2015-03-20, 17:51 | #1 |
Dec 2003
Hopefully Near M48
2×3×293 Posts |
Conceptual Question about Implicit Partial Differentiation
I think I should already know this, but...
Problem: Given an equation of the form f(x,y,z) = 0, find dz/dx. Procedure: Take the partial derivative of both sides wrt x, using the Chain Rule and dx/dx = 1, dy/dx = 0, and dz/dx as the unknown, which must then be solved for algebraically. My question is: Why is dy/dx = 0? I've realized I've never seen an explanation for it... |
2015-03-20, 18:53 | #2 |
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
1C1D_{16} Posts |
If you mean x and y are independent variables, and z is dependent on x and y (I.e. is a function of x and y), then I just said the answer.
Hint: What is a derivative? a derivative measures how one quantity varies [u]as another does[/u]. so dy/dx is asking "if x varies some amount, how does y vary [u]accordingly[/u]?" Last fiddled with by Dubslow on 2015-03-20 at 18:56 |
2015-03-22, 18:52 | #3 |
May 2013
East. Always East.
11·157 Posts |
Make sure to read Dubslow's non-spoiler text first and give it some thought. Then read his spoiler text and think about it some more.
If you don't have it after a good few minutes of thinking, read my spoiler text. It will benefit you to try to figure it out yourself. f(x,y,z) = 0 is probably a re-jiggered equation z = f[SUB]1[/SUB](x,y). For example, the elevation z of the ground at co-ordinates (x,y). The partial derivative dz/dx is asking you how much z changes when you change x (how steep is the climb / descent when you move along the x direction). This will entirely depend on what your "starting point" is so the answer will depend on both x and y. In this case, how does y change when you move in the x direction (or vice versa, for that matter)? It doesn't. Your y coordinate stays the same if you're only moving in x. That's why that partial derivative is zero. This explanation is an example which applies to any two variables that are independent of each other. Note this is an interesting situation where dy/dx = 0 and dx/dy = 0. In a lot of cases you can more or less safely play with dy and dx like they are just normal values (chain rule: dz/dx = dz/dy * dy/dx; what if you "cancel" the dy's?) but in this case you can't: with dy/dx = 0 it's a trap to then assume dx/dy is undefined (i.e. division by 0). Last fiddled with by TheMawn on 2015-03-22 at 18:53 |
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