Prob w/ transposing formula: 5y = (1/5) sqrt[(r/h + h/2)/(r/(3h) - h/(5h))]

[Vulcan]

Trying to transpose this formula to make r the subject.

. . . . .\(\displaystyle 5y\, =\, \dfrac{1}{5}\, \sqrt{\strut \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\,}\)

Got this far but don't think I am heading on the correct path.

. . . . .\(\displaystyle \sqrt{\strut 25y\,}\, =\, \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\)

. . . . .\(\displaystyle \sqrt{\strut 25y\,}\, =\, \left(\dfrac{r}{h}\, +\, \dfrac{h}{2}\right)\, \left(\dfrac{3h}{r}\, -\, \dfrac{5h}{h}\right)\)

. . . . .\(\displaystyle \sqrt{\strut 25y\,}\, =\, \left(\dfrac{r}{h}\, \times\, \dfrac{3h}{r}\right)\, -\, \left(\dfrac{r}{h}\, \times\, \dfrac{5h}{h}\right)\, +\, \left(\dfrac{h}{2}\, \times\, \dfrac{3h}{r}\right)\, -\, \left(\dfrac{h}{2}\, \times\, \dfrac{5h}{h}\right)\)

Any help would be most welcome!

 

[Ishuda]

Trying to transpose this formula to make r the subject.

. . . . .\(\displaystyle 5y\, =\, \dfrac{1}{5}\, \sqrt{\strut \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\,}\)

Got this far but don't think I am heading on the correct path.

. . . . .\(\displaystyle \sqrt{\strut 25y\,}\, =\, \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\)

. . . . .\(\displaystyle \sqrt{\strut 25y\,}\, =\, \left(\dfrac{r}{h}\, +\, \dfrac{h}{2}\right)\, \left(\dfrac{3h}{r}\, -\, \dfrac{5h}{h}\right)\)

. . . . .\(\displaystyle \sqrt{\strut 25y\,}\, =\, \left(\dfrac{r}{h}\, \times\, \dfrac{3h}{r}\right)\, -\, \left(\dfrac{r}{h}\, \times\, \dfrac{5h}{h}\right)\, +\, \left(\dfrac{h}{2}\, \times\, \dfrac{3h}{r}\right)\, -\, \left(\dfrac{h}{2}\, \times\, \dfrac{5h}{h}\right)\)

Any help would be most welcome!

HINT:

\(\displaystyle a\, =\, \dfrac{1}{e}\,\sqrt{\strut \dfrac{a r \,+\, b}{cr\, +\, d}\,}\)

Multiply both sides by e

\(\displaystyle ae\, =\, f\, =\, \sqrt{\strut \dfrac{a r \,+\, b}{cr\,+\,d}\,}\)

Square both sides

\(\displaystyle f^2\, =\, \dfrac{a r\, +\, b}{cr\,+\,d}\)

Now multiply through by cr+d and simplify.

 

[stapel]

Trying to transpose this formula to make r the subject.

. . . . .\(\displaystyle 5y\, =\, \dfrac{1}{5}\, \sqrt{\strut \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\,}\)

Got this far but don't think I am heading on the correct path.

. . . . .\(\displaystyle \sqrt{\strut 25y\,}\, =\, \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\)

By what reasoning (what steps) did you get from the original equation to the line above? It appears that you multiplied through by 5:

. . . . .\(\displaystyle 5y\, =\, \dfrac{1}{5}\, \sqrt{\strut \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\,}\)

. . . . .\(\displaystyle 5(5y)\, =\, 5\left(\dfrac{1}{5}\right)\, \sqrt{\strut \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\,}\)

. . . . .\(\displaystyle 25y\, =\, \sqrt{\strut \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\,}\)

But then the square root somehow magically changed sides...? :shock:

If one were instead to square both sides -- the standard procedure for solving radical equations (here) -- one would instead arrive at this step:

. . . . .\(\displaystyle \left(25y\right)^2\, =\, \left(\sqrt{\strut \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\,}\right)^2\)

. . . . .\(\displaystyle 625y^2\, =\, \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\)

This does not resemble what you've posted, however. Also, going from here:

. . . . .\(\displaystyle \dfrac{\dfrac{r}{h}\, +\, \dfrac{h}{2}}{\dfrac{r}{3h}\, -\, \dfrac{h}{5h}}\)

...to here:

. . . . .\(\displaystyle \left(\dfrac{r}{h}\, +\, \dfrac{h}{2}\right)\, \left(\dfrac{3h}{r}\, -\, \dfrac{5h}{h}\right)\)

...would require that 1/(1/a + 1/b) somehow equalled a + b, which of course is not true:

. . . . .\(\displaystyle \dfrac{1}{\left(\dfrac{1}{a}\, +\, \dfrac{1}{b}\right)}\)

. . . . .\(\displaystyle \dfrac{1}{\left(\dfrac{b}{ab}\, +\, \dfrac{a}{ab}\right)}\)

. . . . .\(\displaystyle \dfrac{1}{\left(\dfrac{b\, +\, a}{ab}\right)}\)

. . . . .\(\displaystyle \dfrac{\dfrac{1}{1}}{\left(\dfrac{b\, +\, a}{ab}\right)}\)

. . . . .\(\displaystyle \left(\dfrac{1}{1}\right)\, \left(\dfrac{ab}{b\, +\, a}\right)\)

. . . . .\(\displaystyle \dfrac{ab}{b\, +\, a}\)

 

[Vulcan]

Redo?

Thanks for your input guys, a couple of brain freezes in what I put originally. I have had another go and have gone over it a couple of times but still think I've missed something?

 

[ksdhart2]

Everything you've done so far seems fine to me, but there's a bit more simplification you can do. Take another look at the denominator. Is there, perhaps, a common term? What happens if you factor that out? Then, don't forget to consider the coefficients out front of the numerator and the denominator (i.e. What's the GCF of 10 and 15?)

 

[Vulcan]

Solved!

There are 2 errors in your attempt:

after squaring both sides: should be 625y^2

after "get both terms with r on LHS:
30r should be -30r, -3750yh^2 should be 3750yh^2

Try again: you should end up with:
r = [3h(250y^2 + h)] / [2(625y^2 - 3)]

Thanks Denis, I knew something wasn't right but sometimes the more you look at it , the worse it gets. There were another couple of errors I spotted as well, but at the end of the day i got there!