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Total differential Problem

  • Thread starter Nanu Nana
  • Start date
  • #1
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Homework Statement


You have two parameters x = 12 and y = 3 set on a machine. The machine generates a function: z (x, y) = 3sin (x ^ 2 + y) y + x ^ 3
Use the total differential of this function in the set point to determine which of the parameters to be set to the most accurate.

Homework Equations


dz = (∂z/∂x) dx + (∂z/∂y) dy
3.Solution
z = 3y sin(x²+y) + x^3
[/B]
dz = (3y cos(x²+y) * 2x + 3x²) dx + (3 sin(x²+y) + 3y cos(x²+y) * 1 + 0) dy
dz = (6xy cos(x²+y) + 3x²) dx + (3 sin(x²+y) + 3y cos(x²+y)) dy

I don't understand why (∂z/∂y) = (3 sin(x²+y) + 3y cos(x²+y) * 1 + 0) dy Where did that zero came from ??? and 1 ??
 

Answers and Replies

  • #2
BvU
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So, NN, where is your attempt at solution ?
What would be your ##\partial z\over \partial y## ?
 
  • #3
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Writing he zero is kind of unnecessary, but it comes from taking the partial derivative of x3 with respect to y. The 1 comes from the y...chain rule.
 
  • #4
BvU
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2019 Award
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Writing he zero is kind of unnecessary, but it comes from taking the partial derivative of x3 with respect to y. The 1 comes from the y...chain rule.
Did you notice the posting in 'homework' ? It is not good for the poster and it is against PF rules to give such a direct answer: it robs the poster from an opportunity to learn from insight.
 
  • #5
95
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So, NN, where is your attempt at solution ?
What would be your ##\partial z\over \partial y## ?
(3 sin(x²+y) + 3y cos(x²+y) * 1 + 0) dy
 
  • #6
95
5
Writing he zero is kind of unnecessary, but it comes from taking the partial derivative of x3 with respect to y. The 1 comes from the y...chain rule.
Oh I see, thank you very much
 
  • #7
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  • #8
32
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Did you notice the posting in 'homework' ? It is not good for the poster and it is against PF rules to give such a direct answer: it robs the poster from an opportunity to learn from insight.
Nope, I didn't notice. I'll likely avoid answering posts in this section from now on.
 

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