9-6 1. (a) By the model: Let

9-6

1. (a) By the model: Let dx

dt = 0 and dy

dt = 0 ⇒ x = 20 , y = 500

x 20 20 30 20 10

y 500 300 500 700 500 dx

dt 0 −0.4 0 0.4 0

dy

dt 0 0 −25 0 25

dy

dx 0 ∞ 0 −∞

figure 1

By the graph , we can know that it start (700,20) with x decreasing , y increases. Up to the level, with x increasing , y decreases. To end up , it get to the equilibrium.

So x is the predator and y is the prey. Moreover, the prey is restricted by the number of the predators. Furthermore, the food about the predators is only y because x decreases with y decreasing .

(b) By the model: Let dx

dt = 0 and dy

dt = 0 ⇒ x = 187.5 , y = 27.08 x 187.5 187.5 87.5 187.5 287.5 y 27.08 17.08 30.42 37.08 23.75 dx

dt 0 11.25 0 −11.25 0

dy

dt 0 0 −0.24 0 0.19

dy

dx 0 −∞ 0 ∞

By the graph , we can know that it start (187.5,17.08) with y increasing , x decreases. Up to the level, with y decreasing , x increases. To end up , it get to the equilibrium.

So y is the predator and x is the prey. Moreover, the prey is restricted by the number of the predators and the resources of environment. Furthermore, the food about the predators is only y because x decreases with y decreasing.

7. By the graphs, the phase trajectory is following as:

10. (a)





 dA

dt = A(2 − 0.01L) = 0 dL

dt = L(−0.5 + 0.0001A) = 0

⇒







(2 − 0.01L) = 0 (−0.5 + 0.0001A) = 0

⇒







L = 200 A = 5000

the point (A, L) = (5000, 200) is inside the curve. If the initial value is the point , the phase trajectory is only a point.

(b) dL

dA = −0.5L + 0.0001AL 2A − 0.01AL

(c)(d) The phase trajectory is the following as:

(e) The graph about the aphids and lady-bug populations as functions of t is following as:

Chapter 9 Problem Plus

1. Solve the equation: dy

dx = y(y − 2)(y − 4) ⇒ 1 8( 1

y − 4 +1 y − 2

y − 2)dy = dx ⇒ y = 2 ± 2

√1 − c₀e^8x

(a) (i) y(0) = −0.3 ⇒ c₀ = 0.2439 ⇒ y(x) = 2 − 2

√1 − 0.2439e^8x

(ii) y(0) = 1 ⇒ c₀ = −3 ⇒ y(x) = 2 − 2

√1 + 3e^8x

(iii) y(0) = 3 ⇒ c₀ = −3 ⇒ y(x) = 2 + 2

√1 + 3e^8x

(iv) y(0) = 4.3 ⇒ c0 = 0.2439 ⇒ y(x) = 2 + 2

√1 + 0.2439e^8x

(b) y₀ = c ⇒ c₀ = 1 − 4

(2 − c)² ⇒ Since y(x) = 2 ± 2

√1 − c₀e^8x , when lim

x→∞y(x) is finite

⇒ 1 − 4

(2 − c)² ≤ 0 ⇒ 0 ≤ c ≤ 4 3.

(a) Up to the graph the question, we sketch y(0)=1 and get the graph as following:

we estimate that when x = 0.3, y = 0.8

(b) h = 0.1, x₀ = 0, y₀ = 1 and Euler’s method:y_n= y_n−1+ hy_n⁰

⇒ y_n= y_n−1+ h(x²_n−1− y²_n−1)

y₁ = 1 + 0.1(0²− 1²) = 0.9, y₂ = 1 + 0.1(0.1²− 0.9²) = 0.82, y₃ = 1 + 0.1(0.2 − 0.82²) = 0.7576

(c) The center of the horizontal line segments of the direction field are field are located on the lines y=x and y=-x. When a solution curve crosses one of these lines, it has a local maximum or minimum.

9. dr

dt + 2tr = r ⇒ dr

dt = r(1 − 2t) ⇒ dr

r = (1 − 2t)dt ⇒ ln(r) = t − t²+ c₀ ⇒ r(t) = c₁e^t−t2 take r(0) = 5 into the equation ⇒ c₁ = 5 ⇒ r(t) = 5 e^t−t2

12. dy

dx = 3 x²e^y ⇒ e^−ydy = 3 x²dx ⇒ −e^−y = c₀+ x³ ⇒ y(x) = − ln(−c₀ − x³)

⇒ Take y(0) = 1 into the equation ⇒ c₀ = −e⁻¹ ⇒ y(x) = − ln(e⁻¹− x³)

14.

y = e^kx Differentiate

==========⇒ dy

dx = ke^kx eliminate k,let k=^{ln y}_x

==============⇒ dy

dx = y ln y Orthogonal x

=========⇒

Trajectories dy

dx = −x y ln y

integrate

=======⇒ 1

2y²ln y −1

4y² = −1 2 x²+ c

=

=⇒ y²ln y − 1

2y²+ x² = C

10-1

3.

4.

9.

(a)

(b)

x =√

t V x² = t . . . (1) , y = 1 − t . . . (2)

(1) + (2) : x²+ y = 1, but x > 0 ⇒ x²+ y = 1 and x > 0

10.

(a)

(b) y = t³ ⇒ t = √³

y, x = t² ======⇒ x = y^{t=sqrt[3]y 23}

13.

(a)

(b) x = sin t . . . (1), y = csc t . . . (2)

(1)×(2):x × y = 1 by(1) and (2) :x > 0 and y > 0 ⇒ xy = 1 , x > 0, y > 0

16.

(a)

(b) x =√

t + 1 ⇒ x² = t + 1 . . . (1) y =√

t − 1 ⇒ y² = t − 1 . . . (2).

