Math 225: Calculus III     Solution 3

Initialization

<< "Calculus`Math225`"

i = {1, 0, 0}; j = {0, 1, 0}; k = {0, 0, 1};

Problem 1

Let f[t_] = ((t^3) - t) i + (3 - (E^(2 t))) j + Log[t] k and
let g[t_] = (t^2) i - t j + Sin[π t] k.
a) Compute f[t].g[t] and Cross[f[t], g[t]].
b) Find the derivative of f[t].g[t] at t==1.
c) Find the integral of Cross[f[t],g[t]] from t==1 to t==2.

Solution

f[t_] = ((t^3) - t) i + (3 - (E^(2 t))) j + Log[t] k; 
    g[t_] = (t^2) i - t j + Sin[π t] k;

f[t].g[t]

f[t]\[Cross]g[t]

(∂_t)(f[t].g[t])

% /. t -> 1

(∫_1^2) f[t]\[Cross]g[t] \[DifferentialD]t

N[%]

Clear[f, g]

Problem 2

Let r[t_] = Sin[(t^2)] i + Cos[(t^2)] j + t k.
a) Find a vector that is tangent to the curve at the point p = {1, 0, (Sqrt[(π/2)])}.
b) Find the equation of the line tangent to the curve r[t] at p.
c) Use ParametricPlot3D[] to plot the curve and the tanget line near p, then 
    combine the plots with Show[].

Solution

p = {1, 0, (Sqrt[(π/2)])};

r[t_] = Sin[(t^2)] i + Cos[(t^2)] j + t k;

a) The value of t that gives r[t] == p is obviously t == (Sqrt[(π/2)]). A tangent vector at
r[(π/2)] == p is given by:

v=r'[(Sqrt[(π/2)])]

b) The equation of the tangent line is

p + t v

c)

tline = 
      ParametricPlot3D[{1, -2 (Sqrt[(π/2)]) t, (Sqrt[(π/2)]) + t, 
          RGBColor[1, 0, 0]}, {t, -1, 1}]

curve = 
      ParametricPlot3D[{Sin[(t^2)], Cos[(t^2)], t, RGBColor[0, 0, 1]}, {t, 
          -2, 3}]

Show[tline, curve, ViewPoint -> {2.372, 1.075, 2.160}]

Clear[p, r, t, tline, curve]

Problem 3

Solve the initial value problem r''[t]==(E^(2 t)) i - Cos[t] j + t k,
r'[0]==j + k, and r[0]==i.
a) Step one: let v[t] be integral of r''[t] with a constant vector added to the result. Then solve the equation v[0]==j+k for the unknown constants. This v[t] is now r'[t].
b) Step two: let r[t] be integral of v[t] with another constant vector added to the result. Then solve the equation r[0]==i for the unknown constants. This gives the final answer r[t].

Solution

Clear[v]

v[t_] = 
      ∫{(E^(2 t)), -Cos[t], t} \[DifferentialD]t + {c1, c2, 
          c3}

Solve[v[0] == j + k]

v[t_] = v[t] /. %[[1]]

r[t_] = ∫v[t] \[DifferentialD]t + {c1, c2, c3}

Solve[r[0] == i]

r[t_] = r[t] /. %[[1]]

Problem 4

The position of a particle in the plane at time t is r[t_] = (i/((t^3) + 1)) - (j/((t^2) + 1)).
a) Find the particle's speed s[t] at any time t≥0. Plot s[t] as a function of t.
b) Find the particle's maximum speed for t≥0.

Solution

r[t_] = (i/((t^3) + 1)) - (j/((t^2) + 1))

a)

s[t_]=Norm[r'[t]]

Plot[s[t], {t, 0, 3}]

b) The plot of s[t] indicates that it has only one maximum for t>0. To find this point we simply solve s'[t]==0.

Simplify[s'[t]]

The maximum seems to be near t==1 so we use this as a starting value in FindRoot[]:

maxpt= FindRoot[s'[t],{t, 1}]

We can also use NSolve[] to find all the roots, including complex numbers. This takes longer.

NSolve[s'[t]==0,t]

The maximum of s[t] is thus:

N[s[t] /. maxpt]

Problem 5

Let r[t_] = t i + (((1 + t))^3) j + (((1 - t))^3) k.
a) Find the unit tangent vector T[t], the unit normal vector n[t] (remember, the letter N is reserved for the function N[] in Mathematica), and the binormal vector
B[t_] = Cross[T[t], n[t]] .
b) Plot T[1], n[1] and B[1] along with the plot of r[t] near t==1.

Solution

r[t_] = t i + (((1 + t))^3) j + (((1 - t))^3) k

a)  The unit tangent vector is given by

T[t_]=((r'[t])/(Norm[r'[t]])) //Simplify

The unit normal vector is given by

n[t_]=((T'[t])/(Norm[T'[t]]))//Simplify

B[t_] = T[t]\[Cross]n[t]  // Simplify

To create the graphics for displaying these vactors at the point r[1]:

v1 = Vector[r[1], r[1] + T[1]]; v2 = Vector[r[1], r[1] + n[1]]; v3 = Vector[r[1], r[1] + B[1]];

curve = ParametricPlot3D[r[t], {t, 0, 1.5}]

Show[curve, v1, v2, v3]