Category: Vector Calculus

Question Number 213045 by MrGaster last updated on 29/Oct/24 $$\stackrel{\rightharpoonup }{a},\stackrel{\rightharpoonup }{bls}theunitvector,\mid \stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}\mid =1$$$$ask\mid \stackrel{\rightharpoonup }{a}-\stackrel{\rightharpoonup }{b}\mid ?$$ Answered by mr W last…

Question Number 210554 by Batmath last updated on 12/Aug/24 Answered by MrGaster last updated on 02/Nov/24 $$\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({px}\right)\right]{x}^{\mathrm{2}{q}−\mathrm{1}} {dx}=\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({tpx}\right)}{{t}}{dt}\right]{x}^{\mathrm{2}{q}−\mathrm{1}} {dx}…

Question Number 207712 by efronzo1 last updated on 24/May/24 Answered by Frix last updated on 24/May/24 $$\mathrm{For}x=q\mathrm{and}A=\left(\begin{array}{c}p\\ 0\end{array}\right)\mathrm{the}\mathrm{area}\mathrm{is}(q-p)\sqrt{p}$$$$\frac{d\left[(q-p)\sqrt{p}\right]}{dp}=0$$$$\frac{q-3p}{2\sqrt{p}}=0\Rightarrow p=\frac{q}{3}$$$$\mathrm{Max}\:\mathrm{area}\:=\frac{\mathrm{2}\sqrt{\mathrm{3}}{q}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{9}} \…

Question Number 204273 by mustafazaheen last updated on 10/Feb/24 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{(\mathrm{x}-1)}^{\ln \left(\frac{2}{4}\right)}}$$$$\mathrm{Domain}\mathrm{f}\left(\mathrm{x}\right)=?$$ Answered by Mathspace last updated on 11/Feb/24 $${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{−{ln}\mathrm{2}} }…

Question Number 201947 by Mingma last updated on 16/Dec/23 Answered by Frix last updated on 18/Dec/23 $$\sin \alpha \sin \beta \cos (\pi -\alpha -\beta )+$$$$+\sin \beta \sin (\pi -\alpha -\beta )\cos \alpha +$$$$+\sin (\pi -\alpha -\beta )\sin \alpha \cos \beta =$$$$=\frac{3-(\cos 2\alpha +\cos 2\beta +\cos 2(\alpha +\beta )}{4}=$$$$\:\:\:\:\:\left[\mathrm{Let}\:\alpha={u}−{v}\wedge\beta={u}+{v}\right]…