Question Number 68315 by 9102176137086 last updated on 08/Sep/19 $$\int\left(\frac{{x}^{−\mathrm{3}} +\mathrm{2}{x}−\mathrm{4}}{{x}}\right) \\ $$ Commented by mathmax by abdo last updated on 10/Sep/19 $$=\int\:\left({x}^{−\mathrm{4}} \:+\mathrm{2}\:−\frac{\mathrm{4}}{{x}}\right){dx}\:=\frac{\mathrm{1}}{−\mathrm{4}+\mathrm{1}}{x}^{−\mathrm{4}+\mathrm{1}} \:+\mathrm{2}{x}−\mathrm{4}{ln}\mid{x}\mid\:+{c}…
Question Number 2163 by Filup last updated on 06/Nov/15 $$\mathrm{For}:\:{y}={f}\left({x}\right)\:\rightarrow\:{x}={g}\left({y}\right) \\ $$$$\mathrm{Therefore}: \\ $$$$\begin{cases}{{x}\left({t}\right)={t}}\\{{y}\left({t}\right)={f}\left({t}\right)}\end{cases} \\ $$$$\mathrm{let}\:\boldsymbol{{r}}\left({t}\right)=\langle{x}\left({t}\right),\:{y}\left({t}\right)\rangle \\ $$$$ \\ $$$$\therefore\int\boldsymbol{{r}}{dt}=\langle\int{tdt},\:\int{f}\left({t}\right){dt}\rangle \\ $$$$ \\ $$$$\mathrm{Does}: \\…
Question Number 2130 by Yozzi last updated on 03/Nov/15 $${Find}\:\boldsymbol{{r}}=\begin{pmatrix}{{x}\left({t}\right)}\\{{y}\left({t}\right)}\end{pmatrix}\:\:\:{satisfying}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{d}\boldsymbol{{r}}}{{dt}}+\boldsymbol{{Ar}}=\mathrm{0} \\ $$$${where}\:\boldsymbol{{A}}=\begin{bmatrix}{−\mathrm{3}\:\:\:\:−\mathrm{1}}\\{\mathrm{8}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{bmatrix}\:{by}\:{using} \\ $$$${a}\:{matrix}\:{integrating}\:{factor}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 133199 by rexford last updated on 19/Feb/21 Answered by mr W last updated on 20/Feb/21 $$\overset{\rightarrow} {\boldsymbol{{AB}}}=\left(−\mathrm{1},\mathrm{5},−\mathrm{3}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{AC}}}=\left(−\mathrm{4},\mathrm{3},\mathrm{3}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{AD}}}=\left(\mathrm{1},\mathrm{7},\lambda+\mathrm{1}\right)…
Question Number 2092 by Filup last updated on 02/Nov/15 $$\mathrm{If}\:\boldsymbol{{A}}=\langle{p}_{{x}} ,\:{p}_{{y}} ,\:{p}_{{z}} \rangle\:\mathrm{is}\:\mathrm{a}\:\mathrm{position}\:\mathrm{vector}\: \\ $$$$\mathrm{in}\:\mathrm{standard}\:\mathrm{position},\:\mathrm{vector}\:\boldsymbol{{r}}=\langle{r}_{{x}} ,\:{r}_{{y}} ,\:{r}_{{z}} \rangle \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{a}\:\mathrm{3}\:\mathrm{dimensional}\:\mathrm{circle} \\ $$$$\mathrm{with}\:\mathrm{focal}\:\mathrm{point}\:\mathrm{at}\:\boldsymbol{{A}},\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{vector}\:\mathrm{equation}\:\boldsymbol{{r}}\left(\theta\right)\:\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is}\:\mathrm{the} \\…
Question Number 132972 by rexford last updated on 17/Feb/21 Answered by Ar Brandon last updated on 17/Feb/21 $$\begin{vmatrix}{\mathrm{i}}&{\mathrm{j}}&{\mathrm{k}}\\{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{3}}\\{\mathrm{1}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}=\mathrm{7i}+\mathrm{5j}+\mathrm{k} \\ $$ Commented by rexford last updated…
Question Number 1851 by 112358 last updated on 14/Oct/15 $${A}\:{plane}\:{has}\:{equation}\:{x}−{z}=\mathrm{4}\sqrt{\mathrm{3}}. \\ $$$${The}\:{line}\:{l}\:{has}\:{vector}\:{equation} \\ $$$$\boldsymbol{{r}}=\lambda\left[\left({cos}\theta+\sqrt{\mathrm{3}}\right)\boldsymbol{{i}}+\left(\sqrt{\mathrm{2}}{sin}\theta\right)\boldsymbol{{j}}+\left({cos}\theta−\sqrt{\mathrm{3}}\right)\boldsymbol{{k}}\right] \\ $$$${where}\:\lambda\:{is}\:{a}\:{scalar}\:{parameter}. \\ $$$${If}\:{l}\:{meets}\:{the}\:{plane}\:{at}\:{P},\:{show}\:{that}, \\ $$$${as}\:\theta\:{varies},\:{P}\:\:{describes}\:{a}\:{circle}.\: \\ $$ Answered by 123456…
Question Number 132920 by liberty last updated on 17/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\mathrm{x}−\mathrm{sin}\:\mathrm{2x}}{\mathrm{x}−\mathrm{sin}\:\mathrm{x}} \\ $$ Answered by bobhans last updated on 17/Feb/21 $$\:{L}'{H}\ddot {{o}pital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}−\mathrm{2cos}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:{x}}\:=…
Question Number 132856 by bramlexs22 last updated on 17/Feb/21 $$\mathrm{Given}\:\mathrm{vector}\:\overset{\rightarrow} {{a}}\:=\:\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}}\:,\:\overset{\rightarrow} {\mathrm{c}}=\hat {\mathrm{j}}−\hat {\mathrm{k}}\:; \\ $$$$\:\overset{\rightarrow} {\mathrm{a}}\:×\:\overset{\rightarrow} {\mathrm{b}}\:=\:\overset{\rightarrow} {\mathrm{c}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\:=\:\mathrm{3}\:\mathrm{then}\:\mid\overset{\rightarrow} {\mathrm{b}}\mid\:=\:?…
Question Number 132853 by EDWIN88 last updated on 17/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{paraboloid}\: \\ $$$$\mathrm{z}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\mathrm{which}\:\mathrm{is}\:\mathrm{closest}\:\mathrm{to}\:\mathrm{the}\:\mathrm{point}\: \\ $$$$\left(\mathrm{3},−\mathrm{6},\mathrm{4}\:\right) \\ $$ Answered by MJS_new last updated on 17/Feb/21…