Unformatted text preview: at a = 0 and b = 0; eliminate the parameter to get (x − h)2 /a2 + (y − k )2 /b2 = 1. If a = b
the curve is a circle with center (h, k ) and radius a; if a = b the curve is an ellipse with center
(h, k ) and major axis parallel to the xaxis when a > b, or major axis parallel to the y axis when
a < b.
(a) ellipses with a ﬁxed center and varying axes of symmetry
(b) (assume a = 0 and b = 0) ellipses with varying center and ﬁxed axes of symmetry
(c) circles of radius 1 with centers on the line y = x − 1 40. Refer to the diagram to get bθ = aφ, θ = aφ/b but θ − α = φ + π/2
so α = θ − φ − π/2 = (a/b − 1)φ − π/2
x = (a − b) cos φ − b sin α = (a − b) sin φ − b sin 41. a−b a−b
b
a−b
b θ φ, bθ
aφ φ. = (a − b) cos φ + b cos
y = (a − b) sin φ − b cos α y φ α x y (a) a a a x a (b) Use b = a/4 in the equations of Exercise 40 to get
1
3
1
3
x = a cos φ + a cos 3φ, y = a sin φ − a sin 3φ;
4
4
4
4
but trigonometric identities yield cos 3φ = 4 cos3 φ − 3 cos φ, sin 3φ = 3 sin φ − 4 sin3 φ,
so x = a cos3 φ, y = a sin3 φ. (c) x2/3 + y 2/3 = a2/3 (cos2 φ + sin2 φ) = a2/3 y 42. y y
1 50 50 0 3 2 1 x x 2 1 1 1 50 50 1 a = −1 a = −2 a=0
y y
50 50 x
1 0 2 1 2 50 50 a=1 a=2 3 x x Supplementary Exercises 1 38 CHAPTER 1 SUPPLEMENTARY EXERCISES
1. 194045; the greatest ﬁveyear slope 2. (a)
(c) f (−1) = 3.3, g (3) = 2
x < −2, x > 3 (b)
(d) (e) the domain is −4 ≤ x ≤ 4.1, the range is
−3 ≤ y ≤ 5 (f ) 3. 4. T x = −3, 3
the domain is −5 ≤ x ≤ 5 and the range is
−5 ≤ y ≤ 4
f (x) = 0 at x = −3, 5; g (x) = 0 at
x = −3, 2 x 70 60 50 40
0 2 4 6 t t 5 8 13 5. If the side has length x and height h, then V = 8 = x2 h, so h = 8/x2 . Then the cost C = 5x2 +
2(4)(xh) = 5x2 + 64/x. 6. Assume that the paint is applied in a thin veneer of uniform thickness, so that the quantity of paint
to be used is proportional to the area covered. If P is the amount of paint to be used, P = kπr2 . The
constant k depends on physical factors, such as the thickness of the paint, absorption of the wood,
etc. 7. y
5 5 1 5 x 8. Suppose the radius of the uncoated ball is r and that of the coated ball is r + h. Then the plastic has
4
4
4
volume equal to the diﬀerence of the volumes, i.e. V = π (r + h)3 − πr3 = πh[3r2 + 3rh + h2 ] in3 .
3
3
3 9. (a) The base has sides (10 − 2x)/2 and 6 − 2x, and the height is x, so V = (6 − 2x)(5 − x)x ft3 . (b) From the picture we see that x < 5 and 2x < 6, so 0 < x < 3.
(c) 3.57 ft ×3.79 ft ×1.21 ft 10. {x = 0} and ∅ (the empty set) 11. impossible; we would have to solve 2(3x − 2) − 5 = 3(2x − 5) − 2, or −9 = −17 12. (a) 13. 1/(2 − x2 ) 15. (3 − x)/x x
−4
f (x)
0
g (x)
3
(f ◦ g )(x)
4
(g ◦ f )(x) −1 (b) no; f (g (x)) can be deﬁned at x = 1,
whereas g , and therefore f ◦ g , requires
x=1
14. g (x) = x2 + 2x −3 −2 −1
0
1
2
3
4
−1
2
1
3 −2 −3
4 −4
2
1 −3 −1 −4
4 −2
0
−3 −2 −1
1
0 −4
2
3
−3
4 −4 −2
1
2
0
3 39 16. Chapter 1 (a) y = x − 1, y = (−x) − 1 = x + 1,
y = 2x + 1, y = 2x + 1 − 3,
y = −2x + 1 + 3 y (b)
3 1
3 1 2 x 1 17. 18. 19. (a) even × odd = odd
(c) even + odd is neither (b) a square is even
(d) odd × odd = even π 3π π 5π 7π 11π
(a) y = cos x − 2 sin x cos x = (1 − 2 sin x) cos x, so x = ± , ± , ,
,− ,−
2
266
6
6
π
3π
π√
5π √
7π √
11π √
(b) (± , 0), (± , 0), ( , 3/2), ( , − 3/2), (− , − 3/2), (−
, 3/2)
2
2
6
6
6
6
If x denotes the distance from A to the base of the tower, and y the distance from B to the
base, then x2 + d2 = y 2 . Moreover h = x tan α = y tan β , so d2 = y 2 − x2 = h2 (cot2 β − cot2 α),
d2
d2 sin2 α sin2 β
. The trigonometric identity
h2 =
2
2 β − cot2 α =
cot
sin α cos2 β − cos2 α sin2 β
d sin α sin β
sin(α + β ) sin(α − β ) = sin2 α cos2 β − cos2 α sin2 β yields h =
.
sin(α + β ) sin(α − β ) (b)
20. (a) 295.72 ft. (a) 3π
2π
(t − 101) =
, or t = 374.75,
when
365
2
which is the same date as t = 9.75, so during the night of January 10th11th (b) y
60
40
20
t
100 200 300 20 (c) from t = 0 to t = 70.58 and from t = 313.92 to t = 365 (the same date as t = 0) , for a total of
about 122 days 21. C is the highest nearby point on the graph; zoom to ﬁnd that the coordinates of C are (2.0944, 1.9132).
Similarly, D is the lowest nearby point, and its coordinates are (4.1888, 1.2284). Since f (x) = 1 x − sin x
2
is an odd function, the coordinates of B are (−2.0944, −1.9132) and those of A are (−4.1888, −1.2284). 22. Let y = A + B sin(at + b). Since the maximum and minimum values of y are 35 and 5, A + B = 35
and A − B = 5, so A = 20, B = 15. The period is 12 hours, so 12a = 2π and a = π/6. The
maximum occurs at t = 2, so 1 = sin(2a + b) = sin(π/3 + b), π/3 + b = π/2, b = π/2 − π/3 = π/6 and
y = 20 + 15 sin(πt/6 + π/6). 23. (a) T...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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