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Question Number 179593 by mathlove last updated on 30/Oct/22

pleas proof the elipse environment formullah p=π(√(2a^2 +2b^2 ))

$${pleas}\:{proof}\:{the}\:{elipse}\:{environment} \\ $$$${formullah} \\ $$$${p}=\pi\sqrt{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} } \\ $$

Commented by mr W last updated on 31/Oct/22

what do you mean with “elipse environment”? do you mean the perimeter of elipse?

$${what}\:{do}\:{you}\:{mean}\:{with}\:``{elipse} \\ $$$${environment}''?\:{do}\:{you}\:{mean}\:{the} \\ $$$${perimeter}\:{of}\:{elipse}? \\ $$

Commented by mathlove last updated on 31/Oct/22

yes

$${yes} \\ $$

Commented by mathlove last updated on 31/Oct/22

ellipse

$${ellipse} \\ $$

Commented by mr W last updated on 31/Oct/22

then you can′t prove, because it′s not true. in fact there is no closed formula to calculate the perimeter of an elipse.

$${then}\:{you}\:{can}'{t}\:{prove},\:{because}\:{it}'{s}\:{not} \\ $$$${true}.\:{in}\:{fact}\:{there}\:{is}\:{no}\:{closed}\:{formula} \\ $$$${to}\:{calculate}\:{the}\:{perimeter}\:{of}\:{an}\:{elipse}. \\ $$

Commented by mathlove last updated on 31/Oct/22

then what is the ellips environment formullah

$${then}\:{what}\:{is}\:{the}\:{ellips}\:{environment}\:{formullah} \\ $$

Commented by mr W last updated on 31/Oct/22

you don′t understand me? i have said, there is no formula for calculating the perimeter of an elipse! i.e. you can′t calculate the perimeter of an elipse exactly through a formula!

$${you}\:{don}'{t}\:{understand}\:{me}?\:{i}\:{have}\:{said}, \\ $$$${there}\:{is}\:{no}\:{formula}\:{for}\:{calculating} \\ $$$${the}\:{perimeter}\:{of}\:{an}\:{elipse}!\:{i}.{e}.\:{you} \\ $$$${can}'{t}\:{calculate}\:{the}\:{perimeter}\:{of}\:{an} \\ $$$${elipse}\:{exactly}\:{through}\:{a}\:{formula}! \\ $$

Commented by mathlove last updated on 31/Oct/22

good thanks

$${good}\:\:{thanks} \\ $$

Commented by MJS_new last updated on 01/Nov/22

for the perimeter of an ellipse we have p=4∫_0 ^(π/2) (√(a^2 cos^2 t +b^2 sin^2 t)) dt which has no exact solution Ramanujan gave this very good approximation p≈π(a+b)(1+((3λ^2 )/(10+(√(4−3λ^2 ))))) with λ=((a−b)/(a+b)) both “easy” formulas p≈π(a+b) p≈π(√(2(a^2 +b^2 ))) are not accurate

$$\mathrm{for}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{an}\:\mathrm{ellipse}\:\mathrm{we}\:\mathrm{have} \\ $$$${p}=\mathrm{4}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\sqrt{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{t}\:+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{t}}\:{dt} \\ $$$$\mathrm{which}\:\mathrm{has}\:\mathrm{no}\:\mathrm{exact}\:\mathrm{solution} \\ $$$$\mathrm{Ramanujan}\:\mathrm{gave}\:\mathrm{this}\:\mathrm{very}\:\mathrm{good}\:\mathrm{approximation} \\ $$$${p}\approx\pi\left({a}+{b}\right)\left(\mathrm{1}+\frac{\mathrm{3}\lambda^{\mathrm{2}} }{\mathrm{10}+\sqrt{\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}} }}\right)\:\mathrm{with}\:\lambda=\frac{{a}−{b}}{{a}+{b}} \\ $$$$ \\ $$$$\mathrm{both}\:``\mathrm{easy}''\:\mathrm{formulas} \\ $$$${p}\approx\pi\left({a}+{b}\right) \\ $$$${p}\approx\pi\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$$\mathrm{are}\:\mathrm{not}\:\mathrm{accurate} \\ $$

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