Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

stamp

  • 2 years ago

[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).

  • This Question is Closed
  1. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    by the chain rule eh \[w=xy\ cos(z)\] \[w(t)=x(t)y(t)\ cos(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))+x(t)y(t)\ cos'(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))-x(t)y(t)z'(t)\ sin(z(t))\]

  3. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So did I goof?

  4. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Or would all that simplify to 4t^3

  5. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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 = \(\frac{1}{\sqrt{1+...}}\) \[w'=cos(cos^{-1}z)(y+2t^2)-t^3z'\ sin(cos^{-1}z)\]

  6. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[w'=z(t^2+2t^2)-t^3z'\ sin(cos^{-1}z)\] \[w'=3t^2z-t^3z'\ sin(cos^{-1}z)\] is that looking right so far?

  7. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it does simplify out to 4t^3

  9. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[w′=cos(z)(x′y+xy′)−xyz′ sin(z)\] 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)

  10. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    provided t not equal -1 or 1 due to that z'

  11. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am not quite seeing the simplification to 4t^3

  12. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  13. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ps thank you for your assistance so far

  14. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i got lost, let me go back to the w' chains :) and sort it back out

  15. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  16. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  17. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1361983895329:dw|

  18. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    we still have to include the caveat of t not equal to -1 or 1

  19. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sin(arccos(t)) = sqrt( 1 - t^2 )

  20. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  21. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    z' = -(1-t^2)^1/2

  22. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i forgot the - part :)

  23. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    w' = (3-1)t^3 w' = 2t^3 ?? :(

  24. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    check out z', its spose to be negative ... not positive

  25. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  26. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    youre welcome

  27. stamp
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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+

  28. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  29. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.