G&R havo B deel 3 C. von Schwartzenberg Toets voorkennis
a b c d
4
12 Differentiaalrekening 1/20 EXTRA: 3 Differentiëren op bladzijde 156 aan het einde van deze uitwerking.
f (x ) = 2x + x + 5x + 7 ⇒ f '(x ) = 8x 3 + 2x + 5. g (x ) = x 2 (2x − 5) = 2x 3 − 5x 2 ⇒ g '(x ) = 6x 2 − 10x . h (x ) = (2x + 1)2 = (2x + 1) ⋅ (2x + 1) = 4x 2 + 2x + 2x + 1 = 4x 2 + 4x + 1 ⇒ h '(x ) = 8x + 4. s (t ) = 1 t 3 + 4t 2 ⇒ s '(t ) = 1 t 2 + 8t . 6
Toets voorkennis
2
2
EXTRA: 4 Extremen berekenen op bladzijde 157 aan het einde van deze uitwerking.
f (x ) = 1 x 3 − x 2 − 8x + 5 ⇒ f '(x ) = x 2 − 2x − 8. 3 2
f '(x ) = 0 ⇒ x − 2x − 8 = 0 (x − 4)(x + 2) = 0 x = 4 ∨ x = −2. Maximum (zie een plot) f ( −2) = 14 1 en minimum (zie een plot) f (4) = −21 2 . 3
3
1a
f (x ) = x 2 ⇒ f '(x ) = 2x en g (x ) = 3x − 7 ⇒ g '(x ) = 3. p (x ) = f (x ) ⋅ g (x ) = x 2 (3x − 7) = 3x 3 − 7 x 2 ⇒ p '(x ) = 9x 2 − 14x .
1b
p '(x ) (zie 1a) = 9x 2 − 14x en f '(x ) ⋅ g '(x ) (zie 1a) = 2x ⋅ 3 = 6x . Dus p ´(x ) ≠ f '(x ) ⋅ g '(x ).
1c
p '(x ) = 9x 2 − 14x en f '(x ) ⋅ g (x ) + f (x ) ⋅ g '(x ) (zie 1a) = 2x ⋅ (3x − 7) + x 2 ⋅ 3 = 6x 2 − 14x + 3x 2 = 9x 2 − 14x . Dus p ´(x ) = f '(x ) ⋅ g (x ) + f (x ) ⋅ g '(x ).
2a
f (x ) = x 2 (2x − 1) ⇒ f '(x ) = 2x (2x − 1) + x 2 ⋅ 2 = 2x (2x − 1) + 2x 2 .
2b
g (x ) = 2x 3(x 2 − 3) ⇒ g '(x ) = 6x 2 (x 2 − 3) + 2x 3 ⋅ 2x = 6x 2 (x 2 − 3) + 4x 4 .
2c
h (x ) = (x 2 − 1)(x 2 + 3) ⇒ h '(x ) = 2x (x 2 + 3) + (x 2 − 1) ⋅ 2x .
2d
j (x ) = (2x 3 + 1)(3x 2 − 1) ⇒ j '(x ) = 6x 2 (3x 2 − 1) + (2x 3 + 1) ⋅ 6x .
3a
f (x ) = (2 − 3x 2 )(2 + 7x ) ⇒ f '(x ) = −6x (2 + 7x ) + (2 − 3x 2 ) ⋅ 7.
3b
g (x ) = (2x − 5)2 = (2x − 5)(2x − 5) ⇒ g '(x ) = 2(2x − 5) + (2x − 5) ⋅ 2 = 4(2x − 5).
3c
h (x ) = (x 2 − 3x )(x 3 + x 2 + x ) ⇒ h '(x ) = (2x − 3)(x 3 + x 2 + x ) + (x 2 − 3x )(3x 2 + 2x + 1).
3d
j (x ) = (3x 2 − 4)2 = (3x 2 − 4)(3x 2 − 4) ⇒ j '(x ) = 6x (3x 2 − 4) + (3x 2 − 4) ⋅ 6x = 12x (3x 2 − 4).
4
g (x ) = c ⋅ f (x ) ⇒ g '(x ) = [c ]'⋅ f (x ) + c ⋅ [f (x )]' = 0 ⋅ f (x ) + c ⋅ f '(x ) = c ⋅ f '(x ).
5a
f (x ) = (4x 2 − 1)(3x + 2) ⇒ f '(x ) = 8x (3x + 2) + (4x 2 − 1) ⋅ 3. yA = f ( −1) = −3 en rc raaklijn = f '( −1) = 17. k : y = 17 x + b door A( −1, − 3) ⇒ 17 ⋅ −1 + b = −3 ⇒ b = −3 + 17 = 14. Dus k : y = 17 x + 14.
5b
B op de y -as (x = 0) ⇒ yB = f (0) = −2 en rc raaklijn = f '(0) = −3. l : y = −3x + b door B (0, − 2) ⇒ −3 ⋅ 0 + b = −2 ⇒ b = −2. Dus l : y = −3x − 2.
6a
f (x ) = (x 2 + 1)(x 2 − 4) = 0 ⇒ x 2 = −1 (kan niet) ∨ x 2 = 4 ⇒ x = −2 ∨ x = 2. Dus A( −2, 0) en B (2, 0). f (x ) = (x 2 + 1)(x 2 − 4) ⇒ f '(x ) = 2x (x 2 − 4) + (x 2 + 1) ⋅ 2x . Dus f '( −2) = −20 en f '(2) = 20. k : y = −20x + b door A( −2, 0) ⇒ −20 ⋅ −2 + b = 0 ⇒ b = −40. Dus k : y = −20x − 40. l : y = 20x + b door B (2, 0) ⇒ 20 ⋅ 2 + b = 0 ⇒ b = −40. Dus l : y = 20x − 40.
6b
f '(1) = −2 ≠ 0 ⇒ de grafiek van f heeft geen extreme waarde voor x = 1.
6c
f '( 1 1 ) = 0 ⇒ de grafiek van f heeft een extreme waarde voor x = 1 1 .
7a
A = O (PQRS ) = PQ ⋅ PS . De grafiek van f snijdt de x -as in O (0, 0) en (6, 0) ( −x 2 + 6x = −x (x − 6) = 0 ⇒ x = 0 ∨ x = 6). PQ = 6 − 2 ⋅ OP = 6 − 2 ⋅ xP = 6 − 2 p en PS = yS = − p 2 + 6 p . Dus A = PQ ⋅ PS = (6 − 2 p )( − p 2 + 6 p ).
7b
A = (6 − 2p )( − p 2 + 6 p ) ⇒ A '( p ) = dA = −2( − p 2 + 6 p ) + (6 − 2 p )( −2 p + 6)
2
2
dp
= 2 p 2 − 12 p − 12 p + 36 + 4 p 2 − 12 p = 6p 2 − 36 p + 36.
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 2/20
7c
dA = 0 ⇒ 6 p 2 − 36 p + 36 = 0 ⇒ p 2 − 6 p + 6 = 0 met D = ( −6)2 − 4 ⋅ 1 ⋅ 6 = 36 − 24 = 12 dp p = 6 − 12 = 6 − 2 3 = 3 − 3 ∨ p = 6 + 12 = 3 + 3 ( > 3 voldoet niet). 2 ⋅1 2 2 We zoeken dus p = 3 − 3. (de enige kandidaat voor het maximum)
8a
f (x ) = ( 1 x 3 − 4)2 − 5 = ( 1 x 3 − 4)( 1 x 3 − 4) − 5 ⇒ f '(x ) = 3 x 2 ( 1 x 3 − 4) + ( 1 x 3 − 4) ⋅ 3 x 2 = 3x 2 ( 1 x 3 − 4). 2 2 2 2 2 2 2 2 yA = f (1) = 7,25 en rc = f '(1) = −10,5. k : y = −10,5x + b door A(1; 7,25) ⇒ −10, 5 ⋅ 1 + b = 7,25 ⇒ b = 17, 75. Dus k : y = −10,5x + 17, 75.
8b
yB = f (0) = 11 en rc = f '(0) = 0. k : y = 0x + b = b door B (0, 11). Dus k : y = 11.
9a 9b
10a
2
x3 =8 x = 2 met f (2) = −5 ⇒T (2, − 5).
O ( ∆ABC ) = 1 ⋅ AC ⋅ AB = 1 ⋅ (OC − OA ) ⋅ f ( p ) = 1 (4 − p )( p 2 − 2p + 3) = (2 − 1 p )( p 2 − 2 p + 3).
2 2 2 2 2 2 d O 1 1 1 3 O '( p ) = = − ( p − 2 p + 3) + (2 − p )(2 p − 2) = − p + p − + 4 p − 4 − p 2 + p = −1 1 p 2 + 6 p − 5 1 . dp 2 2 2 2 2 2 O '( p ) = 0 ⇒ −1 1 p 2 + 6p − 5 1 = 0 (keer − 2) 2 2 3p 2 − 12 p + 11 = 0 met D = ( −12)2 − 4 ⋅ 3 ⋅ 11 = 144 − 132 = 12 p = 12 − 12 (hoort bij een minimum van O ) ∨ p = 12 + 12 ≈ 2, 58. Dus we zoeken p ≈ 2,58. 2⋅3 6
1 = x −3 .
5 = 5 ⋅ 1 = 5x −4 . 4
x3
x4
1 = 1 ⋅ 1 = 1 ⋅ x −2 . 3 x2 3 3x 2
x
x −4 = 14 .
11a
h (x ) = x 2 ⋅ x −2 ⇒ h '(x ) = 2x ⋅ x −2 + x 2 ⋅ [x −2]'.
11d
2
x = 0 (hoort niet bij de top) ∨ 1 x 3 = 4
10b
11b
f '(x ) = 0 ⇒ 3x 2 ( 1 x 3 − 4) = 0
8c
3x −2 = 3 ⋅ 12 = 32 .
x
x
2
h (x ) = x ⋅ x [x
−2
−2
=x
2 + −2
=x
0
1 ⋅ x −6 = 1 ⋅ 1 = 1 . 7 7 x6 7x 6
x
h '(x ) = 0 (zie 1b en gebruik 1a) 2x ⋅ x −2 + x 2 ⋅ [x −2 ]' = 0 2x −1 + x 2 ⋅ [x −2 ]' = 0 x 2 ⋅ [x −2 ]' = −2x −1
11c
= 1 ⇒ h '(x ) = 0.
−1 ]' = −2x2 = −2x −1 − 2 = −2x −3.
x
Dus [x n ]' = ............ = nx n −1 geldt ook voor n = −2. 12a
−1 dus [x −2 ]' = −2x2 .
x
f (x ) = 53 = 5 ⋅ 13 = 5x −3 ⇒ f '(x ) = −15x −4 = −15 ⋅ 14 = − 154 . x
x
1
x
= 1 ⋅ 13 = 1 x −3 ⇒ 5 x 5
x
g '(x ) = − 3 x −4 = − 3 ⋅ 14 = − 3 4 . 5 5 x 5x
12b
g (x ) =
12c
h (x ) = 5x 2 − 52 = 5x 2 − 5 ⋅ 12 = 5x 2 − 5x −2 ⇒ h '(x ) = 10x + 10x −3 = 10x + 10 ⋅ 13 = 10x + 103 .
13a
4 4 f (x ) = x −3 5 = x 3 − 53 = x − 5x −3 ⇒ f '(x ) = 1 + 15x −4 = 1 + 154 .
5x 3
x
x
x
x
x
x
2
x
2
−1
−3
−2
+ 9x −4 = − 22 + 94 .
13b
g (x ) = 2x 3− 3 = 2x3 − 33 = 2x
13c
h (x ) = x + 2 = x + 2 = 1 + 2 x −1 ⇒ h '(x ) = 0 − 2 x −2 = − 2 2 . 3x 3x 3x 3 3 3 3x
14a
f (x ) = 2x −2 1 = 2x2 − 1 2 = 2 x −1 − 1 x −2 ⇒ f '(x ) = − 2 x −2 + 2 x −3 = − 2 2 + 2 3 .
14b
x
x
3x
x
3x
3
3x
2
x
− 3x
3
⇒ g '(x ) = −2x
x
3
3
x
3x
3x
2
g (x ) = 6 − x − 1 = 6 − x + 1 = 6 − x + x −1 ⇒ g '(x ) = 0 − 1 − x −2 = −1 − 12 . x
x
x
x
2
14c
h (x ) = 5 2 − 2x = 5 x −2 − 2 x 2 ⇒ h '(x ) = −5x −3 − 4 x = − 53 − 4 x .
15a
5 ⋅ 2 = 10 en 10 ⋅ 1 = 10, dus 5 ⋅ 2 = 10 ⋅ 1.
15b
5 = x (kruisproducten nemen) 10 2
2x
5 ⋅ 2 = 10 ⋅ x .
5
2
5
5
5
x
10
2
x
5
15c
5 = 3 (kruiselings vermenigvuldigen) 6 x
5x = 18 x = 18 = 3 3 . 5
5
G&R havo B deel 3 C. von Schwartzenberg 16a
12 Differentiaalrekening 3/20
4 = 1 (kruisproducten) 9 2
x = 3 (kruisproducten) x −4 8
16c
x2
3(x − 4) = 8x 3x − 12 = 8x −5x = 12 ⇒ x = 12 = −2 2 .
x = 36 x = −6 ∨ x = 6.
−5
16b
x − 3 = 2 (kruisproducten) x 7
16d
7(x − 3) = 2x 7 x − 21 = 2x 5x = 21 x = 21 = 4 1 . 5
17a
5
3 − 62 = 1 1 2 x 1 1 = 3 = 62 (kruisproducten) 2 2 x 2 3x
= 12
x 2 = 4 ⇒ x = −2 ∨ x = 2.
