Question Number 171908 by ajfour last updated on 21/Jun/22
Commented by ajfour last updated on 21/Jun/22
$${Man}\:{has}\:{to}\:{water}\:{the}\:{tree} \\ $$$${once}\:{with}\:{volume}\:{V}_{\mathrm{0}} \:.\:{His} \\ $$$${bucket}\:{leaks}\:{at}\:{a}\:{rate}\:\frac{{dV}}{{dt}}=−{kV}. \\ $$$${his}\:{speed}={u}\left(\mathrm{2}−\frac{{m}}{{m}+\rho{V}}\right) \\ $$$$\rho{V}<{m}\:.\:{Find}\:{x}\:{such}\:{that}\:{the} \\ $$$${he}\:{can}\:{reach}\:{with}\:{V}_{\mathrm{0}} \:{volume} \\ $$$${of}\:{water}\:{in}\:{bucket}\:{in}\:{min}.\: \\ $$$${time}.\:\:\left({m}\:{is}\:{mass}\:{of}\:{bucket}\right) \\ $$
Commented by mr W last updated on 22/Jun/22
![s_1 =(√(a^2 +x^2 )) v_1 =u t_1 =(s_1 /u)=((√(a^2 +x^2 ))/u) say the volume of water in the bucket at the river is V_1 . s_2 =(√(b^2 +(c−x)^2 )) (dV/dt)=−kV ∫_V_1 ^V (dV/V)=−k∫_0 ^t dt ln (V/V_1 )=−kt V=V_1 e^(−kt) V_0 =V_1 e^(−kt_2 ) (ds/dt)=v=u(2−(m/(m+ρV))) (ds/dt)=u(2−(m/(m+ρV_1 e^(−kt) ))) ∫_0 ^s_2 ds=u∫_0 ^t_2 (2−(1/(1+((ρV_1 )/m)e^(−kt) )))dt s_2 =u[t−((ln (1+((ρV_1 )/m)e^(−kt) ))/k)]_0 ^t_2 (√(b^2 +(c−x)^2 ))=u(t_2 −(1/k)ln ((m+ρV_1 e^(−kt_2 ) )/(m+ρV_1 ))) (√(b^2 +(c−x)^2 ))=u(t_2 −(1/k)ln ((m+ρV_0 )/(m+ρV_0 e^(kt_2 ) ))) total time T=t_1 +t_2 t_2 =T−t_1 =T−((√(a^2 +x^2 ))/u) (√(b^2 +(c−x)^2 ))=u[T−((√(a^2 +x^2 ))/u)−(1/k)ln ((m+ρV_0 )/(m+ρV_0 e^(k(T−((√(a^2 +x^2 ))/u))) ))] (((√(a^2 +x^2 ))+(√(b^2 +(c−x)^2 )))/u)=T−(1/k)ln ((m/(ρV_0 ))/((m/(ρV_0 ))+e^(k(T−((√(a^2 +x^2 ))/u))) )) (((√(a^2 +x^2 ))+(√(b^2 +(c−x)^2 )))/(u/k))=kT−ln ((m/(ρV_0 ))/((m/(ρV_0 ))+e^((kT−((√(a^2 +x^2 ))/(u/k)))) )) let λ=((ρV_0 )/m), Φ=kT, μ=(u/k) (((√(a^2 +x^2 ))+(√(b^2 +(c−x)^2 )))/μ)=Φ+ln(1+λe^(Φ−((√(a^2 +x^2 ))/μ)) ) (dΦ/dx)=0 for minimum T. see example.](https://www.tinkutara.com/question/Q171924.png)
$${s}_{\mathrm{1}} =\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} } \\ $$$${v}_{\mathrm{1}} ={u} \\ $$$${t}_{\mathrm{1}} =\frac{{s}_{\mathrm{1}} }{{u}}=\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{u}} \\ $$$${say}\:{the}\:{volume}\:{of}\:{water}\:{in}\:{the}\:{bucket} \\ $$$${at}\:{the}\:{river}\:{is}\:{V}_{\mathrm{1}} . \\ $$$${s}_{\mathrm{2}} =\sqrt{{b}^{\mathrm{2}} +\left({c}−{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{dV}}{{dt}}=−{kV} \\ $$$$\int_{{V}_{\mathrm{1}} } ^{{V}} \frac{{dV}}{{V}}=−{k}\int_{\mathrm{0}} ^{{t}} {dt} \\ $$$$\mathrm{ln}\:\frac{{V}}{{V}_{\mathrm{1}} }=−{kt} \\ $$$${V}={V}_{\mathrm{1}} {e}^{−{kt}} \\ $$$${V}_{\mathrm{0}} ={V}_{\mathrm{1}} {e}^{−{kt}_{\mathrm{2}} } \\ $$$$\frac{{ds}}{{dt}}={v}={u}\left(\mathrm{2}−\frac{{m}}{{m}+\rho{V}}\right) \\ $$$$\frac{{ds}}{{dt}}={u}\left(\mathrm{2}−\frac{{m}}{{m}+\rho{V}_{\mathrm{1}} {e}^{−{kt}} }\right) \\ $$$$\int_{\mathrm{0}} ^{{s}_{\mathrm{2}} } {ds}={u}\int_{\mathrm{0}} ^{{t}_{\mathrm{2}} } \left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{1}+\frac{\rho{V}_{\mathrm{1}} }{{m}}{e}^{−{kt}} }\right){dt} \\ $$$${s}_{\mathrm{2}} ={u}\left[{t}−\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\rho{V}_{\mathrm{1}} }{{m}}{e}^{−{kt}} \right)}{{k}}\right]_{\mathrm{0}} ^{{t}_{\mathrm{2}} } \\ $$$$\sqrt{{b}^{\mathrm{2}} +\left({c}−{x}\right)^{\mathrm{2}} }={u}\left({t}_{\mathrm{2}} −\frac{\mathrm{1}}{{k}}\mathrm{ln}\:\frac{{m}+\rho{V}_{\mathrm{1}} {e}^{−{kt}_{\mathrm{2}} } }{{m}+\rho{V}_{\mathrm{1}} }\right) \\ $$$$\sqrt{{b}^{\mathrm{2}} +\left({c}−{x}\right)^{\mathrm{2}} }={u}\left({t}_{\mathrm{2}} −\frac{\mathrm{1}}{{k}}\mathrm{ln}\:\frac{{m}+\rho{V}_{\mathrm{0}} }{{m}+\rho{V}_{\mathrm{0}} {e}^{{kt}_{\mathrm{2}} } }\right) \\ $$$${total}\:{time}\:{T}={t}_{\mathrm{1}} +{t}_{\mathrm{2}} \\ $$$${t}_{\mathrm{2}} ={T}−{t}_{\mathrm{1}} ={T}−\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{u}} \\ $$$$\sqrt{{b}^{\mathrm{2}} +\left({c}−{x}\right)^{\mathrm{2}} }={u}\left[{T}−\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{u}}−\frac{\mathrm{1}}{{k}}\mathrm{ln}\:\frac{{m}+\rho{V}_{\mathrm{0}} }{{m}+\rho{V}_{\mathrm{0}} {e}^{{k}\left({T}−\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{u}}\right)} }\right] \\ $$$$\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +\left({c}−{x}\right)^{\mathrm{2}} }}{{u}}={T}−\frac{\mathrm{1}}{{k}}\mathrm{ln}\:\frac{\frac{{m}}{\rho{V}_{\mathrm{0}} }}{\frac{{m}}{\rho{V}_{\mathrm{0}} }+{e}^{{k}\left({T}−\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{u}}\right)} } \\ $$$$\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +\left({c}−{x}\right)^{\mathrm{2}} }}{\frac{{u}}{{k}}}={kT}−\mathrm{ln}\:\frac{\frac{{m}}{\rho{V}_{\mathrm{0}} }}{\frac{{m}}{\rho{V}_{\mathrm{0}} }+{e}^{\left({kT}−\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{\frac{{u}}{{k}}}\right)} } \\ $$$${let}\:\lambda=\frac{\rho{V}_{\mathrm{0}} }{{m}},\:\Phi={kT},\:\mu=\frac{{u}}{{k}} \\ $$$$\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +\left({c}−{x}\right)^{\mathrm{2}} }}{\mu}=\Phi+\mathrm{ln}\left(\mathrm{1}+\lambda{e}^{\Phi−\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{\mu}} \right)\: \\ $$$$\frac{{d}\Phi}{{dx}}=\mathrm{0}\:{for}\:{minimum}\:{T}. \\ $$$${see}\:{example}. \\ $$
Commented by ajfour last updated on 22/Jun/22
$${Thank}\:{you}\:{very}\:{much}\:{sir}… \\ $$
Commented by mr W last updated on 22/Jun/22
