Need help, URGENT maths question: (5 Viewers)

[InteGrand]

Wtf?

I think it's mainly the x and y parts that are confusing you guys? If you want, you can rewrite the partials of f as f₁ and f₂, because that's what they're referring to with those partials, it's just a bit of abuse of notation I guess what they've done.

 

[InteGrand]

Can't you just do it like this?:

Well u and v instead of x and y, with u = tx, and v = ty. And then note that when you're saying ∂ƒ/∂u, what you're referring to is ƒ₁, etc.

It might help if you look at a concrete example of a function ƒ. E.g. ƒ(x,y) = x³y + sin(x + y).

 

[1008]

Well u and v instead of x and y, with u = tx, and v = ty. And then note that when you're saying ∂ƒ/∂u, what you're referring to is ƒ₁, etc.

It might help if you look at a concrete example of a function ƒ. E.g. ƒ(x,y) = x³y + sin(x + y).

Wait, wat. Would my method be valid, or have I got it all wrong?

 

[InteGrand]

Wait, wat. Would my method be valid, or have I got it all wrong?

Well you wrote something like ∂x/∂t, but x and t are independent of each other (so we wouldn't have ∂x/∂t = x. Anyway, if ∂x/∂t equalled x, that'd be like the population growth formulas y' = ky, so ∂x/∂t = x would basically imply something like x = e^t. Like usually when we take partial derivatives, the answer isn't in terms of the original variable we were differentiating unless there's some special relationship between the variables, which there isn't here.).

What you meant was writing u = tx, v = ty, and then f(tx,ty) = f(u,v). Basically the method I showed earlier. Then we can say stuff like ∂u/∂t = x.

 

[InteGrand]

Can't you just do it like this?:

OK I think I get what you're doing now. When you're saying the derivative of ƒ wrt t would be ∂ƒ/∂x ∂x/∂t + etc., by ∂ƒ/∂x, you mean "partial of f wrt its first variable", right? And then ∂x/∂t, you mean "partial of the thing in the first variable's position wrt t", right (which is x, yeah)?

It's just a bit notationally suspect or confusing to write it like that (with x's) because the x appears inside the first variable too. So it's better to just give that thing in the first variable position a name (e.g. u), and so a similar thing for the thing in the second variable's position (e.g. v), and then write something like ƒ₁ to refer to the partial of ƒ wrt its first variable to avoid confusion with letters.

 

[1008]

OK I think I get what you're doing now. When you're saying the derivative of ƒ wrt t would be ∂ƒ/∂x ∂x/∂t + etc., by ∂ƒ/∂x, you mean "partial of f wrt its first variable", right? And then ∂x/∂t, you mean "partial of the thing in the first variable's position wrt t", right (which is x, yeah)?

It's just a bit notationally suspect or confusing to write it like that (with x's) because the x appears inside the first variable too. So it's better to just give that thing in the first variable position a name (e.g. u), and so a similar thing for the thing in the second variable's position (e.g. v), and then write something like ƒ₁ to refer to the partial of ƒ wrt its first variable to avoid confusion with letters.

Yeah, it's just a simple application of the chain rule...I guess

 

[InteGrand]

I think the confusion was more of a notational confusion rather than a mathematical confusion.

 

[1008]

OK I think I get what you're doing now. When you're saying the derivative of ƒ wrt t would be ∂ƒ/∂x ∂x/∂t + etc., by ∂ƒ/∂x, you mean "partial of f wrt its first variable", right? And then ∂x/∂t, you mean "partial of the thing in the first variable's position wrt t", right (which is x, yeah)?

It's just a bit notationally suspect or confusing to write it like that (with x's) because the x appears inside the first variable too. So it's better to just give that thing in the first variable position a name (e.g. u), and so a similar thing for the thing in the second variable's position (e.g. v), and then write something like ƒ₁ to refer to the partial of ƒ wrt its first variable to avoid confusion with letters.

Could you please give an example with ƒ₁?

 

[1008]

OK I think I get what you're doing now. When you're saying the derivative of ƒ wrt t would be ∂ƒ/∂x ∂x/∂t + etc., by ∂ƒ/∂x, you mean "partial of f wrt its first variable", right? And then ∂x/∂t, you mean "partial of the thing in the first variable's position wrt t", right (which is x, yeah)?

It's just a bit notationally suspect or confusing to write it like that (with x's) because the x appears inside the first variable too. So it's better to just give that thing in the first variable position a name (e.g. u), and so a similar thing for the thing in the second variable's position (e.g. v), and then write something like ƒ₁ to refer to the partial of ƒ wrt its first variable to avoid confusion with letters.

Would this be valid then?

