#Irina Mukhametzhanova, Section 24
#This homework is OK to post

#13.3 Homework
#Exercise 3:
#The formula for the length of the curve over the interval [t0,t1] is:
#s = Integral from t0 to t1 (|r'(t)|dt), where r(t) is the position vector
#The derivative of r(t) is:
#r'(t) = <(2t)', (ln(t))', (t^2)'> = <2, 1/t, 2*t>
#The magnitude of r'(t) is:
#|r'(t)| = sqrt(2^2 + (1/t)^2 + (2*t)^2) = sqrt(4 + 1/t^2 + 4*t^2)
#The integral of this over [1,4] is:
int(sqrt(4 + 1/t^2 + 4*t^2), t = 1 .. 4);
#Result: 15 + 2*ln(2)
#ANSWER: The length of this curve over the interval 1<=t<= 4 is 15 + 2*ln(2)

#Exercise 9:
#To find the arc length function, we first need to find r'(t):
#r'(t) = <(t^2)', (2*t^2)', (t^3)'> = <2*t, 4*t, 3*t^2>
#The magnitude of r'(t) is:
#|r'(t)| = sqrt((2*t)^2 + (4*t)^2 + (3*t^2)^2) = sqrt(4*t^2 + 16*t^2 + 9*t^4) = sqrt(20*t^2 + 9*t^4)
#The integral of |r'(t)| is:
int(sqrt(9*t^4 + 20*t^2), t);
#Result: ((9*t^2 + 20)*sqrt(9*t^4 + 20*t^2))/(27*t)
#We can simplify it a little to get rid of a possible 0 in the denominator:
#((9*t^2 + 20)*sqrt(9*t^4 + 20*t^2))/(27*t) = ((9*t^2 + 20)*t*sqrt(9*t^2 + 20))/(27*t) = ((9*t^2 + 20)*sqrt(9*t^2 + 20))/(27)
#We are trying to find the equation of a specific interval, [0,t]. So, our equation turns out to be:
((9*t^2 + 20)*sqrt(9*t^2 + 20))/(27) - ((9*0^2 + 20)*sqrt(9*0^2 + 20))/(27) = ((9*t^2 + 20)*sqrt(9*t^2 + 20))/27 - (20*sqrt(20))/27 = 
# = ((9*t^2 + 20)^(3/2) - 20^(3/2))/27
#ANSWER: The arc length function is equal to ((9*t^2 + 20)^(3/2) - 20^(3/2))/27

#Exercise 11:
#Speed is the magnitude of the position vector's first derivative, or its velocity vector. So, we first find r'(t):
#r'(t) = <(2*t+3)', (4*t-3)', (5-t)'> = <2, 4, -1>
#The velocity vector is always the same at any time t. So, we take the magnitude of this at time t=4 to find the speed:
#speed = |r'(4)| = sqrt(2^2 + 4^2 + (-1)^2) = sqrt(4+16+1) = sqrt(21)
#ANSWER: The speed of the particle at time t=4 is sqrt(21)

#Exercise 13:
#Speed is the magnitude of the position vector's first derivative, or its velocity vector. So, we first find r'(t):
#r'(t) = <(t)', (ln(t))', ((ln(t))^2)'> = <1, 1/t, (2/t)*ln(t)>
#Then, we calculate the velocity vector at time t=1:
#r'(1) = <1, 1/1, (2/1)*ln(1)> = <1,1,0>
#Finally, we take the magnitude of the result to find speed:
#speed = |r'(1)| = sqrt(1^2 + 1^2 + 0^2) = sqrt(2)
#ANSWER: The speed of the particle at time t=1 is sqrt(2)

#Exercise 15:
#Speed is the magnitude of the position vector's first derivative, or its velocity vector. So, we first find r'(t):
#r'(t) = <(sin(3*t))', (cos(4*t))', (cos(5*t))'> = <3*cos(3*t), -4*sin(4*t), -5*sin(5*t)>
#Then, we calculate the velocity vector at time t=pi/2:
#r'(1) = <3*cos(3*pi/2), -4*sin(2*pi), -5*sin(5*pi/2)> = <0,0,-5>
#Finally, we take the magnitude of the result to find speed:
#speed = |r'(1)| = sqrt(0^2 + 0^2 + (-5)^2) = 5
#ANSWER: The speed of the particle at time t=pi/2 is 5


