A calculus how do they get there from here question

[kennykroot]

A calculus "how do they get there from here" question

I'm trying to understand this step in my physics book:

From this (K is a constant.)

(1) $${\frac {{\it dp}}{{\it dy}}}=Kp$$

To this

(2) $${\frac {{\it dp}}{p}}=K{\it dy}$$

and then to this

(3) $$\int \!{p}^{-1}{dp}=\int \!K{dy}$$

How do they get from (1) to (2) How is it that they can break up the notation for the derivative of p with respect to y and treat it like a fraction where dp is divided by dy?

 

[Dick]

It's a sloppy but handy shorthand. More correctly you get (1/p)*(dp/dy)=K. Now integrate both sides dy and do a change of variable y->p(y) on the left side.

 

[HallsofIvy]

I'm trying to understand this step in my physics book:

From this (K is a constant.)

(1) $${\frac {{\it dp}}{{\it dy}}}=Kp$$

To this

(2) $${\frac {{\it dp}}{p}}=K{\it dy}$$

and then to this

(3) $$\int \!{p}^{-1}{dp}=\int \!K{dy}$$

How do they get from (1) to (2) How is it that they can break up the notation for the derivative of p with respect to y and treat it like a fraction where dp is divided by dy?

Well, my guess would be that they actually completed Calculus I and so had learned about "differentials" as opposed to "derivatives"!

 

[kennykroot]

Dick,

Thanks very much.

Kenny