Mathematics: . A 600-gallon tank is lled with 300 gal of pure water. A spigot isopened and a salt solution containing 1 lb of salt per gallon begins owinginto the tank at a rate of 3gal/min. Simultaneously, a drain is opened atthe bottom of the tank allowing the mixture to leave the tank at a rate of1gal/min. What will be the salt content in the tank at the precise momentthat the volume of solution in the tank reaches the tank s capacity?

A rational number with denominator zero is not defined. option 4...

. A 600-gallon tank is lled with 300 gal of pure

lucy8070

28.01.2022

387.87 lb

Step-by-step explanation:

Rate of flow of salt solution, $\frac{dy}{dt}$ = (1 lb/gal * 3 gal/min) - ( $\frac{y}{300 + (3-1)t}$ lb/gal * 1 gal/min)

$\frac{dy}{dt}$ = 3 - $\frac{y}{300 + 2t}$

$\frac{dy}{dt}$ + $\frac{y}{300 + 2t}$ = 3

The linear differential equation with Q = $\frac{1}{2t+300}$ , R = 3

I.F. = e $\int\limits {\frac{1}{2t+300} } \, dt$ = $(t+150)^{\frac{1}{2} }$

y * I.F. = $\int\limits {Q *I.F.} \, dt+c$

y $(t+150)^{\frac{1}{2} }$ = $\int\limits {3*(t+150)^{\frac{1}{2} }} \, dt+c$

At y = 0, t = o, c = $-2(150)^{\frac{3}{2} }$

y $(t+150)^{\frac{1}{2} }$ = $2(t+150)^{\frac{3}{2} }$ $-2(150)^{\frac{3}{2} }$

y = 2(t + 150) - $\frac{2(150)\frac{3}{2} }{(t+150)\frac{1}{2} }$

Net rate of solution accumulation = (3 - 1) gal/min = 2 gal/min

300 gallon was already there when t = 0.

Time taken to fill the gallon = 300 gal / 2 min/gal = 150 min

Salt content at t=150 min is given by

y = 2(150+150) - $\frac{2(150)\frac{3}{2} }{(150+150)\frac{1}{2} }$

y = 387.87 lb

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