Question

How do you derive this formula?

Answer 1

What is the equation in terms in? R? and is g and H constants like gravity and height? if they are, it's not that hard.

Answer 2

yes G is gravity and H is height. I dont know what its in terms of really the question just said to derive that formula.

Answer 3

are you still there?

Answer 4

dv/dr = (1/2)((g/2h)R^2)^(-1/2)((g/H)R)\[\frac{dv}{dR} =\frac{ 1 }{ 2 }((\frac{ g }{ 2H })R ^{2})^{-\frac{ 1 }{ 2 }}((\frac{ g }{ H })R)\]

Answer 5

wait how did you get the last part for (g/H)(R)? I got how you got the derive before that but I dont know how you got that portion that I am talking about

Answer 6

Which works out to be...\[\frac{ dv }{ dR }=\frac{ \frac{ gR }{ H } }{ 2\sqrt{\frac{ gR }{ 2H }} }\]

I'll send an explanation right now.

Answer 7

Yeah I got that part but above you also put in as part of the answer in what you just send me times (g/H)R)

Answer 8

did you do the chain rule and the quotient rule?

Answer 9

Okay now lets assume that that function is...
\[v = \sqrt{f(x)} \]

Because the square root of something is a function this is considers like a function inside of a function, so we can say that this is...

\[v = g(f(x))\]

And when you derive that you take the derivative of the outside function, and everything inside, stays the same, all to 1 less power, then the derivative of the inside. Therefore it becomes...

\[\frac{ dv }{ dR }= g \prime(f(x))\]

then multiplied by the derivative of the inside. So...

\[\frac{ dv }{ dR }= g \prime(f(x))(f \prime(x))\]

Answer 10

No, just chain rule, because if you give values for g and H, it'll be a constant so it's just a funcation to R

Answer 11

Ok but if you are just doing the chain rule shouldnt the answer be