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Question-201421

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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 roof has the shape of a hyperbolic paraboloid with the equation z=(x^2 /(16))−(y^2 /9)+2 (x, y, z in m and z≥0) find the time a water drop on the roof at point A(6, −1, h) takes to flow down along the roof. assume there is no friction or any other resistance.
$${a}\:{roof}\:{has}\:{the}\:{shape}\:{of}\:{a}\:{hyperbolic} \\ $$$${paraboloid}\:{with}\:{the}\:{equation} \\ $$$$\boldsymbol{{z}}=\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{16}}−\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{9}}+\mathrm{2} \\ $$$$\left({x},\:{y},\:{z}\:{in}\:{m}\:{and}\:{z}\geqslant\mathrm{0}\right) \\ $$$${find}\:{the}\:{time}\:{a}\:{water}\:{drop}\:{on}\:{the} \\ $$$${roof}\:{at}\:{point}\:{A}\left(\mathrm{6},\:−\mathrm{1},\:{h}\right)\:{takes}\:{to}\: \\ $$$${flow}\:{down}\:{along}\:{the}\:{roof}. \\ $$$${assume}\:{there}\:{is}\:{no}\:{friction}\:{or}\:{any} \\ $$$${other}\:{resistance}. \\ $$
Commented by ajfour last updated on 08/Dec/23
I shall explain this, its simple but how do we find r^� (t), do tell me guide me..
$${I}\:{shall}\:{explain}\:{this},\:{its}\:{simple}\:{but} \\ $$$${how}\:{do}\:{we}\:{find}\:\bar {{r}}\left({t}\right),\:{do}\:{tell}\:{me} \\ $$$${guide}\:{me}.. \\ $$
Commented by ajfour last updated on 08/Dec/23
(x/4)=p , (y/3)=q z=p^2 −q^2 +2 (∂z/∂p)=2p (∂z/∂q)=−2q N=λ(((i+2pk)/( (√(1+4p^2 )))))×(((j−2qk)/( (√(1+4q^2 ))))) =μ(1k+2qj−2pi) ∣N∣=mgcos θ where cos θ=(1/( (√(1+4q^2 +4p^2 )))) F_(net) ^� =−mgk+(mgcos θ)(((−2pi+2qj+k)/( (√(4p^2 +4q^2 +1))))) (F_(net) ^� /(mg))=−k^� −(((2pi^� −2qj^� −k^� ))/((4p^2 +4q^2 +1))) x=4p, y=3q (F_(net) ^� /(mg))=−k^� −(((8xi^� −6yj^� −k^� ))/( (64x^2 +36y^2 +1)))
$$\frac{{x}}{\mathrm{4}}={p}\:\:,\:\:\frac{{y}}{\mathrm{3}}={q} \\ $$$${z}={p}^{\mathrm{2}} −{q}^{\mathrm{2}} +\mathrm{2} \\ $$$$\frac{\partial{z}}{\partial{p}}=\mathrm{2}{p}\:\:\:\:\:\frac{\partial{z}}{\partial{q}}=−\mathrm{2}{q} \\ $$$${N}=\lambda\left(\frac{{i}+\mathrm{2}{pk}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}\right)×\left(\frac{{j}−\mathrm{2}{qk}}{\:\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} }}\right) \\ $$$$=\mu\left(\mathrm{1}{k}+\mathrm{2}{qj}−\mathrm{2}{pi}\right) \\ $$$$\mid{N}\mid={mg}\mathrm{cos}\:\theta \\ $$$${where}\:\:\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} +\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\bar {{F}}_{{net}} =−{mgk}+\left({mg}\mathrm{cos}\:\theta\right)\left(\frac{−\mathrm{2}{pi}+\mathrm{2}{qj}+{k}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{2}} +\mathrm{1}}}\right) \\ $$$$\frac{\bar {{F}}_{{net}} }{{mg}}=−\hat {{k}}−\frac{\left(\mathrm{2}{p}\hat {{i}}−\mathrm{2}{q}\hat {{j}}−\hat {{k}}\right)}{\left(\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${x}=\mathrm{4}{p},\:\:{y}=\mathrm{3}{q} \\ $$$$\frac{\bar {{F}}_{{net}} }{{mg}}=−\hat {{k}}−\frac{\left(\mathrm{8}{x}\hat {{i}}−\mathrm{6}{y}\hat {{j}}−\hat {{k}}\right)}{\:\left(\mathrm{64}{x}^{\mathrm{2}} +\mathrm{36}{y}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$
Commented by mr W last updated on 08/Dec/23
an object on the surface z=f(x,y) moves in the direction with the largest slope: ▽f=(∂f/∂x)i+(∂f/∂y)j
$${an}\:{object}\:{on}\:{the}\:{surface}\:{z}={f}\left({x},{y}\right) \\ $$$${moves}\:{in}\:{the}\:{direction}\:{with}\:{the} \\ $$$${largest}\:{slope}: \\ $$$$\bigtriangledown{f}=\frac{\partial{f}}{\partial{x}}\boldsymbol{{i}}+\frac{\partial{f}}{\partial{y}}\boldsymbol{{j}} \\ $$
Answered by mr W last updated on 08/Dec/23
in following t is just a parameter, it doesn′t mean the time. (∂z/∂x)=(x/8)=(dx/dt) (dx/x)=(dt/8) ⇒ln x=(t/8)+c_1 ⇒x=C_1 e^(t/8) 6=C_1 ⇒x(t)=6e^(t/8) (∂z/∂y)=−((2y)/9)=(dy/dt) (dy/y)=−((2dt)/9) ⇒ln y=−((2t)/9)+c_2 ⇒y=C_2 e^(−((2t)/9)) −1=C_2 ⇒y(t)=−e^(−((2t)/9)) ⇒z(t)=(9/4)e^(t/4) −(1/9)e^(−((4t)/9)) +2 h=z(0)=(6^2 /(16))−(((−1)^2 )/9)+2=((149)/(36))≈4.14m v(t)=(√(2g(h−z(t)))) =(√(20(((149)/(36))−(9/4)e^(t/4) +(1/9)e^(−((4t)/9)) −2))) =(√(20(((77)/(36))−(9/4)e^(t/4) +(1/9)e^(−((4t)/9)) ))) dl=(√(((dx/dt))^2 +((dy/dt))^2 +((dz/dt))^2 ))dt =(√(((3/4)e^(t/8) )^2 +((2/9)e^(−((2t)/9)) )^2 +((9/(16))e^(t/4) +(4/(81))e^(−((4t)/9)) )^2 ))dt =(√((9/(16))e^(t/4) +(4/(81))e^(−((4t)/9)) +((81)/(256))e^(t/2) +((16)/(6561))e^(−((8t)/9)) +(9/(162))e^(−((7t)/(36))) )) dt dT=(dl/(v(t)))=(((√((9/(16))e^(t/4) +(4/(81))e^(−((4t)/9)) +((81)/(256))e^(t/2) +((16)/(6561))e^(−((8t)/9)) +(9/(162))e^(−((7t)/(36))) ))dt)/( (√(20(((77)/(36))−(9/4)e^(t/4) +(1/9)e^(−((4t)/9)) ))))) (9/4)e^(t_1 /4) −(1/9)e^(−((4t_1 )/9)) +2=0 ⇒t_1 ≈−6.913159307 T_1 =∫_(−6.913159307) ^0 (((√((9/(16))e^(t/4) +(4/(81))e^(−((4t)/9)) +((81)/(256))e^(t/2) +((16)/(6561))e^(−((8t)/9)) +(9/(162))e^(−((7t)/(36))) ))dt)/( (√(20(((77)/(36))−(9/4)e^(t/4) +(1/9)e^(−((4t)/9)) ))))) ≈1.505056 s the trace of the water drop in ground view and in 3D view see diagrams below.
