From the previous section we know that the gradient of a function is orthogonal to the level curves of the graph. This property extends to functions in 3 variables.Their level curves are level surfaces and their gradients are orthogonal to them. This provides a simple way of find the normal to the graph in 3-space and, consequently, its tangent plane. Consider the function z = x2 + y2 and the point ( 1, 1, 2 ). By setting w = x2 + y2 - z = 0, we have defined w as a function of 3 variables the gradient of which is < 2x, 2y, -1 >. This gradient is orthogonal to the level surfaces of w. In particular it is orthogonal to the level surface where w = 0. This is our surface. This vector < 2x, 2y, -1 > is orthogonal to the surface at any point. In particular at ( 1, 1, 2 ), giving the normal vector < 2, 2, -1 >. In addition to being orthogonal to the surface it is orthogonal to the tangent plane there as well. ( Be sure this makes sense to you. ) Thus we have a normal vector, < 2, 2, -1 >, to the tangent plane and a point on the plane. The equation of the tangent plane is: 2( x - 1 ) + 2( y - 1 ) - ( z - 2 ) = 0, or z = 2x + 2y - 2.
The viewpoint is from beneath the surface.
Max - min problems in 3-space
To solve extreme value problems in Calculus I we identified critical points by finding where the derivative is equal to 0 or where the derivative fails to exist. We then test those critical points to determine which are maximums and which are minimums. Sometimes we check endpoints of a closed interval to see if extrema resided there as well. We do virtually the same thing for functions in 2 variables in 3-space. We find critical points where both partial derivatives are 0 or where either partial fails to exist. We test those critical points with a second partial derivative test and perhaps check any endpoints of closed regions. Here is the procedure. Given z = f( x, y ) to find relative and absolute extrema on a closed region, R. First set fx( x, y ) = 0 and fy( x, y ) = 0. Find all ordered pairs that make both first partials zero. Find all ordered pairs that make either first derivative undefined. Then test those critical points with the second partial derivative test. Suppose the point ( h, k ) is a critical point. Let d = fxxfyy - ( fxy )2.
For example, given f( x, y ) = 4sin( x )cos( y ), -p £ x £ p, - p £ y £ p.
fx( x, y ) = 4cos( x )cos( y ) and fy( x, y ) = -4sin( x )sin( y ). fx ( x, y ) = fy( x, y ) = 0 at ( p /2, 0 ), ( - p /2, 0 ). The first partials fail to exist at ( p /2, p ), ( p /2, - p ), ( -p/2, p ) and ( -p/2, p ). These last points are on the border and exist as one-sided derivatives only. Now fxx( x, y ) = -4sin( x )cos( y ) = fyy( x, y ) and fxy( x, y ) = -4cos( x )sin( y ). Let's test the point ( p /2, 0 ): d = ( -4 )( -4 ) - 02 = 16. Since d > 0 and fxx ( p /2, 0 ) < 0, the point ( p /2, 0, 4 ) is a relative maximum. Show that ( -p/2, 0 ) gives a relative minimum value of -4. At the other points along the border we obtain either maximums of 4 or minimums of -4 when substituted into f( x, y ). Clearly the absolute maximum of the function is 4 and its minimum, -4.
A typical applied problem would be one like this. In a corner of the space shuttle there is room for storing an experiment contained in a rectangular solid. If the corner space is bounded by the coordinate axes and the triangular region determined by ( 6, 0, 0 ), ( 0, 3, 0 ) and ( 0, 0, 6 ), find the rectangular solid of maximum volume.
The plane determined by the triangular region is x + 2y + z = 6 and the volume to be maximized is V = xyz.
Both first partials equal 0 at ( 0, 0 ), ( 0, 3 ), ( 6, 0 ) and ( 2, 1 ). The point ( 2, 1, 2 ) gives a maximum volume of 4 cubic units.