5
f (x ) = 2x + 3 = 2x + 3 = 2 + 3x −1 ⇒ f '(x ) = −3x −2 = − 32 . x
x
x
x
Snijden met de x -as ( y = 0) ⇒ f (x ) = 0 ⇒ 2x + 3 = 0 ⇒ 2x + 3 = 0 ⇒ 2x = −3. Dus xA = − 3 . x
2
1
yA = f ( − 3 ) = 0 ( y = 0 was gegeven, zie de regel hierboven) en rc raaklijn = f '( − 3 ) = − 4 . 2
2
3
k : y = − 4 x + b door A( − 3 , 0) ⇒ − 4 ⋅ − 3 + b = 0 ⇒ b = 0 − 4 ⋅ 3 = − 12 = −2. Dus k : y = − 4 x − 2. 3
2
3
2
3 2
6
3
17b
rc raaklijn = f '(x ) = − 3 ⇒ − 32 = − 3 ⇒ 32 = 3 ⇒ 3x 2 = 12 ⇒ x 2 = 4 ⇒ x = −2 ∨ 4 4 4 x x yB = f ( −2) = 1 en yC = f (2) = 3 1 . 2 2 De raakpunten zijn B ( −2, 1 ) en C (2, 3 1 ). 2 2
18a
x + 1 = x + 3 (kruisproducten) x −1 x
18c
x = 3 (kruisproducten) x +2 x −2
18d
x = 1 (kruisproducten) x + 4 2x − 1
x (x + 1) = (x + 3)(x − 1) x 2 + x = x 2 + 2x − 3 −x = −3 x = 3.
18b
x (2x − 1) = x + 4 2x 2 − x = x + 4 2x 2 − 2x − 4 = 0 x2 −x −2 = 0 (x − 2)(x + 1) = 0 x = 2 ∨ x = −1.
4x + 3 = x x +2 4x = x − 3 (kruisproducten) 1 x +2
x (x − 2) = 3(x + 2) x 2 − 2x = 3x + 6 x 2 − 5x − 6 = 0 (x − 6)(x + 1) = 0 x = 6 ∨ x = −1.
19a
x = 2.
(x − 3)(x + 2) = 4x
x 2 − 3x + 2x − 6 = 4x x 2 − 5x − 6 = 0 (x − 6)(x + 1) = 0 x = 6 ∨ x = −1.
2 2 f (x ) = x + 4 = x + 4 = x + 4x −1 ⇒ f '(x ) = 1 − 4x −2 = 1 − 42 .
x
x
x
x
yA = f (3) = 13 en rc raaklijn = f '(3) = 5 . 3
9
k : y = 5 x + b door A(3, 13 ) ⇒ 5 ⋅ 3 + b = 13 ⇒ b = 13 − 5 ⋅ 3 = 13 − 5 = 8 . Dus k : y = 5 x + 8 . 9
19b
3
rc raaklijn = f
9
3
x
yB = f ( −1) = −5 en yC = f (1) = 5. De raakpunten zijn B ( −1, − 5) en C (1, 5). 19c
3
9
3
3
3
9
'(x ) = −3 ⇒ 1 − 42 = −3 ⇒ − 42 = −4 ⇒ −4x 2 = −4 ⇒ x 2 = 1 ⇒ x = −1 ∨ x
1
3
x = 1.
f '(x ) = 0 ⇒ 1 − 42 = 0 ⇒ 1 = 42 ⇒ x 2 = 4 ⇒ x = −2 ∨ x = 2. x
x
1
Maximum (zie een plot) f ( −2) = −4 en minimum (zie een plot)f (2) = 4. 19d
rc raaklijn = f '(x ) = 2 ⇒ 1 − 42 = 2 ⇒ − 42 = 1 ⇒ x 2 = −4 (kan niet) ⇒ geen oplossing. x
Toets voorkennis 1a
x −3 = 13 .
1b
x 6 = x5.
5
6
x
x
1
EXTRA: 5 Machten met positieve en/of gebroken exponenten op bladzijden 158 en 159 aan het einde van deze uitwerking.
1c
x
1 31
1d
x
−1 21
1
= x1 ⋅x 3 = x ⋅3x . = 11 = x
1
2
1 1
x ⋅x
1 2
=
1
x⋅ x
.
G&R havo B deel 3 C. von Schwartzenberg 1
2a 2b
x2 ⋅ x = x2 ⋅x 2 = x
21
2
.
1
2c
−1
x = x −1 − 2 = x −3 . x2
1
20a
12 Differentiaalrekening 4/20
3
x ⋅ x 1
2d
x⋅ x
1 2
x 3⋅ x 1
=
3
1
1
=
1
x ⋅x
−3 = 11 = x 2.
x
=
1 3
1
1
=x
11 x 3
1
1
21c
.
1
2 ⋅x
1
1
− f (x ) = x + x = x + x 2 ⇒ f '(x ) = 1 + 1 x 2 = 1 + 1 1 = 1 + 2
2x
2
1
1 2
− = 1 ⋅ 11 = 1 ⋅ x 2 . 2
2
x2
Uit het hoofd leren:
.
2⋅ x
1
1
[x 2 ]' =
21b
3
1
2 ⋅ x 2 ⋅ [x 2 ]' = 1
21a
−1 1
2 ⋅ x 2 ⋅ [x 2 ]' = 1
20b
x 2 ⋅ [x 2 ]'+ [x 2 ]'⋅ x 2 = 1 1
2
1
1
1
x 2 ⋅ x 2 = x (links en rechts de afgeleide nemen) 1
3
1
− [ x ]' = 1 x 2 = 1 ⋅ 11 = 2
2
3
h (x ) = 1 = 1 1 = x x
x
−1
2
−1 1
⇒ h '(x ) = − 1 x
2
2
2 3 5
3 5
21d
j (x ) = x ⋅ x 3 = x 3 ⋅ x
21e
k (x ) = x 2 ⋅ 4 x = x 2 ⋅ x 4 = x
3 53
=x
1
=− 1
⇒ j '(x ) = 3 3 x ⇒ k '(x ) = 2 1 x
1
=−
11 2x 2
2x ⋅ x
2 53
=−
1 2
1 . 2x ⋅ x
3
5
= 3 3 x 2 ⋅ x 5 = 3 3 x 2 ⋅ x 3.
5
2 41
5
1 41
4
5
= 2 1 x ⋅x
1 4
=21 x ⋅4x.
4
4
2
2 = 2x (1 + x ) + x + 1 . 2⋅ x 2⋅ x
2
1
l (x ) = (x + 1) ⋅ (1 + x ) (productregel) ⇒ g '(x ) = 2x ⋅ (1 + x ) + (x + 1) ⋅
22a
2 2 2 2 −1 − −2 f (x ) = 4x + 1 = 4x +11 = 4x 1+ 1 = 4x 1 + 1 1 = 4x 2 + x 2 ⇒ f '(x ) = 2x 2 − 1 1 x 2 = 21 −
1
x⋅ x
x ⋅x 2
x
1
x
2
1
x
2
2 3
g (x ) = x3 − 4 = x −1 4 = x1 − 41 = x − 4x x
x
3
x
2
x
3
11
1
− 3) = (x
11
23
− 3 f (x ) = x 2 = x 3 ⇒ f '(x ) = 2 x 3 = 2 1 =
2
⇒ g '(x ) = 2 x
− 3)(x
1
3
3x
1
2
− 31
+ 4x
3
h (x ) = (x x − 3) = (x
2
1
2
− 31
22c
2
1
3
2
1 2⋅ x
3
21f
22b
x
1 1 11 g (x ) = x ⋅ 3 x = x ⋅ x 3 = x 3 ⇒ g '(x ) = 1 1 x 3 = 1 1 ⋅ 3 x .
2
3
11
2
−1 31
3
− 3) ⇒ h '(x ) = 1 1 x (x 2
4
= 21 + 3x
1 2
x2
11
2
3x ⋅ x
3
− 3) + (x
11
2
1 3
=
3 2
2x ⋅ x
= 2 −
1 2
x
3 . 2x 2 ⋅ x
4 2 + . 3⋅ 3 x 3x ⋅ 3 x 1
− 3) ⋅ 1 1 x 2 = 3 x (x x − 3). 2
2 . 3⋅ 3 x
yA = f ( 1 ) = 1 en rc raaklijn = f '( 1 ) = 4 . 8
4
8
3
k : y = 4 x + b door A( 1 , 1 ) ⇒ 4 ⋅ 1 + b = 1 ⇒ b = 1 − 4 ⋅ 1 = 1 − 1 = 3 − 2 = 1 . Dus k : y = 4 x + 1 . 3
8
4
3 8
4
yB = f (8) = 4 en rc raaklijn = f '(8) = 1 .
4
3 8
4
6
12
12
12
3
12
3
l : y = 1 x + b door B (8, 4) ⇒ 1 ⋅ 8 + b = 4 ⇒ b = 4 − 8 = 12 − 8 = 4 . Dus l : y = 1 x + 4 . 3
3
3
k snijden met l geeft 4 x + 1 = 1 x + 4 3
12
3
3
3
3
3
3
3
x = 4 − 1 = 16 − 1 = 15 = 5 met y = 1 ⋅ 5 + 4 = 5 + 16 = 21 = 7 . Dus C ( 5 , 7 ). 3
24a
12
12
12
12
4
3 4
1
f (x ) = x x − 3x = x ⋅ x 2 − 3x = x
11
2
3
12
12
12
4
4
4
1
− 3x ⇒ f '(x ) = 1 1 x 2 − 3 = 1 1 x − 3. 2
2
f '(x ) = 0 ⇒ 1 1 x − 3 = 0 ⇒ 1 1 x = 3 ⇒ x = 2 (kwadrateren) ⇒ x = 4 (de enige kandidaat voor een minimum). 2
2
Het minimum is f (4) = 4 ⋅ 4 − 3 ⋅ 4 = 4 ⋅ 2 − 3 ⋅ 4 = −4. 24b
rc raaklijn = f '(0) = −3 geeft k : y = −3x + b door O (0, 0) ⇒ −3 ⋅ 0 + b = 0 ⇒ b = 0. Dus k : y = −3x .
24c
f '(x ) = 3 ⇒ 1 1 x − 3 = 3 ⇒ 1 1 x = 6 ⇒ x = 4 (kwadrateren) ⇒ x = 16 = xA . 2
2
yA = f (16) = 16 ⋅ 16 − 3 ⋅ 16 = 16 ⋅ 4 − 3 ⋅ 16 = 16. Dus A(16, 16). l : y = 3x + b door A(16, 16) ⇒ 3 ⋅ 16 + b = 16 ⇒ b = 16 − 3 ⋅ 16 = −32. Dus l : y = 3x − 32. 25a
1
s (t ) = 10t ⋅ t = 10t ⋅t 2 = 10t
11
2
1
⇒ ds = s '(t ) = 15t 2 = 15 ⋅ t .
Dus ds = 15 ⋅ 1 = 15 ⋅ 1 = 15. dt t =1
dt
25b
De snelheid is ds = 15 ⋅ 8 m/s. d t t =8
G&R havo B deel 3 C. von Schwartzenberg 25c
12 Differentiaalrekening 5/20
108 km/u is 108 ⋅ 1000 = 30 m/s. 60 ⋅ 60 ds = 30
De formule s (t ) = 10t ⋅ t geldt voor 0 ≤ t ≤ 9. Na 9 seconden is s (9) = 10 ⋅ 9 ⋅ 9 = 90 ⋅ 3 = 270 meter afgelegd. De snelheid vanaf t = 9 is s '(9) = 15 ⋅ 9 = 15 ⋅ 3 = 45 m/s. Van t = 9 tot t = 60 legt de trein (60 − 9) ⋅ 45 = 2295 meter af. In de eerste minuut legt de trein 270 + 2295 = 2 565 meter af.
25d
dt
15 ⋅ t = 30 t = 2 ⇒ t = 4. Dus na 4 seconden.
26a
3 3 3 f (x ) = x + 2 = x +1 2 = x 1 +
x
x
x
2
1
x
2
1
1
1
2 = x 2 2 + 2x − 2 ⇒ f '(x ) = 2 1 x 1 2 − x −1 2 = 2 1 x ⋅ 1 2 2
x −
2
yA = f (1) = 3 = 3 en rc raaklijn = f '(1) = 2 1 − 1 = 2 1 − 1 = 1 1 . 1
2
1
2
1 11 x 2
=21x ⋅ x − 2
1
x⋅ x
.
2
k : y = 1 1 x + b door A(1, 3) ⇒ 1 1 ⋅ 1 + b = 3 ⇒ b = 3 − 1 1 = 1 1 . Dus k : y = 1 1 x + 1 1 . 2
26b
2
f '(x ) = 0 ⇒ 2 1 x ⋅ x − 2
2 5
+2
22
1
2
=0⇒
x⋅ x
21 x 2
⋅ x 1
=
1
x⋅ x
2
2
2
⇒ 2 1 x 2 ⋅ x = 1 ⇒ 5 x 3 = 1 ⇒ x 3 = 2 ⇒ x = 3 2 . Dus p = 2 . 2
2
5
5
5
26c
f 3 2 = 5
27a
f (x ) = (x 2 − 5x )2 = (x 2 − 5x )(x 2 − 5x ) ⇒ f '(x ) = (2x − 5)(x 2 − 5x ) + (x 2 − 5x )(2x − 5) = 2(x 2 − 5x )(2x − 5).
27b
f (x ) = (x 2 − 5x )(x 2 − 5x ) ⇒ f '(x ) = [x 2 − 5x ]'⋅ (x 2 − 5x ) + (x 2 − 5x ) ⋅ [x 2 − 5x ]' = 2(x 2 − 5x ) ⋅ [x 2 − 5x ]'.
28
De tabel van y3 = h (x ) valt samen met de tabel van y2 = g '(x ) (de hellingfunctie van g ).
=
32 5
5
62 5
. Dus a = 2 2 , b = 6 en c = 2 . 5
5
Vergeet niet de ketting nog eens extra te differentiëren.
