By Robert Carl Yates

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T These are called critical points. Y Dxy=O DxY,. x xZ Fig. 2 The student should review here the material given previously in Par. 4. t Or at which Dsy fails to exist. Sec. 3 THE DERIVATIVE 48 A critical point is a "high" point if the tangent turns through it from positive to negative slope; a "low" pointif from negative topositive. Such critical points locate relative maxima and minima values of y and offer valuable information in curve sketching. For example, y= x3-3x2+2 (2,-2) with Fig. , at x = 0, 2.

Y = x3- 8 14. ,y= x'-- 8x 49 THE DERIVATIVE Sec. y=x2+1 17. x2y + 4y - 8 = 0 18. (1 + x)y2 = x2(3 - x) 19. x2 + 9y2 - 4x = 0 20. 4. The Derivative in Polar Coordinates Let us apply the delta process to the relation r = f (8) and then seek a geometric interpretation of the derivative Der. Consider the Cardioid r=1+sin0 We have (1) (2) r+Ar=l+sin(B+A8) Ar = sin (8 + AB) - sin 8 = cos 0 sin (AB) + sin 0 cos (AB) - sin 8 cos 9 sin (A9) - sin 0 [1 - cos (A0) ] Ar (3) AB (4) = cos 0 sin (AB) AB Der = (cos 0) (1) - sin 0 1 - cos (A8) AB - (sin 0) (0).

Moreover, since cos0 = A1X2 + µ1F12 -I/T2 +112 , then, if 0 = 90°, A1A2 + µ1µ2 = , and, if this is 0, 0 = 90° or -90°. Thus, two lines are perpendicular if the scalar product of their direction numbers is 0, and conversely. Sec. 5. Slope The angle a is designated as the inclination of P1P2 and tan a as the slope of the segment. ) EXERCISES 2. Find the cosine of the angle between the following pairs of segments of Problem 1 (a), (b) (c), (d) (e), (f) (g), (f) 3. Prove that if tan a1 = -1/tan a2, then the lines with inclinations al and a2 are perpendicular.

### Analytic geometry with calculus by Robert Carl Yates

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