You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**renjer****Member**- Registered: 2006-04-29
- Posts: 50

If x=y=100 and increasing at the rate of 1 per second, z=200, decreasing at the rate of 2 per second. Is R decreasing or increasing and at what rate?

I don't know how to do this question because R is not defined as R(x,y,z) but rather as 1/R. What do I do with this?

Btw I still don't know how to use the math tags. I tried encasing in

tag but the output was [math?] without my equations.*Last edited by renjer (2006-06-07 15:25:34)*

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,316

name g=g(R) a function that g(R)=1/R . Also g=f(x,y,z)=1/x+1/y+1/z

dg=d(1/R)

as well as

dg=d(1/x+1/y+1/z)

solve

d(1/R)=d(1/x+1/y+1/z) out

Or to be more rigorous:

you need to get R[sub]x[/sub] (

To get R[sub]x[/sub]

first get g[sub]x[/sub] and g[sub]R[/sub]

then using a theorem ( I cannot remember its name)

R[sub]x[/sub]=g[sub]x[/sub]/g[sub]R[/sub]

Similarly you can solve R[sub]y[/sub] and R[sub]z[/sub] out and make a dR

The first approach is Leibniz's approach, the second is traditional differential approach, and they get a same answer, some guy may have proved they are equavalent. Leibniz claimed using d notation may simplify algebrac manuplicion, he was great, wasn't he?

*Last edited by George,Y (2006-06-07 16:35:14)*

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,316

Increasing at 0.24 per second.

**X'(y-Xβ)=0**

Offline

Pages: **1**