Answer to Question #329559 in Calculus for Keila

Question #329559

A trough is 75cm long and its ends in the form of an isosceles triangle have an altitude of 20cm 

and a base of 30cm. Water is flowing into the trough at the rate of 100cm3/sec. Find the rate at 

which the water level is rising when the water is 10cm deep?


1
Expert's answer
2022-04-17T08:54:16-0400

Let u=100 cm3/sec, l=75 cm,u=100~cm^3/sec,~l=75\space cm,

H=20 cm, B=30 cm, h=10 cm,H=20~cm,~B=30~cm,~h=10~cm,

xx - the current depth of the water layer.

The current volume of the water:

V=12bxl,V=\frac{1}{2}bxl,

where bb is the base of the triangular cross-section of the water layer; this triangle is similar to the cross-section of the trough with similarity factor of k=xHk=\frac{x}{H}, so b=kB=xBHb=kB=x\frac{B}{H}.

V=12xBHxl=Bl2Hx2u=dVdt=BlHxdxdtdxdt=uHBlxdxdt(x=h)=uHBlh=100203075100.089 (cm/sec)=0.89 (mm/sec)V=\frac{1}{2}x\frac{B}{H}xl=\frac{Bl}{2H}x^2\\ u=\frac{dV}{dt}=\frac{Bl}{H}x\frac{dx}{dt}\Rarr\frac{dx}{dt}=\frac{uH}{Blx}\\ \frac{dx}{dt}(x=h)=\frac{uH}{Blh}=\frac{100\cdot20}{30\cdot75\cdot10}\approx\\ \approx0.089~(cm/sec)=0.89~(mm/sec)


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