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ANTONII [103]
1 year ago
13

A tank originally contains 100 gallon of fresh water. Then water containing 0.5 Lb of salt per gallon is pourd into the tank at

a rate of 2 gal/minute, and the mixture is allowed to leave at the same rate. After 10 minute the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at end of an additional 10 minutes.
Mathematics
1 answer:
Assoli18 [71]1 year ago
7 0

Let S(t) denote the amount of salt (in lbs) in the tank at time t min up to the 10th minute. The tank starts with 100 gal of fresh water, so S(0)=0.

Salt flows into the tank at a rate of

\left(0.5\dfrac{\rm lb}{\rm gal}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = 1\dfrac{\rm lb}{\rm min}

and flows out with rate

\left(\dfrac{S(t)\,\rm lb}{100\,\mathrm{gal} + \left(2\frac{\rm gal}{\rm min} - 2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{S(t)}{50} \dfrac{\rm lb}{\rm min}

Then the net rate of change in the salt content of the mixture is governed by the linear differential equation

\dfrac{dS}{dt} = 1 - \dfrac S{50}

Solving with an integrating factor, we have

\dfrac{dS}{dt} + \dfrac S{50} = 1

\dfrac{dS}{dt} e^{t/50}+ \dfrac1{50}Se^{t/50} = e^{t/50}

\dfrac{d}{dt} \left(S e^{t/50}\right) = e^{t/50}

By the fundamental theorem of calculus, integrating both sides yields

\displaystyle S e^{t/50} = Se^{t/50}\bigg|_{t=0} + \int_0^t e^{u/50}\, du

S e^{t/50} = S(0) + 50(e^{t/50} - 1)

S = 50 - 50e^{-t/50}

After 10 min, the tank contains

S(10) = 50 - 50e^{-10/50} = 50 \dfrac{e^{1/5}-1}{e^{1/5}} \approx 9.063 \,\rm lb

of salt.

Now let \hat S(t) denote the amount of salt in the tank at time t min after the first 10 minutes have elapsed, with initial value \hat S(0)=S(10).

Fresh water is poured into the tank, so there is no salt inflow. The salt that remains in the tank flows out at a rate of

\left(\dfrac{\hat S(t)\,\rm lb}{100\,\mathrm{gal}+\left(2\frac{\rm gal}{\rm min}-2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{\hat S(t)}{50} \dfrac{\rm lb}{\rm min}

so that \hat S is given by the differential equation

\dfrac{d\hat S}{dt} = -\dfrac{\hat S}{50}

We solve this equation in exactly the same way.

\dfrac{d\hat S}{dt} + \dfrac{\hat S}{50} = 0

\dfrac{d\hat S}{dt} e^{t/50} + \dfrac1{50}\hat S e^{t/50} = 0

\dfrac{d}{dt} \left(\hat S e^{t/50}\right) = 0

\hat S e^{t/50} = \hat S(0)

\hat S = 50 \dfrac{e^{1/5}-1}{e^{1/5}} e^{-t/50}

After another 10 min, the tank has

\hat S(10) = 50 \dfrac{e^{1/5}-1}{e^{1/5}} e^{-1/5} = 50 \dfrac{e^{1/5}-1}{e^{2/5}} \approx \boxed{7.421}

lb of salt.

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6 0
3 years ago
In ΔRST, r = 210 inches, t = 550 inches and ∠T=166°. Find all possible values of ∠R, to the nearest degree.
valkas [14]

Answer:

m(∠R) = 5°

Step-by-step explanation:

By applying sine rule in the given triangle RST.

\frac{\text{sin}(\angle R)}{ST}= \frac{\text{sin}(\angle T)}{SR}

\frac{\text{sin}(\angle R)}{r}= \frac{\text{sin}(\angle T)}{t}

\frac{\text{sin}(\angle R)}{210}= \frac{\text{sin}(166)}{550}

sin(∠R) = \frac{210\times \text{sin}(166)}{550}

           = 0.09237

m(∠R) = \text{sin}^{-1}(0.09237)

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4 0
3 years ago
What are the sine cosine and tangent of theta=7pi/4 radians
FrozenT [24]

Answer:

see explanation

Step-by-step explanation:

\frac{7\pi }{4} is in the fourth quadrant

Where sin and tan are < 0 , cos > 0

The related acute angle is 2π - \frac{7\pi }{4} = \frac{\pi }{4}

Hence

sin([\frac{7\pi }{4} ) = - sin(\frac{\pi }{4}) = - \frac{1}{\sqrt{2} } = - \frac{\sqrt{2} }{2}

cos(\frac{7\pi }{4}) = cos(\frac{\pi }{4}) =  \frac{\sqrt{2} }{2}

tan(\frac{7\pi }{4}= - tan(\frac{\pi }{4} = - 1

6 0
3 years ago
Read 2 more answers
In the graph above, the coordinates of the vertices of QPR are Q(3, 0), P(5, 6), and R(7, 0). If AQPR is reflected across
ozzi

Answer:

D. (7, 0)

Step-by-step explanation:

The rule for a reflection over the y-axis is (x, y) → (x, -y)

This means that the x-values stay the same while the y-values change.

Q(x, y) → (x, -y)

Q(3, 0) → (3, 0)

Q'(3, 0)

P(x, y) → (x, -y)

P(5, 6) → (5, -6)

P'(5, -6)

R(x, y) → (x, -y)

R(7, 0) → (7, 0)

R'(7, 0)

Therefore, the correct answer is D.

Hope this helps!

7 0
2 years ago
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