 

[leehuan]

Tsk. Fair enough but I wouldn't get away with it in the exam because that's not what they wanted my final answer to be in.

Plus my lecturer said it'd be a risk to write that unless I explicitly define what I had. But yeah it was mainly a notational error, because that swapping u out for x was just ??? when I first looked at it.

 

[InteGrand]

Could you please give an example with ƒ₁ please?

E.g. Say ƒ(u,v) = u^3 + v^3. So ƒ₁(x,y) = 3u^2 (i.e. "3 times the square of the first variable in f).

Then ƒ(tx,ty) = (tx)^3 + (ty)^3 = t^3 x^3 + t^3 y^3 = t^3 (x^3 + y^3) (incidentally, this is an example of a function ƒ that is homogeneous of degree 3).

Now, ƒ₁(tx,ty) = 3(tx)^2. We wouldn't say this is (∂/∂x)(ƒ(tx,ty)) though, since that'd be 3x^2 * t^3, i.e. partial of the expression " t^3 x^3 + t^3 y^3" wrt x.

So basically, ƒ₁ refers to the partial of the actual function ƒ wrt its first variable, so ƒ₁(tx,ty) is the value of this partial derivative function (the partial derivative is itself a function remember) evaluated at the point (tx,ty).

Meanwhile, (∂/∂x)(ƒ(tx,ty)) means first evaluate ƒ at the point (tx,ty), and then differentiate this expression wrt x.

A kind of single variable analog that may be more familiar from HSC would be as follows.

Say ƒ(x) = sin(x).

Then ƒ'(3x) does not mean (d/dx) sin(3x).

It means evaluate ƒ' at the point 3x.

Now, ƒ' at any point x is ƒ'(x) = cos(x).

So ƒ'(3x) is cos(3x).

So ƒ' refers to the actual derivative of ƒ, whereas (d/dx)(ƒ(3x)) would mean evaluate ƒ at 3x and differentiate the resulting expression wrt x. These are two different things (linked to each other by the chain rule of course).

 

[InteGrand]

Tsk. Fair enough but I wouldn't get away with it in the exam because that's not what they wanted my final answer to be in.

Plus my lecturer said it'd be a risk to write that unless I explicitly define what I had. But yeah it was mainly a notational error, because that swapping u out for x was just ??? when I first looked at it.

What can't you get away with? Writing something like ƒ₁?

 

[leehuan]

What can't you get away? Writing something like ƒ₁?

I mean, they explicitly wanted the answer in that form.

Because I have no idea how fussy UNSW are over exactly identical answers to an equivalent, I don't want to risk leaving it in f₁ notation

 

[InteGrand]

I mean, they explicitly wanted the answer in that form.

Because I have no idea how fussy UNSW are over exactly identical answers to an equivalent, I don't want to risk leaving it in f₁ notation

Well they're just two notations for the same thing. If you get the answer out in the subscript notation form and get penalised, it's kind of like penalising someone for writing y' instead of dy/dx (assuming that you are allowed to use ƒ₁ notation, which should be the case at least if you write a line saying what it means, but I guess you should check with your lecturer for confirmation).

 

[leehuan]

Yeah my lecturer said it was a risk.

But the main problem I had at the time was basically why it was permissible to swap partial f/partial uwith partial f/partial x

 

[1008]

Would this be valid then?

This

 

[InteGrand]

This

Yeah that's valid (remember to say we let t = 1). But do you know why it's valid, i.e. why ∂ƒ/∂x and ∂ƒ/∂u are the same (well I guess this already got answered as leehuan raised this question)?

(Also note here dƒ/dt means dz/dt, where z = ƒ(tx,ty) and x and y are considered fixed.)

 

[leehuan]

Yeah that's valid. But do you know why it's valid, i.e. why ∂ƒ/∂x and ∂ƒ/∂u are the same?

Unfortunately I don't. Please explain; I'm forgetting something really important here.

 

[InteGrand]

Unfortunately I don't. Please explain; I'm forgetting something really important here.

Well it's because the notations are both referring to the partial derivative of ƒ wrt its first designated variable. As seanieg89 said:

Think about what partial f/partial u means, it just means the derivative of the function with respect to the first variable (and you can replace u with x or whatever your favorite greek letter is). You are differentiating f with respect to its first variable and then evaluating it at the coordinates (x,y).

Try to convince yourself that


and


are the same functions, just with different "dummy variables".

Introducing things like u and v can be more hassle than help.

This is also an example of why some people prefer using numerical subscripts for a function to denote differentiation with respect to the j-th variable.

.

 

[leehuan]

Well it's because the notations are both referring to the partial derivative of ƒ wrt its first designated variable. As seanieg89 said:

.

Hm. I see...