Math 225: Calculus III    Solution 4

Initialization

<<Calculus`Math225`

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

Problem 1

Let r[t_] = (3t^5 - 5t^3) i + (t^4+4t) j + (t+1)^3 k.
a) Find the point(s) where the curve defined by r[t] is not smooth.
b) Plot the curve with ParametricPlot3D[] near the singular point(s).
c) Find the length of the Curve from t==-1 to t==1. (Use NIntegrate[]).

Solution

a) Non-smooth points occur where r'[t] == {0,0,0}.

r[t_] = (3t^5 - 5t^3) i + (t^4+4t) j + (t+1)^3 k

r'[t]

Solve[ r'[t] == {0,0,0} ]

The non-smooth point occurs when t == -1.

r[-1]

b)

ParametricPlot3D[r[t], {t,-1.5,0}]

c) Length is the integral of "speed" with respect to "time".

NIntegrate[Norm[r'[t]], {t,-1,1}]

Problem 2

A particle's position is given by
r[t_] = (1 + Sin[t])i + (3 t^2)j + (t + t^3)k.
a) Find the acceleration, a[Pi], of the particle at time t == Pi.
b) Find the tangential and normal components of a[Pi], i.e., find NUMBERS u and v such that  a[Pi] = u T[Pi] + v n[Pi]  without calculating T[Pi] or n[Pi]! (See Section 12.4 in Finney &Thomas or 2.4 in the Lecture Notes).

Solution

a) Acceleration is given by the second derivative.

r[t_] = (1 + Sin[t])i + (3 t^2)j + (t + t^3)k

r'[t]

a[t_] = r''[t]

a[Pi]

b) The tangential component of acceleration is the derivtive of speed:

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

u = s'[Pi]

N[%]

The normal component of acceleration is given by the formula:

v = Sqrt[a[Pi].a[Pi] - u^2] //Simplify

N[%]

Problem 3

Plot the graphs of the following functions, z = f[x,y], with Plot3D[] and plot several level curves of each function with ContourPlot[] (See Section 3.1 in the Lecture Notes and the examples in the Notebook directory.) Compare the graphs to the level curves.
a) z = x + y^2 - y x^3
b) z = (x^3 - y^3)/(x^2 + y^2)
c) z = E^(-x^2-y^2)Sin[x]

Solution

a)

z = x + y^2 - y x^3

Plot3D[z, {x,-2,2}, {y,-2,2}]

ContourPlot[z, {x,-2,2}, {y,-2,2}]

b)

z = (x^3 - y^3)/(x^2 + y^2)

Plot3D[z, {x,-2,2}, {y,-2,2}]

ContourPlot[z, {x,-2,2}, {y,-2,2}]

c)

z = E^(-x^2-y^2)Sin[x]

Plot3D[z,{x,-2,2},{y,-2,2}]

ContourPlot[z,{x,-2,2},{y,-2,2}]

Problem 4

By considering different paths of approach, show that the limit
of the function f[x_,y_] = x^4 y^2/(x^6 + x^2 y^4 + x^3y^3)
as x and y both go to 0 does not exist. Graph the function near {0,0} and explain why this is so based on the appearance of the graph.

Solution

f[x_,y_] = x^4 y^2/(x^6 + x^2 y^4 + x^3y^3)

Restrict f[x,y] to the line y == 2x by substituting  y->2x into f[x,y]. Then take the limit (f[x,y] is actually constant on this line).

Limit[f[x,y] /. {y->2x}, x->0]

Along the line y == 3x:

Limit[f[x,y] /. {y->3x}, x->0]

We don't have to restrict to straight lines. For example, along the curve y == x^2:

f[x,y] /. {y->x^2}

Simplify[%]

Limit[%,x->0]

Since these limits are diufferent (you only need to find two different!) the limit does not exist. The graph of f[x,y] shows this as well:

Plot3D[f[x,y], {x,-1,1}, {y,-1,1}, PlotPoints->30]

You cans see how the graph is "pinched" together over the point {0,0}, indicating that different function values are obtained depending on how one approaches {0,0}.

Problem 5

a) Find the partial derivatives of f[u_,v_] = u^2 Cos[u/v] with respect to u and v.

b) Find the second order partial derivatives of
    f[x_,y_] = x^3 y^2 - y/x^2 + E^(x y)

Solution

a)

Clear[f,u,v]

f[u_,v_] = u^2 Cos[u/v]

D[f[u,v], u]

D[f[u,v], v]

b)

Clear[f,x,y]

f[x_,y_] = x^3 y^2 - y/x^2 + E^(x y)

D[f[x,y],x,x]

D[f[x,y],x,y]

D[f[x,y],y,x]

Notice how D[f[x,y],x,y]] == D[f[x,y],y,x]. This is not a coincindence!

D[f[x,y],y,y]