1
29a
1
− f (x ) = x 2 + 4 = ( x 2 + 4 ) 2 ⇒ f '(x ) = 1 (x 2 + 4) 2 ⋅ 2x = x ⋅ 2
Of korter: f (x ) =
2
x + 4 ⇒ f '(x ) =
1 2
(x + 1
2⋅ x 2+ 4
x
⋅ 2x =
x 2+ 4
.
1 4) 2
Leer van buiten: [ x ]' =
29b
g (x ) = ( 22xx 44 ++ xx 22 )3 ⇒ g '(x ) = 3(2x 4 + x 2 )2 ⋅ (8x 3 + 2x ).
29c
− 3 h (x ) = x 3 + 3x = ( x 3 + 3x ) 3 ⇒ h '(x ) = 1 (x 3 + 3x ) 3 ⋅ (3x 2 + 3) =
1
2
3
j (x ) = ( 2x + 1 ) −2 ⇒ j '(x ) = −2(2x + 1) −3 ⋅ 2 = −4(2x + 1) −3 =
30a
f (x ) =
30b
g (x ) =
30c 30d
2 (x 3 + 3x ) 3
=
1 . 2⋅ x
x 2+1 3
(x 3 + 3x )2
.
−4 . (2x + 1)3
1 = ( 3x + 1 ) −2 ⇒ f '(x ) = −2(3x + 1) −3 ⋅ 3 = −6(3x + 1) −3 = − 6 3 . (3x + 1)2 (3x + 1) 1 4x − 1
1
= (4x
1 − 1) 2
= ( 4x − 1 )
− 21
⇒ g '(x ) = − 1 (4x − 1) 2
1 2
1
−1 21
1
⋅ 4 = −2 ⋅ (4x
11 2
h (x ) = (x 2 + 4) ⋅ x 2 + 4 = (x 2 + 4) ⋅ (x 2 + 4) = ( x 2 + 4 )
11 − 1) 2
Maak een schets van de plot hiernaast.
31b
f (x ) = ( 1 x 2 − 2x )3 ⇒ f '(x ) = 3( 1 x 2 − 2x )2 ⋅ (x − 2). 2 2 2 1 f '(x ) = 0 ⇒ 3( x − 2x ) ⋅ (x − 2) = 0 2 1 x 2 − 2x = 0 ∨ x − 2 = 0 2 2
=
−2 . (4x − 1) ⋅ 4x − 1
1
⇒ h '(x ) = 1 1 (x 2 + 4) 2 ⋅ 2x = 3x ⋅ x 2 + 4. 2
1
2 2 − 1 j (x ) = x + 4 = x + 4 1 = ( x 2 + 4 ) 2 ⇒ j '(x ) = 1 (x 2 + 4) 2 ⋅ 2x = x ⋅ = 1 2 x 2 + 4 (x 2 + 4) 2 2 (x + 4) 2
31a
31c
3x 2 + 3 3⋅
29d
x . x2+ 4
=
x x 2+ 4
.
2
x − 4x = 0 ∨ x = 2 x (x − 4) = 0 ∨ x = 2 x =0 ∨ x =4 ∨ x =2 yA = f (6) = 216 en rc raaklijn = f '(6) = 432. l : y = 432x + b door A(6, 216) ⇒ 432 ⋅ 6 + b = 216 ⇒ b = 216 − 432 ⋅ 6 = −2376. Dus l : y = 432x − 2376.
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 6/20
x 2 + 9 − x 2 + 5x ⇒ f '(x ) =
1
x
32a
f (x ) =
⋅ 2x − 2x + 5 =
− 2x + 5.
32b
f '(3) =
33
f (x ) = x ⋅ 2x + 1 ⇒ f '(x ) = 1 ⋅ 2x + 1 + x ⋅
1 ⋅2 = 2 ⋅ 2x + 1
2x + 1 +
34a
f (x ) = x ⋅ 3x + 1 ⇒ f '(x ) = 1 ⋅ 3x + 1 + x ⋅
1 ⋅3 = 2 ⋅ 3x + 1
3x + 1 +
34b
g (x ) = x ⋅ ( 3x + 1 )3 ⇒ g '(x ) = 1 ⋅ (3x + 1)3 + x ⋅ 3(3x + 1)2 ⋅ 3 = (3x + 1)3 + 9x (3x + 1)2 .
35a
f (x ) = x ⋅ 8 − 2x (BV: 8 − 2x ≥ 0 ⇒ −2x ≥ −8 ⇒ x ≤ 4) ⇒ Df = ←, 4 .
35b
f (x ) = x ⋅ 8 − 2x ⇒ f '(x ) = 1 ⋅ 8 − 2x + x ⋅
35c
f '(x ) = 0 ⇒ 8 − 2x −
35d
−3x = −8 ⇒ x top = −8 = 8 en y top = f ( 8 ) = 8 ⋅ 8 − 2 ⋅ 8 = 8 ⋅ 24 − 16 = 8 ⋅ 8 . Dus de top is ( 8 , 8 8 ). −3 3 3 3 3 3 3 3 3 3 3 3 3
35e
Het extreem in de top is een maximum (zie een plot) ⇒ Bf = ←, 8 8 . 3 3
36a
f (x ) = x ⋅ 2x + 6 (BV: 2x + 6 ≥ 0 ⇒ 2x ≥ −6 ⇒ x ≥ −3) ⇒ Df = −3, → .
36b
f (x ) = x ⋅ 2x + 6 ⇒ f '(x ) = 1 ⋅ 2x + 6 + x ⋅
x 2+ 9 2⋅ x2+ 9 [ x ]' = 1 . yA = f (4) = 9 en rc raaklijn = f '(4) = − 11 . 2⋅ x 5 k : y = − 11 x + b door A(4, 9) ⇒ − 11 ⋅ 4 + b = 9 ⇒ b = 9 + 44 = 89 . Dus k : y = − 11 x + 89 . 5 5 5 5 5 5 3 32 + 9
− 2 ⋅ 3 + 5 = 3 − 1 ≠ 0 ⇒ f heeft geen extreme waarde voor x = 3. 18
x 2x + 1
. (zie Theorie B)
f '(x ) = 0 ⇒ 2x + 6 +
x 8 − 2x
x 2x + 6
1 ⋅ −2 = 2 ⋅ 8 − 2x
= 0 ⇒ 8 − 2x = 1
=0⇒
x 8 − 2x
3x . 2 ⋅ 3x + 1
8 − 2x −
x 8 − 2x
.
⇒ 8 − 2x = x ⋅ 1.
x 1 ⋅ 2 = 2x + 6 + . 2 ⋅ 2x + 6 2x + 6 2x + 6 = −x ⇒ 2x + 6 = −x ⇒ 3x = −6 ⇒ 1 2x + 6
x = −2.
x top = −2 en y top = f ( −2) = −2 ⋅ −4 + 6 = −2 ⋅ 2. Dus de top is ( − 2, − 2 2). 36c
Het extreem in de top is een minimum (zie een plot) ⇒ Bf = −2 2, → .
37a
f (x ) = 2x ⋅
1 ⋅ −2 − 0 = 2 9 − 2x − 2 ⋅ 9 − 2x '(0) = 2 ⋅ 9 − 0 = 2 ⋅ 3 − 0 = 6. 9
9 − 2x − 3 ⇒ f '(x ) = 2 ⋅ 9 − 2x + 2x ⋅
yA = f (0) = 0 ⋅ 9 − 3 = −3 en rc raaklijn = f
2x . 9 − 2x
k : y = 6x + b door A(0, − 3) ⇒ 6 ⋅ 0 + b = −3 ⇒ b = −3 + 0 = −3. Dus k : y = 6x − 3.
2x = 0 ⇒ 2 ⋅ 9 − 2x = 2x ⇒ 2x = 2(9 − 2x ) ⇒ x = 9 − 2x ⇒ 3x = 9 ⇒ x = 3. 1 9 − 2x 9 − 2x x top = 3 en het maximum (zie een plot) is y top = f (3) = 6 ⋅ 9 − 6 − 3 = 6 3 − 3.
37b
f '(x ) = 0 ⇒ 2 ⋅ 9 − 2x −
37c
Df = ←, 4 1 (BV: 9 − 2x ≥ 0 ⇒ −2x ≥ −9 ⇒ x ≤ 4 21 ) en Bf = ←, 6 3 (zie 37b). 2
38a
De hoogtelijn CD deelt AB doormidden ⇒ AD = DB = 1. De stelling van Pythagoras in ∆ADC geeft CD = AC 2 − AD 2 = 22 − 12 = 3.
38b
In ∆ADC :
sin ∠DAC = sin 60° = overst. rz. = DC = 3 = 1 3 en cos ∠DAC = cos 60° = aanl. rz. = AD = 1 . AC
schuine z.
sin ∠DCA
2
2
schuine z.
AC
2
= sin 30° = overst. rz. = AD = 1 en cos ∠DCA = cos 30° = aanl. rz. = DC = 3 = 1 3. schuine z. 2 schuine z. AC 2 2 AC
38c
In ∆ADC :
38d
De stelling van Pythagoras in ∆ABC (fig. 12.12) geeft AC = AB 2 + BC 2 = 12 + 12 = 2. In ∆ABC (fig. 12.12): sin ∠BAC = sin 45° = BC = 1 = 1 ⋅ 2 = 2 = 1 2 en cos ∠BAC = cos 45° = AB = 1 = 1 2. AC
2
2
2
2
2
AC
2
2
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 7/20
39a
sin( 3 π ) = 1 2.
39b
cos( 7 π ) = − 1 3. 6 2
40a
sin(α ) = 1 3 (0 ≤ α ≤ 2π ) ⇒ α = 1 π ∨ α = 2 π .
40b
cos(α ) = − 1 (0 ≤ α ≤ 2π ) ⇒ α = 2 π ∨ 2 3
40c
sin(α ) = − 1 2 (0 ≤ α ≤ 2π ) ⇒ α = 1 1 π ∨ α = 1 3 π .
41
..., − 6 1 π , − 5 1 π , − 4 1 π , − 3 1 π , − 2 1 π , − 1 1 π , − 1 π , 1 π , 1 1 π , 2 1 π , 3 1 π , 4 1 π , 5 1 π , 6 1 π , ...
42a
sin(3x − 1 π ) = 0
4
2
2
sin(1 1 π ) = − 1 3.
39d
cos( 5 π ) = 1 . 3 2
3
3
α
= 1 1 π. 3
4
2
2
2
4
2
2
2
2
39f
sin( − 1 π ) = sin( 7 π ) = − 1 2. 4 4 2
3
6
2
cos(α ) = 0 (0 ≤ α ≤ 2π ) ⇒ α = 1 π ∨ α = 1 1 π .
40e
cos(α ) = 1 3 (0 ≤ α ≤ 2π ) ⇒ α = 1 π ∨ 2 6
40f
cos(α ) = 1 2 (0 ≤ α ≤ 2π ) ⇒ α = 1 π ∨ α = 1 3 π .
2
2
2
2
2
2
α = 1 5 π.
4
2
sin2 (x ) − sin(x ) = 0 sin(x ) ⋅ ( sin(x ) − 1 ) = 0 sin(x ) = 0 ∨ sin(x ) = 1 x = k ⋅ π ∨ x = 1 π + k ⋅ 2π .
2
2
6
4
2
42d cos2 (2x ) + cos(2x ) = 0 cos(2x ) ⋅ ( cos(2x ) + 1 ) = 0 cos(2x ) = 0 ∨ cos(2x ) = −1 x = 1 π + k ⋅ π ∨ x = k ⋅ 2π . 2
2
f (x ) = sin2 (x ) + sin(x ) = 0 sin(x ) ⋅ ( sin(x ) + 1 ) = 0 sin(x ) = 0 ∨ sin(x ) = −1 x = k ⋅ π ∨ x = − 1 π + k ⋅ 2π . 2
f (x ) = 0 (met 0 ≤ x ≤ 2π ) ⇒ x = 0 ∨ x = π ∨ x = 1 1 π ∨ x = 2π . 2
44a
cos(1 1 π ) = − 1 .
40d
2
1 x − 1 π = 1 π + k ⋅π 2 6 2 1 x = 2π + k ⋅π 2 3 x = 4 π + k ⋅ 2π . 3
2 3x = 1 π + k ⋅ π 2 x = 1 π + k ⋅ 1 π. 6 3
39e
42c sin2 (x ) = sin(x )
42b cos( 1 x − 1 π ) = 0
2
2
3
2
3x − 1 π = k ⋅ π
43
39c
f (x ) ≤ 0 (zie een plot) ⇒ x = 0 ∨ π ≤ x ≤ 2π .
sin(x ) cos(x ) − cos(x ) = 0 cos(x ) ⋅ ( sin(x ) − 1 ) = 0 cos(x ) = 0 ∨ sin(x ) = 1 x = 1 π + k ⋅ π ∨ x = 1 π + k ⋅ 2π . 2
2
sin(x ) cos(x ) − cos(x ) = 0 (met 0 ≤ x ≤ 2π ) ⇒ x = 1 π ∨ x = 1 1 π . 2
2
sin(x ) cos(x ) − cos(x ) ≤ 0 (zie een plot) ⇒ 0 ≤ x ≤ 1 π ∨ 1 1 π ≤ x ≤ 2π . 2
44b
2
2
cos (2x ) − cos(2x ) = 0 cos(2x ) ⋅ ( cos(2x ) − 1 ) = 0 cos(2x ) = 0 ∨ cos(2x ) = 1 2x = 1 π + k ⋅ π ∨ 2x = k ⋅ 2π 2
x = 1 π + k ⋅ 1 π ∨ x = k ⋅π. 4
2
cos2 (2x ) − cos(2x ) = 0 (met 0 ≤ x ≤ 2π ) ⇒ x = 0 ∨ x = 1 π ∨ x = 3 π ∨ x = π ∨ x = 1 1 π ∨ x = 1 3 π ∨ x = 2π . 4
4
4
4
cos2 (2x ) − cos(2x ) ≥ 0 (zie een plot) ⇒ x = 0 ∨ 1 π ≤ x ≤ 3 π ∨ x = π ∨ 1 1 π ≤ x ≤ 1 3 π ∨ x = 2π . 4
4
4
4
45a
sin( 1 π ) = 1 ⇒ x = 1 π is een oplossing van sin(x ) = 1 .
