Tank 1 initially contains 50 gals of water with 10 oz of salt in it, while tank 2 initially contains 20 gals of water with 15 oz

of salt in it. water containing 2 oz/gal of salt flows into tank 1 at a rate of 5 gal/min and the well-stirred mixture flows from tank 1 into tank 2 at the same rate of 5 gal/min. the solution in tank 2 flows out to the ground at a rate of 5 gal/min. if x1(t) and x2(t) represent the number of ounces of salt in tank 1 and tank 2, respectively, set up but do not solve an initial value problem describing this system.

Mathematics

1 answer:

Naddik [55]3 years ago

6 0

$\dfrac{\mathrm dx_1}{\mathrm dt}=\dfrac{2\text{ oz}}{1\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}-\dfrac{x_1(t)\text{ oz}}{50\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}$
$\dfrac{\mathrm dx_2}{\mathrm dt}=\dfrac{x_1(t)\text{ oz}}{50\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}-\dfrac{x_2(t)\text{ oz}}{20\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}$

$\implies\begin{cases}\dfrac{\mathrm dx_1}{\mathrm dt}=10-\dfrac1{10}x_1\\\\\dfrac{\mathrm dx_2}{\mathrm dt}=\dfrac1{10}x_1-\dfrac14x_2\\\\x_1(0)=10\\\\x_2(0)=15\end{cases}$

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The length of a rectangle is increasing at a rate of 4 meters per day and the width is increasing at a rate of 1 meter per day.

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Answer:

$\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

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Order of Operations: BPEMDAS

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Area of a Rectangle: A = lw

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Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle l = 10 \ meters$

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$\displaystyle w = 23 \ meters$

$\displaystyle \frac{dw}{dt} = 1 \ m/day$

Step 2: Differentiate

[Area of Rectangle] Product Rule: $\displaystyle \frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

Step 3: Solve

[Rate] Substitute in variables [Derivative]: $\displaystyle \frac{dA}{dt} = (10 \ m)(1 \ m/day) + (23 \ m)(4 \ m/day)$
[Rate] Multiply: $\displaystyle \frac{dA}{dt} = 10 \ m^2/day + 92 \ m^2/day$
[Rate] Add: $\displaystyle \frac{dA}{dt} = 102 \ m^2/day$