#13.4 Homework
#Exercise 1:
#r'(t) = <(4*t^2)', (9*t)'> = <8*t, 9>
#T(t) is the unit tangent vector. So, to find it, we take r'(t) and divide it by its magnitude:
#|r'(t)| = sqrt((8*t)^2 + 9^2) = sqrt(64*t^2 + 81)
#T(t) = r'(t)/|r'(t)| = <(8*t)/sqrt(64*t^2 + 81), 9/sqrt(64*t^2 + 81)>
#T(1) = <(8*1)/sqrt(64*1^2 + 81), 9/sqrt(64*1^2 + 81)> = <8/sqrt(145), 9/sqrt(145)>
#ANSWER: r'(t) = <8*t, 9>, #T(t) = <(8*t)/sqrt(64*t^2 + 81), 9/sqrt(64*t^2 + 81)>, #T(1) = <8/sqrt(145), 9/sqrt(145)>

#Exercise 5:
#r'(t) = <(cos(pi*t))', (sin(pi*t))', (t)'> = <-1*pi*sin(pi*t), pi*cos(pi*t), 1>
#T(t) is the unit tangent vector. So, to find it, we take r'(t) and divide it by its magnitude:
#|r'(t)| = sqrt((-1*pi*sin(pi*t))^2 + (pi*cos(pi*t))^2 + 1^2) = sqrt((pi^2)*(1) + 1) = sqrt(pi^2 + 1)
#T(t) = r'(t)/|r'(t)| = <-1*pi*sin(pi*t)/sqrt(pi^2 + 1), pi*cos(pi*t)/sqrt(pi^2 + 1), 1/sqrt(pi^2 + 1)>
#T(1) = <-1*pi*sin(pi*1)/sqrt(pi^2 + 1), pi*cos(pi*1)/sqrt(pi^2 + 1), 1/sqrt(pi^2 + 1)> = <0, -1*pi/sqrt(pi^2 + 1), 1/sqrt(pi^2 + 1)>
#ANSWER: r'(t) = <-1*pi*sin(pi*t), pi*cos(pi*t), 1>, #T(t) = <-1*pi*sin(pi*t)/sqrt(pi^2 + 1), pi*cos(pi*t)/sqrt(pi^2 + 1), 1/sqrt(pi^2 + 1)>, #T(1) = <0, -1*pi/sqrt(pi^2 + 1), 1/sqrt(pi^2 + 1)>

#Exercise 7:
#The equation for the curvature function kappa(t) is:
#kappa(t) = (|r'(t) x r''(t)|)/|r'(t)|^3
#So first, we need to find r'(t):
#r'(t) = <(1)', (e^t)', (t)'> = <0, e^t, 1>
#Then, we can find r''(t):
#r''(t) = <(0)', (e^t)', (1)'> = <0, e^t, 0>
#We can now find the cross product of the results:
#r'(t) x r''(t) = i*(0-e^t) - j*(0-0) + k*(0-0) = <-e^t,0,0>
#The magnitude of the result is:
#|r'(t) x r''(t)| = e^t
#Now we can find the magnitude of the first derivative cubed:
#|r'(t)|^3 = sqrt(0^2 + e^(2*t) + 1^2)^3 = sqrt(e^(2*t) + 1)^3
#So, out curvature function is:
#ANSWER: kappa(t) = (e^t)/sqrt(e^(2*t) + 1)^3

#Exercise 11:
#The equation for the curvature function kappa(t) at a given point is:
#kappa(t) = (|r'(t) x r''(t)|)/|r'(t)|^3
#Our given value of t is t=-1, so, we need to find r'(-1) and r''(-1) to get to the result:
#r'(t) = <(1/t)', (1/t^2)', (t^2)'> = <-1/t^2, -2/t^3, 2*t>
#r''(t) = <(-1/t^2)', (-2/t^3)', (2*t)'> = <2/t^3, 6/t^4, 2>
#r'(-1) = <-1/(-1)^2, -2/(-1)^3, 2*(-1)> = <-1, 2, -2>
#r''(-1) = <2/(-1)^3, -6/(-1)^4, 2> = <-2, 6, 2>
#The cross product of the result is:
#r'(-1) x r''(-1) = i*(4+12) - j*(-2-4) + k*(-6+4) = <16,-6,-2>
#The magnitude of the cross product is:
#|r'(-1) x r''(-1)| = sqrt(256 + 36 + 4) = sqrt(296)
#The#Now we can find the magnitude of the first derivative cubed at t=-1:
#|r'(-1)|^3 = sqrt(1 + 4 + 4)^3 = sqrt(9)^3 = 3^3 = 27
#ANSWER: The curvature at time t=-1 is:
#kappa(-1) = sqrt(296)/27

