Question Number 224432 by mr W last updated on 10/Sep/25
Commented by mr W last updated on 11/Sep/25
$${if}\:{an}\:{object}\:{takes}\:{the}\:{same}\:{time}\:{to} \\ $$$${reach}\:{the}\:{bottom}\:{of}\:{a}\:{bowl},\:{no}\: \\ $$$${matter}\:{at}\:{which}\:{point}\:{it}\:{is}\:{released} \\ $$$${on}\:{the}\:{interior}\:{of}\:{the}\:{bowl},\:{find}\:{the} \\ $$$${shape}\:{of}\:{the}\:{bowl}.\: \\ $$
Commented by mahdipoor last updated on 11/Sep/25
$$\mathrm{Search}\:\::\:\mathrm{isochronous}\:\mathrm{property} \\ $$
Commented by mr W last updated on 11/Sep/25
Commented by fantastic last updated on 11/Sep/25
Answered by mahdipoor last updated on 11/Sep/25
$${dt}=\frac{{ds}}{{v}}=\frac{\sqrt{\left({dy}\right)^{\mathrm{2}} +\left({dx}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}{g}\left({H}−{y}\right)}}=\sqrt{\frac{\mathrm{1}+{g}^{\mathrm{2}} }{\mathrm{2}{g}\left({H}−{y}\right)}}{dy} \\ $$$${G}\left({y}\right)={x}\:\Rightarrow\:\frac{{dG}}{{dy}}=\frac{{dx}}{{dy}}={g}\left({y}\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:{T}} \sqrt{\mathrm{2}{g}}.{dt}=\int_{\mathrm{0}} ^{\:{H}} \sqrt{\frac{\mathrm{1}+{g}^{\mathrm{2}} }{{H}−{y}}}{dy}={cte}\:\:\left({T}\:\propto!\:{path}\right) \\ $$$$\Rightarrow\forall\mathrm{0}\leqslant{H}\leqslant{limit}\::\:\int_{\mathrm{0}} ^{\:{H}} \sqrt{\frac{\mathrm{1}+{g}^{\mathrm{2}} }{{H}−{y}}}{dy}={cte}\: \\ $$$${what}\:{is}\:{function}\:{G}\left({y}\right)\:\:??? \\ $$