Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

[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
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[w=xy\ cos(z)\]\[w=t(t^2)\ cos(arccos(t))\]\[w=t^3t\]\[w=t^4\]\[dw/dt=4t^3\] Is that all there is to it or am I mistaken?
by the chain rule eh \[w=xy\ cos(z)\] \[w(t)=x(t)y(t)\ cos(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))+x(t)y(t)\ cos'(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))-x(t)y(t)z'(t)\ sin(z(t))\]
So did I goof?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question