#Exercise 17:
#The formula for the curvature of the plane curve at point t is:
#kappa(t) = |f''(t)|/(1 + f'(t)^2)^3/2
#So, we first find f'(t) and f''(t) and find their values at time t=2:
#f'(t) = (t^4)' = 4*t^3
#f''(t) = (4*t^3)' = 12*t^2
#f'(2) = 4*2^3 = 32
#f''(2) = 12*2^2 = 48
#kappa(2) = |48|/(1 + 32^2)^3/2 ≈ 0.001462700599
#ANSWER: The curvature of the curve on the plane at t=2 is about 0.001462700599

#Exercise 21:
#The equation for the curvature function kappa(t) at t is:
#kappa(t) = (|r'(t) x r''(t)|)/|r'(t)|^3
#So, we first need to find r'(t) and r''(t)
diff(t - tanh(t), t);
#Result: tanh(t)^2
diff(sech(t), t);
#Result: -sech(t)*tanh(t)
#r'(t) = <tanh(t)^2, -sech(t)*tanh(t)>
diff(tanh(t)^2, t)
#Result: 2*tanh(t)*(1 - tanh(t)^2) = 2*tanh(t)*sech(t)^2
diff(-sech(t)*tanh(t), t);
#Result: sech(t)*tanh(t)^2 - sech(t)*(1 - tanh(t)^2) = sech(t)*tanh(t)^2 - sech(t)^3 = sech(t)*(tanh(t)^2 - sech(t)^2) = sech(t)*(tanh(t)^2 - sech(t)^2)
#r''(t) = <2*tanh(t)*sech(t)^2, sech(t)*(tanh(t)^2 - sech(t)^2)>
#Now, we find the cross product of the results:
#r'(t)x r''(t) = k*(tanh(t)^2*sech(t)*(1-2*sech(t)^2) + 2*tanh(t)*sech(t)^2*sech(t)*tanh(t)) =
# = k*(tanh(t)^2*sech(t) - 2*tanh(t)^2*sech(t)^3 + 2*tanh(t)^2*sech(t)^3) = 
# = k*(tanh(t)^2*sech(t)*(1 - 2*sech(t) + 2*sech(t))) = k*(tanh(t)^2*sech(t))
#The magnitude of the cross product would be:
#|r'(t) x r''(t)| = tanh(t)^2*sech(t)
#Next, we find the magnitude of r'(t) and cube it:
#|r'(t)|^3 = sqrt(tanh(t)^4 + sech(t)^2*tanh(t)^2)^3 = sqrt(tanh(t)^2*(tanh(t)^2 + sech(t)^2))^3 =
# = tanh(t)^3
#The curvature equation for the vector would be:
#kappa(t) = tanh(t)^2*sech(t)/tanh(t)^3 = sech(t)/tanh(t) = 1/sinh(t) = csch(t)


#13.5 Homework
#Exercise 3:
#The velocity vector is the first derivative of the position vector, and acceleration vector is the second derivative.
#v(t) = r'(t) = <(t^3)', (1-t)', (4*t^2)'> = <3*t^2, -1, 8*t>
#a(t) = r''(t) = v'(t) = <(3*t^2)', (-1)', (8*t)'> = <6*t, 0, 8>
#Their values at time t=1 are:
#v(1) = <3, -1, 8>
#a(1) = <6, 0, 8>
#Speed is the magnitude of the velocity vector, so:
#speed(1) = |v(1)| = sqrt(9 + 1 + 64) = sqrt(74)
#ANSWER: The velocity vector at time t=1 is <3, -1, 8>, the acceleration vector is <6, 0, 8> and the speed is sqrt(74)

