Engineering Calculus Notes 269

# Engineering Calculus Notes 269 - 257 3.4 LEVEL CURVES...

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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 x-value 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 x-value 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.

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