Fangyao- solution 7.sagews

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####################pbset 7
# Exercise 1
#Exercise 1. Make a CAS function riemann_sum(f, a, b, n, c) that takes five inputs:
#•end points a,b, a<b,of interval of integration
#•function f continuous on the interval
#•positive integer n, the number of subintervals of equal length in partition of interval [a,b]
#•and parameter p.
#The function returns
#• left Riemann sum if c = 0;
#• middle Riemann sum if c = 1;
#• right Riemann sum if c = 2
#Test your function on example of your choice.

def riemann_sum(f,a,b,n,p):#p takes only one of the values 0, 1, 2
    #Define the partition of the interval [a,b] into n subintervals of equal length:
    delt=(b-a)/n #encode the length of a subinterval #delt=(b-a)/n
    L=[a+delt*j for j in range(n+1)] # L is a list of end points of subintervals   #a+n*delt=b
    # For each case below S is the sum of the given function values at appropriate x-values
    if p==0:
        S=sum(a+k*delt for k in range(n))
    elif p==1:
        S=sum(((a+b)/2)+k*delt for k in range(n))
    else:
        S=sum(b-k*delt for k in range(n))
    return S*delt

#My example:
riemann_sum(x^3-3*x^2,0,2,30,0).n()

1.93333333333333

#Your example:
riemann_sum(2*x^3-3*x^2+x,0,3,40,0).n()

4.38750000000000

#Exercise 2
#Make a CAS function integral_MVT(f, a, b) that takes the end points of a closed interval and a continuous function defined on this interval and returns a figure with the graph of the function and the line y=f__ave. Here f_ave is the average value of the function f on the interval [a,b].
def integral_MVT(f,a,b):
    G=Graphics()
    #Make a plot of f on [a,b]:
    fcn_plot=plot(f,a,b,legend_label="f(x)",figsize=[4,2],thickness=2)
    G+=fcn_plot
    #Find the average value of f on [a,b]:
    f_ave=f.integral(x,a,b)/(b-a)
    #Find location of the point c s.t. f(c)=f_ave
    c=find_root(f-f_ave,a,b)
    #Define the constant function equal the f_ave on [a,b] and make a plot of the function
    f_c(x)=f(c)
    f_ave_plot=plot(f_ave,a,b,thickness=2,legend_label="line y=f_ave",linestyle="dashed")
    G+=f_ave_plot
    # Make a plot of two points, c on the x-axis and corresponding point on the graph of f
    pt_plot=point([[c,0],[c,f(c)]],size=40,color='red')
    G+=pt_plot
    return G.show()
#My example:
f(x)=3*sin(x)
integral_MVT(f,0,pi)# Include your own example

# Your example:
f(x)=x^2+2
integral_MVT(f,0,2*pi)

# Exercise 3
#Consider the parametric curve x=t^2-4,  y=t-t^3, -2<=t<=2.
#Plot the curve. You will see two parts of the area bounded by the curve, a curvilinear triangle and a loop.
#Consider only the area bounded by the curvilinear triangle. Use CAS assistance to find the vertical line that halves the area bounded by this curve. Make a figure with the curve plot and the vertical line.  Show your work.
var('t')
x(t)=t^2-4
y(t)=t-t^3

t

f_plot=plot(x(t),-3,t,3,figsize=[4,3])
x_plot=plot(y(t),-3,t,3,figsize=[4,3])

(f_plot+x_plot).show()

part1=parametric_plot((x(t),y(t)),(t,-2,1));part1

#Mini project: Area of the tilted ellipse

#Part 1. Find the general formula for the area of ellipse that is defined by the formula
#A*x^2+2*B*x*y+C*y^2 = 1  with B^2-A*C < 0.
#Hint: Apply the completing the square technique to the first two terms in the lhs of the given equation. This allows you to rewrite the given equation in the form ((x+k)^(2))/(a^(2))+(y^(2))/(b^(2))=1 with k, a, b depending on parameters A, B, C. Then think about relation between the area of standard ellipse and the area of its image under shear transformations

#Part 2. Apply the formula derived in Part 1 to the ellipse with parameters A = 1/4, B = 2/5, C = 1.
#Plot the given (sheared) ellipse and the corresponding standard ellipse in one figure.

