1. ## Comparing two formulas...

I have two formulas for a velocity displayed below

Vn = 175 / ( p^0.43 ) eqn 1

and

Ve = 135 / ( p^0.5 ) eqn 2

I know that for any value of p i choose, Equation 2 will give me a lower velocity.

I know this from a simple trial and error method, by substituting different values for p.

Besides this trial and error method, is there any way that i can prove that Ve<Vn all the time?

2. Originally Posted by Rush
I have two formulas for a velocity displayed below

Vn = 175 / ( p^0.43 ) eqn 1

and

Ve = 135 / ( p^0.5 ) eqn 2

I know that for any value of p i choose, Equation 2 will give me a lower velocity.

I know this from a simple trial and error method, by substituting different values for p.

Besides this trial and error method, is there any way that i can prove that Ve<Vn all the time?
If Vn> Ve, then $\displaystyle \frac{175}{p^{0.43}}> \frac{135}{p^{0.5}}$

Assuming p is positive, that is the same as $\displaystyle 175p^{0.5}> 135p^{0.43}$ which leads to [tex]\frac{p^{0.5}{p^.43}= p^{.07}> \frac{135}{175}= \frac{4}{5}[tex] so that $\displaystyle p> \left(\frac{4}{5}\)^{25}= .0038$, approximately. So this is NOT true for [b]all[b] p but is true for p> .004.

3. Originally Posted by HallsofIvy
If Vn> Ve, then $\displaystyle \frac{175}{p^{0.43}}> \frac{135}{p^{0.5}}$

Assuming p is positive, that is the same as $\displaystyle 175p^{0.5}> 135p^{0.43}$ which leads to [tex]\frac{p^{0.5}{p^.43}= p^{.07}> \frac{135}{175}= \frac{4}{5}[tex] so that $\displaystyle p> \left(\frac{4}{5}\)^{25}= .0038$, approximately. So this is NOT true for all p but is true for p> .004.

Sorry, but i cannot read ur equations properly,

can u retype it for me?
(esp the below line)

[b][b][b][b][tex]\frac{p^{0.5}{p^.43}= p^{.07}> \frac{135}{175}= \frac{4}{5}[tex] so that $\displaystyle p> \left(\frac{4}{5}\)^{25}= .0038$, approximately. So this is NOT true for all p but is true for p> .004.

4. i specifically cannot understand how u got, p^0.5 x p^0.43 = p^0.07