# A-ball-lies-on-the-function-z-xy-at-the-point-1-2-2-Find-the-point-in-the-xy-plane-where-the-ball-will-touch-it-an-unsolved-old-question-Q200929-

Question Number 201214 by mr W last updated on 02/Dec/23
$$\mathrm{A}\:\mathrm{ball}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{function}\:{z}={xy}\:\mathrm{at} \\$$$$\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{2},\mathrm{2}\right).\:\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in} \\$$$$\mathrm{the}\:{xy}−\mathrm{plane}\:\mathrm{where}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{will} \\$$$$\mathrm{touch}\:\mathrm{it}. \\$$$$\\$$$$\left({an}\:{unsolved}\:{old}\:{question}\:{Q}\mathrm{200929}\right) \\$$
Answered by mr W last updated on 03/Dec/23
$${the}\:{ball}\:{lies}\:{on}\:{the}\:{surface}\:\boldsymbol{{z}}=\boldsymbol{{xy}} \\$$$${and}\:{follows}\:{the}\:{line}\:{with}\:{maximum} \\$$$${slope}\:{when}\:{it}\:{goes}\:{down}. \\$$$${initial}\:{position}\:{is}\:\left(\mathrm{1},\mathrm{2},\mathrm{2}\right). \\$$$$\frac{\partial{z}}{\partial{x}}={y}={x}'\left({t}\right) \\$$$$\frac{\partial{z}}{\partial{y}}={x}={y}'\left({t}\right) \\$$$$\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix} \\$$$$\begin{vmatrix}{\mathrm{0}−\lambda}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}−\lambda}\end{vmatrix}=\lambda^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\$$$$\lambda_{\mathrm{1}} =\mathrm{1},\:\lambda_{\mathrm{2}} =−\mathrm{1} \\$$$$\Rightarrow\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:={c}_{\mathrm{1}} \begin{pmatrix}{\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix}\:{e}^{{t}} +{c}_{\mathrm{2}} \begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}{e}^{−{t}} \\$$$${at}\:{t}=\mathrm{0}: \\$$$${x}={c}_{\mathrm{1}} +{c}_{\mathrm{2}} =\mathrm{1} \\$$$${y}=−{c}_{\mathrm{1}} +{c}_{\mathrm{2}} =\mathrm{2} \\$$$$\Rightarrow{c}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}},\:{c}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}} \\$$$${equation}\:{of}\:{the}\:{curve}\:{traced}\:{by}\:{the}\: \\$$$${ball}\:{is}\:{thus} \\$$$$\begin{cases}{{x}\left({t}\right)=\frac{−{e}^{{t}} +\mathrm{3}{e}^{−{t}} }{\mathrm{2}}}\\{{y}\left({t}\right)=\frac{{e}^{{t}} +\mathrm{3}{e}^{−{t}} }{\mathrm{2}}}\\{{z}\left({t}\right)=\frac{−{e}^{\mathrm{2}{t}} +\mathrm{9}{e}^{−\mathrm{2}{t}} }{\mathrm{4}}}\end{cases} \\$$$${when}\:{the}\:{ball}\:{touches}\:{the}\:{xy}−{plane}: \\$$$${z}\left({t}\right)=\frac{−{e}^{\mathrm{2}{t}} +\mathrm{9}{e}^{−\mathrm{2}{t}} }{\mathrm{4}}=\mathrm{0} \\$$$$\Rightarrow{e}^{{t}} =\sqrt{\mathrm{3}} \\$$$$\Rightarrow{x}\left({t}\right)=\frac{−\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{0} \\$$$$\Rightarrow{y}\left({t}\right)=\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}} \\$$$${i}.{e}.\:{the}\:{ball}\:{touches}\:{the}\:{xy}−{plane} \\$$$${at}\:\left(\mathrm{0},\:\sqrt{\mathrm{3}},\:\mathrm{0}\right). \\$$
Commented by mr W last updated on 02/Dec/23
$${if}\:{the}\:{ball}\:{is}\:{initially}\:{at}\:\left(\sqrt{\mathrm{2}},\:\sqrt{\mathrm{2}},\:\mathrm{2}\right), \\$$$${c}_{\mathrm{1}} +{c}_{\mathrm{2}} =\sqrt{\mathrm{2}} \\$$$$−{c}_{\mathrm{1}} +{c}_{\mathrm{2}} =\sqrt{\mathrm{2}} \\$$$$\Rightarrow{c}_{\mathrm{1}} =\mathrm{0},\:{c}_{\mathrm{2}} =\sqrt{\mathrm{2}} \\$$$${equation}\:{of}\:{the}\:{curve}\:{traced}\:{by}\:{the}\: \\$$$${ball}\:{is}\:{then} \\$$$$\begin{cases}{{x}\left({t}\right)=\sqrt{\mathrm{2}}{e}^{−{t}} }\\{{y}\left({t}\right)=\sqrt{\mathrm{2}}{e}^{−{t}} }\\{{z}\left({t}\right)=\mathrm{2}{e}^{−\mathrm{2}{t}} }\end{cases} \\$$
Commented by mr W last updated on 02/Dec/23
Commented by mr W last updated on 02/Dec/23
Commented by Akira181 last updated on 05/Mar/24
$$\mathrm{It}\:\mathrm{is}\:\mathrm{pretty}\:\mathrm{sure},\:\mathrm{congratulations}! \\$$$$\mathrm{I}\:\mathrm{took}\:\mathrm{other}\:\mathrm{way}\:\mathrm{by}\:\mathrm{first}\:\mathrm{parametrizing} \\$$$$\mathrm{and}\:\mathrm{simplifying}\:\mathrm{as}\:\mathrm{square}\:\mathrm{roots}. \\$$