(b) P(F) = 1 2 × 1 4 + 1 2 × 1 = 5 8 P(E) = 1 2 P(F

1. (a) This is the pedigree of the family together with their possible genotypes.

X Y X X orX X X Y

X Y X Y X X orX X X X

X Y

(b)

P(F) = 1 2× 1

4 +1

2 ×1 = 5 8 P(E) = 1

2 P(F|E) = 1

2× 1

2 ×1 = 1 4 P(E|F) = P(EFˆ)

P(F) = P(F|E)·P(E) P(F) =

1 4 × ¹2

5 8

= 1 5

2. (a)

P(X =k) =C_k⁵⁰(1 4)^k(3

4)^50−k (b)

P(X ≥30) =C30⁵⁰(1 4)³⁰(3

4)²⁰+C31⁵⁰(1 4)³¹(3

4)¹⁹+· · ·+C50⁵⁰(1 4)⁵⁰(3

4)⁰ 3. (a)

0×0.31 + 1×0.21 + 2×0.19 + 3×0.17 + 4×0.12 = 1.58 (b)

Var = 0.21−1.581²+ 0.19(2−1.58)²+ 0.17(3−1.58)²+ 0.12(4−1.58)²

= 1.9236 σ = √

Var = 1.387 4.

k 0 1 2 3 4 5 6 7

P(X =k) 0.13536 0.27073 0.27073 0.18048 0.09024 0.0361 0.01203 0.003447 P(Y =k) 0.66667 0.22222 0.074074 0.0247 0.00823 0.00274 0.00091

k 8 9 10 11

P(X =k) 0.00086 0.00019 0.000038 0.000006 P(Y =k) 0.0003 0.0001016 0.00003387 0.0000112901 P(X =k)> P(Y =k) when k= 2,3, . . . ,10.

5. (a) The vector that goes from (2,1) to (-3,4) has the form

−3−2 4−1

= −5

3

. This vector has length p

(−5)² + 3² =√ 34.

Normalizing this vector yields

v=

−5/√ 34 3/√

34

and the directional derivative is

D^vf(2,1) = (▽f(2,1))·v= 4/5

2/5

·

−5/√ 34 3/√

34

= −14 5√

34.

(b) The gradient vector of f at (-3,4) is perpendicular to the level curve through (-3,4).

Evaluating ▽f at (-3,4), we find

▽f(−3,4) =







− 6 825 25





.

Therefore, the line that is perpendicular to the level curve passing through (-3,4) has the form

−3 4

+t







− 6 825 25





, fort∈R.

Z dy y(y−4) =

Z

1/2 dx.

We use the partial fraction method to integrate the left-hand side.

1

y(y−4) = A

y + B y−4

= (A+B)y−4A y(y−4) Comparing the last term to the integrand, we find

A+B = 0 and −4A= 1 and thus

A= −1

4 and B = 1 4. Using the partial fraction decomposition, we must integrate

1 4

Z ( 1

y−4 − 1

y) dy= Z

1/2dx which yields

1

4(ln|y−4| −ln|y|) = 1/2x+C¹. Simplifying this results in

ln|y−4

y | = 2x+ 4C¹

|y−4

y | = e^4C1e^2x y−4

y = ±e^4C1e^2x y−4

y = Ce^2x. Using the initial conditony(0) =−3, we find

C = −7

−3. The solution is therefore

y−4

y = 7

3e^2x.

If we want the solution in the formy=f(x), we must solve for y. We find

y = 12

3−7e^1/2x.

7. (a)

rN(1− N

K)−H = 2N(1− N

1000)−100 = 0 when N(1− N

1000) = 50 N − 1

1000N² = 50 N²−1000N + 50000 = 0

N = 500 + 200√

5 is locally stable.

N = 500−200√

5 is unstable.

(b)

g(N) = 2N(1− N

1000)−100

= − 1

500N²+ 2N −100

= − 1

500(N²−1000N)−100

= − 1

500(N −500)²+ 400

The maximal harvesting rate that maintains a positive population size is 400.

8. (a) p= 0 is unstable.

p= 1−D− ^m_c is stable.

(b)

1−pˆ−D= 1−(1−D− m

c )−D= m c. (c)

c(1−pˆ−D) =c·m c =m.

9. Denote the area of the copper disc by A={(x, y)|x²+y² ≤1}. (a)

D¹T(x, y) = 2x−1 D2T(x, y) =−4y D¹¹T(x, y) = 2 D¹²T(x, y) = 0 D²¹T(x, y) = 0 D²²T(x, y) = −4

Solve

D¹T(x, y) = 2x−1 D²T(x, y) =−4y , we find (x, y) = (1/2,0). But

△(1/2,0) = det

D¹¹T(1/2,0) D¹²T(1/2,0) D²¹T(1/2,0) D²²T(1/2,0)

= det

2 00 −4

=−8<0 Therefore there is no extrema on the inner part of A.

F(x, y, λ) =x²2y²−x+λ(x²+y²−1).

Then

∂F

∂x(x, y, λ) = −2x−1 + 2xλ

∂F

∂y(x, y, λ) = −4y+ 2yλ

∂F

∂λ(x, y, λ) = x²+y²−1

Solve







∂F

∂x(x, y, λ) = 0

∂F

∂y(x, y, λ) = 0

∂F

∂λ(x, y, λ) = 0

, we find y = 0 or λ= 2. and x=±p

1−y² =±1 or x= ¹6. Substituting x, we findy =±^√635. Now,

T(1,0) = 0 T(−1,0) = 2 T(1

6,

√35

6 ) = −25 12

, therefore the maximam temperature on the boundary of A is 2 and the minimum temperature on the boundary of A is ⁻₁₂²⁵.