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Question Number 34231 by byaw last updated on 03/May/18 Answered by MJS last updated on 03/May/18 $$\mathrm{136}°−\mathrm{46}°=\mathrm{90}°\:\Rightarrow\:\mathrm{cos}\:\mathrm{136}°=−\mathrm{sin}\:\mathrm{46}°;\:\mathrm{sin}\:\mathrm{136}°=\mathrm{cos}\:\mathrm{46}° \\ $$$${v}_{\mathrm{1}} =\left(\mathrm{20}\:\angle\mathrm{136}°\right)=\begin{pmatrix}{\mathrm{20cos}\:\mathrm{136}°}\\{\mathrm{20sin}\:\mathrm{136}°}\end{pmatrix}=\begin{pmatrix}{−\mathrm{20sin}\:\mathrm{46}°}\\{\mathrm{20cos}\:\mathrm{46}°}\end{pmatrix} \\ $$$${v}_{\mathrm{2}} =\left(\mathrm{5}\:\angle\mathrm{46}°\right)=\begin{pmatrix}{\mathrm{5cos}\:\mathrm{46}°}\\{\mathrm{5sin}\:\mathrm{46}°}\end{pmatrix} \\ $$$${v}={v}_{\mathrm{1}}…
Question Number 34230 by byaw last updated on 03/May/18 Answered by MJS last updated on 03/May/18 $$\mathrm{aerial}: \\ $$$$\mathrm{bottom}\:{C}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{top}\:{B}=\begin{pmatrix}{\mathrm{0}}\\{{a}}\end{pmatrix} \\ $$$$\mathrm{surveyor}: \\ $$$$\mathrm{position}\:\mathrm{1}\:{A}_{\mathrm{1}}…
Question Number 164639 by cortano1 last updated on 20/Jan/22 Answered by mr W last updated on 20/Jan/22 $${x}={a}\:\mathrm{cos}\:\mathrm{2}\pi{t}\:\Rightarrow\frac{{dx}}{{dt}}=−\mathrm{2}\pi{a}\:\mathrm{sin}\:\mathrm{2}\pi{t} \\ $$$${y}={a}\:\mathrm{sin}\:\mathrm{2}\pi{t}\:\Rightarrow\frac{{dy}}{{dt}}=\mathrm{2}\pi{a}\:\mathrm{cos}\:\mathrm{2}\pi{t} \\ $$$${z}={bt}\:\Rightarrow\frac{{dz}}{{dt}}={b} \\ $$$${at}\:{t}=\mathrm{1}: \\…
Question Number 164367 by Zaynal last updated on 16/Jan/22 $$\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}, \\…