Question

how to get R in

\[\frac{30}{e} = e^{-5/R} (-\frac{250}{R} + \frac{100}{R^2})\]

Answer 1

Graph them, I suppose...

Answer 2

This is a transcendental function with a linear function... It wouldn't be possible to solve unless you use Newton's Approximation. http://www.math.brown.edu/UTRA/linapprox.html

Answer 3

lol isnt this just algebra

Answer 4

i just need to get R for my diff eq thingy

Answer 5

http://www.wolframalpha.com/input/?i=30%2Fe+%3D+e%5E%28-5%2Fr%29*%28-250%2Fr+%2B+100%2Fr%5E2%29 suggests -27.55 :|

Answer 6

ugh what did i do wrong :/

Answer 7

dyslexic i am today... sorry...

Answer 8

what is the solution for \[\frac{dq}{dt} + \frac{q}{0.02R} = \frac{100}{R}\]

Answer 9

\[q= Ae^{\frac{-50}{R}t}+2\]

Answer 10

Unless q is just a function of R.

Answer 11

how did you get this?

Answer 12

Pretty much the same thing we did before. I reasoned, If q were constant, what would it have to be to get 100/R :: answer 2.

Answer 13

what i got was \[\LARGE q = \frac 2R + Ae^{-\frac{t}{0.02R}}\]

Answer 14

Of course 50t/R = t/0.02R so we got that part the same. But I think you have q/.02R in your equation so if you plug in your answer you will have a R^2 in the denominator.

Answer 15

\[\frac{2}{.02R}=\frac{100}{R}\]
\[\frac{2/R}{.02R}<>\frac{100}{R}\]

Answer 16

Essentially, set A=0 and see if it solves the ODE.

Answer 17

oh lol R cancels out *facepalm*

Answer 18

wait lemme redo this

Answer 19

ugh my end answer is still wrong >.<

Answer 20

okay...how to solve for R in
\[\large \frac{30}{e} = -\frac{3e^{-\frac 5R}}{R}\]

Answer 21

this is where it came from...maybe i just did something wrong

http://openstudy.com/updates/4ffd2fc8e4b00c7a70c5dc86