Starting with a tossing surface (red):
z = f(x,y) = -(1/4)x² + 16
Consider the 4D surface given by the function:
F(x,y,z) = f(x,y) - z = -(1/4)x² + 16 - z
grad F(x,y,z) = - (1/2)x i + 0 j - k
So at given point (2.788, 5.576, 14.070):
grad F(2.788,5.576,14.070) = -1.394i + 0j - k
From which we can write the equation of the
normal line to the tangent plane:
z = .71736 x + 12.070 in the plane parallel to the z-x plane through the given point.
This normal line is not shown.
Computing grad F(2.788, 5.576, 14.070) dot (x-2.788, y-5.576, z-14.070) set equal to zero is one way to find the tangent plane at the given point.
Tangent Plane, light blue, at the given point is: z = -1.394 x + 17.956472 (y = anything)
The angle, A, of inclination of this tangent plane is computed by finding the cos A =
abs( grad F(2.788, 5.576, 14.070) dot k) divided by the magnitude of grad F.
So angle A = 54.34585 degrees
If we use the direction of the vector from (0,0) to (2.788, 5.576),
to compute a unit vector u = .447213595 i + .894427191 j then:
v = -.894427191 i + .447213595 j is perpendicular to u and
v dot (x-2.788, y-5.576, z-14.070) set equal to zero gives the
equation of the plane of the toss.
Toss Plane, light green, through the given point is: y = 2 x (z = anything)
An equation reprsenting the parabola of the toss "IN" this light green plane is:
z = -(1/20)x² + 16 which may be observed by someone in the crowd
while the marcher throwing the toss looks at z = -(1/4)x² + 16.
Notes about grad f(x,y):
1. grad f(x,y) = -(1/2) x i + 0 j and grad f(2.788,5.576) = -1.394 i + 0 j.
2. Magnitude of grad f(2.788,5.576) = 1.384 = maximum increase value but
the directional derivative in the direction of the toss is:
grad f(2.788,5.576) dot u = -.623415751.
The point (2.788, 5.576, 14.070) represents the position of the centroid
of the tossed object 1/10 second after the peek, vertex, of the throw.
Through this point verify that the line given by
(.5 t, t, -.697 t + 17.956472) is the intersection of the tangent plane
and the tossing plane.
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