Question

partial derivatives, how to tackle this?
http://puu.sh/h2cAZ/b4e2f1a365.png

Answer 1

@Hero @dan815 @hartnn @perl

Answer 2

@hartnn gave up, who's next

Answer 3

@jim_thompson5910 @iambatman @Callisto @CGGURUMANJUNATH

Answer 4

You should begin to solve this by finding

\[\frac{\delta u}{\delta s}\]

and

\[\frac{\delta u}{\delta t}\]

You can then use the chain rule, and use some of algebra to solve for g(s, t) and h(s, t)

Will probably take a lot of effort, but doable none the less, the problem isnt nearly as hard as it initially seems.

Answer 5

"You can then use the chain rule, and use some of algebra to solve for g(s, t) and h(s, t)"
can you explain more in detail what to do here?

Answer 6

By chain rule and also use of product rule, you know that:

\[\frac{\delta u}{\delta s}=\frac{\delta u}{\delta x}*\frac{\delta x}{\delta s}+\frac{\delta u}{\delta y}*\frac{\delta y}{\delta s}\]

This can also be applied to find du/dt

\[\frac{\delta u}{\delta t}=\frac{\delta u}{\delta x}*\frac{\delta x}{\delta t}+\frac{\delta u}{\delta y}*\frac{\delta y}{\delta t}\]

Answer 7

but how do u find du/dx?

Answer 8

can u write u .

Answer 9

∂u/∂s = (∂u/∂x)*(2e^(2s)cos(5t)) + (∂u/∂y)*(2e^(2s)sin(5t)) EQN(1)
∂u/∂s = (∂u/∂x)*(-5e^(2s)sin(5t)) + (∂u/∂y)*(5e^(2s)cos(5t)) EQN(2)

Square equation 1 and equation 2, to make it look like in the equation you are given, it requires a fair bit of algebra, but you can get there in the end

(∂u/∂s)^2 = 4{(∂u/∂x)^2*(e^(2s)*cos(5t))^2 + 2*(∂u/∂x)*(∂u/∂y)*(e^(4s)cos(5t)sin(5t))+ (∂u/∂y)^2*(e^(2s)sin(5t))^2}

This can be reduced to:

(∂u/∂s)^2/{4e^(4s)} = (∂u/∂x)^2*(cos^2(5t)) + 2*(∂u/∂x)*(∂u/∂y)((cos(5t)sin(5t))+ (∂u/∂y)^2*(sin^2(5t)) EQN(3)

This can also be done for EQN(2) to give EQN(4)
(∂u/∂t)^2 = 25*{(∂u/∂x)^2*(e^(2s)*sin(5t))^2 - 2*(∂u/∂x)*(∂u/∂y)*(e^(4s)(cos(5t)*sin(5t)+ (∂u/∂t)^2*(e^(2s)* cos(5t))^2}

This can be reduced to:
(∂u/∂t)^2/{25e^(4s)} = (∂u/∂x)^2*(sin^2(5t)) - 2*(∂u/∂x)*(∂u/∂y)*(cos(5t)*sin(5t))+ (∂u/∂t)^2*(cos^2(5t)) EQN(4)

Now we can add equation 3 and equation 4, to give:

1/{4e^(4s)}*25*(∂u/∂s)^2 + 1/{25*e^(4s)}(∂u/∂t)^2 = g(s,t)(∂u/∂s)^2+h(s,t)(∂u/∂t)^2
(∂u/∂x)^2*(cos^2(5t) + sin^2(5t)) + (∂u/∂y)^2*(sin^2(5t) + cos^2(5t)) = g(s,t)(∂u/∂s)^2+h(s,t)(∂u/∂t)^2
(∂u/∂x)^2 + (∂u/∂y)^2 = g(s,t)(∂u/∂s)^2+h(s,t)(∂u/∂t)^2

So g = 1/{4e^(4s)} and h = 1/{25e^(4s)}.

Answer 10

Uhm, well that is awkward, I guess let me re-type this out in word and take a picture.. I guess openstudy does't like me

Answer 11

There we go.

Answer 12

thankuuuuu :D

Answer 13

@chaise can i do something like this:
since we know (du/dx)^2 = g(s,t) * (du/ds)^2, then
g(s,t)
= (du/dx)^2 / (du/ds)^2
= 1/(dx/ds)^2

Answer 14

yes

Answer 15

hmm well lemme think if its really okaay to just do this
(du/dx)^2 = g(s,t) * (du/ds)^2,
u might have to worry about some dependant variables

Answer 16

but solution adds a dy/ds as well, it says
g(s,t)
= 1/((dx/ds)^2 + (dy/ds)^2)

Answer 17

du/ds= du/dx*dx/ds + du/dy*dy/ds
du/dt= du/dx*dx/dt + du/dy*dy/dt

Answer 18

yea, that is true, but how does this explain g(s,t) is not 1/(dx/ds)^2 ?

Answer 19

if u do some algebra ull see that du/dx is dependant on both du/ds and dy/dt

Answer 20

" can i do something like this:
since we know (du/dx)^2 = g(s,t) * (du/ds)^2, then
g(s,t)
= (du/dx)^2 / (du/ds)^2
= 1/(dx/ds)^2

27 minutes ago"

here you are makign an assumption that du/ds and g(s,t) is only coming out of the du/dx

Answer 21

which necessarily doesnt have to be true