Question

find dr/du, dr/dv, dr/dt
r=xlny
x=3u+vt
y=uvt

Answer 1

I dont understand the question

Answer 2

It may just be me though.

Answer 3

find \[\delta r/\delta u\] \[\delta r/\delta v\] \[\delta r/\delta t\]

Answer 4

does that make it more understandable?

Answer 5

Nope! probably cause its out of my league :D Haven't learned that yet.

Answer 6

haha thanks for trying, how about you @KingGeorge ?

Answer 7

What subject is that? Just curious:) I'm gonna end up taking it im sure

Answer 8

its like.. calc 3? i think.. it depends on where you're taking it.. for my its either math 252, or calc 20c, but i think the general subject is calc 3, specifically, multivariable differential equations, and partial fraction chain rule decomposition

Answer 9

Yup I need calc 3 for my comp science major. Im only in trig right now.

Answer 10

good luck to you! i dunno about you but i'm terrriibbblleee at math, and i have a really bad prof, so i'm struggling to get a D -_-"

Answer 11

Its been my strong subject I guess. I'm almost done with my trig class and I have 90,97,98 test grades so far.

But yea, Professors make the biggest difference.

Answer 12

I have the ones who teach straight from a powerpoint word by word..

Answer 13

for most topics it's not that bad, for math its terrible.. she lectures straight out of the book.. doesn't teach anything conceptual, just tells you how to do stuff, and then doesn't really let us ask questions

Answer 14

Thats dumb haha

Answer 15

first get r in terms of u,v,t
then take derivative with respect to each variable, holding the rest as constants

r=(3u+vt) ln(uvt)
i'll do dr/du, v and t will act as constants
use product rule

dr/du = 3ln(uvt) + (3u+vt)(1/u)

Answer 16

You're a smart cow!:D

Answer 17

let me tell you what i got, cuz it's wrong, and maybe you could see what i'm doing wrong?
for dr/du
dr/dx = lny
dy/du = 3
dr/dy = x/y
dx/du = xvt

3ln(uvt) + (3u+vt)/v

Answer 18

and, this is in the chapter of the chain rule, so i think i have to use it haha

Answer 19

@eliassaab i'm not quite sure how you got o.O

Answer 20

oh i see, well you get same answer
how do you end up with a v on bottom?

Answer 21

dx/du = 3
dy/du = vt

Answer 22

so using the formula dr/du=
(dr/dx)(dx/du) + (dr/dy)(dy/du)

dr/dy = x/y so plugging that in you get
[(3u+vt) / uvt] x [vt] so the vt on the bottom cancels, i think?

Answer 23

sorry dy/du is vt

Answer 24

correct, leaving u on the bottom

Answer 25

sorry that was a typo!! -_- i have a hard time telling my u's and v's apart...
3ln(uvt) + (3u+vt)/u

Answer 26

so did we get the same wrong answer ?

Answer 27

the right answer is
3ln(uvt)+3+vt/u and i'm not sure whats different, what i did wrong

Answer 28

no we are correct, they just split the fraction up

(3u+vt)/u = 3 + vt/u

Answer 29

but there's a u at the top in the first part, and there sin't one in the second.. i'm so confused

Answer 30

eliasaab, how are you obtaining this equation by substitution of x,y
r=lmt2uv2+3lmtu2v

Answer 31

itz, ??
3u/u = 3

Answer 32

but.. its part of the equation.. are you splitting it like 3u/u + vt/u ?! you can do that?

Answer 33

Replacing x and y in r, we obtain
\[ r=3 u \ln (t u v)+t v \ln (t u v)
\]

\[
\frac {\partial r }{du}=\frac{t v}{u}+3 \ln (t u v)+3
\]
\[
\frac {\partial r }{dv}=t \ln (t u v)+t+\frac{3 u}{v}
\]
\[
\frac {\partial r }{dt}=v \ln (t u v)+\frac{3 u}{t}+v
\]

Answer 34

could you explain how you did the derivations for that? if you don't mind?

Answer 35

You have r as a function of u , v and t,
You take the partial derivatives of r with respect to u, then to v, then to t