Mathmom

Four times I tried to answer this question, but question keeps getting deleted. First few times, I thought perhaps poster deleted question because he wanted to repost in hopes of getting an answer, but last time question had only been posted for about 20 minutes.

Question:

A leaky bucket that weighs 6 lb and a rope of neglible weight are used to draw water from a well that is 373 ft deep. The bucket is filled with 35 lb of water and is pulled up at a rate of 5 ft/sec, but water leaks out of the bucket at a rate of 0.5 lb/sec. Find the work done in pulling the bucket to the top of the well (in ft-lb).

Additional Details (included last time question posted):

This is what I've done:

Force at time t: 6 lbs + 35 lbs - 0.5 lbs/sec * t sec = (41 - 0.5t) lbs

F(t) = 41 - 0.5t

Now we will express F in terms of distance x:

v = x/t = 5

t = x/5

F(x) = 41 - 0.5t x/5

F(x) = 41 - 0.1 x

Integrating from 0 to 373, I got:

W = ∫ (44 - 0.1x) dx from 0 to 373

W = 8336.55 ft-lbs

It says its not the correct answer. What am I doing wrong?

_______________________________

My answer:

What you are doing wrong is forgetting to take into account that eventually, bucket becomes empty

Here's my answer, which I tried to post before:

Let t = time (in sec)

Let h = height of bucket (in feet) from bottom of well -----> h varies from 0 to 373

Bucket is pulled up at a rate of 5 ft/sec -----> dh/dt = 5 -----> h = 5t -----> t = h/5

Let w = weight of bucket and water (in lbs), w(0) = 6 + 35 = 41

Water leaks out of bucket at a rate of 0.5 lb/sec -----> dw/dt = -0.5

w = 41 - 0.5t

w = 41 - 0.5 (h/5)

w = 41 - 0.1 h

Since water leaks out at a rate of 0.1 lb/ft, and initial amount of water = 35lb,

it will take 35lbs/(0.1 lb/ft) = 350 ft for bucket to be completely empty

So we have piecewise function:

w(h) = 41 - 0.1 h . . . . . 0 ≤ h ≤ 350 <--------- this matches your equation

w(h) = 6 . . . . . . . . . . . . 350 < h ≤ 373 <----- weight of empty bucket only

Work required to lift bucket and water weighing w(h) lbs a distance of ∆h feet

W = w(h) ∆h ft-lb

Integrating from h = 0 to h = 373, we get:

W = ∫₀³⁷³ w(h) dh

W = ∫₀³⁵⁰ (41 - 0.1 h) dh + ∫₃₅₀³⁷³ (6) dh

W = (41h - 0.05h²) |₀³⁵⁰ + 6h |₃₅₀³⁷³

W = (41 * (350 - 0) - 0.05 * (350² - 0)) + 6 (373 - 350)

W = 14350 - 6125 + 138

W = 8,363 ft-lb

_______________________________

I guess I'll just send email to person who posted question, with a link to this question and answer. I'm done trying to answer question, but didn't want my answer to go to waste.

Mαthmφm

Again, someone posts 5 math questions within a few minutes then deletes them all. In the interest of not letting my work go to waste, here is the question and my answer. Any comments or alternate solutions are welcome.

Question:

Find the point (in the first quadrant) on the lemniscate 2(x^2+y^2)2=25(x^2−y^2) where the tangent is horizontal.

how do you solve for x and y?

______________________________

My answer:

First we find dy/dx by implicit differentiation:

2(x²+y²)² = 25(x²-y²)

4(x² + y²) (2x + 2y dy/dx) = 25 (2x - 2y dy/dx)

2x (4x² + 4y²) + 2y (4x² + 4y²) dy/dx = 50x - 50y dy/dx

2y (4x² + 4y²) dy/dx + 50y dy/dx = 50x - 2x(4x² + 4y²)

2y (4x² + 4y² + 25) dy/dx = 2x (25 - 4x² - 4y²)

dy/dx = x (25 - 4x² - 4y²) / [y (4x² + 4y² + 25)]

Tangent is horizontal when dy/dx = 0

x (25 - 4x² - 4y²) / [y (4x² + 4y² + 25)] = 0

x (25 - 4x² - 4y²) = 0

x = 0

25 - 4x² - 4y² = 0 -----> x² + y² = 25/4

Now we plug these values into equation of lemniscate: 2(x²+y²)² = 25(x²-y²)

x = 0

2(0+y²)² = 25(0-y²)

2y⁴ + 25y² = 0

y² (2y² + 25) = 0

y = 0

But when x = 0 and y = 0, then dy/dx = 0/0 which is indeterminate.