(1)−(2):x²− y² = 2 by (1) and (2): x > 0 and y > 0.

So x²− y² = 2 where x > 0 and y > 0

21. x = 5 sin t , y = 2 cos t ⇒ sin t = x

5, cos t = y

2 ⇒ sin²t + cos²t = 1 ⇒ (x

5)²+ (y

2)² = 1.

The motion is from (0, −2) to (0, −2) and moves clockwise around the ellipse ( (x

5)²+ (y

2)² = 1 ) 3 times. So the matched graph is V.

28. (a) x = t⁴− t + 1 = (t⁴+ 1) − t > 0 ∀t ∈ R and y = t² ≥ 0 ⇒ the graph is in the first quadrant.

(b) x = t²− 2t = t(t − 2) and y =√

t ≥ 0. x is negative and y is positive for 0 ≤ t ≤ 2 So the matched graph is I.

(c) x = sin 2t has period π. y(t + 2π) = sin(t + 2π + sin(2(t + 2π))) = sin(t + sin(2t)) = y(t) has period 2π. These equations match graph II since x cycles through the values -1 to 1 twice as y cycles through those values once.

(d) x = cos 5t has the period 2π

5 and y = sin 2t has the period π. And the graph is through (1, 0) for t = 2kπ and (−1, 0) for t = (2k + 1)π. So the matched graph is VI.

(e) x = t + sin 4t and y = t²+ cos 3t. As t becomes large , t and t² is dominant term. So the graph looks like the graph of y = x² with the oscillations. So the matched graph is IV.

(f) x = sin 2t

4 + t² and y = cos 2t

4 + t². As t → ∞, x → 0 and y → 0 So the matched graph is III.

10-2

2. x = 1

t ⇒ dx dt = −1

t² and y =√

te^−t ⇒ dy

dt = e^−tt⁻¹² (1

2 − t). So dy

dx = e^−tt³²(t −1 2) 5. x = t cos t ⇒ dx

dt = cos t − t sin t and y = t sin t ⇒ dy

dt = sin t + t cos t. So t = π ,dy

dx = −π

−1 = π Moreover, the tangent line is y − y(π) = π(x − x(π) ⇒ y = πx + π²

6. x = cos θ + sin 2θ ⇒ x(0) = 1 and y = sin θ + cos 2θ ⇒ y(0) = 1.

dx

dθ = − sin θ + 2 cos θ and dy

dt = cos θ − 2 sin 2θ. So t = 0 ,dy dx = 1

2. So the tangent line is y = 1

2x + 1 2. 11. x = t²+ 1 and y = t²+ t ⇒ dx

dt = 2t and dy

dt = 2t + 1. So dy

dx = 2t + 1 2t . Moreover, d^dy_dx

dt = −1 2t², d²y

dx² =

d^dy_dx dt dx dt

=

−1 2t²

2t = −1

4t³. So the curve is concave upward as d²y

dx² > 0 ⇒ t < 0 16. x = cos 2t and y = cos t ⇒ dx

dt = −2 sin 2t and dy

dt = − sin t. So dy

dx = 1

4 cos t = sec t 4 . Moreover, d^dy_dx

dt = sec t tan t 4 , d²y

dx² =

d^dy_dx dt dx dt

=

sec t tan t 4

−2 sin 2t = − sec³t

16 . So the curve is concave upward as d²y

dx² > 0 ⇒ π

2 < t < π

30. x = 3t²+ 1 and y = 2t³+ 1 ⇒ dx

dt = 6t and dy

dt = 6t² ⇒ dy dx = t

The tangent line is y − (2t³+ 1) = t(x − (3t²+ 1)) and passes through x = 4, y = 3.

So 2 − 2t³ = 3t − 3t³ ⇒ t³− 3t + 2 = 0 ⇒ (t + 2)(t − 1)² = 0 ⇒ t = −2 or t = 1.

The tangent line is y − 3 = x − 4 or y − 3 = −2(x − 4).

31. The area is 4 Z a

0

y dx x=a cos θ,y=b sin θ

===========⇒ 4 Z 0

π 2

b sin θa(− sin θ) dθ = 4ab Z ^π₂

0

sin²θdθ = abπ

41. dx

dt = 6t , dy

dt = 6t², Arc length=

Z 1 0

p(6t)² + (6t²)²dt ⇒ 6 Z 1

0

t√

t²+ 1 dt = 2(t²+1)(³²) 

1 0 = 4√

2−2

44. dx

dt = −3 sin t + 3 sin 3t and dy

dt = 3 cos t − 3 cos 3t, Arc length =

Z π 0

3p

(sin 3t − sin t)²+ (cos 3t − cos t)²dt = Z π

0

3√

2 − 2 sin 3t sin t − 2 cos 3t cos t dt

= Z π

0

3p

2 + cos 4t − cos 2t − (cos 4t + cos 2t) dt = 6 Z π

0

r1 − cos 2t 2

= 6 Z π

sin t dt = −6 cos t 

π = 12

62. By the formula:

The area of surface = Z 1

0

2π 3t²p

(3 − 3t²)² + (6t)²dt = 18π Z 1

0

t²p

(t²− 1)²+ (2t)²dt

= 18π Z 1

0

t²(t²+ 1) dt = 18π(1 5 +1

3) = 48 5 π 63. By the formula:

The area of surface = Z ^π₂

0

2π a sin³θ q

(−3a cos²θ sin θ)²+ (3a cos θ sin²θ)²dθ

= 2πa Z ^π₂

0

sin³θ q

(3a cos θ sin θ)(cos θ²+ sin²θ) dθ

= 6πa² Z ^π₂

0

sin⁴θ cos θ dθ = 6 5πa²