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Category: Differentiation

bobhans-How-do-you-find-a-point-on-the-curve-y-x-2-closest-to-the-point-0-18-

Question Number 106779 by bobhans last updated on 07/Aug/20 $$\:\:\:\:\:\:\:\:\:^{\succ\mathrm{bobhans}\prec} \\ $$$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{find}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{closest}\:\mathrm{to}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{0},\mathrm{18}\right)\:? \\ $$ Answered by john santu last updated on 07/Aug/20 $$\:\:\:\:\:\:\:^{@\mathrm{JS}@}…

Max-and-min-P-2-x-3-y-subject-to-constraint-x-2-9-y-2-25-1-x-2-y-2-

Question Number 172253 by cortano1 last updated on 25/Jun/22 $$\:\:{Max}\:{and}\:{min}\:{P}=\sqrt{\mathrm{2}}\:{x}+\:\sqrt{\mathrm{3}}\:{y} \\ $$$${subject}\:{to}\:{constraint}\: \\ $$$$\:\frac{{x}^{\mathrm{2}} }{\mathrm{9}}\:+\frac{{y}^{\mathrm{2}} }{\mathrm{25}}\:\leqslant\:\mathrm{1}\:\leqslant\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$ Answered by mr W last updated…

Proof-that-d-n-dx-n-cos-x-cos-x-npi-2-d-n-dx-n-sin-x-sin-x-npi-2-where-n-Z-

Question Number 41151 by rahul 19 last updated on 02/Aug/18 $$\mathrm{Proof}\:\mathrm{that}\::\:\frac{\mathrm{d}^{\mathrm{n}} }{\mathrm{d}{x}^{{n}} }\left(\mathrm{cos}\:{x}\right)\:=\:\mathrm{cos}\:\left({x}+\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{d}^{\mathrm{n}} }{\mathrm{d}{x}^{{n}} }\left(\mathrm{sin}\:{x}\right)\:=\:\mathrm{sin}\:\left({x}+\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{where}\:\mathrm{n}\in\mathbb{Z}. \\ $$ Commented by prof Abdo…

Question-41117

Question Number 41117 by ajfour last updated on 02/Aug/18 Answered by MJS last updated on 02/Aug/18 $${r}=\mathrm{1} \\ $$$${P}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{A}=\begin{pmatrix}{\mathrm{cos}\:\alpha}\\{\mathrm{1}+\mathrm{sin}\:\alpha}\end{pmatrix}\:\:{F}=\begin{pmatrix}{\mathrm{cos}\:\alpha}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mid{AP}\mid=\sqrt{\mathrm{2}\left(\mathrm{1}+\mathrm{sin}\:\alpha\right)} \\ $$$$\mid{AF}\mid=\mathrm{1}+\mathrm{sin}\:\alpha \\ $$$$\mid{FP}\mid=\mathrm{cos}\:\alpha…