#Exercise 5:
#The velocity vector is the first derivative of the position vector, and acceleration vector is the second derivative.
#v(theta) = r'(theta) = <(sin(theta))', (cos(theta))', (cos(3*theta))'> = <cos(theta), -sin(theta), -3*sin(3*theta)>
#a(theta) = r''(theta) = v'(theta) = <(cos(theta))', (-sin(theta))', (-3*sin(3*theta))'> = <-sin(theta), -cos(theta), -9*cos(3*theta)>
#Their values at time theta=pi/3 are:
#v(1) = <cos(pi/3), -sin(pi/3), -3*sin(pi)> = <1/2, -sqrt(3)/2, 0>
#a(1) = <-sin(pi/3), -cos(pi/3), -9*cos(pi)> = <-sqrt(3)/2, -1/2, 9>
#Speed is the magnitude of the velocity vector, so:
#speed(1) = |v(1)| = sqrt((1/2)^2 + (-sqrt(3)/2)^2) = sqrt(1) = 1
#ANSWER: The velocity vector at time theta=pi/3 is <1/2, -sqrt(3)/2, 0>, the acceleration vector is <-sqrt(3)/2, -1/2, 9> and the speed is 1

#Exercise 15:
#We can find the velocity vector by integrating the acceleration vector:
#v(t) = int(a(t)dt) = i*(t^2/2) + j*(4*t) + C
#We can use the fact that v(0) = <3,-2> to find the arbitrary constant C:
#v(0) = 0 + 0 + C = C = <3,-2>
#So, the velocity vector is:
#v(t) = i*(t^2/2) + j*(4*t) + 3*i - 2*j = <t^2/2 + 3, 4*t - 2>
#We can find the position vector by integrating the velocity vector:
#r(t) = int(v(t)dt) = i*(t^3/6 + 3*t) + j(2*t^2 - 2*t) + C
#We can use the fact that r(0) = <0,0> to find C:
#r(0) = 0 + 0 + C = C = <0,0>
#So, the position vector is:
#r(t) = i*(t^3/6 + 3*t) + j(2*t^2 - 2*t) + C = <t^3/6 + 3*t, 2*t^2 - 2*t>
#ANSWER: v(t) = <t^2/2 + 3, 4*t - 2> and r(t) = <t^3/6 + 3*t, 2*t^2 - 2*t>

#Exercise 17:
#We can find the velocity vector by integrating the acceleration vector:
#v(t) = int(a(t)dt) = k*(t^2/2) + C
#We can use the fact that v(0) = i to find the arbitrary constant C:
#v(0) = 0 + C = C = i
#So, the velocity vector is:
#v(t) = i + k*(t^2/2)
#We can find the position vector by integrating the velocity vector:
#r(t) = int(v(t)dt) = t*i + (t^3/6)*k + C
#We can use the fact that r(0) = j to find C:
#r(0) = 0 + 0 + C = C = j
#So, the position vector is:
#r(t) = t*i + j + (t^3/6)*k
#ANSWER: v(t) = i + k*(t^2/2) and r(t) = t*i + j + (t^3/6)*k

#Exercise 31:
#The particle is slowing down because, when we plot the two vectors:
with(VectorCalculus);
PlotVector([<12, 20, 20>, <2, 1, -3>]);
#The acceleration's z-value is negative, and is pointing downward, unlike the velocity vector


#14.1 Homework
#Exercise 1:
#To evaluate the function at specified points, we plug in saif points into the function:
#f(2,2) = 2 + 2*2^3 = 2 + 2*8 = 2 + 16 = 18
#f(-1,4) = -1 + 4*(-1)^3 = -1 + 4*(-1) = -5

#Exercise 3:
#To evaluate the function at specified points, we plug in saif points into the function:
#h(3,8,2) = 3*8*2^(-2) = 24*(1/4) = 6
#h(3,-2,-6) = 3*(-2)*(-6)^(-2) = (-6)*(1/36) = -1/6

#Exercise 7:
#The only restriction fot the domain is that 4*x^2-y has to be greater than 0, otherwise the logarithm would be undefined. Our y has to be less than 4*x^2 (y < 4*x^2)
#So, our sketch would be coloring everything below, and including, the parabola y = 4*x^2
#(Sketch is attached in the separate file)