Checking graph of 2(x²+y²)² = 25(x²-y²), we see that tangent is not horizontal at (0,0)

http://www2.wolframalpha.com/input/?i=2%28x%C2%B2%...

x² + y² = 25/4

2(x²+y²)² = 25(x²-y²)

2(x²+y²)² = 25(x²+y²-2y²)

2(25/4)² = 25(25/4-2y²)

625/8 = 625/4 - 50y²

50y² = 625/4 - 625/8

50y² = 625/8

y² = 25/16

y = 5/4 (we eliminate y = -5/4 ---> not in first quadrant)

x² + y² = 25/4

x² = 25/4 - y²

x² = 25/4 - 25/16

x² = 75/16

x = 5√3/4 (we eliminate x = -5√3/4 ---> not in first quadrant)

Point in first quadrant where tangent is horizontal: (5√3/4, 5/4)

Mαthmφm

So I answered a question about differential equations, and put in quite a bit of effort.

Then, when I went to preview my answer, I was told that the question was deleted, so I went to this user's page and found no questions listed. So I checked to see if he/she had previously posted any other questions (there are ways to do this) and lo and behold, this person had actually asked and deleted SIX questions.

I will not name this person (lest my question gets reported and deleted), and there is no way to actually report this person (you can only report questions - but they keep getting deleted).

Instead, I will post the question and my answer. Perhaps this can help some other person.

___________________________________

Question:

Differential equations solve these impossible problems...for me at least... please?

1. y ' ' + y = tanx

Use variation of parameters I guess?

2. Undetermined coefficients?

y ' ' + 6y ' +9 = x²+6x+9

___________________________________

My answer:

First we find solutions to homogeneous differential equation: y'' + y = 0

r² + 1 = 0

r = ±i

y₁ = e^0 cos(x) = cos(x)

y₂ = e^0 sin(x) = sin(x)

Particular solution: yp = u₁(x) y₁(x) + u₂(x) y₂(x), where

u₁'y₁ + u₂'y₂ = 0

u₁'y₁' + u₂'y₂' = tan(x)

u₁' cos(x) + u₂' sin(x) = 0 . . . . . . . . [1]

−u₁' sin(x) + u₂' cos(x) = tan(x) . . . . [2]

Multiply equation [1] by sin(x) and [2] by cos(x) and solve for u₂'

u₁' sin(x)cos(x) + u₂' sin²(x) = 0

−u₁' sin(x)cos(x) + u₂' cos²(x) = sin(x)

------------------------ ------------------------

u₂' (sin²x + cos²x) = sin(x)

u₂' = sin(x)

u₂ = -cos(x)

Multiply equation [1] by cos(x) and [2] by -sin(x) and solve for u₁'

u₁' cos²(x) + u₂' sin(x)cos(x) = 0

u₁' sin²(x) − u₂' sin(x)cos(x) = −sin²x/cos(x)

---------------------------- ----------------------------

u₁' (sin²x + cos²x) = −sin²x/cos(x)

u₁' = −sin²x/cos(x)

u₁' = (cos²x−1)/cos(x)

u₁' = cos(x) − sec(x)

u₁ = sin(x) − ln|tan(x) + sec(x)|

yp = cos(x) (sin(x) − ln|tan(x) + sec(x))| − sin(x) cos(x)

yp = −cos(x) ln|tan(x) + sec(x)|

General solution:

y = A cos(x) + B sin(x) − cos(x) ln|tan(x) + sec(x)|

==============================

Is differential equation really y'' + 6y' + 9 = x² + 6x + 9

or should it actually be y'' + 6y' + 9y = x² + 6x + 9 ?

I'll assume the latter (since I don't see much need to have +9 on both sides)

Find solution to homogeneous differential equation y'' + 6y' + 9y = 0

r² + 6r + 9 = 0

(r + 3)² = 0

r = -3 (double root)

yh = c₁ e^(-3x) + c₂ x e^(-3x)

Use method of undetermined coefficients to find particular solution

yp = Ax² + Bx + C

yp' = 2Ax + B

yp'' = 2A

y'' + 6y' + 9y = x² + 6x + 9

(2A) + 6(2Ax + B) + 9(Ax² + Bx + C) = x² + 6x + 9

(9A)x² + (12A + 9B)x + (2A + 6B + 9C) = x² + 6x + 9

Matching coefficients we get

9A = 1

12A + 9B = 6

2A + 6B + 9C = 9

Solving, we get: A = 1/9, B = 14/27, C = 17/27

yp = 1/9 x² + 14/27 x + 17/27

yp = 1/27 (3x² + 14x + 17)

General solution:

y = c₁ e^(-3x) + c₂ x e^(-3x) + 1/27 (3x² + 14x + 17)

Ματπmφm

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How does that even come close to "Contact us"?

Here's my problem: I recently changed internet provider, so I decided to update my contact information with Yahoo. I changed my email address, but I could not not change my Yahoo ID (which was the same as my old email address). Unfortunately, my old Yahoo account was just deactivated, and I have lost all access to my Yahoo!Answers information. I've been answering questions for 4 years, and have accumulated 61,410 pts by answering over 8,000 question (mostly in math).

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