$${in}\:{following}\:{t}\:{is}\:{just}\:{a}\:{parameter},\: \\ $$$${it}\:{doesn}'{t}\:{mean}\:{the}\:{time}. \\ $$$$\frac{\partial{z}}{\partial{x}}=\frac{{x}}{\mathrm{8}}=\frac{{dx}}{{dt}} \\ $$$$\frac{{dx}}{{x}}=\frac{{dt}}{\mathrm{8}}\:\Rightarrow\mathrm{ln}\:{x}=\frac{{t}}{\mathrm{8}}+{c}_{\mathrm{1}} \:\Rightarrow{x}={C}_{\mathrm{1}} {e}^{\frac{{t}}{\mathrm{8}}} \\ $$$$\mathrm{6}={C}_{\mathrm{1}} \\ $$$$\Rightarrow{x}\left({t}\right)=\mathrm{6}{e}^{\frac{{t}}{\mathrm{8}}} \\ $$$$\frac{\partial{z}}{\partial{y}}=−\frac{\mathrm{2}{y}}{\mathrm{9}}=\frac{{dy}}{{dt}} \\ $$$$\frac{{dy}}{{y}}=−\frac{\mathrm{2}{dt}}{\mathrm{9}}\:\Rightarrow\mathrm{ln}\:{y}=−\frac{\mathrm{2}{t}}{\mathrm{9}}+{c}_{\mathrm{2}} \Rightarrow{y}={C}_{\mathrm{2}} {e}^{−\frac{\mathrm{2}{t}}{\mathrm{9}}} \\ $$$$−\mathrm{1}={C}_{\mathrm{2}} \\ $$$$\Rightarrow{y}\left({t}\right)=−{e}^{−\frac{\mathrm{2}{t}}{\mathrm{9}}} \\ $$$$\Rightarrow{z}\left({t}\right)=\frac{\mathrm{9}}{\mathrm{4}}{e}^{\frac{{t}}{\mathrm{4}}} −\frac{\mathrm{1}}{\mathrm{9}}{e}^{−\frac{\mathrm{4}{t}}{\mathrm{9}}} +\mathrm{2} \\ $$$${h}={z}\left(\mathrm{0}\right)=\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{16}}−\frac{\left(−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{9}}+\mathrm{2}=\frac{\mathrm{149}}{\mathrm{36}}\approx\mathrm{4}.\mathrm{14}{m} \\ $$$${v}\left({t}\right)=\sqrt{\mathrm{2}{g}\left({h}−{z}\left({t}\right)\right)} \\ $$$$\:\:=\sqrt{\mathrm{20}\left(\frac{\mathrm{149}}{\mathrm{36}}−\frac{\mathrm{9}}{\mathrm{4}}{e}^{\frac{{t}}{\mathrm{4}}} +\frac{\mathrm{1}}{\mathrm{9}}{e}^{−\frac{\mathrm{4}{t}}{\mathrm{9}}} −\mathrm{2}\right)} \\ $$$$\:\:=\sqrt{\mathrm{20}\left(\frac{\mathrm{77}}{\mathrm{36}}−\frac{\mathrm{9}}{\mathrm{4}}{e}^{\frac{{t}}{\mathrm{4}}} +\frac{\mathrm{1}}{\mathrm{9}}{e}^{−\frac{\mathrm{4}{t}}{\mathrm{9}}} \right)} \\ $$$${dl}=\sqrt{\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} +\left(\frac{{dy}}{{dt}}\right)^{\mathrm{2}} +\left(\frac{{dz}}{{dt}}\right)^{\mathrm{2}} }{dt} \\ $$$$\:\:=\sqrt{\left(\frac{\mathrm{3}}{\mathrm{4}}{e}^{\frac{{t}}{\mathrm{8}}} \right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{9}}{e}^{−\frac{\mathrm{2}{t}}{\mathrm{9}}} \right)^{\mathrm{2}} +\left(\frac{\mathrm{9}}{\mathrm{16}}{e}^{\frac{{t}}{\mathrm{4}}} +\frac{\mathrm{4}}{\mathrm{81}}{e}^{−\frac{\mathrm{4}{t}}{\mathrm{9}}} \right)^{\mathrm{2}} }{dt} \\ $$$$\:\:=\sqrt{\frac{\mathrm{9}}{\mathrm{16}}{e}^{\frac{{t}}{\mathrm{4}}} +\frac{\mathrm{4}}{\mathrm{81}}{e}^{−\frac{\mathrm{4}{t}}{\mathrm{9}}} +\frac{\mathrm{81}}{\mathrm{256}}{e}^{\frac{{t}}{\mathrm{2}}} +\frac{\mathrm{16}}{\mathrm{6561}}{e}^{−\frac{\mathrm{8}{t}}{\mathrm{9}}} +\frac{\mathrm{9}}{\mathrm{162}}{e}^{−\frac{\mathrm{7}{t}}{\mathrm{36}}} }\:{dt} \\ $$$${dT}=\frac{{dl}}{{v}\left({t}\right)}=\frac{\sqrt{\frac{\mathrm{9}}{\mathrm{16}}{e}^{\frac{{t}}{\mathrm{4}}} +\frac{\mathrm{4}}{\mathrm{81}}{e}^{−\frac{\mathrm{4}{t}}{\mathrm{9}}} +\frac{\mathrm{81}}{\mathrm{256}}{e}^{\frac{{t}}{\mathrm{2}}} +\frac{\mathrm{16}}{\mathrm{6561}}{e}^{−\frac{\mathrm{8}{t}}{\mathrm{9}}} +\frac{\mathrm{9}}{\mathrm{162}}{e}^{−\frac{\mathrm{7}{t}}{\mathrm{36}}} }{dt}}{\:\sqrt{\mathrm{20}\left(\frac{\mathrm{77}}{\mathrm{36}}−\frac{\mathrm{9}}{\mathrm{4}}{e}^{\frac{{t}}{\mathrm{4}}} +\frac{\mathrm{1}}{\mathrm{9}}{e}^{−\frac{\mathrm{4}{t}}{\mathrm{9}}} \right)}} \\ $$$$\frac{\mathrm{9}}{\mathrm{4}}{e}^{\frac{{t}_{\mathrm{1}} }{\mathrm{4}}} −\frac{\mathrm{1}}{\mathrm{9}}{e}^{−\frac{\mathrm{4}{t}_{\mathrm{1}} }{\mathrm{9}}} +\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{t}_{\mathrm{1}} \approx−\mathrm{6}.\mathrm{913159307} \\ $$$${T}_{\mathrm{1}} =\int_{−\mathrm{6}.\mathrm{913159307}} ^{\mathrm{0}} \frac{\sqrt{\frac{\mathrm{9}}{\mathrm{16}}{e}^{\frac{{t}}{\mathrm{4}}} +\frac{\mathrm{4}}{\mathrm{81}}{e}^{−\frac{\mathrm{4}{t}}{\mathrm{9}}} +\frac{\mathrm{81}}{\mathrm{256}}{e}^{\frac{{t}}{\mathrm{2}}} +\frac{\mathrm{16}}{\mathrm{6561}}{e}^{−\frac{\mathrm{8}{t}}{\mathrm{9}}} +\frac{\mathrm{9}}{\mathrm{162}}{e}^{−\frac{\mathrm{7}{t}}{\mathrm{36}}} }{dt}}{\:\sqrt{\mathrm{20}\left(\frac{\mathrm{77}}{\mathrm{36}}−\frac{\mathrm{9}}{\mathrm{4}}{e}^{\frac{{t}}{\mathrm{4}}} +\frac{\mathrm{1}}{\mathrm{9}}{e}^{−\frac{\mathrm{4}{t}}{\mathrm{9}}} \right)}} \\ $$$$\:\:\:\approx\mathrm{1}.\mathrm{505056}\:{s} \\ $$$$ \\ $$$${the}\:{trace}\:{of}\:{the}\:{water}\:{drop}\:{in}\:{ground} \\ $$$${view}\:{and}\:{in}\:\mathrm{3}{D}\:{view}\:{see}\:{diagrams} \\ $$$${below}. \\ $$
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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