### Steve's Blog Assignment

*The town of Calcuville has a water tower whose tank is a circular ellipsoid, formed by rotating an ellipse around its minor axis. The tank is 20 feet tall and 50 feet wide.*

*a) If there are 7.46 gallons of water per cubic foot, what is the capacity of the tank to the nearest 1000 gallons?*

*b)Calcuville imposes water rationing regulations whenever the tank is only one quarter full. How deep is the water in the tank when water rationing becomes necessary?*

*c) During peak usage, 5000 gallons of water are used per hour. How fast is the water level dropping if it is 12 feet deep when peak usage begins?*

Equation for an ellipse ((x-h)²)/a² + ((y-k)²)/b² = 1

width is related to the x axis, and height is related to the y axis

50 = 2a 20 = 2b

a = 25 b = 10

Substitute the values of

**a**and

**b**into the equation and the coordinate (0,0) because no specific coordinate was mentioned and it is also easier to work with.

((x-0)²)/25² + ((y-0)²)/10² = 1

x²/625 + y²/100 = 1

Find volume by using the technique of slicing. Solve for x because ellipse is rotated around the minor axis which happens to be the y axis.

x² + 625y²/100 = 625

x² = 625 - 625y²/100

x = (+/-)√(625-(25y²/4))

You need both the positive and the negative of the square root because it is an ellipse and it exists in both the first and fourth quadrants.

V(y) = πΣ(0,10) √(625-(25y²/4))²dy

Σ takes the place of the integral sign and the numbers after it are the (a,b) of the integral.

That part covers only the top half or the first quadrant of the ellipse, but the bottom or fourth quadrant

**MUST**be included in order to find the volume of the whole ellipse.

V(y) = πΣ(0,10)(√(625-(25y²/4)))² + πΣ(-10,0)(-√(625-(25y²/4)))²

Because the inner function is squared, it becomes positive which means that it becomes the same as the positive one and it can become one integral.

V(y) = πΣ(-10,10)(√(625-(25y²/4)))Â²

The square root and the squared cancel each other out. Leaving a parabola.

V(y) = πΣ(-10,10)(625-(25y²/4))

Antidifferentiate then evaluate

= π(625y-(25y^3/12) from -10 to 10

= π{[625(10) -(25(10^3)/12)]-[625(-10)-(25(-10^3)/12)]}

= π[(4166.6667)-(-4166.6667)]

= π(8333.3333)

= 26179.9388 ft^3

Use the constant of 7.46 gallons of water per cubic foot to calculate how many gallons of water there are.

V= (26179.9388 ft^3)(7.46 gal/ft^3)

V= 195302.3433 gallons

**Answer to Part a) 196000 gallons of water**

Part b) Requires changing volume from gallons to cubic feet and take one quarter of that amount.

0.25V= (1/4)(195302.3433 gallons/7.46 gallons/ft^3)

0.25V= 6544.9847 ft^3

Now take the integral from the bottom of the tank to y and set it equal to 0.25V

πΣ(-10,y)(625-(25y²/4)) = 6544.9847

Divide by π

Σ(-10,y)(625-(25y²/4)) = 2083.3333

Antidifferentiate the left side and evaluate it while still equaling 2083.3333.

625y-(25y^3/12) from -10 to y

[625y-(25y^3)/12]-[625(-10)-(25(-10^3)/12)] = 2083.3333

625y-((25y^3)/12)+4166.6667 = 2083.3333

625y-(25y^3)/12) = -2083.3333

Solve for y to find the depth.

Multiply by 12.

7500y -25y^3 = -25000

0 = 25y^3 -7500y -25000

0 = 25(y^3 -300y -1000)

Using a calculator, I found y to equal -3.472964, in terms of depth, -3.472964 is not the depth, but it is the y coordinate on the graph of the ellipse. For example, when y=-10, depth is 0 ft. If y =0, depth is 10 ft, and finally if y=10, depth is 20 ft, the max.

depth = 10-3.472964

**Answer Part b) depth = 6.5270 ft**

Part c) Given dV/dt = -5000 gallons/hour and the depth of the water is 12 feet, I can find the y value needed to find the rate of how fast the water level is dropping. y=2.

V = 7.46π[(625y-(25y^3)/12)-4166.6667]

Differentiate implicitly

dV/dt = 7.46π[625y'-(25y²y'/4) -0]

dV/dt = 7.46π[625y'-(25/4)y²y']

Substitute the values for dV/dt, and y to solve for y'.

-5000 = 7.46π[625y'-(25/4)(2²)y']

-213.3444 = 625y' -25y'

-213.3444 = 600y'

-.3556 = y'

**Answer Part c) y' = -0.3556 feet/hour, the depth is decreasing at a rate of 0.3556 feet per hour when the depth is 12 feet**.

## 2 Comments:

At 8:56 AM CDT, Mr. Kuropatwa said…

(a) This looks like a rounding error. The correct answer is 196 000 gallons of water, rounded to the nearest 1000 gallons.

(b) Correct!

(c) Correct!

At 7:12 PM CDT, Mr. Kuropatwa said…

I'm glad you edited your work! The thing is though, your work doesn't support a solution of 196 000 gallons. Acording to the work you show the answer is 195 000 gallons. Hmmm ..... I wonder if you can sort this out? It may mean going over your work one more time. ;-)

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