CoCalc -- MTH 333 Roller Coaster Part I (Brandi Williams, Jaimie James, Jammie Johnson, Skyler Self).sagews

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MIDNIGHT SHOCK ROLLER COASTER

The vectors that make up our roller coaster are labelled r1-r11. These are the vectors that we used to calculate the curvature of our roller coaster for all times, t.

var('t',domain='real')
r1=vector([61*cos(t), 0,61*sin(2*t)] ) #red hill
r2=vector([t,0,0]) #blue line
r3=vector([-15*sin(t)-50,t,(-30*cos(t)+30)]) #Orange Loop
r4=vector([t,6.3,0]) #black line
r5=vector([(-(30*cos(t/7.5)+120.5)), -(30*sin(t/sqrt(55))-7), 3.2*t/sqrt(50)])#green half loop connect to black spiral
r6=vector([-150,-(-70*sin(0.5*t)-7), (45*sin(1.2*t))])#red half hill
r7=vector([(t-91), (5*sin(t)+5), (5*cos(t))+5])
r8=vector([-5*cos(t)-144.9, -5*sin(t)+56.8, t+30.2])
r9=vector([(170*cos(t))-100,(cos(t)+67.5),(12*sin(10*t)+12)])
r10=vector([(2*t)-144.9,-(10.5*sin(t/5)-57),(10.2*cos(-t/5)+15)]) #green loop connects to blue spiral/black wave
r11=vector([(35*sin(t/sqrt(50))+70),-((35*cos(t/8))-35), 3.2*t/sqrt(50)])

t

For each vector, we found the curvature in terms of t. The curvature, as we know, is: $K= [[r' X r'']]/[[r']]^3$ . So, we found the curvatures using this expression and that work is shown below. Once we calculated that curvatures, we plotted each one.

a1=vector([-61*sin(t),0,122*cos(2*t)]); #Derivative of Vector (r1)
a2=vector([-61*cos(t), 0, -244*sin(2*t)]); #Second Derivative of Vector
crossA= a1.cross_product(a2); #cross product
topA=norm(crossA) # Length of r' X r''
bottomA=((norm(a1))^3) #(Length of r')^3
K1=(topA)/(bottomA) #Curvature
show(K1)

\displaystyle \frac{7442 \, \sqrt{{\left(\cos\left(2 \, t\right) \cos\left(t\right) + 2 \, \sin\left(2 \, t\right) \sin\left(t\right)\right)}^{2}}}{{\left(14884 \, \cos\left(2 \, t\right)^{2} + 3721 \, \sin\left(t\right)^{2}\right)}^{\frac{3}{2}}}

b1=vector([1,0,0]);
b2=vector([0,0,0]);

The vector r2 has a curvature of 0. We know this because the cross product of the first and second derivative is 0. Also, it's a line.

c1=vector([-15*cos(t),t,(30*sin(t))]); #Derivative of vector
c2=vector([15*sin(t),t,(30*cos(t))]); #Second Derivative of vector
crossC= c1.cross_product(c2); #cross product
topC=norm(crossC) #Length of r'X r''
bottomC=((norm(c1))^3) #(Length of r')^3
K3=(topC)/(bottomC) #Curvature
show(K3)

\displaystyle \frac{\sqrt{225 \, {\left(t \cos\left(t\right) + t \sin\left(t\right)\right)}^{2} + 900 \, {\left(t \cos\left(t\right) - t \sin\left(t\right)\right)}^{2} + 202500 \, {\left(\cos\left(t\right)^{2} + \sin\left(t\right)^{2}\right)}^{2}}}{{\left(t^{2} + 225 \, \cos\left(t\right)^{2} + 900 \, \sin\left(t\right)^{2}\right)}^{\frac{3}{2}}}

d1=vector([1,0,0])
d2=vector([0,0,0])

The vector r4 has a curvature of 0. We know this because the cross product of the first and second derivative is 0. Also, it's a line like before.

