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Question Number 201421 by mr W last updated on 06/Dec/23

Commented by mr W last updated on 06/Dec/23

Commented by mr W last updated on 06/Dec/23

$a r o o f h a s t h e s h a p e o f a h y p e r b o l i c$ $p a r a b o l o i d w i t h t h e e q u a t i o n$ $z = \frac{x^{2}}{16} - \frac{y^{2}}{9} + 2$ $(x, y, z i n m a n d z ⩾ 0)$ $f i n d t h e t i m e a w a t e r d r o p o n t h e$ $r o o f a t p o i n t A (6, - 1, h) t a k e s t o$ $f l o w d o w n a l o n g t h e r o o f .$ $a s s u m e t h e r e i s n o f r i c t i o n o r a n y$ $o t h e r r e s i s t a n c e .$
Commented by ajfour last updated on 08/Dec/23

$I s h a l l e x p l a i n t h i s, i t s s i m p l e b u t$ $h o w d o w e f i n d \bar{r} (t), d o t e l l m e$ $g u i d e m e . .$
Commented by ajfour last updated on 08/Dec/23

$\frac{x}{4} = p, \frac{y}{3} = q$ $z = p^{2} - q^{2} + 2$ $\frac{\partial z}{\partial p} = 2 p \frac{\partial z}{\partial q} = - 2 q$ $N = λ (\frac{i + 2 p k}{\sqrt{1 + 4 p^{2}}}) \times (\frac{j - 2 q k}{\sqrt{1 + 4 q^{2}}})$ $= μ (1 k + 2 q j - 2 p i)$ $∣ N ∣= m g \cos θ$ $w h e r e \cos θ = \frac{1}{\sqrt{1 + 4 q^{2} + 4 p^{2}}}$ ${\bar{F}}_{n e t} = - m g k + (m g \cos θ) (\frac{- 2 p i + 2 q j + k}{\sqrt{4 p^{2} + 4 q^{2} + 1}})$ $\frac{{\bar{F}}_{n e t}}{m g} = - \hat{k} - \frac{(2 p \hat{i} - 2 q \hat{j} - \hat{k})}{(4 p^{2} + 4 q^{2} + 1)}$ $x = 4 p, y = 3 q$ $\frac{{\bar{F}}_{n e t}}{m g} = - \hat{k} - \frac{(8 x \hat{i} - 6 y \hat{j} - \hat{k})}{(64 x^{2} + 36 y^{2} + 1)}$
Commented by mr W last updated on 08/Dec/23

$a n o b j e c t o n t h e s u r f a c e z = f (x, y)$ $m o v e s i n t h e d i r e c t i o n w i t h t h e$ $l a r g e s t s l o p e :$ $▽ f = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j$
	Answered by mr W last updated on 08/Dec/23

	$i n f o l l o w i n g t i s j u s t a p a r a m e t e r,$ $i t {d o e s n}^{'} t m e a n t h e t i m e .$ $\frac{\partial z}{\partial x} = \frac{x}{8} = \frac{d x}{d t}$ $\frac{d x}{x} = \frac{d t}{8} \Rightarrow \ln x = \frac{t}{8} + c_{1} \Rightarrow x = C_{1} e^{\frac{t}{8}}$ $6 = C_{1}$ $\Rightarrow x (t) = 6 e^{\frac{t}{8}}$ $\frac{\partial z}{\partial y} = - \frac{2 y}{9} = \frac{d y}{d t}$ $\frac{d y}{y} = - \frac{2 d t}{9} \Rightarrow \ln y = - \frac{2 t}{9} + c_{2} \Rightarrow y = C_{2} e^{- \frac{2 t}{9}}$ $- 1 = C_{2}$ $\Rightarrow y (t) = - e^{- \frac{2 t}{9}}$ $\Rightarrow z (t) = \frac{9}{4} e^{\frac{t}{4}} - \frac{1}{9} e^{- \frac{4 t}{9}} + 2$ $h = z (0) = \frac{6^{2}}{16} - \frac{{(- 1)}^{2}}{9} + 2 = \frac{149}{36} \approx 4.14 m$ $v (t) = \sqrt{2 g (h - z (t))}$ $= \sqrt{20 (\frac{149}{36} - \frac{9}{4} e^{\frac{t}{4}} + \frac{1}{9} e^{- \frac{4 t}{9}} - 2)}$ $= \sqrt{20 (\frac{77}{36} - \frac{9}{4} e^{\frac{t}{4}} + \frac{1}{9} e^{- \frac{4 t}{9}})}$ $d l = \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} + {(\frac{d z}{d t})}^{2}} d t$ $= \sqrt{{(\frac{3}{4} e^{\frac{t}{8}})}^{2} + {(\frac{2}{9} e^{- \frac{2 t}{9}})}^{2} + {(\frac{9}{16} e^{\frac{t}{4}} + \frac{4}{81} e^{- \frac{4 t}{9}})}^{2}} d t$ $= \sqrt{\frac{9}{16} e^{\frac{t}{4}} + \frac{4}{81} e^{- \frac{4 t}{9}} + \frac{81}{256} e^{\frac{t}{2}} + \frac{16}{6561} e^{- \frac{8 t}{9}} + \frac{9}{162} e^{- \frac{7 t}{36}}} d t$ $d T = \frac{d l}{v (t)} = \frac{\sqrt{\frac{9}{16} e^{\frac{t}{4}} + \frac{4}{81} e^{- \frac{4 t}{9}} + \frac{81}{256} e^{\frac{t}{2}} + \frac{16}{6561} e^{- \frac{8 t}{9}} + \frac{9}{162} e^{- \frac{7 t}{36}}} d t}{\sqrt{20 (\frac{77}{36} - \frac{9}{4} e^{\frac{t}{4}} + \frac{1}{9} e^{- \frac{4 t}{9}})}}$ $\frac{9}{4} e^{\frac{t_{1}}{4}} - \frac{1}{9} e^{- \frac{4 t_{1}}{9}} + 2 = 0$ $\Rightarrow t_{1} \approx - 6.913159307$ $T_{1} = \int_{- 6.913159307}^{0} \frac{\sqrt{\frac{9}{16} e^{\frac{t}{4}} + \frac{4}{81} e^{- \frac{4 t}{9}} + \frac{81}{256} e^{\frac{t}{2}} + \frac{16}{6561} e^{- \frac{8 t}{9}} + \frac{9}{162} e^{- \frac{7 t}{36}}} d t}{\sqrt{20 (\frac{77}{36} - \frac{9}{4} e^{\frac{t}{4}} + \frac{1}{9} e^{- \frac{4 t}{9}})}}$ $\approx 1.505056 s$ $t h e t r a c e o f t h e w a t e r d r o p i n g r o u n d$ $v i e w a n d i n 3 D v i e w s e e d i a g r a m s$ $b e l o w .$
	Commented by mr W last updated on 07/Dec/23

	Commented by mr W last updated on 08/Dec/23

	Commented by mr W last updated on 08/Dec/23

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