# AP Calculus by Shirley O. Hockett, David Bock

By Shirley O. Hockett, David Bock

Either Calculus AB and Calculus BC are lined during this entire AP attempt practise handbook. potential attempt takers will locate 4 perform tests in Calculus AB and 4 extra in Calculus BC, with all questions spoke back and suggestions defined. The guide additionally offers a close 10-chapter evaluate protecting subject matters for either assessments. The authors additionally provide an outline of the AP Calculus tests, which include suggestion to scholars on making most sensible use in their graphing calculators.

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Extra info for AP Calculus

Sample text

As x Æ 0ϩ, this l/x function has the indeterminate form ϱ/ϱ. Apply L’Hôpital’s rule: lim+ ln y = lim+ x→0 x→0 1/ x −1 / x 2 = lim+ ( − x ) = 0. x→0 0 So y Æ e or 1. 12. (A) Use the Parts Formula with u = x and dv = ex dx. Then du = dx and v = ex, and ( xe 15. (B) x − ∫ e x dx ) = ( xe 1 0 x − ex The arc length is given by the integral Ύ 1 0 ) 1 0 = ( e − e ) − ( 0 − 1) . 1 + ( 3 x ) dx which is 1 2 1 1 1/ 2 3/ 2 1 + 9 x ) ( 9dx ) = i (1 + 9 x ) ( ∫ 9 3 9 0 16. (E) Separating variables yields 1 2 2 1 0 = ( ) 2 103 / 2 − 1 .

40. 547 . a 2 0 (D) If x = 2t ϩ 1, then t = x − 1 , so dt = 1 dx . When t = 0, x = 1; when t = 3, x = 7. 2 2 (E) Use A(t) = A0ekt, where the initial amount A0 is 50. Then A(t) = 50ekt. Since . 9 . 3 . 9 t 35 Diagnostic Test Calculus AB 7_4324_APCalc_01DiagnosticAB Diagnostic Test Calculus AB 7_3679_APCalc_01DiagnosticAB 36 10/3/08 4:18 PM Page 36 AP Calculus 26 y x 41. (C) See the figure above. Since x2 ϩ y2 = 262, it follows that dy 2 x dx + 2 y = 0 dt dt at any time t. When x = 10, then y = 24 and it is given that Hence, 2(10)(3) + 2(24) 42.

6 Note that the distance covered in 6 seconds is 6 ∫0 v(t ) dt , the area between the velocity curve and the t-axis. −2 − 0 . 5− 4 21. (C) Acceleration is the slope of the velocity curve, 22. (D) Particular solutions appear to be branches of hyperbolas. See page 21. 23. (A) Differentiating implicitly yields 2xyyЈ + y2 – 2yЈ + 12y2yЈ = 0. When y = 1, x = 4. Substitute to find yЈ. 24. (C) 25. (D) 26. (B) Separate to get x ∫a g(t ) dt − x ∫b g(t ) dt dy y 2 = x ∫a g(t ) dt = 2 x dx , − + b ∫x g(t ) dt = b ∫a g(t ) dt.