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stamp
 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).
stamp
 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).

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stamp
 one year ago
Best 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?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1by 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
 one year ago
Best ResponseYou've already chosen the best response.0Or would all that simplify to 4t^3

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we 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
 one year ago
Best 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?

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1it does simplify out to 4t^3

stamp
 one year ago
Best 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)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1provided t not equal 1 or 1 due to that z'

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

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i got lost, let me go back to the w' chains :) and sort it back out

amistre64
 one year ago
Best 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1361983895329:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we still have to include the caveat of t not equal to 1 or 1

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1so that reduces to w' = (31) t^3 which means i prolly missed a sign in there someplace along the way :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i forgot the  part :)

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1check out z', its spose to be negative ... not positive

stamp
 one year ago
Best ResponseYou've already chosen the best response.0Ok. I will ponder all of this and do some work on paper to verify everything. Thank you again for your time.

stamp
 one year ago
Best 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+

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1yes, but z' is negative so that changes the sign .... and carries it to the end as (3+1) not (31)
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