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var('A B C')

<function reset at 0x7f7926cb30c8> (A, B, C)

a=1/sqrt(A);b=sqrt(A/A*C-B^2)

pi/sqrt(1/4-4/25)

10/3*pi

A=0.25;B=0.4;C=1
a=1/sqrt(A);b=sqrt(A/A*C-B^2)
a

2.00000000000000

b

0.916515138991168

x(t)=(2*cos(t))
y(t)=((5/3)*sin(t))
p=parametric_plot((x(t),y(t)),(t,0,2*pi),figsize=[4,3],thickness=2);p

x(t)=(2*cos(t)-(8/3)*sin(t))
y(t)=((5/3)*sin(t))
q=parametric_plot((x(t),y(t)),(t,0,2*pi),color='red',figsize=[4,3],thickness=2);q+p

## E1 incorrect, -3; E2: 2/2.Extra points: +2. Total: 4/5

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x(t)=2*cos(t)-(8/3)*sin(t)
y(t)=(5/3)*sin(t)
p=parametric_plot(((x(t),y(t)),(t,0,2*pi)),figsize=[4,3],thickness=2);p

Error in lines 3-3 Traceback (most recent call last): File "/cocalc/lib/python2.7/site-packages/smc_sagews/sage_server.py", line 1191, in execute flags=compile_flags), namespace, locals) File "", line 1, in <module> File "/ext/sage/sage-8.9_1804/local/lib/python2.7/site-packages/sage/misc/decorators.py", line 493, in wrapper return func(*args, **options) File "/ext/sage/sage-8.9_1804/local/lib/python2.7/site-packages/sage/plot/plot.py", line 2625, in parametric_plot raise ValueError("the number of functions and the number of variable ranges is not a supported combination for a 2d or 3d parametric plots") ValueError: the number of functions and the number of variable ranges is not a supported combination for a 2d or 3d parametric plots

#####################################################
#Mini project: Projectile motion
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### Your solution for Step 1 goes here
x0=0; y0=2; l=20; h=3.5; V=15;H=2;g=9.81
var('alpha')
# Governing equations: m*x'=0, m*y'=-m*g.
# Initial conditions:x0=0,y0=H,vx=V*cos(alpha),vy=V*sin(alpha)
# Solution:
x(t)=V*cos(alpha)*t
y(t)=-g*(t^2/2)+V*sin(alpha)*t+H

alpha

# You code goes here
# Write the equation x(tfence)=(fence location) using solution x(t) found on Step 1
#Solve the equaiton for tfence
tfence=l/(V*cos(alpha));tfence
# tfence involves the unknown parameter $\alpha$.

4/3/cos(alpha)

# Write the equation y(tfence)=h using th solution for y(t) found on Step 1.

eq=y(tfence)==h;eq

20*sin(alpha)/cos(alpha) - 8.72000000000000/cos(alpha)^2 + 2 == 3.50000000000000

#Use the identity 1/cos(s)^2=1+tan(s)^2 to rewrite the equation as quadratic with respect to $z=tan(alpha)$ and solve for $z$. Use the dictionary type for your solution.
var('z')
eq_z=20*z-8.72*(1+z^2)-1.5==0
tan_alpha=solve(eq_z,z,solution_dict=True);tan_alpha

z [{z: -1/218*sqrt(6801) + 125/109}, {z: 1/218*sqrt(6801) + 125/109}]

# Now use the inverse of the function and find alpha1 and alpha2 from equation tan(alpha)=z
alpha1=(arctan(tan_alpha[0][z])).n()
alpha2=(arctan(tan_alpha[1][z])).n()

#Define parametric equations of the two trajectories
#First trajectory:
y1(t)=y(t).subs(alpha=alpha1);x1(t)=x(t).subs(alpha=alpha1)
#Second trajectory
y2(t)=y(t).subs(alpha=alpha2);x2(t)=x(t).subs(alpha=alpha2)
#For each trajectory find the time when projectile hits the ground. Experiment with time range to bracket the solution or just play with plots to find the needed range
tfinal1=find_root(y1(t),0,4)
tfinal2=find_root(y2(t),0,4)

plot1=parametric_plot((x1(t),y1(t)),(t,0,tfinal1),color='green')
plot2=parametric_plot((x2(t),y2(t)),(t,0,tfinal2),color='blue')
(plot1+plot2).show()