Math 225: Calculus III	Quiz 5    2/25&2/27/96

<< "Calculus`Math225`"; i = {1, 0, 0}; j = {0, 1, 0}; k = {0, 0, 1};

1. Find the equation of the plane tangent to the surface x^3 y - z^3 x == 2 at the point {1,3,1}.

Solution

The normal vector to the surface at {1,3,1} is given by the gradient:.

Grad[x^3 y - z^3 x]

n = % /. {x->1, y->3, z->1}

The equation of thetangent plane is then given by

n.{x,y,z} == n.{1,3,1}

2. Show that the points {1,1} and {-1,-1} are critical points of the
function f[x_,y_] = 2y^3 - 3 y + 2 x^3 - 3x^2 y and determine whether
these points are local maxima, local minima, or saddle points.

Solution

f[x_,y_] = 2y^3 - 3 y + 2 x^3 - 3x^2 y;

We must show that {±1,±1} are solutions of

eqns = { D[f[x,y],x] == 0,
         D[f[x,y],y] == 0 }

eqns /. {x->1, y->1}

eqns /. {x->-1, y->-1}

We can also see this by solving the equations...

Solve[eqns]

or using the function CriticalPoints[]:

cp = CriticalPoints[f[x,y]]

To detect which are max/min or saddle points we use the second derivative test.

d = D[f[x,y],x,x]D[f[x,y],y,y] - D[f[x,y],x,y]^2

Evaluate d at the critical points {±1,±1}:

d /. {x->1, y->1}

d /. {x->-1, y->-1}

Since  d > 0, the graph is uniformly curved at these points. To see if it is curved up or down, check a second partial derivative:

D[f[x,y],x,x] /. {x->1, y->1}

Positive means curved up at {1,1}, so f[x,y] has a local minimum at {1,1}.

D[f[x,y],x,x] /. {x->-1, y->-1}

Negative means curved down at {-1,-1}, so f[x,y] has a local maximum at {-1,-1}.

Plot3D[f[x,y], {x,-2,2}, {y,-2,2},
BoxRatios->{1,1,1},ViewPoint->{2.560, 2.094, 0.713}]