Question

Anyone got any ideas?
Attachment to follow,

Answer 1

I had dz/dv = dz/dx*dx/dv + dz/dy*dy/dv

Answer 2

so wouldn't dz/dx =2 in this case?
and dx/dv = 4
and dz/dy = c and dy/dv = d?

Answer 3

so the answer should be 8 + cd

Answer 4

which it says is wrong...

Answer 5

@ParthKohli

Answer 6

@calculusfunctions

Answer 7

@tukajo You have z = f(x, y), x = x(u, v), y = y(u, v) and you wish to evaluate ∂z/∂v at (0, 3). Thus\[\frac{∂z }{∂v }=\frac{ ∂z }{ ∂x }\frac{∂x }{∂v }+\frac{∂z }{ ∂y }\frac{ ∂y }{∂v }\]

Do you understand thus far?

Answer 8

@tukajo this is where we have to start. Do you understand so far? Yes or no?

Answer 9

It's very simple from here. Do you know what to do next? Yes or no?

Answer 10

I think he gets that part @calculusfunctions , since he already typed it.

Answer 11

Well if he gets that then the rest really is simple because it requires substituting the values given in the chart, in to the that equation.

Answer 12

sorry my internet crashed on me, yes I get it.

Answer 13

So what is your answer then?

Answer 14

dz/dx = 2
dx/dv = 4
dz/dy = c
dy/dv = d
so 2*4 + c*d = 8 + cd?

Answer 15

it says that is incorrect though?

Answer 16

rfl

Answer 17

dx/dv (0,3)= 4 means at u=0.,v=3
but u need at x=0,y=3
so u can't use dx/dv = 4

Answer 18

it's also funny that you'd ask that.

Answer 19

hartnn can you explain a little further?

Answer 20

xv (0,3) = 4
means partial derivative of x w.r.t. v when u=0,v=3 equals 4
but u don't need that info while calculating
\(\large\frac{∂z (0,3) }{∂v }=\frac{ ∂z(0,3) }{ ∂x }\frac{∂x }{∂v }+\frac{∂z(0,3) }{ ∂y }\frac{ ∂y }{∂v }\)
even idk how u find ∂x /∂v and ∂y/∂v

Answer 21

thanks for trying haha

Answer 22

@hartnn, what?

Answer 23

i may be wrong though....sorry for confusion

Answer 24

could you go into that algebraic

Answer 25

now i think zv(0,3) means the question is asking partial derivative of z w.r.t v when u= 0,v=3
hence from the table xv(0,3) = 4 will be there
now to find zx(0,3) means partial derivative of z w.r.t x when u= 0,v=3
but x(0,3)=2 and y(0,3)=1
so zx(2,1) needs to be found which is p
so u get 1st term as 4p
same for 2nd term.,....

Answer 26

ah I got you.

Answer 27

@hartnn actually I think you maybe correct! I didn't observe that chart properly.

Answer 28

@tukajo let's do this again because @hartnn is correct. I made a mistake. Sorry.

Answer 29

Okie dokie

Answer 30

I'm ready, sorry. No more distractions, let's do this :)

Answer 31

OK we have z = f(x, y), x = x(u, v), y = y(u, v) with x(0, 3) = 2 and y(0, 3) = 1. This implies that u = 0 and v =3. Correct?

Answer 32

Yes

Answer 33

OK so then it should follow that x = 2 since x(u, v) = 2 because x(0, 3) = 2. Similarly y = 1 since y(u, v) = 1 because y(0, 3) = 1. Thus we need to evaluate\[\frac{ ∂z }{∂x }|_{(2, 1)}\]and\[\frac{∂z }{∂y }|_{(2, 1)}\]Do you understand so far?

Answer 34

yes I got you.

Answer 35

OK so then so far tell me what these values are.

Answer 36

dz/dy|(2,1) = a
dz/dx(2,1) = p

Answer 37

dz/dx(2,1) * dx/dv(0,3) = 4p
dz/dy(2,1)*dy/dv(0,3) = ad.
dz/dv = 4p + ad
correct?

Answer 38

well yea, it is correct. But I mean I am logically following it correctly, I think.

Answer 39

I worked it out and yes, finally! Sorry for the mistake earlier. Even teachers are human!

Answer 40

it's all good, thanks for your help :)

Answer 41

Welcome!

Answer 42

Now I have to go work on some programming. Lol C++ here I come.