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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
[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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stampBest ResponseYou've already chosen the best response.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?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
by the chain rule eh \[w=xy\ cos(z)\] \[w(t)=x(t)y(t)\ cos(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))+x(t)y(t)\ cos'(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))x(t)y(t)z'(t)\ sin(z(t))\]
 one year ago

stampBest ResponseYou've already chosen the best response.0
Or would all that simplify to 4t^3
 one year ago

amistre64Best ResponseYou've already chosen the best response.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 = \(\frac{1}{\sqrt{1+...}}\) \[w'=cos(cos^{1}z)(y+2t^2)t^3z'\ sin(cos^{1}z)\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[w'=z(t^2+2t^2)t^3z'\ sin(cos^{1}z)\] \[w'=3t^2zt^3z'\ sin(cos^{1}z)\] is that looking right so far?
 one year ago

stampBest ResponseYou've already chosen the best response.0
You forgot your x' and y' and z' on your derivatives but it is ok because I recognize what you did.
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
it does simplify out to 4t^3
 one year ago

stampBest ResponseYou've already chosen the best response.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)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
provided t not equal 1 or 1 due to that z'
 one year ago

stampBest ResponseYou've already chosen the best response.0
I am not quite seeing the simplification to 4t^3
 one year ago

stampBest ResponseYou've already chosen the best response.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.
 one year ago

stampBest ResponseYou've already chosen the best response.0
ps thank you for your assistance so far
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
i got lost, let me go back to the w' chains :) and sort it back out
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[w'=x'y\cos(z)+xy'\cos(z)+xy~\cos'(z(t))\] \[w'=x'y\cos(z)+xy'\cos(z)xyz'~\sin(z)\] \[w'=x'y\cos(z)+xy'\cos(z)xyz'~\sin(z)\] \[x = t,~ y = t^2, ~z = cos^{1}(t)\]\[x' = 1, ~y' = 2t, ~z' =(1t^2)^{(1/2)}\] \[w'=t^2t+t~2t~ttt^2(1t^2)^{(1/2)}~\sin(\cos^{1}(t))\] \[w'=t^3+2t^3t^3(1t^2)^{(1/2)}~\sin(\cos^{1}(t))\] \[w'=t^3(3(1t^2)^{(1/2)}~\sin(\cos^{1}(t))\] the key here is in definine sin(cos^1 (t)) better
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1361983895329:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
we still have to include the caveat of t not equal to 1 or 1
 one year ago

stampBest ResponseYou've already chosen the best response.0
sin(arccos(t)) = sqrt( 1  t^2 )
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
so that reduces to w' = (31) t^3 which means i prolly missed a sign in there someplace along the way :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
i forgot the  part :)
 one year ago

stampBest ResponseYou've already chosen the best response.0
w' = (31)t^3 w' = 2t^3 ?? :(
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
check out z', its spose to be negative ... not positive
 one year ago

stampBest ResponseYou've already chosen the best response.0
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

stampBest ResponseYou've already chosen the best response.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+
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
yes, but z' is negative so that changes the sign .... and carries it to the end as (3+1) not (31)
 one year ago
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