45b
2 1 π en 4 1 π liggen op dezelfde plaats als 1 π op de eenheidscirkel. (precies één of twee rondgangen verder) 6 6 6
45c
sin( 5 π ) = 1 ⇒ x = 5 π is een oplossing van sin(x ) = 1 .
45d
2 5 π en 6
46a
2 sin( 1 x ) = 1
6
2
6
2
6
6
2
2
− 1 1 π liggen op dezelfde plaats als 5 π op de eenheidscirkel. (precies één rondgang verder of terug) 6 6
2
sin( 1 x ) = 1
2 2 1 x = 1 π + k ⋅ 2π ∨ 1 x = 5 π + k ⋅ 2π (keer 2) 2 6 2 6 x = 1 π + k ⋅ 4π ∨ x = 5 π + k ⋅ 4π . 3 3
46b
2cos(x − 1 π ) = 1 3
cos(x − 1 π ) = 1 3
2
x − 1 π = 1 π + k ⋅ 2π ∨ x − 1 π = − 1 π + k ⋅ 2π 3
3
3
x = 2 π + k ⋅ 2π ∨ x = k ⋅ 2π . 3
3
G&R havo B deel 3 C. von Schwartzenberg 46c
12 Differentiaalrekening 8/20
2 sin(2x − 1 π ) = − 3 sin(2x − 1 π ) = − 1 3
2cos(3x − π ) = − 1 cos(3x − π ) = − 1
2x − 1 π = 4 π + k ⋅ 2π ∨ 2x − 1 π = − 1 π + k ⋅ 2π
3x − π = 2 π + k ⋅ 2π ∨ 3x − π = − 2 π + k ⋅ 2π
46d
4
4
2
2
3
4 3 4 3 2x = 19 π + k ⋅ 2π ∨ 2x = − 1 π + k ⋅ 2π 12 12 x = 19 π + k ⋅ π ∨ x = − 1 π + k ⋅ π . 24 24
47a
3
9
47c
6
sin(2x − 1 π ) = 1 2 2
2x − 1 π = 1 π + k ⋅ 2π ∨ 2x − 1 π = 3 π + k ⋅ 2π 6
4
6
4
2x = 5 π + k ⋅ 2π ∨ 2x = 11 π + k ⋅ 2π 12
12
x = 5 π + k ⋅ π ∨ x = 11 π + k ⋅ π . (met x op [0, 2π ]) x 47b
48a
24 = 5 π ∨ 24
3
x = 5 π + k ⋅ 2π ∨ x = 1 π + k ⋅ 2π.
2 sin(2x − 1 π ) = 2 6
3
3x = 5 π + k ⋅ 2π ∨ 3x = 1 π + k ⋅ 2π 3
9
sin( 2 x ) = − 1 2
3 2 2 x = − 1 π + k ⋅ 2π ∨ 2 x = 5 π + k ⋅ 2π (keer 3 ) 2 3 4 3 4 x = − 3 π + k ⋅ 3π ∨ x = 15 π + k ⋅ 3π . (met x op [0, 2π ]) 8 8 x = 15 π . 8
24
x = 1 5 π ∨ x = 11 π ∨ x = 1 11 π . 24
24
24
2cos(3x − 1 π ) = 3
2 cos(3x − 1 π ) = 1 3 2 2 3x − 1 π = 1 π + k ⋅ 2π ∨ 3x − 1 π = − 1 π + k ⋅ 2π 2 6 2 6 3x = 2 π + k ⋅ 2π ∨ 3x = 1 π + k ⋅ 2π 3 3 x = 2 π + k ⋅ 2 π ∨ x = 1 π + k ⋅ 2 π . (met x op [0, 2π ]) 9 3 9 3 x = 2 π ∨ x = 8 π ∨ x = 14 π ∨ x = 1 π ∨ x = 7 π ∨ 9 9 9 9 9
(
47d cos( 1 x ) = − 1 3
2 2 1 x = 5 π + k ⋅ 2π ∨ 1 x = − 5 π + k ⋅ 2π (keer 2) 2 6 2 6 x = 5 π + k ⋅ 4π ∨ x = − 5 π + k ⋅ 4π . (met x op [0, 2π ]) 3 3 x = 5π. 3
x = 13 π . 9
)
sin(x ) ⋅ cos(x ) − 1 3 = 0 2
sin(x ) = 0 ∨ cos(x ) = 1 3 2
x = k ⋅ π ∨ x = 1 π + k ⋅ 2π ∨ x = − 1 π + k ⋅ 2π . 6
( (
6
) )
sin(x ) ⋅ cos(x ) − 1 3 = 0 (met 0 ≤ x ≤ 2π ) ⇒ x = 0 ∨ x = π ∨ x = 2π ∨ x = 1 π ∨ x = 11 π . 2
6
sin(x ) ⋅ cos(x ) − 1 3 ≥ 0 (zie een plot) ⇒ 0 ≤ x ≤ 1 π ∨ π ≤ x ≤ 11 π ∨ x = 2π . 48b
49a 49b 49c
2
6
6
2
sin (x ) − 1 1 sin(x ) − 1 = 0 2 sin(x ) − 2 ⋅ sin(x ) + 1 = 0 2 sin(x ) = 2 ∨ sin(x ) = − 1 2 x = kan niet ∨ x = 7 π + k ⋅ 2π ∨ x = − 1 π + k ⋅ 2π . 6 6 sin2 (x ) − 1 1 sin(x ) − 1 = 0 (met 0 ≤ x ≤ 2π ) ⇒ x = 7 π ∨ x = 11 π . 2 6 6 sin2 (x ) − 1 1 sin(x ) − 1 ≤ 0 (zie een plot) ⇒ 0 ≤ x ≤ 7 π ∨ 11 π ≤ x ≤ 2π . 2 6 6
(
)
(
)
Zie de eerste drie schermen hiernaast. Vermoedelijk: f (x ) = sin(x ) ⇒ f '(x ) = cos(x ). TABLE doet het vermoeden versterken. Zie de eerste twee schermen hieronder. Vermoedelijk: g (x ) = cos(x ) ⇒ g '(x ) = − sin(x ). TABLE doet het vermoeden weer versterken. cos(x )
49d
3
Vermoedelijk: h (x ) = sin( x − 2 ) ⇒ h '(x ) = cos(x − 2) ⋅ 1 = cos(x − 2). TABLE doet het vermoeden alleen maar versterken. Zie de eerste twee schermen hiernaast. Vermoedelijk: j (x ) = cos( x + 1 ) ⇒ j '(x ) = − sin(x + 1) ⋅ 1 = − sin(x + 1). TABLE doet het vermoeden versterken. Zie de laatste twee schermen hiernaast.
sin(x )
6
G&R havo B deel 3 C. von Schwartzenberg 50ac
12 Differentiaalrekening 9/20
Zie de eerste drie schermen hiernaast. Vermoedelijk f (x ) = sin( 2x ) ⇒ f '(x ) = cos(2x ) ⋅ 2 = 2cos(2x ) . periode π amplitude 2
periode 22π =π amplitude 1
50bc
Controle met TABLE doet het vermoeden versterken. Zie de eerste twee schermen hiernaast. Vermoedelijk g (x ) = sin( 3x ) ⇒ g '(x ) = cos(3x ) ⋅ 3 = 3cos(3x ) . periode 2 π 3 amplitude 3
periode 23π = 23 π amplitude 1
Controle met TABLE doet het vermoeden versterken. 51
Vermoedelijk k (x ) = cos( 3x ) ⇒ k '(x ) = − sin(3x ) ⋅ 3 = −3 sin(3x ). Controle met TABLE doet het vermoeden versterken.
52a
f (x ) = cos( 2x ) ⇒ f '(x ) = − sin(2x ) ⋅ 2 = −2 sin(2x ).
52b
g (x ) = x ⋅ cos(x ) ⇒ g '(x ) = 1 ⋅ cos(x ) + x ⋅ − sin(x ) = cos(x ) − x sin(x ).
52c
h (x ) = 3 + 4 sin( 2x − 1 π ) ⇒ h '(x ) = 0 + 4 cos(2x − 1 π ) ⋅ 2 = 8cos(2x − 1 π ).
52d
j (x ) = 10 + 16 sin( 1 (x − 1)) = 10 + 16 sin( 1 (xx −−11) ) ⇒ j '(x ) = 0 + 16cos( 1 x − 1 ) ⋅ 1 = 8 cos( 1 x − 1 ).
3
3
2
2
Of
3
2
2
2
2
2
2
j (x ) = 10 + 16 sin( 1 (x − 1) ) ⇒ j '(x ) = 0 + 16cos( 1 x − 1 ) ⋅ 1 ⋅ 1 = 8cos( 1 x − 1 ). 2
2
2
2
2
2
53a
f (x ) = sin( ax + b ) ⇒ f '(x ) = cos(ax + b ) ⋅ a = a cos(ax + b ).
53b
g (x ) = cos( ax + b ) ⇒ g '(x ) = − sin(ax + b ) ⋅ a = −a sin(ax + b ).
54ab
I: II:
55a
f (x ) = x cos( 2x ) ⇒ f '(x ) = 1 ⋅ cos(2x ) + x ⋅ − sin(2x ) ⋅ 2 = cos(2x ) − 2x sin(2x ).
55b
g (x ) = x 2 sin( 3x ) ⇒ g '(x ) = 2x ⋅ sin(3x ) + x 2 ⋅ cos(3x ) ⋅ 3 = 2x sin(3x ) + 3x 2 cos(3x ).
55c
h (x ) = 2x sin( 3x − 1 ) ⇒ h '(x ) = 2 ⋅ sin(3x − 1) + 2x ⋅ cos(3x − 1) ⋅ 3 = 2 sin(3x − 1) + 6x cos(3x − 1).
55d
j (x ) = 1 + 3x cos( 1 x ) ⇒ j '(x ) = 0 + 3 ⋅ cos( 1 x ) + 3x ⋅ − sin( 1 x ) ⋅ 1 = 3cos( 1 x ) − 1 1 x sin( 1 x ).
56ab
I:
f (x ) = sin2 (x ) = sin(x ) ⋅ sin(x ) ⇒ f '(x ) = cos(x ) ⋅ sin(x ) + sin(x ) ⋅ cos(x ) = 2 sin(x ) ⋅ cos(x ).
II:
f (x ) = sin2 (x ) = sin(x )
f (x ) = x sin( 2x ) ⇒ f '(x ) = 1 ⋅ sin(2x ) + x ⋅ cos(2x ) ⋅ 2 = sin(2x ) + 2x cos(2x ). (mijn voorkeur) f (x ) = x sin( 2x ) ⇒ f '(x ) = 1 ⋅ sin(2x ) + x ⋅ 2cos(2x ) = sin(2x ) + 2x cos(2x ).
2
2
(
2
(
)
2
)
2
2
2
2
⇒ f '(x ) = 2 sin(x ) ⋅ cos(x ). (mijn voorkeur, maar bepaal je eigen voorkeur).
57a
f (x ) = cos2 (x ) = cos(x )
57b
g (x ) = 2 sin2 (x ) = 2 sin(x )
57c
h (x ) = 1 + 2cos2 (x ) = 1 + 2 cos(x )
57d
j (x ) = x + 3sin2 (x ) = x + 3 sin(x )
58a
f (x ) = sin3(x ) = sin(x )
58b
g (x ) = x sin2 (x ) = x ⋅ sin(x )
58c
h (x ) =
58d
j (x ) = 2x ⋅ cos( x 2 ) ⇒ j '(x ) = 2 ⋅ cos(x 2 ) + 2x ⋅ − sin(x 2 ) ⋅ 2x = 2cos(x 2 ) − 4x 2 sin(x 2 ).
(
⇒ f '(x ) = 2cos(x ) ⋅ − sin(x ) = −2 sin(x ) ⋅ cos(x ). 2
)
⇒ g '(x ) = 2 ⋅ 2 sin(x ) ⋅ cos(x ) = 4 sin(x ) ⋅ cos(x ).
(
(
(
3
)
2
2
)
⇒ h '(x ) = 0 + 2 ⋅ 2cos(x ) ⋅ − sin(x ) = −4 sin(x ) ⋅ cos(x ).
2
)
⇒ j '(x ) = 1 + 3 ⋅ 2 sin(x ) ⋅ cos(x ) = 1 + 6 sin(x ) ⋅ cos(x ). 2
⇒ f '(x ) = 3 ( sin(x ) ) ⋅ cos(x ) = 3sin2 (x ) ⋅ cos(x ).
(
2 + sin(x ) ⇒ h '(x ) =
2
)
⇒ g '(x ) = 1 ⋅ sin2 (x ) + x ⋅ 2 sin(x ) ⋅ cos(x ) = sin2 (x ) + 2x sin(x ) ⋅ cos(x ).
cos(x ) 1 ⋅ cos(x ) = . 2 ⋅ 2 + sin(x ) 2 ⋅ 2 + sin(x )
Gebruik: [ x ]' =
1 . 2⋅ x
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 10/20 2
(
)
59a
f (x ) = cos2 (x ) − cos(x ) = cos(x ) − cos(x ) ⇒ f '(x ) = 2cos(x ) ⋅ − sin(x ) + sin(x ) = −2 sin(x ) ⋅ cos(x ) + sin(x ). f '(x ) = 0 ⇒ −2 sin(x ) ⋅ cos(x ) + sin(x ) = 0 sin(x ) ⋅ ( −2cos(x ) + 1 ) = 0 max. (zie plot) f (0) = 0 sin(x ) = 0 ∨ − 2cos(x ) = −1 min. f ( 1 π ) = − 1 3 4 x = k ⋅ π ∨ cos(x ) = 1 max. f (π ) = 2 2 x = k ⋅ π ∨ x = 1 π + k ⋅ 2π ∨ x = − 1 π + k ⋅ 2π min. f ( 5 π ) = − 1 3 3 3 4 5 1 x op [0, 2π ] geeft x = 0 ∨ x = π ∨ x = 2π ∨ x = π ∨ x = π . max. f (2π ) = 0.
