Question Number 164367 by Zaynal last updated on 16/Jan/22 $$\int\:\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \:\boldsymbol{{dx}} \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{{A}}\right\} \\ $$ Answered by puissant last updated on 16/Jan/22 $$\Omega=\int{e}^{{tanx}} {dx}\:;\:{t}={tanx}\:\rightarrow\:{dx}=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}…
Question Number 33199 by 786786AM last updated on 12/Apr/18 Commented by 786786AM last updated on 15/Apr/18 $$\mathrm{pl}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sir}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 98408 by bemath last updated on 13/Jun/20 Answered by smridha last updated on 13/Jun/20 $$\boldsymbol{{for}}\:\boldsymbol{{L}}_{\mathrm{1}} \:\boldsymbol{{any}}\:\boldsymbol{{point}}\:\left(\boldsymbol{{p}}_{\mathrm{1}} \right)\boldsymbol{{can}}\:\boldsymbol{{be}}\:\boldsymbol{{described}}\:\boldsymbol{{as}} \\ $$$$\left(−\mathrm{1}−\mathrm{3}\boldsymbol{{t}},\mathrm{3}+\mathrm{2}\boldsymbol{{t}},−\mathrm{2}+\boldsymbol{{t}}\right). \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{L}}_{\mathrm{2}} \boldsymbol{{any}}\:\boldsymbol{{point}}\:\left(\boldsymbol{{p}}_{\mathrm{2}} \right)\:\boldsymbol{{can}}\:\boldsymbol{{be}}\:\boldsymbol{{described}}…
Question Number 32872 by byaw last updated on 04/Apr/18 Answered by byaw last updated on 02/Aug/18 $${please}\:{help} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 98325 by bemath last updated on 13/Jun/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{curvature}\:\mathrm{vector}\:\mathrm{and} \\ $$$$\mathrm{its}\:\mathrm{magnitude}\:\mathrm{at}\:\mathrm{any}\:\mathrm{point}\: \\ $$$$\overset{\rightarrow} {\mathrm{r}}\:=\:\left(\theta\right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\overset{\rightarrow} {\mathrm{r}}=\:\left(\mathrm{acos}\:\theta,\mathrm{asin}\:\theta,\mathrm{a}\theta\right) \\ $$$$.\mathrm{Show}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{feet}\:\mathrm{of}\:\mathrm{the} \\ $$$$\bot\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{to}\:\mathrm{the}\:\mathrm{tangent}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{curve}\:\mathrm{that}\:\mathrm{completely}\:\mathrm{lies} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{hyperbolic}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…
Question Number 32713 by byaw last updated on 31/Mar/18 $$\mathrm{A}\:\mathrm{iniform}\:\mathrm{rod}\:\mathrm{of}\:\mathrm{length}\:\mathrm{2}{a}\:\mathrm{and}\:\mathrm{weight}\:{W} \\ $$$$\mathrm{is}\:\mathrm{smoothly}\:\mathrm{pivoted}\:\mathrm{to}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{point} \\ $$$$\mathrm{at}\:{A}.\:\mathrm{A}\:\mathrm{load}\:\mathrm{of}\:\mathrm{weight}\:\mathrm{2}{W}\:\mathrm{is}\:\mathrm{attached} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{end}\:{B}.\:\mathrm{The}\:\mathrm{rod}\:\mathrm{is}\:\mathrm{kept}\:\mathrm{horizontally} \\ $$$$\mathrm{by}\:\mathrm{a}\:\mathrm{string}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{the}\:\mathrm{midpoint} \\ $$$${D}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rod}\:\mathrm{and}\:\mathrm{to}\:\mathrm{a}\:\mathrm{point}\:{C}\:\mathrm{vertically} \\ $$$$\mathrm{above}\:{A}.\: \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string}\:\mathrm{is}\:\mathrm{2}{a},\:\mathrm{find}, \\…
Question Number 32529 by byaw last updated on 27/Mar/18 Commented by byaw last updated on 27/Mar/18 $$\mathrm{Please}\:\mathrm{help} \\ $$ Commented by mrW2 last updated on…
Question Number 98018 by hardylanes last updated on 11/Jun/20 $${the}\:{vector}\:{equations}\:{of}\:{two}\:{lines}\:{l}_{\mathrm{1}} \:{and}\:{l}_{\mathrm{2}} \:{are}\:{given}\:{by} \\ $$$${l}_{\mathrm{1}} :{r}=\mathrm{5}{i}−{j}+{k}+\lambda\left(−\mathrm{3}{i}+\mathrm{2}{j}\right) \\ $$$${l}_{\mathrm{2}} :{r}=\mathrm{2}{i}+\mathrm{3}{j}+\mathrm{2}{k}+\mu\left(\mathrm{2}{j}+{k}\right) \\ $$$${find} \\ $$$${thd}\:{position}\:{vdctor}\:{of}\:{intersection}\:{of}\:{thf}\:{linds}\:{l}_{\mathrm{1}} \:{and}\:{l}_{\mathrm{2}} \: \\…
Question Number 31264 by ajfour last updated on 04/Mar/18 $${Find}\:{the}\:{shortest}\:{distance}\:{from}\:{the} \\ $$$${plane}\:\bar {\boldsymbol{{r}}}.\bar {\boldsymbol{{n}}}=\boldsymbol{{q}}\:\:{to}\:{the}\:{sphere} \\ $$$$\:\mid\bar {\boldsymbol{{r}}}−\bar {\boldsymbol{{r}}}_{\mathrm{0}} \mid=\boldsymbol{{R}}\:\:. \\ $$ Commented by ajfour last…
Question Number 96240 by mr W last updated on 30/May/20 $${Find}\:{the}\:{shortest}\:{distance}\:{from}\:{the} \\ $$$${point}\:{P}\left(\mathrm{2},−\mathrm{3},\mathrm{5}\right)\:{to}\:{the}\:{line}\:{L} \\ $$$$\frac{{x}+\mathrm{3}}{\mathrm{2}}=\frac{{y}−\mathrm{1}}{−\mathrm{3}}=\frac{{z}−\mathrm{2}}{\mathrm{4}} \\ $$ Answered by 1549442205 last updated on 31/May/20 $$\mathrm{Denote}\:\mathrm{Q}\:\mathrm{be}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{L}\:\mathrm{such}\:\mathrm{that}…