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Question An underground storage tank is being filled with liquid as shown in the diagram.Initially the tank is empty.At time t hours after filling begins,the volume of liquid is v m^3 & the depth of liquid is h m.It is given that v=4/3(h^3).The liquid is poured in at a rate of 20m^3 per hour,but owing to leakage,liquid is lost at a rate proportional to h^2.When h=1,dh/dt=4.95. i)show that h satisfies the differential equation dh/dt=5/h^2-1/(20). ii)verify that (20h^2)/(100-h^2)= -20 + (2000)/((10-h)(10+h)). Many thanks
Answer Ok , let's review our given facts :
1. At any given time t , the tank has an amount of liquid v(t).
This amount is due to Pouring & Leaking. That means :
v(t)=P(t)-L(t) .
2. P'(t)=20 m^3/h & L'(t)=a[h(t)]^2 . Where a is a constant.
3. v(t)=(4/3)[h(t)]^3 .
First of all let's find a, using the information ,
When h=1,dh/dt=4.95 :
1st : If h=1 then v=(4/3)[1]^3=4/3 . & v'=20-a*1=20-a .
2nd : We know that v(t)=(4/3)[h(t)]^3 which means
h(t)=[3v(t)/4]^(1/3). In this case,
h'(t)=(3/4)^(1/3)*(1/3)[v(t)]^(-2/3)*v'(t)
h'(t)=(3/4)^(1/3)*(1/3)[v(t)]^(-2/3)*[P'(t)-L'(t)]
LEt's use our data :
4.95=(3/4)^(1/3)*(1/3)[4/3]^(-2/3)*[20-a]=0.247(20-a)
(20-a)=19.8 -> a=0.2
Now let's write the equation of h'(t) :
v(t)=(4/3)[h(t)]^3
v'(t)=4[h(t)]^2*h'(t)
P'(t)-L(t)=4[h(t)]^2*h'(t)
20-a[h(t)]^2=4[h(t)]^2*h'(t)
20-(0.2)[h(t)]^2=4[h(t)]^2*h'(t)
20/h(t)^2 - 0.2 = 4h'(t)
5/h(t)^2 - 1/20 = h'(t) Q.E.D
