The area of the geometry is the sum of the area of the triangle and the semicircle will be 49.12 square feet.
<h3>What is Geometry?</h3>It deals with the size of geometry, region, and density of the different forms both 2D and 3D.
The geometry is the combination of the semicircle and right triangle.
Then the area of the geometry will be
Area = 1/2 x 6 x 8 + π/8 x 8²
Area = 4 x 6 + 3.14 x 8
Area = 24 + 25.12
Area = 49.12 square feet
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Answer:
(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is .
(b) The concentration (in lbs per gallon) when it is at the point of overflowing is .
(c) The theoretical limiting concentration if the tank has infinite capacity is .
Step-by-step explanation:
This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If <em>Q(t)</em> gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for <em>Q(t)</em>.
The main equation that we’ll be using to model this situation is:
Rate of change of <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> exits the tank
where,
Rate at which <em>Q(t)</em> enters the tank = (flow rate of liquid entering) x (concentration of substance in liquid entering)
Rate at which <em>Q(t)</em> exits the tank = (flow rate of liquid exiting) x (concentration of substance in liquid exiting)
Let be the concentration of salt water solution in the tank (in
) and
the time (in minutes).
Since the solution being pumped in has concentration and it is being pumped in at a rate of
, this tells us that the rate of the salt entering the tank is
But this describes the amount of salt entering the system. We need the concentration. To get this, we need to divide the amount of salt entering the tank by the volume of water already in the tank.
is the volume of brine in the tank at time t. To find it we know that at t = 0 there were 200 gallons, 3 gallons are added and 2 are drained, and the net increase is 1 gallons per second. So,
Therefore,
The rate at which enters the tank is
The rate of the amount of salt leaving the tank is
and the rate at which exits the tank is
Plugging this information in the main equation, our differential equation model is:
Since we are told that the tank starts out with 200 gal of solution, containing 100 lb of salt, the initial concentration is
Next, we solve the initial value problem
We solve for C(t)
D is the constant of integration, to find it we use the initial condition
So the concentration of the solution in the tank at any time t (before the tank overflows) is
(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is just the concentration of the solution times its volume
(b) Since the tank can hold 500 gallons, it will begin to overflow when the volume is exactly 500 gal. We noticed before that the volume of the solution at time t is . Solving the equation
tells us that the tank will begin to overflow at 300 minutes. Thus the concentration at that time is
(c) If the tank had infinite capacity the concentration would then converge to,
The theoretical limiting concentration if the tank has infinite capacity is