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

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In summary: So it would be:∂f/∂r = ∂f/∂F*∂F/∂r + ∂f/∂G*∂G/∂r + ∂f/∂H*∂H/∂rIn summary, the conversation discusses finding the partial derivative of f with respect to t using the tree diagram method. The correct formula is ∂f/∂x* ∂x/∂t + ∂f/∂y* ∂y/∂t and in a hypothetical scenario with three variables, the formula would be ∂f/∂F*∂F/∂r + ∂f
  • #1
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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  • #2
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]
 
  • #3
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
 
  • #4
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 ?
 
  • #5
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.
 
  • #6
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 ?
 
  • #7
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.
 

1. What is "Partial F/ Partial T" in this context?

In this context, "Partial F/ Partial T" refers to the partial derivative of the function F with respect to the variable T. This means taking the derivative of F while holding all other variables constant.

2. What does "F= (x,y)" mean?

The notation "F= (x,y)" means that the function F is defined in terms of two variables, x and y. This means that the output of F will depend on the values of x and y.

3. What does it mean for x and y to be functions of s and t?

This means that the variables x and y are defined in terms of the variables s and t. In other words, the values of x and y are dependent on the values of s and t.

4. Why is only step 1 needed in this situation?

In this situation, only step 1 is needed because we are only interested in finding the partial derivative of F with respect to T. Since all other variables are held constant, we do not need to perform any additional steps.

5. Can you provide an example of this concept in action?

Sure, let's say we have the function F(x,y) = 2x + 3y. If we want to find the partial derivative of F with respect to x, we would use the notation "Partial F/ Partial x" and the result would be 2. This means that for any given value of y, the rate of change of F with respect to x is always 2.

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