e1=vector([4*sin(0.133333*t),-6*(sqrt(5/11)*cos(t/sqrt(55))),0.4525])
e2=vector([0.53332*cos(0.133333*t),(6/11)*sin(t/(sqrt(55))),0])
crossE= e1.cross_product(e2); #cross product
topE=norm(crossE) # Length of r' X r''
bottomE=((norm(e1))^3) #(Length of r')^3
K5=(topE)/(bottomE) #Curvature
show(K5)

\displaystyle \frac{\sqrt{{\left(3.19992000000000 \, \sqrt{\frac{5}{11}} \cos\left(\frac{1}{55} \, \sqrt{55} t\right) \cos\left(0.133333000000000 \, t\right) + \frac{24}{11} \, \sin\left(\frac{1}{55} \, \sqrt{55} t\right) \sin\left(0.133333000000000 \, t\right)\right)}^{2} + 0.0582388657252900 \, \cos\left(0.133333000000000 \, t\right)^{2} + 0.0609192148760331 \, \sin\left(\frac{1}{55} \, \sqrt{55} t\right)^{2}}}{{\left(\frac{180}{11} \, \cos\left(\frac{1}{55} \, \sqrt{55} t\right)^{2} + 16 \, \sin\left(0.133333000000000 \, t\right)^{2} + 0.204756250000000\right)}^{\frac{3}{2}}}

f1=vector([0,35*cos(0.5*t),54*cos(1.2*t)]);
f2=vector([0,-17.5*sin(0.5*t),-64.8*sin(1.2*t)])
crossF= f1.cross_product(f2);
topF= norm(crossF)
bottomF= ((norm(f1))^3)
K6=(topF)/(bottomF)
show(K6)

\displaystyle \frac{\sqrt{{\left(-2268.00000000000 \, \cos\left(0.500000000000000 \, t\right) \sin\left(1.20000000000000 \, t\right) + 945.000000000000 \, \cos\left(1.20000000000000 \, t\right) \sin\left(0.500000000000000 \, t\right)\right)}^{2}}}{{\left(2916 \, \cos\left(1.20000000000000 \, t\right)^{2} + 1225 \, \cos\left(0.500000000000000 \, t\right)^{2}\right)}^{\frac{3}{2}}}

g1=vector([0,(5*cos(t)),(-5*sin(t))])
g2=vector([0,(-5*sin(t)),(-5*cos(t))])
crossG= g1.cross_product(g2);
topG=norm(crossG)
bottomG=((norm(g1))^3)
K7=(topG)/(bottomG)
show(K7)

\displaystyle \frac{25 \, \sqrt{{\left(\cos\left(t\right)^{2} + \sin\left(t\right)^{2}\right)}^{2}}}{{\left(25 \, \cos\left(t\right)^{2} + 25 \, \sin\left(t\right)^{2}\right)}^{\frac{3}{2}}}

h1=vector([5*sin(t),-5*cos(t),1])
h2=vector([5*cos(t),5*sin(t),0])
crossH= h1.cross_product(h2)
topH= norm(crossH)
bottomH= ((norm(h1))^3)
K8= (topH)/(bottomH)
show(K8)

\displaystyle \frac{\sqrt{625 \, {\left(\cos\left(t\right)^{2} + \sin\left(t\right)^{2}\right)}^{2} + 25 \, \cos\left(t\right)^{2} + 25 \, \sin\left(t\right)^{2}}}{{\left(25 \, \cos\left(t\right)^{2} + 25 \, \sin\left(t\right)^{2} + 1\right)}^{\frac{3}{2}}}

i1=vector([-170*sin(t),-sin(t),120*cos(10*t)])
i2=vector([-170*cos(t),-cos(t),-1200*sin(10*t)])
crossI= i1.cross_product(i2);
topI=norm(crossI)
bottomI=((norm(i1))^3)
K9=(topI)/(bottomI) #Curvature
show(K9)

