By James Stewart

This handbook comprises worked-out suggestions to each odd-numbered workout in Multivariable Calculus, 8e (Chapters 1-11 of Calculus, 8e).

**Read Online or Download Student Solutions Manual, Chapters 10-17 for Stewart’s Multivariable Calculus, 8th PDF**

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**Extra resources for Student Solutions Manual, Chapters 10-17 for Stewart’s Multivariable Calculus, 8th**

**Example text**

The foci are at ± 2 − 1 1 = (±1 1). 2 1 29. 32 − 6 − 2 = 1 ⇔ 32 − 6 = 2 + 1 ⇔ 3(2 − 2 + 1) = 2 + 1 + 3 ⇔ 3( − 1)2 = 2 + 4 ⇔ ( − 1)2 = 23 ( + 2). This is an equation of a parabola with 4 = 23 , so = 16 . The vertex is (1 −2) and the focus is . 1 −2 + 16 = 1 − 11 6 31. The parabola with vertex (0 0) and focus (1 0) opens to the right and has = 1, so its equation is 2 = 4, or 2 = 4. 33. The distance from the focus (−4 0) to the directrix = 2 is 2 − (−4) = 6, so the distance from the focus to the vertex is 1 (6) 2 = 3 and the vertex is (−1 0).

92 + 4 2 = 36 is given by ⇔ 2 2 + = 1. We use the parametrization = 2 cos , = 3 sin , 0 ≤ ≤ 2. The circumference 4 9 2 2 ()2 + ()2 = 0 (−2 sin )2 + (3 cos )2 0 2 2 √ = 0 4 sin2 + 9 cos2 = 0 4 + 5 cos2 = √ 2 − 0 = , and () = 4 + 5 cos2 to get 8 4 (0) + 4 4 + 2 2 + 4 3 + 2 () + 4 5 + 2 3 + 4 7 + (2) ≈ 159. ≈ 8 = 4 3 4 4 2 4 Now use Simpson’s Rule with = 8, ∆ = Copyright 2016 Cengage Learning.

3 , or 5 . 3 = 2 cos − 2 cos 2 = 2 1 + cos − 2 cos2 = 2(1 − cos )(1 + 2 cos ) = 0 ⇒ Thus the graph has vertical tangents where = , 3 and 5 , 3 and horizontal tangents where = →0 3 2 3 3 2 1 −2 3 2 √ 3 3 2 − 12 −323 −3 3 2 4 . 3 To determine 0 4 3 5 3 and = 0, so there is a horizontal tangent there. what the slope is where = 0, we use l’Hospital’s Rule to evaluate lim 2 3 0 √ 0 √ − √ 3 2 31. ” For instance, for − 10 ≤≤ , 10 there are two petals, one with 0 and one with 0.