stamp
  • stamp
[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).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
stamp
  • stamp
\[w=xy\ cos(z)\]\[w=t(t^2)\ cos(arccos(t))\]\[w=t^3t\]\[w=t^4\]\[dw/dt=4t^3\] Is that all there is to it or am I mistaken?
amistre64
  • amistre64
by the chain rule eh \[w=xy\ cos(z)\] \[w(t)=x(t)y(t)\ cos(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))+x(t)y(t)\ cos'(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))-x(t)y(t)z'(t)\ sin(z(t))\]
stamp
  • stamp
So did I goof?

Looking for something else?

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

More answers

stamp
  • stamp
Or would all that simplify to 4t^3
amistre64
  • amistre64
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 = \(\frac{1}{\sqrt{1+...}}\) \[w'=cos(cos^{-1}z)(y+2t^2)-t^3z'\ sin(cos^{-1}z)\]
amistre64
  • amistre64
\[w'=z(t^2+2t^2)-t^3z'\ sin(cos^{-1}z)\] \[w'=3t^2z-t^3z'\ sin(cos^{-1}z)\] is that looking right so far?
stamp
  • stamp
You forgot your x' and y' and z' on your derivatives but it is ok because I recognize what you did.
amistre64
  • amistre64
it does simplify out to 4t^3
stamp
  • stamp
\[w′=cos(z)(x′y+xy′)−xyz′ sin(z)\] 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)
amistre64
  • amistre64
provided t not equal -1 or 1 due to that z'
stamp
  • stamp
I am not quite seeing the simplification to 4t^3
stamp
  • stamp
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.
stamp
  • stamp
ps thank you for your assistance so far
amistre64
  • amistre64
i got lost, let me go back to the w' chains :) and sort it back out
stamp
  • stamp
ok
amistre64
  • amistre64
\[w'=x'y\cos(z)+xy'\cos(z)+xy~\cos'(z(t))\] \[w'=x'y\cos(z)+xy'\cos(z)-xyz'~\sin(z)\] \[w'=x'y\cos(z)+xy'\cos(z)-xyz'~\sin(z)\] \[x = t,~ y = t^2, ~z = cos^{-1}(t)\]\[x' = 1, ~y' = 2t, ~z' =(1-t^2)^{(-1/2)}\] \[w'=t^2t+t~2t~t-tt^2(1-t^2)^{(-1/2)}~\sin(\cos^{-1}(t))\] \[w'=t^3+2t^3-t^3(1-t^2)^{(-1/2)}~\sin(\cos^{-1}(t))\] \[w'=t^3(3-(1-t^2)^{(-1/2)}~\sin(\cos^{-1}(t))\] the key here is in definine sin(cos^-1 (t)) better
amistre64
  • amistre64
|dw:1361983895329:dw|
amistre64
  • amistre64
we still have to include the caveat of t not equal to -1 or 1
stamp
  • stamp
sin(arccos(t)) = sqrt( 1 - t^2 )
amistre64
  • amistre64
so that reduces to w' = (3-1) t^3 which means i prolly missed a sign in there someplace along the way :)
amistre64
  • amistre64
z' = -(1-t^2)^1/2
amistre64
  • amistre64
i forgot the - part :)
stamp
  • stamp
w' = (3-1)t^3 w' = 2t^3 ?? :(
amistre64
  • amistre64
check out z', its spose to be negative ... not positive
stamp
  • stamp
Ok. I will ponder all of this and do some work on paper to verify everything. Thank you again for your time.
amistre64
  • amistre64
youre welcome
stamp
  • stamp
@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+
amistre64
  • amistre64
yes, but z' is negative so that changes the sign .... and carries it to the end as (3+1) not (3-1)

Looking for something else?

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