Partial derivatives of implicitly defined functions

[compliant]

Homework Statement

If the equations

x^2 - 2(y^2)(s^2)t - 2st^2 = 1
x^2 + 2(y^2)(s^2)t + 5st^2 = 1

define s and t as functions of x and y, find \partial^2 t / \partial y^2

The Attempt at a Solution

Equating the two, we get 4y^2*s^2*t = -7s*t^2. My main problem is, as simple as this sounds, how do I implicitly differentiate a single term with three variables? (i.e. 4y^2*s^2*t) Because if I can implicitly differentiate it in terms of y, I can probably find \partial t / \partial y in terms of the other partial derivatives.

 

[djeitnstine]

Partial differentiation is not implicit differentiation if that's what you mean. You simply treat all other variables as constant. In this case you get 8y*s^2 etc...

 

[compliant]

But how would I get the partial derivative of t wrt y from there? I mean, aren't we assuming t is a function of x and y (and with s as well?) ?So let me try this out:

8y*s^2*t + 4y^2(2 \partial s / \partial y)t + 4y^2*s^2*(\partial t / \partial y) = -7(\partial s / \partial y)*t^2 - 7s(2 \partial t / \partial y)

 

[gabbagabbahey]

But how would I get the partial derivative of t wrt y from there? I mean, aren't we assuming t is a function of x and y (and with s as well?) ?

Don't combine your two original equations; instead differentiate each one with respect to y--- you will end up with two equations, each involving \frac{\partial s}{\partial y} and \frac{\partial t}{\partial y}, which you can then solve for \frac{\partial t}{\partial y}.

 

[compliant]

Don't combine your two original equations; instead differentiate each one with respect to y--- you will end up with two equations, each involving \frac{\partial s}{\partial y} and \frac{\partial t}{\partial y}, which you can then solve for \frac{\partial t}{\partial y}.

Sounds like a plan. So let me try this again...[-4ys^2t - 2y^2(2s\frac{\partial s}{\partial y})t - 2y^2 s^2(\frac{\partial t}{\partial y})] + [-2(\frac{\partial s}{\partial y})t^2 - 2s(2 \frac{\partial t}{\partial y})] = 1

[4ys^2t + 2y^2(2s\frac{\partial s}{\partial y})t + 2y^2 s^2(\frac{\partial t}{\partial y})] + [5(\frac{\partial s}{\partial y})t^2 + 5s(2 \frac{\partial t}{\partial y})] = 1[3(\frac{\partial s}{\partial y})t^2 + 3s(2 \frac{\partial t}{\partial y})] = 2

Is there a way to eliminate the \frac{\partial s}{\partial y} term?

 

[gabbagabbahey]

Sounds like a plan. So let me try this again...


[-4ys^2t - 2y^2(2s\frac{\partial s}{\partial y})t - 2y^2 s^2(\frac{\partial t}{\partial y})] + [-2(\frac{\partial s}{\partial y})t^2 - 2s(2 \frac{\partial t}{\partial y})] = 1

[4ys^2t + 2y^2(2s\frac{\partial s}{\partial y})t + 2y^2 s^2(\frac{\partial t}{\partial y})] + [5(\frac{\partial s}{\partial y})t^2 + 5s(2 \frac{\partial t}{\partial y})] = 1

Careful, when you differentiate an equation, you need to differentiate both sides of it.

\frac{\partial}{\partial y} (1)\neq 1

Is there a way to eliminate the \frac{\partial s}{\partial y} term?

Sure...you have 2 equations and 2 unkowns; think back to your high school algebra

 

[compliant]

Careful, when you differentiate an equation, you need to differentiate both sides of it.

\frac{\partial}{\partial y} (1)\neq 1



Sure...you have 2 equations and 2 unkowns; think back to your high school algebra

Ok, so now I have

3t^2 \frac {\partial s}{\partial y} = -6s \frac{\partial t}{\partial y}
\frac {\partial s}{\partial y} = \frac{-2s}{t^2} \frac{\partial t}{\partial y}

I sub this back into the first equations to get \frac{\partial t}{\partial y}. However, I get a different value for \frac{\partial t}{\partial y} when I plug it into the second equation. If I'm going to write the answer, would I write them both down? Because I feel there really should only be one answer...