Menu Close

Author: Tinku Tara

Question-138336

Question Number 138336 by mr W last updated on 12/Apr/21 Commented by mr W last updated on 12/Apr/21 $${uniform}\:{rod}\:{with}\:{length}\:{L}\:\left(>{s}\right)\:{and}\: \\ $$$${mass}\:{m},\:{cube}\:{with}\:{side}\:{length}\:{s}\:{and}\: \\ $$$${mass}\:{M}.\:{one}\:{end}\:{of}\:{the}\:{rod}\:{is}\:{hinged}.\: \\ $$$${no}\:{friction}.…

Question-72803

Question Number 72803 by mr W last updated on 03/Nov/19 Commented by mr W last updated on 03/Nov/19 $${a}\:{ball}\:{is}\:{thrown}\:{from}\:{point}\:{A}\:{with}\:{a} \\ $$$${speed}\:{u}.\:{after}\:{striking}\:{a}\:{point}\:{on}\:{the} \\ $$$${inclined}\:{plane},\:{it}\:{returns}\:{back}\:{to}\:{its} \\ $$$${original}\:{position}.\:{if}\:{the}\:{coefficient}\:{of}…

Q1-A-balloon-in-the-shape-of-cone-surrmo-surmounted-by-hemispherical-top-the-daimeter-of-balloon-is-equal-to-hieght-of-cone-find-the-rate-of-change-of-volume-of-the

Question Number 7265 by 2402@gmail.com last updated on 20/Aug/16 $${Q}\mathrm{1}.\:{A}\:{balloon}\:{in}\:{the}\:{shape}\:{of}\:{cone}\:{surrmo} \\ $$$$\:\:\:\:\:\:\:\:\:\:{surmounted}\:{by}\:{hemispherical} \\ $$$$\:\:\:\:\:\:\:\:\:{top}\:.\:{the}\:{daimeter}\:{of}\:{balloon}\:{is} \\ $$$$\:\:\:\:\:\:\:{equal}\:{to}\:{hieght}\:{of}\:{cone}\:.\:{find}\:{the} \\ $$$$\:\:\:\:\:{rate}\:{of}\:{change}\:{of}\:{volume}\:{of}\:{the} \\ $$$$\:\:\:{balloone}\:{with}\:{respect}\:{to}\:{its}\:{total} \\ $$$$\:\:\:\:{height}\:\:{h}\:. \\ $$ Commented…

Question-7263

Question Number 7263 by Tawakalitu. last updated on 19/Aug/16 Commented by Rasheed Soomro last updated on 20/Aug/16 $${I}\:{have}\:{answered}\:{your}\:{question}#\mathrm{6852} \\ $$$${which}\:{is}\:{very}\:{resembling}\:{to}\:{this}\:{question}. \\ $$$${Please}\:{see}\:{if}\:{you}\:{have}\:{not}\:{seen}\:{yet}. \\ $$ Commented…

Integrate-f-x-y-1-1-x-2-y-2-2-over-the-triangle-with-vertices-0-0-1-0-1-3-after-changing-it-to-polar-form-

Question Number 72796 by Learner-123 last updated on 03/Nov/19 $${Integrate}\:{f}\left({x},{y}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{over} \\ $$$${the}\:{triangle}\:{with}\:{vertices}\:\left(\mathrm{0},\mathrm{0}\right)\:,\left(\mathrm{1},\mathrm{0}\right), \\ $$$$\left(\mathrm{1},\sqrt{\mathrm{3}}\right)\:{after}\:{changing}\:{it}\:{to}\:{polar}\:{form}. \\ $$ Answered by mind is power last…

Question-138329

Question Number 138329 by JulioCesar last updated on 12/Apr/21 Answered by Ñï= last updated on 12/Apr/21 $$\int\frac{{dx}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}=\int\frac{{dx}}{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}}=\int\frac{{dx}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}} \\ $$$$=\int\frac{{csc}^{\mathrm{2}} \mathrm{2}{xd}}{{csc}^{\mathrm{2}}…

Question-7258

Question Number 7258 by Tawakalitu. last updated on 19/Aug/16 Commented by Yozzia last updated on 19/Aug/16 $${y}'=\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}\Rightarrow{y}=\int\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}{dx}. \\ $$$${Let}\:{u}=\mathrm{5}−{x}^{\mathrm{2}} \Rightarrow{du}=−\mathrm{2}{xdx}\Rightarrow{xdx}=\frac{−\mathrm{1}}{\mathrm{2}}{du}. \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−{du}}{\:\sqrt{{u}}}=\frac{−\mathrm{1}}{\mathrm{2}}\int{u}^{−\mathrm{1}/\mathrm{2}} {du}=\frac{−\mathrm{1}}{\mathrm{2}}×\mathrm{2}{u}^{\mathrm{1}/\mathrm{2}}…