Unformatted text preview: Quiz 1 1. Describe fully the geometry of the surface or region in R3 given by the set of points
(1:, y, 2) satisfying the equation a72+y2+22+32=4x. Show that the intersection of this geornetric object With the plane 2 = —l is a circle,
and determine its centre and radius. Solution:
We rearrange and complete the square: 172—4a:+y2+22+32=0 3 2 9 25 5 2 this is a sphere with centre at (2,0, —3/2) and radius 5/2. Substituting 2 = —1 into the original equation gives 1:2 —— 3/2 +1— 3 = 41:, or :52—4:c+y2=2 => (:v—2)2——y2=2+4=6, which in the gay—coordinates represents a circle centred at (a:,y) = (2,0) and with
radius x/é. In three dimensions, it is a circle lying in the plane 2 = —l, with centre (a:,y,2) = (2,0, —1) and radius x/E. 2. Find the unit vectors that are parallel to the tangent line to the parabola y = $2 at
the point (2, 4). Solution:
The slope of the tangent line to the curve 3/ = 1:2 is found by taking the derivative; thus
. change in y
the tangent line at a: = 2 has slope 29119,:2 = 4 = —. .
change In a: Two vectors parallel to the tangent line are (1,4) and —(1,4> = (—1,—4>. To find the
corresponding unit vectors, we divide by the magnitude v12 + 42 = V17- Hence we find: Two unit vectors parallel to the tangent line are ...
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