... and carries it to the end as (3+1) not (3-1)

    • one year ago
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