Question

Let \((u,v)=\mathbf{f}(x,y,x)=(2x+y^3,xe^{5y-7z})\)

Compute \(D\mathbf{f}(x,y,z),\text{ and }\partial (u,v)/\partial(x,y)\)

Answer 1

so I have \[D\mathbf{f}(x,y,z)=\left[\begin{matrix} \partial u / \partial x & \partial u / \partial y & \partial u / \partial z \\ \partial v / \partial x & \partial v / \partial y & \partial v/ \partial z \end{matrix}\right]\]

Answer 2

but I am struggling with how to find \[\frac{\partial(u,v)}{\partial(x,y)}=(10x-3y^2)e^{5y-7z}\]

Answer 3

@amistre64 if you could take a look at this I would be really stoked!

Answer 4

are we are talking about vector differentiation here?

Answer 5

yes

Answer 6

ok so \[D\mathbf{F}(x,y,z)=\left[\begin{matrix} 2 & 3y^2 & 0 \\ e^{5y-7z} & 5xe^{5y-7z} & -7xe^{5y-7z}\end{matrix}\right]\]

Answer 7

so it's quite clear that the answer is the determinant if I ignored the third column.

Answer 8

If the understanding of notations is correct, that is your \(\partial(u,v)\over\partial(x,y)\). D will be its Jacobian, a vector

Answer 9

did I understand it correctly?

Answer 10

\(\partial(u,v)\over\partial(x,y)\) is scalar

Answer 11

\[D=\left({\partial\over\partial x}\hat{i}+{\partial\over\partial y}\hat{j}+{\partial\over\partial z}\hat{k}\right)\]

Answer 12

if you take dot product of D with (u,v), you get a scalar
if you take cross product of D with (u,v), you get a vector

Answer 13

wait, why is the jacobian a vector?

Answer 14

isn't it the matrix I wrote down?

Answer 15

yes.. now it needs the i, j, k row on the top to classify it as a vector.

Answer 16

\[D=\left[\begin{matrix} i & j & k \\2 & 3y^2 & 0 \\ e^{5y-7z} & 5xe^{5y-7z} & -7xe^{5y-7z}\end{matrix}\right]\]

Answer 17

now, make it a determinant and evaluate
you'd get an (i,j,k) vector

Answer 18

which represents what?

Answer 19

rate of change of "f" along each axis

Answer 20

one is "gradient" -> scalar
one is "curl" -> vector

Answer 21

\[\det(D)=\left\langle 10\,x\,{e}^{5\,y-7\,z}-3\,{y}^{2}\,{e}^{5\,y-7\,z},-21\,x\,{y}^{2}\,{e}^{5\,y-7\,z}, -14\,x\,{e}^{5\,y-7\,z}\right\rangle\]

Answer 22

yes

Answer 23

so my answer is just the one component of that vector.

Answer 24

the entire vector would be the answer

Answer 25

it's not though.

Answer 26

you i,j,k components in the vector notation are not in proper order

Answer 27

\[\left<-21xy^2e^{5y-7z},-14x.....\right>\]

Answer 28

oh yeah I see that mistake I just threw it in from maxima, still, my book tells me that
\[\frac{\partial (u,v)}{\partial (x,y)} = (10-3y^2)e^{5y-7z}\]\[\frac{\partial (u,v)}{\partial (y,z)} = -21xy^2e^{5y-7z}\]\[\frac{\partial (u,v)}{\partial (x,z)} = -14xe^{5y-7z}\]

Answer 29

exactly.. k, i and j components respectively

Answer 30

so I can see that I take the term that is not in the partial derivative is the answer. but honestly I have no idea what I am doing. very frustrating.

Answer 31

you mean the maxima interpreter?

Answer 32

I am lost with the symbols you are using. If you can provide me with their meaning/ book sentence .. I can help

Answer 33

no sorry I meant I just used maxima to compute the determinant and I copied it down in the wrong order.

Answer 34

and btw I really appreciate your help electrokid. I just find that less people look at questions when you "medal" them and I still would like others to look at this. I wish we could medal more than one person...