Mathematics, 12.03.2020 21:32 jamarlittles9132

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A water tank at a filtration plant is built in the shape of an inverted cone with height 5.2 m and diameter 5 m at the top. Water is being pumped into the tank at a rate of 1.2 m³/min. Find the rate at which the water level is rising when there is 8π m³ of water

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Answer from: que81hhh

The rate at which the water level is rising when there is 8π m³ of water in the tank is; dh/dt = 0.075 m /min

Implicit Differentiation

The formula for Volume of a cone is:

V = ¹/₃πr²h

We are given:

diameter; d = 5 m Thus; radius; r = d/2 = 5/2 = 2.5 m height; h = 5.2 m

Thus;

r/h = 2.5/5.2

Thus; r = (25/52)h

Now, when V = 8π m³/min., the height will be gotten from;

V = (¹/₃)π((25/52)h)²h

V = (¹/₃)π(625/2704)h³

V = π(625/8112)h³

Plugging in 8π for V gives us;

8π = π(625/8112)h³

Simplifying this gives;

h³ = 103.8336

h = ∛103.8336

h ≈ 4.7

Now using implicit differentiation, let us differentiate both sides of V = (¹/₃)π(625/2704)h³ to get;

dV/dt = π(625/2704)h²(dh/dt)

We are given;

dV/dt = 1.2 m³/min

To get dh/dt which is the rate of water level rising, we will plug in 1.2 for dV/dt and 4.7 for h to get;

1.2 = π(625/2704)(4.7)²(dh/dt)

Thus;

dh/dt ≈ 0.075 m /min

Read more about implicit differentiation at; link

Answer from: 1536468

0.075 m/min

Explanation:

You need to use derivatives which is an advanced concept used in calculus.

1. Write the equation for the volume of the cone:

$V=\dfrac{1}{3}\pi r^2h$

2. Find the relation between the radius and the height:

r = diameter/2 = 5m/2 = 2.5mh = 5.2mh/r =5.2 / 2.5 = 2.08

3. Filling the tank:

Call y the height of water and x the horizontal distance from the axis of symmetry of the cone to the wall for the surface of water, when the cone is being filled.

The ratio x/y is the same r/h

x/y=r/hy = x . h / r

The volume of water inside the cone is:

$V=\dfrac{1}{3}\pi x^2y$

$V=\dfrac{1}{3}\pi x^2(2.08)\cdot x\\\\\\V=\dfrac{2.08}{3}\pi x^3$

4. Find the derivative of the volume of water with respect to time:

$\dfrac{dV}{dt}=2.08\pi x^2\dfrac{dx}{dt}$

5. Find x² when the volume of water is 8π m³:

$V=\dfrac{2.08}{3}\pi x^3\\\\\\8\pi=\dfrac{2.08}{3}\pi x^3\\\\\\ 11.53846=x^3\\ \\ \\ x=2.25969\\ \\ \\ x^2=5.1062$ m²

6. Solve for dx/dt:

$1.2m^3/min=2.08\pi(5.1062m^2)\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=0.03596m/min$

7. Find dh/dt:

From y/x = h/r = 2.08:

$y=2.08x\\\\\\\dfrac{dy}{dx}=2.08\dfrac{dx}{dt}\\\\\\\dfrac{dy}{dt}=2.08(0.035964m/min)=0.0748m/min\approx0.075m/min$

That is the rate at which the water level is rising when there is 8π m³ of water.

Answer from: Quest

5x+10 is the correct one

hope this

Answer from: Quest

what statement? is there an image?

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