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)


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Question Number 201214 by mr W last updated on 02/Dec/23

$A ball lies on the function z = x y at$ $the point (1, 2, 2) . Find the point in$ $the x y - plane where the ball will$ $touch it .$ $(a n u n s o l v e d o l d q u e s t i o n Q 200929)$
	Answered by mr W last updated on 03/Dec/23
	$the ball lies on the surface z=xy and follows the line with maximum slope when it goes down. initial position is (1,2,2). (∂z/∂x)=y=x′(t) (∂z/∂y)=x=y′(t) (((x′)),((y′)) ) = ((0,1),(1,0) ) ((x),(y) ) determinant (((0−λ),1),(1,(0−λ)))=λ^2 −1=0 λ_1 =1, λ_2 =−1 ⇒ ((x),(y) ) =c_1 ((1),((−1)) ) e^t +c_2 ((1),(1) )e^(−t) at t=0: x=c_1 +c_2 =1 y=−c_1 +c_2 =2 ⇒c_1 =−(1/2), c_2 =(3/2) equation of the curve traced by the ball is thus { ((x(t)=((−e^t +3e^(−t) )/2))),((y(t)=((e^t +3e^(−t) )/2))),((z(t)=((−e^(2t) +9e^(−2t) )/4))) :} when the ball touches the xy−plane: z(t)=((−e^(2t) +9e^(−2t) )/4)=0 ⇒e^t =(√3) ⇒x(t)=((−(√3)+(√3))/2)=0 ⇒y(t)=(((√3)+(√3))/2)=(√3) i.e. the ball touches the xy−plane at (0, (√3), 0).$
	$t h e b a l l l i e s o n t h e s u r f a c e z = x y$ $a n d f o l l o w s t h e l i n e w i t h m a x i m u m$ $s l o p e w h e n i t g o e s d o w n .$ $i n i t i a l p o s i t i o n i s (1, 2, 2) .$ $\frac{\partial z}{\partial x} = y = x^{'} (t)$ $\frac{\partial z}{\partial y} = x = y^{'} (t)$ $(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix})$ $\| \begin{matrix} 0 - λ & 1 \\ 1 & 0 - λ \end{matrix} \| = λ^{2} - 1 = 0$ $λ_{1} = 1, λ_{2} = - 1$ $\Rightarrow (\begin{matrix} x \\ y \end{matrix}) = c_{1} (\begin{matrix} 1 \\ - 1 \end{matrix}) e^{t} + c_{2} (\begin{matrix} 1 \\ 1 \end{matrix}) e^{- t}$ $a t t = 0 :$ $x = c_{1} + c_{2} = 1$ $y = - c_{1} + c_{2} = 2$ $\Rightarrow c_{1} = - \frac{1}{2}, c_{2} = \frac{3}{2}$ $e q u a t i o n o f t h e c u r v e t r a c e d b y t h e$ $b a l l i s t h u s$ ${\begin{cases} x (t) = \frac{- e^{t} + 3 e^{- t}}{2} \\ y (t) = \frac{e^{t} + 3 e^{- t}}{2} \\ z (t) = \frac{- e^{2 t} + 9 e^{- 2 t}}{4} \end{cases}$ $w h e n t h e b a l l t o u c h e s t h e x y - p l a n e :$ $z (t) = \frac{- e^{2 t} + 9 e^{- 2 t}}{4} = 0$ $\Rightarrow e^{t} = \sqrt{3}$ $\Rightarrow x (t) = \frac{- \sqrt{3} + \sqrt{3}}{2} = 0$ $\Rightarrow y (t) = \frac{\sqrt{3} + \sqrt{3}}{2} = \sqrt{3}$ $i . e . t h e b a l l t o u c h e s t h e x y - p l a n e$ $a t (0, \sqrt{3}, 0) .$
	Commented by mr W last updated on 02/Dec/23
	$if the ball is initially at ((√2), (√2), 2), c_1 +c_2 =(√2) −c_1 +c_2 =(√2) ⇒c_1 =0, c_2 =(√2) equation of the curve traced by the ball is then { ((x(t)=(√2)e^(−t) )),((y(t)=(√2)e^(−t) )),((z(t)=2e^(−2t) )) :}$
	$i f t h e b a l l i s i n i t i a l l y a t (\sqrt{2}, \sqrt{2}, 2),$ $c_{1} + c_{2} = \sqrt{2}$ $- c_{1} + c_{2} = \sqrt{2}$ $\Rightarrow c_{1} = 0, c_{2} = \sqrt{2}$ $e q u a t i o n o f t h e c u r v e t r a c e d b y t h e$ $b a l l i s t h e n$ ${\begin{cases} x (t) = \sqrt{2} e^{- t} \\ y (t) = \sqrt{2} e^{- t} \\ z (t) = 2 e^{- 2 t} \end{cases}$
	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

	$It is pretty sure, congratulations!$ $I took other way by first parametrizing$ $and simplifying as square roots .$
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