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larisa86 [58]
3 years ago
14

A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

and is lost through evaporation at a rate proportional to the surface area.(a) Show that the volume V(t) of water in the pond at time t satisfies the differential equation dV/dt=k−απ(3a/πh)2/3V2/3,whereαis the coefficient of evaporation.(b) Find the equilibrium depth of water in the pond. Is the equilibrium asymptotically stable?(c) Find a condition that must be satisfied if the pond is not to overflow.
Mathematics
1 answer:
Scrat [10]3 years ago
5 0

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

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is vector v with an initial point of (-5,22) and a terminal point of (20,60) equal to vector u with an intintal point of (50,120
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Answer:

Yes, vectors u and v are equal.

Step-by-step explanation:

We need to check whether vectors u and v are equal or not.

If the initial point is (x_1,y_1) and terminal point is (x_2,y_2), then the vector is

Vector=(x_2-x_1)i+(y_2-y_1)j

Vector v with an initial point of (-5,22) and a terminal point of (20,60).

\overrightarrow v=(20-(-5))i+(60-22)j

\overrightarrow v=25i+38j       ..... (1)

Vector u with an initial point of (50,120) and a terminal point of (75,158).

\overrightarrow u=(75-50)i+(158-120)j

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\overrightarrow u=\overrightarrow v

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5 0
3 years ago
PLEASE HELP ME WITH IT.
scoundrel [369]

OK.  I did it.  Now let's see if I can go through it without
getting too complicated.

I think the key to the whole thing is this fact:

     A radius drawn perpendicular to a chord bisects the chord.

That tells us several things:

-- OM bisects AB. 
   'M' is the midpoint of AB.
   AM is half of AB.

-- ON bisects AC.
    'N' is the midpoint of AC.
   AN is half of AC.

--  Since AC is half of AB,
     AN is half of AM.
     a = b/2 

Now look at the right triangle inside the rectangle.
'r' is the hypotenuse, so

                                            a² + b² = r²

But  a = b/2, so             (b/2)² + b² = r²

(b/2)² = b²/4                   b²/4   + b² = r²

Multiply each side by 4:     b² + 4b² = 4r²
                                       -  -  -  -  -  -  -  -  -  -  -
                                            0  + 5b² = 4r²  
Repeat the
original equation:                a² +  b² =  r²

Subtract the last
two equations:                  -a² + 4b² = 3r² 

Add  a²  to each side:              4b²  =  a² + 3r² .    <=== ! ! !
 
7 0
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