Lab Assignment 1 Student Solutions

Problem #1

Derivative Practice ... pg 175 #35

D[Sin[Tan[Sqrt[1 + x^3]]],x]

        2                   3                  3  2
     3 x  Cos[Tan[Sqrt[1 + x ]]] Sec[Sqrt[1 + x ]]
     ----------------------------------------------
                                 3
                     2 Sqrt[1 + x ]

Derivative Practice...pg 497 #23

Integrate[1/(Sqrt[1 + 2 x] + 3),x]

     Sqrt[1 + 2 x] - 3 Log[3 + Sqrt[1 + 2 x]]

Problem #2

     -SurfaceGraphics-

This is the graph of the saddle shape.

     -Graphics-

The hyperbolas are apparent in these traces (representing the planes z=1,z=4,and z=10, respectively).

     -Graphics-

The traces in the planes x=1,x=2,and x=4 reveal the parabolic character of the graph planely.Similar traces (opening upward,since x^2>0),would be seen in planes y=k,where k is a real number.

Since the graph contains traces of hyperbolas and parabolas, depending on the trace, the name "hyperbolic paraboloid" makes sense.

Problem #3

Plot3D[x y^3/(x^2+y^6),{x,-2,2},{y,-2,2},PlotPoints->40]

     -SurfaceGraphics-

From this graph we can tell that the function does not exist at zero. The question is how to know the equation of the curve to approach along to prove it. By looking at the graph, it can be seen that it will be a complex graph of ax^3+bx^2+cx+d. The first guess is simply x=y^3. This proves to be correct and is proved when done on paper comes up with an answer of one-half.

Problem #4

Plot3D[Cos[x]*Cos[y],{x,-Pi,Pi},{y,-Pi,Pi}]

     -SurfaceGraphics-

Plot3D[Cos[x]+Cos[y],{x,-Pi,Pi},{y,-Pi,Pi}]

     -SurfaceGraphics-

You can tell by these graphs that they are a different shape and that cos[x]+cos[y] has a larger range (smaller and bigger z values)

ContourPlot[Cos[x]*Cos[y],{x,-Pi,Pi},{y,-Pi,Pi}]

     -ContourGraphics-

ContourPlot[Cos[x]+Cos[y],{x,-Pi,Pi},{y,-Pi,Pi}]

     -ContourGraphics-

>From these contour plots, one can see that they are similar around y=0 and x=0, but as you go out to values getting closer to values of Pi and -Pi, you see that z=cos[x] + cos[y] goes downward and z= cos[x]*cos[y] goes upward.

Problem #5

5. The funtional form g(x,y)=ln(ax^2 +y^2 +c)

     -Graphics-

This functional form is a combination of logarithms in two dimensions (x and y are independent of each other), so beginning by looking at a natural log first is useful. The natural log increases very rapidly near x=0, then more slowly when x >> 0. It always increases and is undefined for x < 0.

'A' affects the rate at which the trace in the x-z plane 'goes through' the log function. At small values for a, the curve increases rapidly and the levels off quickly, although it continuels to increase. For small values of a, the curve increases much more gradually. 'A' affects the graph only in tghe x direction.

"C' is the smallest number whose log will be taken--at (0,0). If z is negative, Mathematica goes nuts (because Log[x] where x< 0 is undefined). Because c dictates where on the natural log curve the traces in both axes must begin, it also affects how steep the curve is near the origin--small values of z lead to a more sharply curving surface near the origin, while larger values of z lead to a more gently curing surface.

In the following three examples, I have plotted the 3D plot, then the x-axis, then the y-axis.