Why Can't I Derive r/R as √3 - 1 in This Chemistry Problem?

  • Thread starter Thread starter Puchinita5
  • Start date Start date
  • Tags Tags
    Figure
AI Thread Summary
The discussion centers on a chemistry problem involving the relationship between variables r and R, where the user struggles to derive r/R as √3 - 1. Despite having the correct answer, they cannot simplify their equation to reach this conclusion. They simplify the equation down to 2R^2 = 2Rr + r^2 but are unsure how to proceed. A key suggestion is to treat the equation as a quadratic, leading to the realization that the correct solution for r/R is √3 - 1, while the other solution is unphysical. The user is encouraged to simplify only one side of the equation and then take the square root to find the desired relationship.
Puchinita5
Messages
178
Reaction score
0
simple math i can't figure out!

so I'm trying to do this chemistry problem...i have the answer, but for some reason i can't do the math myself... I've taken calculus but for some reason i just can't get this!

i have from my answer sheet:

(2 R)^2 + (2√2 R)^2 = (2 r + 2 R)^2

therefore
(√3 -1) R = r

r/R = √3 -1 = 0.73

---------------------------------------------

i cannot seem to figure out how to get to r/R ?

when i do the math, i get

4R^2 +8R^2= 4R^2+ 4Rr+ 4Rr+4r^2

which i simplify to

8R^2=8Rr+4r^2

then to 2R^2= 2Rr+r^2

then what?? where on Earth does √3 -1 come from?

please help!
 
Physics news on Phys.org
two variables, simple math that i cannot solve!

Homework Statement



I have been able to figure out this problem until I had to do some math simplification. I have the answer in front of me but i cannot seem to understand how they got to the answer that they did.

(2 R)^2 + (2R√2)^2= (2r + 2R)^2

therefore
(√3 -1) R = r

r/R = √3 -1 = 0.73

Homework Equations





The Attempt at a Solution



I have no idea how the heck they solved for r/R to get √3 -1 = 0.73

the furthest i can simplify the expression is down to 2R^2=2Rr+r^2

what other steps do i need to do?! I'm desperate to figure this silly problem out!
 


If you divide that by R^2, you get

(r/R)^2 + 2 (r/R) - 2 = 0

which is a quadratic equation for r/R with two solutions, √3 -1 and -√3 -1.

If r and R are both positive, the second solution is unphysical.
 


Sometimes it doesn't pay to multiply everything out and combine terms. The variable r appears in only one place in your original equation, and if that pesky 2 in the exponent wasn't there on the RHS, you could easily solve for it. To get rid of the square, you want to take the square root of both sides. So what you want to do is simplify only the left side and then take the square root of both sides of the equation.
 


Please don't double-post your problems!
Puchinita5 said:

Homework Statement



I have been able to figure out this problem until I had to do some math simplification. I have the answer in front of me but i cannot seem to understand how they got to the answer that they did.

(2 R)^2 + (2R√2)^2= (2r + 2R)^2

therefore
(√3 -1) R = r

r/R = √3 -1 = 0.73

Homework Equations





The Attempt at a Solution



I have no idea how the heck they solved for r/R to get √3 -1 = 0.73

the furthest i can simplify the expression is down to 2R^2=2Rr+r^2

what other steps do i need to do?! I'm desperate to figure this silly problem out!
 


(two threads merged)
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top