Question

z = f(u) + g(v) and u = x - 4t, v = x + 4t
dz/dx = 3, g'(v) = 2
find {partial derivative} dz/dt

Answer 1

picture included

Answer 2

not exactly sure where to begin with this

Answer 3

I get 4

Answer 4

how do you set it up?

Answer 5

\[\frac{dz}{dt}=\frac{d f}{du}\frac{\partial u}{\partial t} +\frac{d g}{dv}\frac{\partial v}{\partial t}\]

Answer 6

that's kinda what i was thinking but i'm not quite sure where partial(dz/dx) and g'(v) would fit into that

Answer 7

bc if z = f(u) + g(v), then the partial dz/dx would be related to both functions put together, right?

Answer 8

\[\frac{dz}{dx}=\frac{d f}{du}\frac{\partial u}{\partial x} +\frac{d g}{dv}\frac{\partial v}{\partial x}=3\]

\[\frac{\partial u}{\partial x}=1\]

\[\frac{\partial v}{\partial x}=1\]

\[\frac{d f}{du} +\frac{d g}{dv}=3\]
\[\frac{d f}{du} +2=3\]
\[\frac{d f}{du} =1\]

Answer 9

see what I did?

Answer 10

ah

Answer 11

ok so you do the original z -> dz/dx = each one individually + added together, which will give you more values that you can use to figure out what partial dz/dt is

Answer 12

yes...now plug in what you know to get

\[\frac{dz}{dt}=\frac{d f}{du}\frac{\partial u}{\partial t} +\frac{d g}{dv}\frac{\partial v}{\partial t}\]

Answer 13

so dz/dt = (1) ( -4) + (2)(4) = -4 + 8 = 4
but this is dz/dt and not partial dz/dt right?

Answer 14

that is it..I should have written \(\displaystyle\frac{\partial z}{\partial t}\)

Answer 15

ah ok
bleh well that makes a lot of sense :0
i feel dumb now

Answer 16

\[\frac{\partial z}{\partial t}=\frac{d f}{du}\frac{\partial u}{\partial t} +\frac{d g}{dv}\frac{\partial v}{\partial t}\]

Answer 17

i was looking at it all like it was greek thinking where are all these extra symbols coming from

Answer 18

ic

Answer 19

Thank you very much! I really appreciate it!

Answer 20

no problem