Mathematical modelling question

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lektor
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Hey,

So, In a question about finding volume in a dome we were

Given that V = pi.k(k^2/r - k^2/3r^3 - r/8)
And K = m/2pi.p.t

Firstly we are asked to obtain an expression for the value of r that maximises the volume of air and rearrange to obtain an equation of the form ar^4 + br^2 + c = 0

V' => r^4/8 + kr^2 - k^2 = 0

So far I am confident that is correct, but next it asks you to solve the equation for r^2 and hence find r, approximate squareroot of 96 to be 10.

Therefore pi^2 = 10

and substituting the value of k into this equation i obtained,

r^4/8 + mr^2/2pi.p.t - m^2/4pi^2.p^2.t^2 = 0

That is where i get stuck and i hope someone can help me out, thnx :\
 
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lektor said:
Hey,

So, In a question about finding volume in a dome we were

Given that V = pi.k(k^2/r - k^2/3r^3 - r/8)
And K = m/2pi.p.t
Is that [itex]V= \pi k(\frac{k^2}{r}-\frac{k^2}{3r^3}- \frac{r}{8}[/itex] and [itex]k= \frac{m}{2\pi pt}[/itex]? If you don't use LaTex, use lots of parentheses. Also do not use "k" and "K" to mean the same thing.

Firstly we are asked to obtain an expression for the value of r that maximises the volume of air and rearrange to obtain an equation of the form ar^4 + br^2 + c = 0

V' => r^4/8 + kr^2 - k^2 = 0

So far I am confident that is correct, but next it asks you to solve the equation for r^2 and hence find r, approximate squareroot of 96 to be 10.

Therefore pi^2 = 10
What?? Not the [itex]\pi[/itex] I know! "Approximate square root of 96 to be 10"? Why?

and substituting the value of k into this equation i obtained,

r^4/8 + mr^2/2pi.p.t - m^2/4pi^2.p^2.t^2 = 0

That is where i get stuck and i hope someone can help me out, thnx :\
You were told to solve for r2 first- use the quadratic formula. That's fairly straight forward and gives a relatively simple formula for r2. Since you still have unknowns m, p, t in the formula, I see no reason to "approximate" [itex]\pi[/itex] by 10.
 
lektor said:
Hey,

So, In a question about finding volume in a dome we were

Given that V = pi.k(k^2/r - k^2/3r^3 - r/8)
And K = m/2pi.p.t
Is that [itex]V= \pi k(\frac{k^2}{r}-\frac{k^2}{3r^3}- \frac{r}{8}[/itex] and [itex]k= \frac{m}{2\pi pt}[/itex]? If you don't use LaTex, use lots of parentheses. Also do not use "k" and "K" to mean the same thing.

Firstly we are asked to obtain an expression for the value of r that maximises the volume of air and rearrange to obtain an equation of the form ar^4 + br^2 + c = 0

V' => r^4/8 + kr^2 - k^2 = 0

So far I am confident that is correct, but next it asks you to solve the equation for r^2 and hence find r, approximate squareroot of 96 to be 10.

Therefore pi^2 = 10
What?? Not the [itex]\pi[/itex] I know! "Approximate square root of 96 to be 10"? Why?

and substituting the value of k into this equation i obtained,

r^4/8 + mr^2/2pi.p.t - m^2/4pi^2.p^2.t^2 = 0

That is where i get stuck and i hope someone can help me out, thnx :\
You were told to solve for r2 first- use the quadratic formula that gives a fairly straight forward expression for r. Since you still have unknowns m, p, t, I see no reason for "approximating" [itex]\pi[/itex] by 10. You aren't going to get a numerical answer anyway.