59b
yA = f ( 2 π ) = 3 en rc raaklijn = f '( 2 π ) = 3.
3
3
4
3
3
k : y = 3 ⋅ x + b door A( 2 π , 3 ) ⇒ 3 ⋅ 2 π + b = 3 ⇒ b = 3 − 2 π 3. Dus k : y = x 3 + 3 − 2 π 3. 3
60a
4
3
4
4
3
4
3
f (x ) = x ⋅ cos(x ) ⇒ f '(x ) = 1 ⋅ cos(x ) + x ⋅ − sin(x ) = cos(x ) − x sin(x ). yA = f ( 1 π ) = 1 π ⋅ 0 = 0 en rc raaklijn = f '( 2 π ) = 0 − 1 π ⋅ 1 = − 1 π . 2
2
3
2
2
k : y = − 1 π x + b door A( 1 π , 0) ⇒ − 1 π ⋅ 1 π + b = 0 ⇒ b = 1 π 2 . Dus k : y = − 1 π x + 1 π 2 . 2
2
2
2
4
2
4
60b
yB = f (π ) = π ⋅ −1 = −π en rc raaklijn = f '(π ) = −1 − π ⋅ 0 = −1. l : y = −x + b door B (π , − π ) ⇒ −π + b = −π ⇒ b = 0 ⇒ l gaat door de oorsprong.
60c
f '(1) = cos(1) − 1 ⋅ sin(1) ≠ 0 ⇒ f heeft geen top voor x = 1.
61a
f (x ) = sin2 (x ) − 3 ⋅ sin(x ) = sin(x )
(
2
)
− 3 ⋅ sin(x ) ⇒ f '(x ) = 2 sin(x ) ⋅ cos(x ) − 3 ⋅ cos(x ).
f '(x ) = 0 ⇒ 2 sin x ⋅ cos(x ) − 3 ⋅ cos(x ) = 0 cos(x ) ⋅ 2 sin(x ) − 3 = 0
(
)
min. (zie plot) f ( 1 π ) = − 3 3
x
2 = 1 π + k ⋅π ∨ 2
2
min. f ( 2 π ) = − 3
2
3
x = 1 π + k ⋅ 2π ∨ x = 2 π + k ⋅ 2π 3
2
2
3
4
max. f (1 1 π ) = ( −1)2 − 3 ⋅ −1 = 1 + 3.
3
2
x op [0, 2π ] geeft x = 1 π ∨ x = 1 1 π ∨ x = 1 π ∨ x = 2 π . 61b
4
max. f ( 1 π ) = 12 − 3 ⋅ 1 = 1 − 3
cos(x ) = 0 ∨ 2 sin(x ) = 3 x = 1 π + k ⋅ π ∨ sin(x ) = 1 3
3
yA = f ( 1 π ) = sin2 ( 1 π ) − 3 ⋅ sin( 1 π ) = ( 1 )2 − 3 ⋅ 1 = 1 − 1 3 en 6
6
6
2
2
4
2
rc raaklijn = f '( 1 π ) = 2 sin( 1 π ) ⋅ cos( 1 π ) − 3 ⋅ cos( 1 π ) = 2 ⋅ 1 ⋅ 1 3 − 3 ⋅ 1 3 = 1 3 − 3 . 6
6
6
6
2 2
2
2
k : y = ( 1 3 − 3 )x + b door A( 1 π , 1 − 1 3) ⇒ ( 1 3 − 3 ) ⋅ 1 π + b = 1 − 1 3. 2
2
6
4
2
6
2
2
2
Dus k : y = ( 1 3 − 3 )x + 1 − 1 3 − 1 π ( 1 3 − 3 ). 2
61c
2
4
2
6
4
2
2
2
f '(x ) = 3 ⇒ 2 sin(x ) ⋅ cos(x ) − 3 ⋅ cos(x ) = 3 (niet algebraïsch op te lossen) intersect geeft
x ≈ 3,1415... = π ∨ x ≈ 4,1887... = 4 π . 3
Controle: f '(π ) = 2 sin(π ) ⋅ cos(π ) − 3 ⋅ cos(π ) = 2 ⋅ 0 ⋅ −1 − 3 ⋅ −1 = 3. f '( 4 π ) = 2 sin( 4 π ) ⋅ cos( 4 π ) − 3 ⋅ cos( 4 π ) = 2 ⋅ − 1 3 ⋅ − 1 − 3 ⋅ − 1 = 1 3 + 1 3 = 3. en 3
62
3
Sinusoïde y = sin(x ) 1
verm. t.o.v. de y -as met
verm.t.o.v. de x -as met b
translatie (d , a )
y = sin(cx )
3
3
2
toppen (binnen één periode) ( 21 π , 1) en ( 23 π , − 1)
c
1
1 2
1
2
verm.t.o.v. de x -as met b
translatie (d , a )
y = a + b cosn(c (x − d ))
Geen opgave 63 te vinden.
64a
I = 2x ⋅ x ⋅ h = 40 ⇒ h = 402 = 202 ;
64b
x = 4 (dm) ⇒ h = 202 = 20 = 5 = 1,25 (dm) en M = 8 ⋅ 4 + 2 ⋅ 8 ⋅ 1,25 + 2 ⋅ 4 ⋅ 1,25 = 62 (dm2 ).
64c
M = 2 ⋅ 2x ⋅ h + 2 ⋅ x ⋅ h + 1 ⋅ 2x ⋅ x = 4xh + 2xh + 2x 2 = 6xh + 2x 2 (dm2 ).
4
16
1
verm. t.o.v. de y -as met
63
x
toppen (binnen één periode) (0, 1) en (π , − 1)
y = b cos(cx )
( 21c π + d , a + b ) en ( 23c π + d , a − b )
2x
2
y = cos(cx )
3 2
( 21c π , b ) en ( 23c π , − b )
y = a + b sin(c (x − d ))
2
Sinusoïde y = cos(x )
( c ⋅ π , 1) en ( c ⋅ π , − 1)
y = b sin(cx )
2
c
(0, 1) en ( c1 ⋅ π , − 1) (0, b ) en ( c1 π , − b ) (d , a + b ) en ( c1 π + d , a − b )
x = 2 (dm) ⇒ h = 202 = 20 = 5 (dm) en M = 4 ⋅ 2 + 2 ⋅ 4 ⋅ 5 + 2 ⋅ 2 ⋅ 5 = 68 (dm2 ). 2
4
4
G&R havo B deel 3 C. von Schwartzenberg 65a
12 Differentiaalrekening 11/20
I = 2x ⋅ x ⋅ h = 72 (dm3 ) ⇒ h = 722 = 362 (dm). 2x
x
K = 0, 4 ⋅ 2x ⋅ x + 0,2(2 ⋅ 2x ⋅ h + 2 ⋅ x ⋅ h ) = 0, 8x 2 + 1,2xh = 0,8x 2 + 1,2x ⋅ 362 = 0, 8x 2 + 43,2 (€). x
43,2
x
65b
43,2 K = 0,8x + = 0, 8x + 43,2x ⇒ dK = K ' = 1, 6x − 43,2x −2 = 1, 6x − 2 dx x x dK = 0 ⇒ 1, 6x − 43,2 = 0 ⇒ 1,6x = 43,2 ⇒ 1, 6x 3 = 43,2 ⇒ x 3 = 43,2 = 27 ⇒ x = 3 27 = 3 (dm). dx 1 1,6 x2 x2 De materiaalkosten zijn minimaal (er is slechts 1 kandidaat) bij de afmetingen van 3 bij 6 bij 4 dm.
66a
I = x ⋅ x ⋅ h = 16 (dm3 ) ⇒ h = 162 (h is de hoogte in dm).
2
2
−1
x
O = x ⋅ x + 4 ⋅ x ⋅ h = x 2 + 4xh = x 2 + 4x ⋅ 162 = x 2 + 64 (dm2 ). x
x
2
66b
O = x + 64 = x 2 + 64x −1 ⇒ dO = O ' = 2x − 64x −2 = 2x − 642 . dx x x dO = 0 ⇒ 2x − 64 = 0 ⇒ 2x = 64 ⇒ 2x 3 = 64 ⇒ x 3 = 32 ⇒ x = 3 32 ≈ 3,17 (dm). dx 1 x2 x2 De oppervlakte O is minimaal (er is slechts 1 kandidaat) bij de afmetingen van 3,17 bij 3,17 bij 1,59 dm.
67
O = x ⋅ y = 75 (m2 ) ⇒ y = 75 . x
K = 10x + 20(x + 2y ) = 30x + 40y = 30x + 40 ⋅ 75 = 30x + 3000 (€). x
x
. K = 30x + 3000 = 30x + 3 000x −1 ⇒ dK = K ' = 30 − 3 000x −2 = 30 − 3000 2
dx x x dK = 0 ⇒ 30 − 3000 = 0 ⇒ 30 = 3000 ⇒ 30x 2 = 3 000 ⇒ x 2 = 100 (met x > 0) ⇒ x = 100 = 10 (m). dx 1 x2 x2 De kosten K zijn minimaal (er is slechts 1 kandidaat) bij de afmetingen 10 (voor de vierde zijde) bij 7,5 m.
68a
O = x ⋅ y = 1200 (m2 ) ⇒ y = 1200 . x
K = 60y + 15(x + y ) = 15x + 75y = 15x + 75 ⋅ 1200 = 15x + 90000 (€). x
68b
68c
x
K = 15x + 90000 = 15x + 90 000x −1 ⇒ dK = K ' = 15 − 90 000x −2 = 15 − 90000 . x dx x2 dK = 0 ⇒ 15 − 90000 = 0 ⇒ 15 = 90000 ⇒ 15x 2 = 90 000 ⇒ x 2 = 6 000 (met x > 0) ⇒ x = dx 1 x2 x2 De minimale kosten (1 kandidaat) zijn € 2323, 79 bij de afmetingen 77,5 (langs de beek) bij 15,5 m.
6 000 ≈ 77, 5 (m).
K = 15x + 90000 = 2 500 intersect geeft x ≈ 52, 6 (minder lang). x
De afmetingen zijn 52,6 bij 22,8 meter. 69a
O = 2 ⋅ π r 2 (bodem en deksel) + 2π r ⋅ h (mantel) = 2π r 2 + 2π rh (cm2 ).
69b
I = π r 2 ⋅ h = 1 000 (cm3 ) ⇒ h = 1000 (cm). 2 πr
2
2
69c
O = 2π r + 2π rh = 2π r + 2π r ⋅ 1000 = 2π r 2 + 2000 (cm2 ). 2
69d
O = 2π r 2 + 2000 = 2π r 2 + 2 000r −1 ⇒ dO = O ' = 4π r − 2 000r −2 = 4π r − 2000 . 2
r
πr
r dr r dO = 0 ⇒ 4π r − 2000 = 0 ⇒ 4π r = 2000 ⇒ 4π r 3 = 2 000 ⇒ r 3 = 500 ⇒ r = 3 500 ≈ 5, 4 (cm). π π dr 1 r2 r2 De hoeveelheid materiaal is minimaal (er is slechts 1 kandidaat) bij r = 3 500 ≈ 5, 4 cm en h = 1000 ≈ 10,8 cm. 2 π
70a
I = π r 2 ⋅ h = 500 (cm3 ) ⇒ h = 5002 (cm). πr
K = 1 ⋅ π r 2 (bodem) + 1 ⋅ 2π rh (mantel) + 2 ⋅ π r 2 (deksel) + 2 ⋅ 2π r ⋅ 1 (rand met hoogte 1 cm) = 3π r 2 + 4π r + 2π rh = 3π r 2 + 4π r + 2π r ⋅ 5002 = 3π r 2 + 4π r + 1000 . πr
70b
r
De materiaalkosten zijn minimaal (optie minimum is toegestaan) bij r ≈ 3, 5 cm en h ≈ 12, 6 cm. (oplossen met de afgeleide geeft een derdegraadsvergelijking die niet algebraïsch kan worden opgelost, deze vergelijking kan vervolgens dan met intersect grafisch-numeriek worden opgelost)
πr
G&R havo B deel 3 C. von Schwartzenberg 71a
12 Differentiaalrekening 12/20
I = π r 2 ⋅ h ⇒ h = I 2 . (hierin is I een bepaalde inhoud dus voor te stellen als een of ander vast getal) πr
2
71b
O = 2 ⋅ π r (bodem en deksel) + 2π r ⋅ h (mantel) = 2π r 2 + 2π r ⋅ I 2 = 2I + 2π r 2 .
71c
O = 2I + 2π r 2 = 2I ⋅ r −1 + 2π r 2 ⇒ dO = O ' = −2I ⋅ r −2 + 4π r = 4π r − 2I2 . r dr r dO = 0 ⇒ 4π r − 2I = 0 ⇒ 4π r = 2I ⇒ 4π r 3 = 2I ⇒ r 3 = I ⇒ r = 3 I . dr 2π 2π 1 r2 r2
r
πr
Dus O is minimaal (er is slechts 1 kandidaat) bij r = 3 I . 2π
71d
2
r = 3 I = 3 π r h (alles tot de derde macht nemen) 2π 2π r 3 = π r 2h (kruiselings vermenigvuldigen) 1 2π
2π r 3 = π r 2h (links en rechts delen door π r 2 ) 2r = h . 72a
AB + AC + BC = 12 x + AC + AC = 12 2AC = 12 − x AC = 12 − x AC AC
CD 2 = AC 2 − AD 2 = (6 − 1 x )2 − ( 1 x )2 = (6 − 1 x )(6 − 1 x ) − ( 1 x )2
72b
2
2
2
2
2
= 36 − 3x − 3x + 1 x 2 − 1 x 2 = 36 − 6x . 4
2 = 12 − x 2 2 = 6 − 1 x. 2
4
CD = 36 − 6x . O (ABC ) = 1 ⋅ AB ⋅ CD = 1 x ⋅ 36 − 6x .
