Partial F/ Partial T F= (x,y) x and y = functs of s and t Only step 1 needed

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Homework Help Overview

The discussion revolves around finding the partial derivative of a function f(x,y) with respect to t, where f is defined as f(x,y) = Sqrt[x^4 + y], and both x and y are functions of s and t. Participants are exploring the chain rule in the context of multivariable calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the chain rule, questioning the inclusion of variables in their calculations. Some express uncertainty about their previous methods and seek clarification on the correct approach to differentiate f with respect to t.

Discussion Status

There is an ongoing exploration of the correct formula for the partial derivative, with some participants confirming the correct application of the chain rule. Multiple interpretations of the problem are being discussed, and guidance has been offered regarding the differentiation process.

Contextual Notes

Participants are reflecting on their previous test experiences and the constraints of homework rules that allow for redoing problems. There is an acknowledgment of potential confusion due to fatigue and the complexity of the functions involved.

Yezman
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If f(x,y) = Sqrt[x^4 + y] and x = s^2 + t^2 + 1 and y = t^2 +t*Cos(2s)

How do i find

Partial f
--------
Partial tI made the tree diagram how f depends on x and y, which both depend on s and t... so on my test I said it was

(Partials of all of these)f/x*x/s*x/t + f/y*y/s*y/tLooking back now I wonder why I put the s's in there, seems like I don't need them. (this was from a recent test and we get to redo 1 problem). My new idea is

f/x*x/t + f/x*x/t ? That seems to simple though

Can someone here point in the right direction?

Thanks
 
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Think of this as:

[tex]f(x(s),y(t)) = \sqrt{(x(s,t))^4 + y(s,t)}\,,\text{ where } x(s,t) = s^2 + t^2 + 1\text{ and }y(s,t) = t^2 +t\,\,\cos(2s)[/tex]
 
Yezman said:
If f(x,y) = Sqrt[x^4 + y] and x = s^2 + t^2 + 1 and y = t^2 +t*Cos(2s)

How do i find

Partial f
--------
Partial t


I made the tree diagram how f depends on x and y, which both depend on s and t... so on my test I said it was

(Partials of all of these)


f/x*x/s*x/t + f/y*y/s*y/t


Looking back now I wonder why I put the s's in there, seems like I don't need them. (this was from a recent test and we get to redo 1 problem). My new idea is

f/x*x/t + f/x*x/t ? That seems to simple though

Can someone here point in the right direction?

Thanks

If what you mean by that last equation is

f'(t) = (fx)*x'(t) + (fy)*y'(t)

that is the correct formula to work the problem
 
Is that an f prime? I am looking for ∂f/∂t

I re-read the section (stewart Calc)

Wouldn't it just be


∂f/∂x* ∂x/∂t + ∂f/∂y* ∂y/∂t ?
 
Yezman said:
Is that an f prime? I am looking for ∂f/∂t

I re-read the section (stewart Calc)

Wouldn't it just be


∂f/∂x* ∂x/∂t + ∂f/∂y* ∂y/∂t ?

Yes, exactly. I didn't notice that x and y were also functions of s in addition to t.
 
Ok... Idk what i was thinking when i had to multiply three things at a time (lack of sleep probably)Just to check... if you had

(I made this up so something might be weird)
∂f/∂r = ?

f(x,y,z) & x =(s,t,r) & y =(s,t,r) & z =(s,t,r)

∂f/∂r = ∂f/∂x*∂x/∂r + ∂f/∂y*∂y/∂r + ∂f/∂z*∂z/∂r ?
 
Yezman said:
Ok... Idk what i was thinking when i had to multiply three things at a time (lack of sleep probably)


Just to check... if you had

(I made this up so something might be weird)
∂f/∂r = ?

f(x,y,z) & x =F(s,t,r) & y =G(s,t,r) & z =H(s,t,r)

∂f/∂r = ∂f/∂x*∂x/∂r + ∂f/∂y*∂y/∂r + ∂f/∂z*∂z/∂r ?

Yes, you have the idea. I inserted functions for you.
 

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