Download e-book for kindle: AP Calculus by Shirley O. Hockett, David Bock

By Shirley O. Hockett, David Bock

ISBN-10: 0764143247

ISBN-13: 9780764143243

Either Calculus AB and Calculus BC are coated during this complete AP try instruction handbook. potential try out takers will locate 4 perform assessments in Calculus AB and 4 extra in Calculus BC, with all questions replied and recommendations defined. The guide additionally presents a close 10-chapter assessment protecting themes for either assessments. The authors additionally supply an outline of the AP Calculus tests, together with recommendation to scholars on making top use in their graphing calculators.

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Part B 29. (B) Set x −1 x −1 A B = = + . + +2 x 1 x x + 1 x 2 + ( ) ( ) x + 3x + 2 x − 1 = A ( x + 2 ) + B ( x + 1) ; 2 x = −2 implies − 3 = − B, or B = 3; x = −1 implies − 2 = A, or A = −2. 30. (E) S is the region bounded by y = sec x, the y-axis, and y = 4. y (x,4) R r 0 (x,y) x 7_3679_APCalc_02DiagnosticBC 10/3/08 4:19 PM Page 59 Diagnostic Test Calculus BC 59 We send region S about the x-axis. Using washers, ⌬V = p(R2 Ϫ r 2) ⌬x. Symmetry allows us to double the volume generated by the first-quadrant portion of S.

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.

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. 5 dy dy = 0, so =– . 4 dt dt (E) Let u = 2x and note that F ′( x ) = 1 . 1 − u3 Then F ′( x ) = F ′(u)u ′( x ) = 2 F ′(u) = 2 i 43. (D) v(5) Ϫ v(0) = ∫ 5 0 dx = 3 . dt 1 i . 1 − (2 x )3 a(t ) dt = − 1 π i 22 + 1 (3) (2) = − π + 3 . Since v(5) = 0, 4 2 Ϫv(0) = Ϫπ ϩ 3; so v(0) = π Ϫ 3. 102 5 44. (C) 45. (D) Let y = (x3 Ϫ 4x2 ϩ 8)ecos(x ).

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AP Calculus by Shirley O. Hockett, David Bock


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