\displaystyle \frac{120 \, \sqrt{28901} \sqrt{{\left(\cos\left(10 \, t\right) \cos\left(t\right) + 10 \, \sin\left(10 \, t\right) \sin\left(t\right)\right)}^{2}}}{{\left(14400 \, \cos\left(10 \, t\right)^{2} + 28901 \, \sin\left(t\right)^{2}\right)}^{\frac{3}{2}}}

j1=vector([2,(-2.1*cos(t/5)),(-2.04*sin(t/5))])
j2=vector([0, (0.42*sin(t/5)),(-0.408*cos(t/5))])
crossJ= j1.cross_product(j2);
topJ=norm(crossJ)
bottomJ=((norm(j1))^3)
K10=(topJ)/(bottomJ) #Curvature
show(K10)

\displaystyle \frac{\sqrt{{\left(0.856800000000000 \, \cos\left(\frac{1}{5} \, t\right)^{2} + 0.856800000000000 \, \sin\left(\frac{1}{5} \, t\right)^{2}\right)}^{2} + 0.665856000000000 \, \cos\left(\frac{1}{5} \, t\right)^{2} + 0.705600000000000 \, \sin\left(\frac{1}{5} \, t\right)^{2}}}{{\left(4.41000000000000 \, \cos\left(\frac{1}{5} \, t\right)^{2} + 4.16160000000000 \, \sin\left(\frac{1}{5} \, t\right)^{2} + 4\right)}^{\frac{3}{2}}}


k1=vector([((7*cos(t/(5*sqrt(2)))/(sqrt(2)))),(35/8)*sin(t/8), 0.4525])
k2=vector([(-7/10)*sin(t/(5*sqrt(2))),(35/64)*cos(t/8),0])
crossK= k1.cross_product(k2);
topK=norm(crossK)
bottomK=((norm(k1))^3)
K11(t)=(topK)/(bottomK) #Curvature
show(K11(t))

\displaystyle \frac{\sqrt{\frac{2401}{16384} \, {\left(5 \, \sqrt{2} \cos\left(\frac{1}{10} \, \sqrt{2} t\right) \cos\left(\frac{1}{8} \, t\right) + 8 \, \sin\left(\frac{1}{10} \, \sqrt{2} t\right) \sin\left(\frac{1}{8} \, t\right)\right)}^{2} + 0.0612369155883789 \, \cos\left(\frac{1}{8} \, t\right)^{2} + 0.100330562500000 \, \sin\left(\frac{1}{10} \, \sqrt{2} t\right)^{2}}}{{\left(\frac{49}{2} \, \cos\left(\frac{1}{10} \, \sqrt{2} t\right)^{2} + \frac{1225}{64} \, \sin\left(\frac{1}{8} \, t\right)^{2} + 0.204756250000000\right)}^{\frac{3}{2}}}

A=K1(2)
show(A)

\displaystyle \frac{7442 \, \sqrt{{\left(\cos\left(4\right) \cos\left(2\right) + 2 \, \sin\left(4\right) \sin\left(2\right)\right)}^{2}}}{{\left(14884 \, \cos\left(4\right)^{2} + 3721 \, \sin\left(2\right)^{2}\right)}^{\frac{3}{2}}}

#B=K2(2) <--Curvature for r2 is 0.
#D=K4(2) <--Curvature for r4 is 0.

C=K3(2)
show(C)

\displaystyle \frac{\sqrt{202500 \, {\left(\cos\left(2\right)^{2} + \sin\left(2\right)^{2}\right)}^{2} + 900 \, {\left(\cos\left(2\right) + \sin\left(2\right)\right)}^{2} + 3600 \, {\left(\cos\left(2\right) - \sin\left(2\right)\right)}^{2}}}{{\left(225 \, \cos\left(2\right)^{2} + 900 \, \sin\left(2\right)^{2} + 4\right)}^{\frac{3}{2}}}

E=K5(2)
show(E)