72c
2
2
73a
K = 12 ⋅ (200 − x ) + 10 ⋅ x 2 + 3 600 ⋅ 2 = 2 400 − 12x + 20 ⋅ x 2 + 3 600 (€).
73b
K = 2 400 − 12x + 20 ⋅
x2 +3 600 ⇒ dK = K ' = −12 + 20 ⋅ 3600 dx
dK = 0 ⇒ −12 + 20x =0⇒ dx x 2 + 3600 2 2
20x
x 2 + 3600
1 2 ⋅ x 2 + 3600 2
20x
⋅ 2x = −12 +
x 2 + 3600
[ x ]' =
.
1 . 2⋅ x
+ 3 600 ⇒ 400x 2 = 144(x 2 + 3 600)
= 12 ⇒ 20x = 12 ⋅ x 1
400x = 144x + 144 ⋅ 3 600 ⇒ 256x 2 = 518 400 ⇒ x 2 = 2 025 (met x > 0) ⇒ x = 45 (m). De minimale kosten (er is slechts 1 kandidaat) zijn € 3360 bij x = 45 m. 74a
K = a ⋅ (200 − x ) + 10 ⋅ x 2 + 3 600 ⋅ 2 = 200a − ax + 20 ⋅ x 2 + 3 600 (€). (hierbij is a een constante!!!) K = 200a − ax + 20 ⋅
x2 +3 600 ⇒ dK = K ' = −a + 20 ⋅ 3600 dx
1 2 ⋅ x 2 + 3600
⋅ 2x = −a +
20x
x 2 + 3600
.
74b
dK = 0 ⇒ dK = 0 ⇒ −a + 20 ⋅ 60 = 0 ⇒ 1200 = a ≈ 14,14 (€). dx x = 200 − AP dx x = 200 − 140 = 60 7200 602 + 3600
75a
AB ' = 5002 + 2002 = 250 000 + 40 000 = 290 000 (m). De totale kosten via het traject AB 'B zijn 100 ⋅ 290 000 + 150 ⋅ 100 ≈ 68 852 (€).
75b
AB = 5002 + 3002 = 340 000 (m). Verder is BC = 100 ⋅ 340 000 (m) en AC = 200 ⋅ 340 000 (m). 300
300
Een kabel in een rechte lijn van A naar B kost 100 ⋅ 2 ⋅ 340 000 + 150 ⋅ 1 ⋅ 340 000 ≈ 68 852 (€). 3
75c
2
2
2
3
2
2
AP = x + 200 = x + 40 000 (m) en PB = (500 − x ) + 100 = (500 − x )(500 − x ) + 1002 = 250 000 − 500x − 500x + x 2 + 10 000 = x 2 − 1 000x + 260 000 (m). De kosten voor een kabel via punt P zijn K = 100 ⋅ x 2 + 40 000 + a ⋅ x 2 − 1 000x + 260 000 (€).
K = 100 ⋅
x 2 + 40 000 + a ⋅
x 2 − 1 000x + 260 000 heeft als afgeleide functie
dK = K ' = 100 ⋅ 1 1 ⋅ 2x + a ⋅ ⋅ (2x − 1 000) = dx 2 ⋅ x 2 + 40000 2 ⋅ x 2 − 1000x + 260000
75d
a (400 −500) 100 ⋅ 400 dK =0⇒ + =0 dx x = 400 4002 + 40000 4002 − 1000 ⋅ 400 + 260000 −100a = − 100 ⋅ 400 4002 − 1000 ⋅ 400 + 260000 4002 + 40000 2 400 − 1000 ⋅ 400 + 260000 ≈ 126, 49 (€/m). a = 100 ⋅ 400 ⋅ 4002 + 40000
100
100x
x 2 + 40000
+
a (x −500) x 2 − 1000x + 260000
.
G&R havo B deel 3 C. von Schwartzenberg 76a
12 Differentiaalrekening 13/20
AP = x 2 + 0,12 = x 2 + 0, 01 (km) en BP = (0, 4 − x )2 + 0,22 = (0, 4 − x )(0, 4 − x ) + 0,22 snelheid =
afstand tijd
⇒ tijd =
= 0,16 − 0, 4x − 0, 4x + x 2 + 0, 04 = x 2 − 0, 8x + 0,20 (km).
afstand snelheid
De totale tijd t is gelijk aan t = 76b
2
x + 0,01 18
+
x 2 − 0,8x + 0,2 12
= 1 ⋅ x 2 + 0, 01 + 1 ⋅ x 2 − 0,8x + 0,2 (uur). 18
12
76c
Maak een schets van jouw plot. Zie een voorbeeld hiernaast. Vermeld het WINDOW-scherm. De optie minimum geeft als top het punt (x , t ) = (0,243...; 0, 035...).
76d
Frits heeft minimaal 0, 035... uur = 0, 035... ⋅ 60 ⋅ 60 ≈ 129 seconden nodig.
77a
Van S (TART) naar aankomst L(and) is SL = x 2 + 22 = x 2 + 4 (km) en van L naar F (INISH) is LF = 10 − x (km). 2 De totale tijd t is gelijk aan t = x + 4 + 10 − x = 1 ⋅ x 2 + 4 + 1 ⋅ (10 − x ) (uur).
4
77b
12
12
t = 1 ⋅ x 2 + 4 + 1 ⋅ (10 − x ) ⇒ dt = t ' = 1 ⋅ 4
dx
12
x dt = 0 ⇒ = 1 ⇒ 12x = 4 ⋅ dx 12 4⋅ x 2+ 4
x 1 ⋅ 2x + 1 ⋅ −1 = − 1 . 4 2⋅ x 2+ 4 12 4 ⋅ x 2 + 4 12 2 2 2 2 2
x + 4 ⇒ 3x = x + 4 ⇒ 9x = x + 4 ⇒ 8x = 4 ⇒ x 2 = 1 (x > 0) ⇒ x = 2
Dus t is minimaal (er is slechts 1 kandidaat) voor x = 78a
4
1 ( ≈ 0, 707) km. 2
AB + BC = x + BC = 60 ⇒ BC = 60 − x . (nu de stelling van Pythagoras in ∆ABC ) AC = BC 2 − AB 2 = (60 − x )2 − x 2 = (60 − x )(60 − x ) − x 2 = 3 600 − 60x − 60x + x 2 − x 2 = 3 600 − 120x . OABC = 1 ⋅ AB ⋅ AC = 1 x ⋅ 3 600 − 120x . 2
2
1 ⋅ −120 = 3600 − 120x − dx 2 2 2 2 ⋅ 3600 − 120x dO = 0 ⇒ 3600 − 120x = 30x ⇒ 60x = 3 600 − 120x ⇒ 180x = 3 600 ⇒ x = 3600 = 20. dx 2 180 3600 − 120x De maximale (er is slechts 1 kandidaat) oppervlakte O (20) ≈ 346, 41.
= 1x ⋅ 2
3600 3 600 − 120x ⇒ dO = O ' = 1 ⋅ 3 600 − 120x + 1 x ⋅
78b
O
79a
AP + PD = x + PD = 20 ⇒ PD = 20 − x . (nu de stelling van Pythagoras in ∆ADP )
30x . 3600 − 120x
AD = PD 2 − AP 2 = (20 − x )2 − x 2 = (20 − x )(20 − x ) − x 2 = 400 − 20x − 20x + x 2 − x 2 = 400 − 40x . OABC = 1 ⋅ AD ⋅ AP = 1 ⋅ 400 − 40x ⋅ x = 1 x ⋅ 400 − 40x . 2
79b
2
2
10x 1 400 − 40x ⇒ dO = O ' = 1 ⋅ 400 − 40x + 1 x ⋅ O ⋅ −40 = 400 − 40x − . dx 2 2 2 2 ⋅ 400 − 40x 400 − 40x dO = 0 ⇒ 400 − 40x = 10x ⇒ 20x = 400 − 40x ⇒ 60x = 400 ⇒ x = 400 = 40 = 20 (cm). dx 2 60 6 3 400 − 40x De maximale (er is slechts 1 kandidaat) oppervlakte O ( 20 ) ≈ 38, 49 cm2 . 3
= 1x ⋅ 2
1. 2
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 14/20
Diagnostische toets D1a
f (x ) = x 3(2x + 1) ⇒ f '(x ) = 3x 2 (2x + 1) + x 3 ⋅ 2 == 3x 2 (2x + 1) + 2x 3.
D1b
g (x ) = (x 2 − 2)(3x 2 + 4) ⇒ g '(x ) = 2x (3x 2 + 4) + (x 2 − 2) ⋅ 6x .
D1c
h (x ) = (x 2 − 4)2 = (x 2 − 4)(x 2 − 4) ⇒ h '(x ) = 2x (x 2 − 4) + (x 2 − 4) ⋅ 2x = 4x (x 2 − 4).
D2a f (x ) = x 2 (3x − 4) ⇒ f '(x ) = 2x (3x − 4) + x 2 ⋅ 3 = 2x (3x − 4) + 3x 2 . yA = f ( −1) = −7 en rc raaklijn = f '( −1) = 17.
k : y = 17x + b door A( −1, − 7) ⇒ 17 ⋅ −1 + b = −7 ⇒ b = −7 + 17 = 10. Dus k : y = 17 x + 10.
D2b f '( −1) = 2(3 − 4) + 3 ⋅ 1 = −2 + 3 ≠ 0. Dus de grafiek van f heeft geen top voor x = 1. D2c
Snijden met de x -as ( y = 0) ⇒ f (x ) = 0 ⇒ x 2 (3x − 4) = 0 ⇒ x = 0 ∨ 3x = 4. Dus xB = 4 . 3 yB = f ( 4 ) = 0 ( y = 0) en rc raaklijn = f '( 4 ) = 16 . 3
3
3
k : y = 16 x + b door B ( 4 , 0) ⇒ 16 ⋅ 4 + b = 0 ⇒ b = 0 − 16 ⋅ 4 = − 64 . Dus k : y = 16 x − 64 . 3
3
3
3
3
3
9
3
9
D3a f (x ) = −x 2 + 3x + 4 = 0 ⇒ x 2 − 3x − 4 = 0 ⇒ (x − 4)(x + 1) = 0 ⇒ x = 4 ∨ xA = −1. Nu is: AB = xB − xA = p − −1 = p + 1 en BC = yC = − p 2 + 3p + 4. O ( ∆ABC ) = 1 ⋅ AB ⋅ BC = 1 ⋅ ( p + 1) ⋅ ( − p 2 + 3p + 4) = ( 1 p + 1 )( − p 2 + 3p + 4). 2
D3b
2
2
2
O ( p ) = ( 1 p + 1 )( − p 2 + 3p + 4) ⇒ O '( p ) = 1 ( − p 2 + 3p + 4) + ( 1 p + 1 ) ⋅ ( −2 p + 3) 2 2 2 2 2 = − 1 p 2 + 1 1 p + 2 − p 2 + 1 1 p − p + 1 1 = −1 1 2 2 2 2 2 x top = −1 + 4 = 3 en O '( 3 ) = 3,125 ≠ 0. Dus O is niet maximaal voor p = 3 . 2 2 2 2
p 2 + 2p + 3 1 . 2
D3c O '( p ) = 0 ⇒ −1 1 p 2 + 2 p + 3 1 = 0 (keer − 2) 2
2
3p 2 − 4 p − 7 = 0 met D = ( −4)2 − 4 ⋅ 3 ⋅ −7 = 16 + 84 = 100 p = 4 − 10 = −6 = −1 ( ≤ −1 voldoet niet) ∨ p = 4 + 10 = 14 = 7 . Dus O maximaal voor p = 7 = 2 1 . 2⋅3 6 6 6 3 3 3 D4a f (x ) = 34 = 3 ⋅ 14 = 3x −4 ⇒ f '(x ) = −12x −5 = −12 ⋅ 15 = − 125 . x
x
3
x
− 33 = 4x 3 − 3x −3 ⇒
2
= 12x 2 + 94 .
D4b
g ( x ) = 4x
D4c
3 3 h (x ) = 2x 3− 3 = 2x3 − 33 = 2 − 3x −3 ⇒ h '(x ) = 9x −4 = 94 .
D4d
2 2 k (x ) = 6 − x = 6 − x = 6x −1 − x ⇒ k '(x ) = −6x −2 − 1 = − 62 − 1.
x
x
x
x
D4e l (x ) =
1 3x 6
x
g '(x ) = 12x + 9x
x
−4
x
x
x
x
x
= 1 ⋅ 16 = 1 x −6 ⇒ l '(x ) = −2x −7 = − 27 . 3 x 3 x
D4f
2 2 m (x ) = x + 2x + 1 = x + 2x + 1 = x + 2 + x −1 ⇒ m '(x ) = 1 − x −2 = 1 − 12 .
D5a
x +1 = 2 3 x
x
x
x
D5b
3(x + 1) = 2x 3x + 3 = 2x x = −3.
x
x
6 = 2x 9 3
D5c
x2
2x
= 54
x 3 = 27 = 33 x = 3.
−3 = − 1 3 3 = 1 x2 3 2
D5d
x2
(x − 3)(x + 26) = (x + 1)(x + 2)
x =9 x = 3 ∨ x = −3.
D6a f (x ) = 2x − 4 = 2x − 4 = 2 − 4x −1 ⇒ f '(x ) = 4x −2 = 42 . x
x
x
yA = f (4) = 1 en rc raaklijn = f '(4) = 1 .
x
4
k : y = 1 x + b door A(4, 1) ⇒ 1 ⋅ 4 + b = 1 ⇒ b = 1 − 1 = 0. Dus k : y = 1 x . 4
4
'(x ) = 1 ⇒ 42 = 1 ⇒ x 2 = 4 ⇒ x = −2 ∨
D6b rc raaklijn = f x yB = f ( −2) = 4 en yC = f (2) = 0. De raakpunten zijn B ( −2, 4) en C (2, 0).