#Exercise 21:
#At x=0 and y=0, z=12
#At x=0 and z=0, y=3
#At y=0 and z=0, x=4
#So, we draw the plane that crosses those three points on the axes.
#(Sketch is attached in the separate file)
#The horizontal traces (if we use a plane z=a to cut through the plane) would have the equation 3*x + 4*y = 12-c
#The vertical traces (if we use planes x=a and y=a to cut through the plane) would have equations z = (12 - 3*a) - 4*y and z = -3*x + (12 - 4*a)

#Exercise 23:
#You draw z = x^2, then draw z = 4*y^2, then combine the two parabolas together.
#(Sketch is attached in the separate file)
#The horizontal traces (if we use a plane z=a to cut through the figure) are all ellipses
#The vertical traces (if we use planes x=a and y=a to cut through the figure) would be parabolas with equations z = a^2 + 4*y^2 and z = 4*a^2 + x^2

#Exercise 33:
#(Sketch is attached in the separate file)

#Exercise 35:
#(Sketch is attached in the separate file)


#14.2 Homework
#Exercise 9:
#lim(x,y -> 2,5) (g(x,y) - 2*f(x,y)) = lim(x,y -> 2,5) (g(x,y)) - lim(x,y -> 2,5) (2*f(x,y)) = 7 - 2*lim(x,y -> 2,5) (f(x,y)) = 7 - 2*3 = 1
#ANSWER: 1

#Exercise 11:
#lim(x,y -> 2,5) (exp(f(x,y)^2-g(x,y))) = exp(lim(x,y -> 2,5) (f(x,y)^2-g(x,y))) = 
# = exp(lim(x,y -> 2,5) (f(x,y)^2) - lim(x,y -> 2,5) (g(x,y)) = 
# = exp(lim(x,y -> 2,5) (f(x,y))^2 - 7) = exp(3^2 - 7) = 
# = exp(9-7) = exp(2)
#ANSWER: e^2

#Exercise 15:
#When we try to plug in (0,0) into the function, we end up with 0/0. So, we need to see if the limit is the same from all directions using a line y = mx:
#lim(x,y -> 0,0) ((x^3 + y^3)/x*y^2) = lim(x,y -> 0,0) ((x^3 + m^3*x^3)/m^2*x^3) = lim(x,y -> 0,0) (x^3*(1+m^3)/m^2*x^3) = lim(x,y -> 0,0) ((1+m^3)/m^2) = (1+m^3)/m^2
#The limit changes with the slope, so, the limit does not exist

#Exercise 21:
#First, we try to plug in (0,0) into the function to see if we can just find the limit by plugging in the point:
#The result turns out to be 0/0, so, we need another method to find the limit.
#We can use the y=mx method to see if the limit is the same from all directions:
#lim(x,y -> 0,0) (x*y/(3*x^2 + 3*y^2)) = lim(x,y -> 0,0) (m*x^2/(3*x^2 + 3*m^2*x^4)) = lim(x,y -> 0,0) (m/(3 + 3*m*x^2)) = m/3
#The limit DOES depend on the slope, so, the limit does not exist

#Exercise 23:
#We try to approach the origin by following the x-axis (so, y=z=0):
#lim(x,y,z -> 0,0,0) ((x + 0 + 0)/(x^2 + 0^2 + 0^2) = lim(x,y,z -> 0,0,0) (1/x) = 1/0
#Now we follow the y-axis (so, x=z=0):
#lim(x,y,z -> 0,0,0) ((0 + y + 0)/(0^2 + y^2 + 0^2)) = lim(x,y,z -> 0,0,0) (1/y) = 1/0
#ANSWER: Because the two limits do not exist, the limit at the origin does not exist as well

#Exercise 27:
#First, we try to plug in the values to see if limit can be easily found:
#(-2)^4*cos(1*pi)/exp(-2+1) = 16*(-1)/exp(-1) = -16*e
#ANSWER: The limit is equal to -16*e

#Exercise 31:
#First, we try to plug in the values to see if limit can be easily found:
#1/sqrt(3^2 + 4^2) = 1/sqrt(9+16) = 1/sqrt(25) = 1/5
#ANSWER: The limit is equal to 1/5

#Exercise 35:
#First, we try to plug in the values to see if limit can be easily found:
#(-3)^2*(-2)^3 + 4*(-3)*(-2) = 9*(-8) + 4*6 = -72 + 24 = -48
#ANSWER: The limit is equal to -48