\displaystyle \frac{\sqrt{{\left(3.08681825452188 \, \sqrt{\frac{5}{11}} \cos\left(\frac{2}{55} \, \sqrt{55}\right) + 0.574945632088101 \, \sin\left(\frac{2}{55} \, \sqrt{55}\right)\right)}^{2} + 0.0609192148760331 \, \sin\left(\frac{2}{55} \, \sqrt{55}\right)^{2} + 0.0541946961953164}}{{\left(\frac{180}{11} \, \cos\left(\frac{2}{55} \, \sqrt{55}\right)^{2} + 1.31581347396443\right)}^{\frac{3}{2}}}

F=K6(2)
show(F)

\displaystyle 0.0165083865620941

G=K7(2)
show(G)

\displaystyle \frac{25 \, \sqrt{{\left(\cos\left(2\right)^{2} + \sin\left(2\right)^{2}\right)}^{2}}}{{\left(25 \, \cos\left(2\right)^{2} + 25 \, \sin\left(2\right)^{2}\right)}^{\frac{3}{2}}}

H=K8(2)
show(H)

\displaystyle \frac{\sqrt{625 \, {\left(\cos\left(2\right)^{2} + \sin\left(2\right)^{2}\right)}^{2} + 25 \, \cos\left(2\right)^{2} + 25 \, \sin\left(2\right)^{2}}}{{\left(25 \, \cos\left(2\right)^{2} + 25 \, \sin\left(2\right)^{2} + 1\right)}^{\frac{3}{2}}}

I=K9(2)
show(I)

\displaystyle \frac{120 \, \sqrt{28901} \sqrt{{\left(\cos\left(20\right) \cos\left(2\right) + 10 \, \sin\left(20\right) \sin\left(2\right)\right)}^{2}}}{{\left(14400 \, \cos\left(20\right)^{2} + 28901 \, \sin\left(2\right)^{2}\right)}^{\frac{3}{2}}}

J=K10(2)
show(J)

\displaystyle \frac{\sqrt{{\left(0.856800000000000 \, \cos\left(\frac{2}{5}\right)^{2} + 0.856800000000000 \, \sin\left(\frac{2}{5}\right)^{2}\right)}^{2} + 0.665856000000000 \, \cos\left(\frac{2}{5}\right)^{2} + 0.705600000000000 \, \sin\left(\frac{2}{5}\right)^{2}}}{{\left(4.41000000000000 \, \cos\left(\frac{2}{5}\right)^{2} + 4.16160000000000 \, \sin\left(\frac{2}{5}\right)^{2} + 4\right)}^{\frac{3}{2}}}

K=K11(2)
show(K)

\displaystyle \frac{\sqrt{\frac{2401}{16384} \, {\left(5 \, \sqrt{2} \cos\left(\frac{1}{4}\right) \cos\left(\frac{1}{5} \, \sqrt{2}\right) + 8 \, \sin\left(\frac{1}{4}\right) \sin\left(\frac{1}{5} \, \sqrt{2}\right)\right)}^{2} + 0.0612369155883789 \, \cos\left(\frac{1}{4}\right)^{2} + 0.100330562500000 \, \sin\left(\frac{1}{5} \, \sqrt{2}\right)^{2}}}{{\left(\frac{49}{2} \, \cos\left(\frac{1}{5} \, \sqrt{2}\right)^{2} + \frac{1225}{64} \, \sin\left(\frac{1}{4}\right)^{2} + 0.204756250000000\right)}^{\frac{3}{2}}}

Curvature at t=2

K1(2) = 0.00415

K2(2) = 0

K3(2) = 0.13051

K4(2) = 0

K5(2) = 74.3356

K6(2) = 0.01651

K7(2) = 0.2

K8(2) = 0.1923

K9(2) = 8.25 * 10^-9

K10(2) = 0.0485

K11(2) = 0.02203

From these calculations, we can conclude that the MAXIMUM curvature occurs at r5 because when we plugged in 2, we calculated 74.3356. Compared to the other values, this was the greatest value. We also can conclude that the MINIMUM curvature occurs at r9 because when we plugged in 2, we calculated 8.225*10^-9.

