Question

[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).

Answer 1

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

Answer 2

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

Answer 3

So did I goof?

Answer 4

Or would all that simplify to 4t^3

Answer 5

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 = \(\frac{1}{\sqrt{1+...}}\)

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

Answer 6

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

Answer 7

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

Answer 8

it does simplify out to 4t^3

Answer 9

\[w′=cos(z)(x′y+xy′)−xyz′ sin(z)\]

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)

Answer 10

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

Answer 11

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

Answer 12

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.

Answer 13

ps thank you for your assistance so far

Answer 14

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

Answer 15

ok

Answer 16

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