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

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

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

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 .