This is the plot of the curvatures of (K1-K11)

A=plot(K1,(t,0,pi/2), color='red') #use parameters from plots
C=plot(K3,(t,0,2.05*pi), color='green')
E=plot(K5,(t,0,sqrt(52.91)*pi),color='orange')
F=plot(K6,(t,0,pi/2),color='blue')
G=plot(K7, (t, 0, 4.027 * pi),color='black')
H =plot(K8, (t, 0, 4.027 * pi),color='red')
I=plot(K9, (t,0,pi/2),color='blue')
J=plot(K10, (t,0,sqrt(50)*pi),color='red')
show(A+C+E+F+H+G+I+J)

u=r1(t=1).n(digits=5)
r1prime=diff(r1,t)
p=norm(r1prime)
v=(r1prime)/(p)
h=(u)+(v)
show(h)

\displaystyle \left(-\frac{61 \, \sin\left(t\right)}{\sqrt{14884 \, \cos\left(2 \, t\right)^{2} + 3721 \, \sin\left(t\right)^{2}}} + 32.958,\,0.00000,\,\frac{122 \, \cos\left(2 \, t\right)}{\sqrt{14884 \, \cos\left(2 \, t\right)^{2} + 3721 \, \sin\left(t\right)^{2}}} + 55.467\right)

h(t=1) + (u) simplified is <65.925,0,111.934>

A=plot1=parametric_plot3d(r1,(t,0,pi/2),color='purple',thickness=2)
W=arrow3d((32.958,0.0,55.467),(65.925,0,111.934),10,color='red')
show(W+A)

3D rendering not yet implemented

This is the plot of the Unit Tangent Vector for r1 at t=1.

u3=r3(t=1).n(digits=5)
r3prime=diff(r3,t)
p3=norm(r3prime)
v3=(r3prime)/(p3)
h3=(u3)+(v3)
q=((h3(t=1))+(u3))
show(q)

\displaystyle \left(-\frac{15 \, \cos\left(1\right)}{\sqrt{225 \, \cos\left(1\right)^{2} + 900 \, \sin\left(1\right)^{2} + 1}} - 125.24,\,\frac{1}{\sqrt{225 \, \cos\left(1\right)^{2} + 900 \, \sin\left(1\right)^{2} + 1}} + 2.0000,\,\frac{30 \, \sin\left(1\right)}{\sqrt{225 \, \cos\left(1\right)^{2} + 900 \, \sin\left(1\right)^{2} + 1}} + 27.582\right)

B=plot3=parametric_plot3d(r3,(t, 0, 2.05 * pi),thickness=2, color='purple')
W2=arrow3d((-62.622,1.0,13.791),(q),10,color='red')
show(W2+B)

3D rendering not yet implemented

This is the plot of the Unit Tangent Vector for r3 at t=1.

u5=r7(t=1).n(digits=5)
r7prime=diff(r7,t)
p5=norm(r7prime)
v5=(r7prime)/(p5)
h5=(u5)+(v5)
q5=((h5(t=1))+(u5))
show(q5)

\displaystyle \left(\frac{1}{\sqrt{25 \, \cos\left(1\right)^{2} + 25 \, \sin\left(1\right)^{2} + 1}} - 180.00,\,\frac{5 \, \cos\left(1\right)}{\sqrt{25 \, \cos\left(1\right)^{2} + 25 \, \sin\left(1\right)^{2} + 1}} + 18.415,\,-\frac{5 \, \sin\left(1\right)}{\sqrt{25 \, \cos\left(1\right)^{2} + 25 \, \sin\left(1\right)^{2} + 1}} + 15.403\right)

C=plot5=parametric_plot3d(((t-91), (5*sin(t)+5), (5*cos(t))+5), (t, 0, 6.94 * pi), thickness=2, color='purple')
W5=arrow3d((u5),(q5),10,color='red')
show(W5+C)

3D rendering not yet implemented

This is the plot of the Unit Tangent Vector for r7 at t=1.

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