1047.sagews

#Part (a)

t = var('t')

g = var('g')

h1 = function('h1',t)

DE = diff(h1,t,2) + g

desolve(DE, [h1,t])

\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{2} \, g t^{2} + k_{2} t + k_{1}

#k1 is assumed to be the starting height h(0)

#k2 is assumed to be the starting velocity vh(0)

h(t,k1,k2) = -1/2*g*t^2 + k2*t + k1

vh(t,k2) = diff(h(t,k1,k2),t)

vh(t,k2)

\newcommand{\Bold}[1]{\mathbf{#1}}-g t + k_{2}

solve( h(t) == 0,t)

\newcommand{\Bold}[1]{\mathbf{#1}}\left[t = \frac{{(k_{2} - \sqrt{2 \, g k_{1} + k_{2}^{2}})}}{g}, t = \frac{{(k_{2} + \sqrt{2 \, g k_{1} + k_{2}^{2}})}}{g}\right]

t0 = (k2 - sqrt(2*g*k1 + k2^2))/g

t0i = (k2 + sqrt(2*g*k1 + k2^2))/g

vh(t = t0, k2 = 0)

\newcommand{\Bold}[1]{\mathbf{#1}}-k_{2} + \sqrt{2 \, g k_{1} + k_{2}^{2}}

a(t,k2) = vh(t = t0i,k2 = 0)

a(t,0)

\newcommand{\Bold}[1]{\mathbf{#1}}-\sqrt{g k_{1}} \sqrt{2}

#Because k2 = 0, we can assume that vh(t = t0i,k2 = 0) =  - sqrt(2*g*k1), where k1 is h(0)

##Part (b)

#Rate of falling water is given by diff(h,t),  If we imagine every tiny little piece of water to be falling from the very top at height h, we can use the standard gravitation differential equation to show it's velocity. We solved this already, and found the solution - sqrt(2*g*k1), where k1 is h. Torticelli's law tells us that multiplying this rate by the area of the small hole at the bottom through which the water drains out is equivalent to the change of the height at the top multiplied by it's change in height. this gives:
#    
#    A(h)*diff(h,t) = -a*sqrt(2*g*k1)

#Part(c)

m = .50/.195

x = var('x')

c1=plot(.50-m*x,x,(0,.195))

c2=plot(m*(x-1/200),x,(0,.20))

c2

c1

var('y r')

\newcommand{\Bold}[1]{\mathbf{#1}}\left(y, r\right)

y == m*(r-1/200)

\newcommand{\Bold}[1]{\mathbf{#1}}y = 2.56410256410256 \, r - 0.0128205128205128

solve(_, r )

\newcommand{\Bold}[1]{\mathbf{#1}}\left[r = \frac{39}{100} \, y + \frac{1}{200}\right]

#This tells us that the radius r varies by r == 39/100*y + 1/200

R(y) = 39/100*y + 1/200

n(R(.50))

\newcommand{\Bold}[1]{\mathbf{#1}}0.200000000000000

n(R(0))

\newcommand{\Bold}[1]{\mathbf{#1}}0.00500000000000000

A(y) = R(y)^2*pi

A(y)

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{40000} \, {(78 \, y + 1)}^{2} \pi

y=function('y',t)

y

\newcommand{\Bold}[1]{\mathbf{#1}}y\left(t\right)

#a = (1/200)^2*pi

a = var('a')

#Substituting all the parts for the equation, we get A(y)* diff(y,t) = -a * sqrt(2*g*y). We now solve:

DEiii = A(y)* diff(y,t)  + a * sqrt(2*g*y)

DEiii

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{40000} \, {(78 \, y\left(t\right) + 1)}^{2} \pi D[0]\left(y\right)\left(t\right) + \sqrt{g y\left(t\right)} \sqrt{2} a

desolve(DEiii,[y,t])

\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{200000} \, \frac{{(5 \, \pi + 6084 \, \pi y\left(t\right)^{2} + 260 \, \pi y\left(t\right))} \sqrt{g y\left(t\right)} \sqrt{2}}{a g} = c + t

## Part (d)

## Because there is no easy solution for this constant we see that c is actually a time constant, and we can find it by letting t = 0 and y = .5

ciii(y,t,a,g) =  -1/200000*(5*pi + 6084*pi*y^2 +260*pi)*sqrt(g*y)*sqrt(2)/(a*g) - t

ciii(y,t,a,g)

\newcommand{\Bold}[1]{\mathbf{#1}}-t - \frac{1}{200000} \, \frac{{(265 \, \pi + 6084 \, \pi y^{2})} \sqrt{g y} \sqrt{2}}{a g}

n(ciii(.5,0,(1/200)^2*pi,9.8))

\newcommand{\Bold}[1]{\mathbf{#1}}-114.103354508989

## HAHAHA!!! By letting y = 0, we quickly see that this means that the equation gives t= 114.10 [sec] when y = 0. HAHA!

#part e

#The equation remains somewhat related, with a having the same value. However, we find that the radius function now changes. The radius must equal .2 m when the height reaches zero, and .005 m when the height reaches .5 m.