4
x = 2.
x −3 = x +1 x + 2 x + 26
x 2 + 23x − 78 = x 2 + 3x + 2 20x = 80 x = 4.
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 15/20
1
21
D7a f (x ) = 2x 2 ⋅ x = 2x 2 ⋅ x 2 = 2x D7b
⇒ f '(x ) = 5x
2
1
1
g (x ) = (x + 4) ⋅ x = x ⋅ x 2 + 4x 2 = x
11
2
11
1
2
= 5x 1 ⋅ x 2 = 5x ⋅ x .
1
1
+ 4x 2 ⇒ g '(x ) = 1 1 x 2 + 2x
−1
2
2
= 1 1 ⋅ x + 21 = 1 1 ⋅ x + 2 . 2
x
1 2
2
2
Of g (x ) = (x + 4) ⋅ x (productregel) ⇒ g '(x ) = 1 ⋅ x + (x + 4) ⋅ [x ]' = 1 ⋅ x + (x + 4) ⋅ 1 x
− 21
2
2
1
1
1
− − −1 D7c h (x ) = x3 + 4 = x +1 4 = x1 + 41 = x 3 + 4x 3 ⇒ h '(x ) = 2 x 3 − 4 x 3 = 2 1 −
x
x3
x3
3
x3
3
3x 3
D7d k (x ) = (x 3 + x )(1 − x ) ⇒ k '(x ) = (3x 2 + 1)(1 − x ) + (x 3 + x ) ⋅ − 1 x
−1
2
2
4 3x ⋅ x
=
1 3
x
= x + x +4 . 2⋅ x
4 2 − . 3⋅ 3 x 3x ⋅ 3 x
3 = (3x 2 + 1)(1 − x ) − x + x . 2⋅ x
D8a f (x ) = 3( x 2 + 4x ) 4 ⇒ f '(x ) = 12(x 2 + 4x )3 ⋅ (2x + 4). D8b
x 2 + 2 ⇒ g '(x ) =
g (x ) =
1 2⋅ x 2+ 2
⋅ 2x =
x x2+2
.
Gebruik: [ x ]' =
1 . 2⋅ x
D8c h (x ) = x 2 ( 2x − 1 ) 4 ⇒ h '(x ) = 2x (2x − 1) 4 + x 2 ⋅ 4(2x − 1)3 ⋅ 2 = 2x (2x − 1) 4 + 8x 2 (2x − 1)3. 1 ⋅ −1 = 2⋅ 2 −x
D8d k (x ) = x ⋅
2 − x ⇒ k '(x ) = 1 ⋅ 2 − x + x ⋅
D9a f (x ) = x ⋅
50 − x 2 ⇒ f '(x ) = 1 ⋅ 50 − x 2 + x ⋅
f '(x ) = 0 ⇒ 50 − x 2 −
x2 50 − x
2
2−x −
1 2 ⋅ 50 − x
2 = 0 ⇒ 50 1− x =
2
x2 50 − x
2
x 2⋅ 2 −x
.
⋅ −2x = 50 − x 2 −
x2 50 − x 2 2
⇒ 1 ⋅ x 2 = 50 − x 2 ⇒ x
.
= 50 ⇒ x = ±5.
f ( −5) = −5 ⋅ 25 = −25 en f (5) = 5 ⋅ 25 = 25. De toppen zijn ( −5, − 25) en (5, 25). D9b yA = f (1) = 1 ⋅ 49 = 7 en rc raaklijn = f '(1) = 49 − 1 = 7 − 1 = 48 . 7 7 49 48 48 48 1 k: y = x + b door A(1, 7) ⇒ ⋅1 + b = 7 ⇒ b = 7 − = . Dus k : y = 48 x + 1 . 7
7
7
7
7
7
D9c Df = − 50, 50 (BV: 50 − x 2 ≥ 0 ⇒ −x 2 ≥ −50 ⇒ x 2 ≤ 50 ⇒ − 50 ≤ x ≤ 50). Bf = −25, 25 (zie D9a en een plot). D10a sin(α ) = − 1 3 (0 ≤ α ≤ 2π ) ⇒ α = 1 1 π ∨ α = 1 2 π .
D10c sin(α ) = 1 (0 ≤ α ≤ 2π ) ⇒ α = 1 π ∨ α = 5 π .
D10b cos(α ) = − 1 3 (0 ≤ α ≤ 2π ) ⇒ α = 3 π ∨ 2 4
D10d cos(α ) = 1 (0 ≤ α ≤ 2π ) ⇒ α = 1 π ∨ 2 3
2
3
3
α
D11a 4 sin(2x + 1 π ) = 2 2
sin(2x + 1 π ) = 1 2 2
2x + 1 π = 1 π + k ⋅ 2π ∨ 2x + 1 π = 3 π + k ⋅ 2π 2
4
2
4
2x = − 1 π + k ⋅ 2π ∨ 2x = 1 π + k ⋅ 2π 4
4
x = − 1 π + k ⋅ π ∨ x = 1 π + k ⋅ π . (met x op [0, 2π ]) 8
8
x = 7 π ∨ x = 1 7 π ∨ x = 1 π ∨ x = 1 1 π. 8
8
6
6
α = 1 2 π. 3
D11c cos(1 1 x ) = −1
2
2
2
= 1 1 π. 4
8
8
D11b cos(x − 1 π ) = − 1 3 6 2 x − 1 π = 5 π + k ⋅ 2π ∨ x − 1 π = − 5 π + k ⋅ 2π 6 6 6 6 x = π + k ⋅ 2π ∨ x = − 2 π + k ⋅ 2π . (met x op [0, 2π ]) 3 x = π ∨ x = 4 π. 3
2 3 x = π + k ⋅ 2π (keer 2 ) 3 2 x = 2 π + k ⋅ 4 π . (met x op [0, 2π ]) 3 3 x = 2 π ∨ x = 2π . 3 D11d 2 sin(3x + 1 π ) = − 3 2 sin(3x + 1 π ) = − 1 3 2 2 3x + 1 π = 4 π + k ⋅ 2π ∨ 3x + 1 π = − 1 π + k ⋅ 2π 2 3 2 3 3x = 5 π + k ⋅ 2π ∨ 3x = − 5 π + k ⋅ 2π 6 6 x = 5 π + k ⋅ 2 π ∨ x = − 5 π + k ⋅ 2 π . (met x op [0, 2π ]) 18 3 18 3 x = 5 π ∨ x = 17 π ∨ x = 29 π ∨ x = 7 π ∨ x = 19 π ∨ 18 18 18 18 18
x = 31 π . 18
D12a f (x ) = x 2 + 2cos(x ) ⇒ f '(x ) = 2x + 2 sin(x ). D12b g (x ) = 2x 2 ⋅ cos(x ) ⇒ g '(x ) = 4x ⋅ cos(x ) + 2x 2 ⋅ − sin(x ) = 4x cos(x ) − 2x 2 ⋅ sin(x ). D12c h (x ) = cos( x 2 ) ⇒ h '(x ) = − sin(x 2 ) ⋅ 2x = −2x sin(x 2 ). D12d j (x ) = cos2 (2x ) = cos( 2x ) ⋅ cos( 2x ) ⇒ j '(x ) = − sin(2x ) ⋅ 2 ⋅ cos(2x ) + cos(2x ) ⋅ − sin(2x ) ⋅ 2 = −4 sin(2x ) ⋅ cos(2x ). 2
Of j (x ) = cos2 (2x ) = cos( 2x ) ⇒ j '(x ) = 2cos(2x ) ⋅ − sin(2x ) ⋅ 2 = −4 sin(2x ) ⋅ cos(2x )
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 16/20
D13a h (t ) = 0,1 sin( 100πt ) ⇒ h '(t ) = 0,1cos(100πt ) ⋅ 100π = 10π cos(100πt ). D13b u (t ) = 4 sin( 80πt ) + 5 cos( 81πt ) ⇒ u '(t ) = 4 cos(80πt ) ⋅ 80π + 5 ⋅ − sin(81πt ) ⋅ 81π = 320π cos(80πt ) − 405π sin(81πt ). D13c f (x ) = 5x ⋅ sin( 3x ) ⇒ f '(x ) = 5 ⋅ sin(3x ) + 5x ⋅ cos(3x ) ⋅ 3 = 5 sins(3x ) + 15x cos(3x ). − sin(x ) 1 ⋅ − sin(x ) = . 2 ⋅ 2 + cos(x ) 2 ⋅ 2 + cos(x )
2 2 ++ cos( cos(xx )) ⇒ g '(x ) =
D13d g (x ) =
(
D14a f (x ) = x cos2 (x ) = x ⋅ cos(x )
2
)
Gebruik: [ x ]' =
1 . 2⋅ x
⇒ f '(x ) = 1 ⋅ cos2 (x ) + x ⋅ 2cos(x ) ⋅ − sin(x ) = cos2 (x ) − 2x sin(x ) ⋅ cos(x ).
D14b yA = f (π ) = π cos2 (π ) = π ⋅ ( −1)2 = π en rc raaklijn = f '(π ) = cos2 (π ) − 2π sin(π ) ⋅ cos(π ) = ( −1)2 − 2π ⋅ 0 = 1. k : y = x + b door A(π , π ) ⇒ 1 ⋅ π + b = π ⇒ b = π − π = 0. Dus k : y = x . D15a AB + AC = x + AC = 20 ⇒ AC = 20 − x . 2
2
2
(nu de stelling van Pythagoras in ∆ABC )
2
BC = AC + AC = x + (20 − x ) = x 2 + (20 − x )(20 − x ) = x 2 + 400 − 20x − 20x + x 2 = 2x 2 − 40x + 400. OABEDC = OABC + OBEDC = 1 ⋅ AB ⋅ AC + BC 2 2
= 1 ⋅ x ⋅ (20 − x ) + 2x 2 − 40x + 400 = 10x − 1 x 2 + 2x 2 − 40x + 400 = 1 1 x 2 − 30x + 400. 2
2
2
2
D15b O = 1 1 x − 30x + 400 ⇒ dO = O ' = 3x − 30. 2 dO = 0 ⇒ 3x = 30 ⇒ dx
dx
x = 10. De oppervlakte is minimaal (er is slechts 1 kandidaat) voor x = 10.
(cm). D16a I = x ⋅ 1 x ⋅ h = 3 000 (cm3 ) ⇒ x 2h = 9 000 (cm3 ) ⇒ h = 9000 2 3
x
O = x ⋅ 1 x + 2 ⋅ x ⋅ h + 2 ⋅ 1 x ⋅ h = 1 x 2 + 2 2 xh = 1 x 2 + 8 x ⋅ 9000 = 1 x 2 + 24000 (cm2 ). 2 3
3
3
3
3
3
x
3
x
D16b O = 1 x 2 + 24000 = 1 x 2 + 24 000x −1 ⇒ dO = O ' = 2 x − 24 000x −2 = 2x − 24000 3 3 dx 3 3 x x2 dO = 0 ⇒ 2x = 24000 ⇒ 2x 3 = 3 ⋅ 24 000 ⇒ x 3 = 3 ⋅ 12 000 ⇒ x = 3 36 000 ≈ 33, 0 (cm). dx 3 x2 De oppervlakte is minimaal (er is slechts 1 kandidaat) bij de afmetingen van 33,0 bij 11,0 bij 8,3 cm. (de optie minimum is hier ook geoorloofd en geeft met van een een geschikt venster dezelfde x -waarde) D17a O = x ⋅ y = 1200 (m2 ) ⇒ y = 1200 . x
De totale lengte van de afrastering is L = 4x + y = 4x + 1200 (m). x
L is minimaal (de optie minimum mag toegepast worden) voor x ≈ 17,3 (m). De afmetingen van het stuk land zijn 17,3 bij 69,3 m. D17b De kosten voor de afrastering zijn K = 60(2x + y ) + 20 ⋅ 2x = 160x + 60y = 160x + 60 ⋅ 1200 (€).
K is minimaal (optie minimum mag) voor x ≈ 21,2 (m). De afmetingen van het stuk land zijn 21,2 bij 56,6 m.
x
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 17/20
Gemengde opgaven 12. Differentiaalrekening G31a l =
(xP )2 + ( yP )2 =
G31b l =
p 4 − 7 p 2 + 16 ⇒ dl =
(
p2 + 4 − p2 dp
2p 3 − 7 p
dl = 0 ⇒ dp
4
2
p − 7 p + 16
2
)
=
1 4
(
)(
)
p 2 + 4 − p 2 4 − p 2 = p 2 + 16 − 4 p 2 − 4 p 2 + p 4 = p 4 − 7 p 2 + 16.
2
2 ⋅ p − 7 p + 16
⋅ (4 p 3 − 14 p ) =
2p 3 − 7 p
p 4 − 7 p 2 + 16
.
= 0 (teller = 0) ⇒ 2 p 3 − 7 p = 0 ⇒ 2 p ( p 2 − 3 1 ) = 0 ⇒ 2 p = 0 ∨ p 2 = 3 1 (met p > 0) ⇒ p = 3 1 . 2
1
Dus l is minimaal (er is slechts 1 kandidaat) voor p =
2
2
31. 2
G31c Deze cirkel heeft als straal de minimale lengte l uit G31b (omdat de cirkel de parabool raakt). 4
2
2
r = 3 1 − 7 3 1 + 16 = 2 2
(3 21 )
− 7 ⋅ 3 1 + 16 = 12 1 − 24 1 + 16 = 28 1 − 24 1 = 3 3 . 2 4 2 4 2 4 Dus de oppervlakte van deze cirkel is O = π r 2 = π ⋅ 3 3 = 3 3 π . 4
4
G32c 3sin(2x − 1 π ) = −1 1 3
G32a sin(x ) = − 1 2
3
2
4
2
sin(2x − 1 π ) = − 1 3
x = − 1 π + k ⋅ 2π ∨ x = 1 1 π + k ⋅ 2π .
