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Parabola in a plane where y is held to a constant value c The plane is perpendicular to the y axis, and intersects that axis at the point (0,c,0)
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Parabola in a plane where z is held to a constant value c The plane is perpendicular to the z axis, and intersects that axis at the point (0,0,c)
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Imagine a parabola in the plane z = c whose axis is parallel to the y axis as shown in Fig 18-5 Here, x follows t while y follows f (t) We therefore have the parametric system x=t y = f (t) = a1t2 + a2t + a3 z=c For a parabola in the plane z = c whose axis is parallel to the x axis instead of the y axis, the value of y follows t while the value of x follows f (t), so we have x = f (t) = a1t2 + a2t + a3 y=t z=c
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An example Consider a quadratic function in the plane x = 2 Suppose that the parametric equations are
x=2 y=t z = t2 3t + 2
Lines and Curves in Three-Space
Using the knowledge we ve gained so far in this chapter, along with our existing knowledge of algebra (such as we got from Algebra Know-It-All or a comparable algebra book), let s draw a graph of this function Imagine that we re gazing broadside at the plane x = 2 from some distant point on the positive x axis We ve been told that y = t If we stay in the plane x = 2, we can therefore write the quadratic function by direct substitution as z = y2 3y + 2 The coefficient of y2 is positive, so the parabola opens in the positive z direction The above polynomial equation factors into z = (y 1)(y 2) so we can see that z = 0 when y = 1, and also that z = 0 when y = 2 Because x is always equal to 2, we know that the points (2,1,0) and (2,2,0) are on the parabola The curve opens in the positive z direction, so we know that the parabola must have an absolute minimum The y-value at the point, ymin, is the average of the y-values of the points where z = 0 Therefore ymin = (1 + 2)/2 = 3/2 To find the z-value at this point, we plug 3/2 into the quadratic function and get zmin = (3/2)2 3 3/2 + 2 = 9/4 9/2 + 2 = 9/4 18/4 + 8/4 = (9 18 + 8)/4 = 1/4 We ve determined that the coordinates of the absolute minimum are (2,3/2, 1/4) We also know that the points (2,1,0) and (2,2,0) lie on the parabola Figure 18-6 shows these points They re close together, so it s difficult to get a clear picture of the parabola based on their locations But we can find another point to help us draw the curve When we plug in 0 for y, we get z = y2 3y + 2 = 02 3 0 + 2 =0 0+2=2 This tells us that the point (2,0,2) is on the curve It s also shown in Fig 18-6
Another example Now let s look at a quadratic function in the plane where y = 5 Suppose that the parametric equations are
x=t y=5 z = 2t2 + 4t + 3
Parabolas
(2, 0, 2)
Each axis increment is 1/4 unit
(2, 1, 0) +y (2, 2, 0) (2, 3/2, 1/4)
Figure 18-6
Graph of a parabola in a plane parallel to the yz plane, such that x has a constant value of 2 On both axes, each increment represents 1/4 unit
Imagine that we re gazing broadside at the plane y = 5 from somewhere on the negative y axis We have been told that x = t, so we can write the quadratic function as z = 2x2 + 4x + 3 This parabola opens in the positive z direction, because the coefficient of x2 is positive That means this parabola attains an absolute minimum for some value of x Let s call it xmin When x is the independent variable and z is the dependent variable, the general polynomial form for a quadratic function is z = a1x2 + a2x + a3 where a1, a2, and a3 are constants From our algebra courses, we know that xmin = a2/(2a1) In this situation, we have xmin = 4/(2 2) = ( 4)/4 = 1
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