Question Number 221050 by MrGaster last updated on 23/May/25 Commented by MrGaster last updated on 23/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 221044 by fantastic last updated on 23/May/25 $${If}\:{V}\:{be}\:{a}\:{function}\:{of}\:{x}\:{and}\:{y},\:{prove}\:{that} \\ $$$$\frac{\partial^{\mathrm{2}} {V}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {V}}{\partial{y}^{\mathrm{2}} }=\frac{\partial^{\mathrm{2}} {V}}{\partial{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}}\:\frac{\partial{V}}{\partial{r}}+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{\partial^{\mathrm{2}} {V}}{\partial\theta^{\mathrm{2}} }, \\ $$$${where}\:{x}={r}\:\mathrm{cos}\:\theta\:,\:{y}={r}\mathrm{sin}\:\theta \\ $$…
Question Number 221047 by universe last updated on 23/May/25 Answered by breniam last updated on 24/May/25 $$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{r}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \left({k}−\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)!}−\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{k}=\mathrm{0}}…
Question Number 221036 by SdC355 last updated on 23/May/25 $$\mathrm{prove} \\ $$$$\int_{{P}\in\left[\epsilon,\epsilon+\delta\right]} \sqrt{\left(\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}{f}\left({t}\right)\right)^{\mathrm{2}} +\left(\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\mathrm{g}\left({t}\right)\right)^{\mathrm{2}} }\:\mathrm{d}{t}\leq\int_{{P}\in\left[\epsilon,\epsilon+\delta\right]} \:\frac{{C}_{\mathrm{1}} {f}^{\left(\mathrm{1}\right)} \left({t}\right)+{C}_{\mathrm{2}} \mathrm{g}^{\left(\mathrm{1}\right)} \left({t}\right)}{\:\sqrt{{C}_{\mathrm{1}} ^{\mathrm{2}} +{C}_{\mathrm{2}} ^{\mathrm{2}} }}\:\mathrm{d}{t} \\…
Question Number 221034 by Nicholas666 last updated on 23/May/25 Answered by vnm last updated on 23/May/25 $${let}\:{lim}={a} \\ $$$$\mathrm{sin}\frac{{a}}{\:\sqrt{{n}}}=\frac{{a}}{\:\sqrt{{n}}}−\frac{\mathrm{1}}{\mathrm{6}}\frac{{a}^{\mathrm{3}} }{{n}\sqrt{{n}}}+{o}\left(\frac{\mathrm{1}}{{n}\sqrt{{n}}}\right)=\frac{{a}}{\:\sqrt{{n}}}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{6}{n}}+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$$\sqrt{{n}+\mathrm{1}}=\sqrt{{n}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}}=\sqrt{{n}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$$\mathrm{sin}\frac{{a}}{\:\sqrt{{n}}}\centerdot\sqrt{{n}+\mathrm{1}}=\frac{{a}}{\:\sqrt{{n}}}\sqrt{{n}}\left(\left(\mathrm{1}−\frac{{a}^{\mathrm{2}}…
Question Number 221060 by Tawa11 last updated on 23/May/25 Please what’s TSA of a frustrum of a cone? Any easy method? Answered by mr W…
Question Number 221056 by fantastic last updated on 23/May/25 Answered by Frix last updated on 24/May/25 $$\mathrm{Area}\:\mathrm{of}\:\mathrm{trapezoid}\:=\frac{\mathrm{16}+\mathrm{22}}{\mathrm{2}}{h}=\mathrm{19}{h} \\ $$$${h}=\sqrt{\mathrm{4}^{\mathrm{2}} −\left(\frac{\mathrm{22}−\mathrm{16}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{7}} \\ $$$${r}=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mathrm{Shaded}\:\mathrm{area}\:=\:\mathrm{19}\sqrt{\mathrm{7}}−\frac{\mathrm{7}\pi}{\mathrm{4}}…
Question Number 221057 by fantastic last updated on 23/May/25 Commented by fantastic last updated on 23/May/25 $${Please}\:{find}\:{the}\:{area}\:{of}\:{shaded}\:{region}\:{using}\:{calculus} \\ $$ Answered by mr W last updated…
Question Number 221017 by Rojarani last updated on 22/May/25 Answered by Ghisom last updated on 23/May/25 $$\mathrm{min}\:\left(\frac{{a}^{\mathrm{2}} }{{b}}+\frac{{b}^{\mathrm{2}} }{{c}}+\frac{{c}^{\mathrm{2}} }{{a}}\right)\:=\mathrm{1}\:\mathrm{when}\:{a}={b}={c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{min}\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:=\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{when}\:{a}={b}={c}=\frac{\mathrm{1}}{\mathrm{3}}…
Question Number 221012 by mr W last updated on 22/May/25 Commented by mr W last updated on 24/May/25 $$\cancel{{yes}}\:{it}'{s}\:{not}\:{given}\:{that}\:{it}\:{is}\:{a}\:{square}. \\ $$ Commented by fantastic last…