Unformatted text preview: 257 3.4. LEVEL CURVES
expressed as the function of x
y=−
=− 8 − x2
4
x2
2− .
4 (We need the minus sign to get y = −1 when x = 2.) Note that this
solution is local : near (2, 1) we would need to use the positive root. Near
√
(2 2, 0), we cannot solve for y in terms of x, because the “vertical line
√
test” fails: for any xvalue slightly below x = 2 2, there are two distinct
points with this abcissa (corresponding to the two signs for the square
root). However, near this point, the “horizontal line test” works: to each
√
y value near y = 0, there corresponds a unique xvalue near x = 2 2
yielding a point on the ellipse, given by
8 − 4y 2 . x= While we are able in this particular case to determine what works and
what doesn’t, in other situations an explicit solution for one variable in
terms of the other is not so easy. For example, the curve
x3 + xy + y 3 = 13 (3.14) contains the point (3, −2). We cannot easily solve this for y in terms of x,
but implicit diﬀerentiation yields
3x2 + y + x dy
dy
+ 3y 2
=0
dx
dx (3.15) and substituting x = 3, y = −2 we get the equation
27 − 2 + 3 dy
dy
+ 12
=0
dx
dx which is easily solved for dy/dx:
15 dy
= −25
dx
dy
5
=− .
dx
3 It seems that we have found the slope of the line tangent to the locus of
Equation (3.14) at the point (3, −2); but how do we know that this line
even exists? Figure 3.3 illustrates what we think we have found. ...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
 Fall '08
 ALL
 Calculus

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