m2 = ((5/1000)-(2/10))/((1/2)-0)

m2

\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{39}{100}

r2(y) = m2*y+(2/10)

r2(0)

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{5}

r2(.5)

\newcommand{\Bold}[1]{\mathbf{#1}}0.00500000000000000

A2(y) = pi*r2(y)^2

A2(y)

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{10000} \, {(39 \, y - 20)}^{2} \pi

a2 = pi*(.005)^2

# We now substitute these values into the new equation:

y=function('y',t)

DEe = A2(y)* diff(y,t)  + a * sqrt(2*g*y)

DEe

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{10000} \, {(39 \, y\left(t\right) - 20)}^{2} \pi D[0]\left(y\right)\left(t\right) + \sqrt{g y\left(t\right)} \sqrt{2} a

desolve(DEe,[y,t])

\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{50000} \, \frac{{(2000 \, \pi + 1521 \, \pi y\left(t\right)^{2} - 2600 \, \pi y\left(t\right))} \sqrt{g y\left(t\right)} \sqrt{2}}{a g} = c + t

#One again, we have to find c by letting y=.5 and t = 0;

ce(y,t,a,g) = -1/50000*(2000*pi + 1521*pi*y^2 -2600*pi*y)*sqrt(g*y)*sqrt(2)/(a*g) - t

n(ce(.5,0,a2,9.8),digits=7)

\newcommand{\Bold}[1]{\mathbf{#1}}-276.0586

#This then gives us that the time constant comes out to be -276.1. If we let y = 0, we quickly see that  t = 276.1 [s]

#Part (f)

#I'll be anylizing (with my little sister!) a sphere like lamp cover with a upper diameter of 0.282575 m and a lower one of 0.0381. The total height is 0.123825. I estimated a .3 m diameter, so thats approximately a .15 m radius.

plot(-sqrt(.15^2-t^2)+0.15,t,0,.15)

-sqrt(.15^2-(0.0381/2)^2)+0.15

\newcommand{\Bold}[1]{\mathbf{#1}}0.00121459244939376

plot(-sqrt(.15^2-t^2)+0.15-0.0012,t,0,.15)

#The '-.00125' is to account for the cut shape along the bottom, which turns out to be about an eighth of an inch. Thus, the radius with respect to h is given by: r = sqrt(h-(0.15-0.0012))^2 - (.15^2))

plot(sqrt(-((-t+(0.15-0.00121779550295673))^2 - (.15^2))),t,0,.15)

y = function('y',t)

l = var('l')

A(y) = sqrt(-((-y+l)^2 - (r^2)))

A(y)

\newcommand{\Bold}[1]{\mathbf{#1}}\sqrt{-{(l - y)}^{2} + r^{2}}

ac = var('ac')

y =function('y',t)

DEc = pi*(A(y))^2 * diff(y,t) + ac*sqrt(2*g*y)

DEc

\newcommand{\Bold}[1]{\mathbf{#1}}-{({(l - y\left(t\right))}^{2} - r^{2})} \pi D[0]\left(y\right)\left(t\right) + \sqrt{g y\left(t\right)} \sqrt{2} \mbox{ac}

desolve(DEc, [y,t])

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{15} \, \frac{{(15 \, \pi \sqrt{2} l^{2} - 10 \, \pi \sqrt{2} l y\left(t\right) - 15 \, \pi \sqrt{2} r^{2} + 3 \, \pi \sqrt{2} y\left(t\right)^{2})} \sqrt{g y\left(t\right)}}{\mbox{ac} g} = c + t

# Again, even though we cannot easily solve for y, we can find the time constan c by setting t = 0 and y(0) = 0.123825. Also for l we substitute 0.15-0.0012, for r we substitute .15, and pi*(0.0381/2)^2 for ac.

c_time_c(t,y,ac,g,l,r) = 1/15*(15*pi*sqrt(2)*l^2 - 10*pi*sqrt(2)*l*y - 15*pi*sqrt(2)*r^2+3*pi*sqrt(2)*y^2)*sqrt(g*y)/(ac*g) - t

c_time_c(t,y,ac,g,l,r)

\newcommand{\Bold}[1]{\mathbf{#1}}-t + \frac{1}{15} \, \frac{{(15 \, \pi \sqrt{2} l^{2} - 10 \, \pi \sqrt{2} l y - 15 \, \pi \sqrt{2} r^{2} + 3 \, \pi \sqrt{2} y^{2})} \sqrt{g y}}{\mbox{ac} g}

n(c_time_c(0,0.127,pi*(0.0381/2)^2,9.8,0.15-0.0012,.15))

\newcommand{\Bold}[1]{\mathbf{#1}}-4.31696446452679

#This time constant is 4 seconds... let me go try it.

#Well... we tried it out. It came out to a disappointing 5.2 seconds. That's not too bad... but my little sister was not impressed. I have failed...