3
4
2
2x − 1 π = − 1 π + k ⋅ 2π ∨ 2x − 1 π = 4 π + k ⋅ 2π
G32b sin(2x − 1 π ) ⋅ cos(x + 1 π ) = 0 4 3 sin(2x − 1 π ) = 0 ∨ cos(x + 1 π ) = 0 4 3 2x − 1 π = k ⋅ π ∨ x + 1 π = 1 π + k ⋅ π 4 3 2 2x = 1 π + k ⋅ π ∨ x = 1 π + k ⋅ π 4 6 x = 1 π + k ⋅ 1 π ∨ x = 1 π + k ⋅π. 8 2 6
3
3
3
2x = k ⋅ 2π ∨ 2x = 5 π + k ⋅ 2π
3
3
x = k ⋅π ∨ x = 5 π + k ⋅π. 6
G33a f (x ) = x 2 ⋅ cos( 2x ) ⇒ f '(x ) = 2x ⋅ cos(2x ) + x 2 ⋅ − sin(2x ) ⋅ 2 = 2x cos(2x ) − 2x 2 sin(2x ).
sin(x ) + x ⇒ g '(x ) =
G33b g (x ) =
(
G33c h (x ) = cos(x )
2
)
cos(x ) + 1 1 ⋅ (cos(x ) + 1) = . 2 ⋅ sin(x ) + x 2 ⋅ sin(x ) + x
Gebruik: [ x ]' =
1 . 2⋅ x
− 2 sin( 3x ) ⇒ h '(x ) = 2cos(x ) ⋅ − sin(x ) − 2 cos(3x ) ⋅ 3 = −2 sin(x ) cos(x ) − 6cos(3x ).
G34a f (x ) = sin2 (x ) − sin(x ) = 0 sin(x ) ⋅ ( sin(x ) − 1 ) = 0 sin(x ) = 0 ∨ sin(x ) = 1 x = k ⋅ π ∨ x = 1 π + k ⋅ 2π . x op [0, 2π ] geeft x = 0 ∨ x = 1 π ∨ x = π ∨ x = 2π . 2
2
(
G34b f (x ) = sin2 (x ) − sin(x ) = sin(x )
2
)
− sin(x ) ⇒ f '(x ) = 2 sin(x ) ⋅ cos(x ) − cos(x ).
f '(x ) = 0 ⇒ 2 sin(x ) ⋅ cos(x ) − cos(x ) = 0 cos(x ) (2 sin(x ) − 1 ) = 0 cos(x ) = 0 ∨ 2 sin(x ) = 1 x = 1 π + k ⋅ π ∨ sin(x ) = 1
min. (zie plot) f ( 1 π ) = − 1
x = 1 π + k ⋅ π ∨ x = 1 π + k ⋅ 2 ∨ x = 5 π + k ⋅ 2π .
max. f (1 1 π ) = 2.
2
max. f ( 1 π ) = 2 2
min. f ( 5 π ) = − 1
2
2
6
6
6
6
4
2
f '(x ) = 0 (met 0 ≤ x ≤ 2π ) ⇒ x = 1 π ∨ x = 1 π ∨ x = 5 π ∨ x = 1 1 π . (de extreme waarden hierboven) 6
2
6
2
2
G34c yA = f (π ) = sin (π ) − sin(π ) = 0 en rc raaklijn = f '(π ) = 2 sin(π ) ⋅ cos(π ) − cos(π ) = 0 − −1 = 1.
k : y = x + b door A(π , 0) ⇒ 1 ⋅ π + b = 0 ⇒ b = −π . Dus k : y = x − π .
G35a I = x ⋅ 2x ⋅ h = 12 (m3 ) ⇒ h = 122 = 62 (m). 2x
x
K = 120 ⋅ x ⋅ 2x + 80 ⋅ (2 ⋅ x ⋅ h + 2x ⋅ h ) = 240x 2 + 80 ⋅ 4xh = 240x 2 + 320x ⋅ 62 = 240x 2 + 1920 (€). x
x
G35b K = 240x + 1920 = 240x 2 + 1 920x −1 ⇒ dK = K ' = 480x − 1 920x −2 = 480x − 1920 . x dx x2 dK = 0 ⇒ 480x = 1920 ⇒ 480x 3 = 1 920 ⇒ x 3 = 1920 = 4 ⇒ x = 3 4 ≈ 1,59 (m). dx 480 1 x2 De kosten zijn minimaal (er is slechts 1 kandidaat) bij de afmetingen van 1,59 bij 3,17 bij 2,38 m. 2
4
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 18/20
G35c O zijaanzicht = x ⋅ h + 1 x ⋅ x ⇒ I = (xh + 1 x 2 ) ⋅ 2x = 2x 2h + x 3. 2 2
x 2 x
3 3 G35d I = 12 (m3 ) ⇒ 2x 2h = 12 − x 3 ⇒ h = 12 − x2 = 122 − x 2 = 62 − x (m).
2x
2x
2x
x
2
K = 120 ⋅ 2x ⋅ x ⋅ 2 + 80 ⋅ 2x ⋅ h + 80 ⋅ (xh + 1 x 2 ) ⋅ 2 = 240x 2 ⋅ 2 + 160xh + 160xh + 80x 2 2
45°
x
= 240x 2 ⋅ 2 + 320x ⋅ ( 62 − x ) + 80x 2 = 240x 2 ⋅ 2 + 1920 − 160x 2 + 80x 2 = 240x 2 ⋅ 2 − 80x 2 + 1920 (€). x
2
x
x
G35e K is minimaal (optie minimum geoorloofd) voor x ≈ 1, 55 (m). De afmetingen van het vloeroppervlak zijn 1,55 bij 3,09 m. De hoogte aan de voorkant is m en de hoogte tegen de gevel is 3,28 m. G36a f (x ) = 0 ⇒ 27 x − x 4 = 0
O∆OST = 6. Dus 1 ⋅ OS ⋅ yT = 1 ⋅ 3 ⋅ yT = 1 1 ⋅ yT = 6 ⇒ yT = 61 = 12 = 4.
27 x − x 4 = 0
2
x (27 − x 3 ) = 0 x = 0 ∨ 27 = x 3 x = 0 ∨ x = 3 27 = 3. Dus OS = 3.
2
2
1
3
2
4
y = 27x − x = 4 (niet algebraïsch op te lossen ⇒ ) intersect geeft x ≈ 0, 60 ∨ x ≈ 2, 77. Dus T (0, 60; 4) en U (2, 77; 4).
G36b AB = f ( p ) − g ( p ) = 3 ⇒ 27 x − x 4 − 8x − x 4 = 3. Intersect geeft p ≈ 1,34. G36c h (x ) = cx − x 4 met domein [0,10]. Dus h (0) = 0 en h (10) = 0. h (10) = 10c − 10 000 = 0 10c − 10 000 = 0 10c = 10 000 c = 1 000. G36d h (x ) =
cx − x 4 ⇒ h '(x ) =
Dus h (x ) = 1 000x − x 4 op het domein [0,10]. Optie maximum geeft x ≈ 6,30 en y ≈ 68, 74. Dus het bereik van h is [0;68,74].
1 2 ⋅ cx − x
4
⋅ (c − 4x 3 ) =
h heeft een maximum voor x = 1,5 ⇒ h '(1, 5) = 0 ⇒
c − 4x 3 . 2 ⋅ cx − x 4 c − 4 ⋅ 1,53 2 ⋅ 1,5c − 1,5 4
= 0 ⇒ c − 4 ⋅ 1, 53 = 0 ⋅ ... ⇒ c = 4 ⋅ 1,53 = 13,5. 1
G37a f (3) = f ( −3) = 34 − 16 = 81 − 16 = 65. Dus de grafiek van f is 65 omlaag verschoven. G37b f (x ) = x 4 − 16 ⇒ f '(x ) = 4x 3 rcm = f '(2) = 4 ⋅ 23 = 4 ⋅ 8 = 32 m: y = 32x + b door ( −2, 0) ⇒ 32 ⋅ −2 + b = 0 ⇒ b = 64. Dus m: y = 32x + 64. G37c g (x ) = x 3(x 4 − 16) (productregel of) = x 7 − 16x 3 ⇒ g '(x ) = 7x 6 − 48x 2 .
g '(x ) = 7x 6 − 48x 2 = 0 ⇒ 7x 2 (x 4 −
48 ) ⇒ x 2 = 0 ∨ x 4 = 48 ⇒ x = 0 ∨ x = ± 4 48 . 7 7 7 De x -coördinaten van de toppen zijn − 4 48 en 4 48 (bij x = 0 geen top maar een buigpunt). 7 7
G38a A balk = 2 ⋅ 7, 5 ⋅ 4 + 2 ⋅ 7,5 ⋅ 10 + 2 ⋅ 4 ⋅ 10 = 60 + 150 + 80 = 290 (cm2 ). Acilinder = 2 ⋅ π ⋅ 32 + 2π ⋅ 3 ⋅ 10, 6 ≈ 256 (cm2 ). Omdat V balk = V cilinder hoort de kleinste F bij de verpakking met de kleinste oppervlakte. Dus de cilindervormige verpakking heeft de kleinste F -waarde. 2 G38c F = 2 + π r = 2r −1 +
G38b 20 < h < 40 ⇒ 20 < 8000 < 40 2
r
πr
20 < 8000 én 8000 < 40 2 2 πr
πr
20π r 2 < 8 000 én 40π r 2 > 8 000
r 2 < 8000 én r 2 > 8000 20π
(0 <) r <
40π 8000 ≈ 11,3 én 20π
Dus 8, 0 < r < 11,3
r > 8000 ≈ 8, 0. 40π
4000
π 4000
r 2 geeft
dF = F ' = −2r −2 + π ⋅ 2r = − 2 + π r . dr 4000 r 2 2000 dF = − 2 + π r = 0 dr r 2 2000 πr = 2 2000 r 2 3
π r = 4 000
r 3 = 4000 ⇒ r = 3 4000 ≈ 10,8 (cm). π
π
G&R havo B deel 3 C. von Schwartzenberg
12 Differentiaalrekening 19/20 20
G39a Bestudeer het vooraanzicht hiernaast. 90 − 10 = 80 en 80 : 2 = 40 (cm).
h
Pythagoras: h = 402 − 52 ≈ 39, 7 (cm). G39b Zie de hiernaast op schaal 1 : 10 (het vooraanzicht en) zijaanzicht.
40 50
5 10
G39c Maak een schets met de gegevens. Zie hieronder.
40
50
2x
h
90 − x 2
1 2
x
35
35
20 vooraanzicht
x
Pythagoras:
h=
(
90 − x 2
2
2
) −( x) 1 2
=
2
( 45 − 21 x )
− 1 x2 = 4
50 zijaanzicht
( 45 − 21 x )( 45 − 21 x ) − 41 x 2
= 2 025 − 45 ⋅ 1 x − 1 x ⋅ 45 + 1 x 2 − 1 x 2 = 2 025 − 45x . 2
2
4
4
G39d I = 0, 075x ⋅ 2 025 − 45x optie maximum (is toegestaan) ⇒ x = 30 (cm). De inhoud is maximaal (er is slechts 1 kandidaat) bij x = 30 en h = 2 025 − 45 ⋅ 30 ≈ 26 cm. Of (algebraisch met de afgeleide)
I = 0, 075x ⋅ 2 025 − 45x −45 ⋅ 0,075x dI = 0, 075 ⋅ 2 025 − 45x + 0, 075x ⋅ 1 ⋅ −45 = 0, 075 ⋅ 2 025 − 45x + . dx 2 ⋅ 2025 − 45x 2 ⋅ 2025 − 45x dI = 0 ⇒ 0,075 ⋅ 2025 − 45x = 45 ⋅ 0,075x ⇒ 2 ⋅ 0, 075 ⋅ (2 025 − 45x ) = 45 ⋅ 0, 075 x dx 1 2 ⋅ 2025 − 45x 4 050 − 90x = 45x ⇒ 4 050 = 135x ⇒ x = 4050 = 30 (cm). Dit geeft dan (zoals hierboven) 135
39e
= 0, 075x ⋅ 2 025 − 45x ≥ 30 (dm3 ). = 30 intersect ⇒ x ≈ 10,1 ∨ x ≈ 43,1 (cm). ≥ 30 (zie een plot) ⇒ 10,1 ≤ x ≤ 43,1. = 2 025 − 45x ≥ 20 (cm) intersect of 2 025 − 45x ≥ 400 −45x ≥ −1 625 x ≤ 1625 ≈ 36,1 (cm).
I I I h
45
h ≥ 20 én I ≥ 30 ⇒ x ≤ 36,1 én 10,1 ≤ x ≤ 43,1. Dus 10,1 ≤ x ≤ 36,1 (cm).
h ≈ 26 (cm).
G&R havo B deel 3 C. von Schwartzenberg
Voorkennis
12 Differentiaalrekening 20/20
3 Differentiëren (bladzijde 156)
7a
f (x ) = 2x 3 + 3x 2 ⇒ f '(x ) = 6x 2 + 6x .
7b
g (x ) = 1 x 4 − x 3 + 1 ⇒ g '(x ) = 2x 3 − 3x 2 .
7c
h (x ) = x 2 (x 2 − 2) = x 4 − 2x 2 ⇒ h '(x ) = 4x 3 − 4x .
7d
k (x ) = (4x 2 + 1)2 = (4x 2 + 1)(4x 2 + 1) = 16x 4 + 4x 2 + 4x 2 + 1 = 16x 4 + 8x 2 + 1 ⇒ k '(x ) = 64x 3 + 16x .
8a
f ( p ) = 4 p 3 − 1 p ⇒ f '( p ) = 12 p 2 − 1 .
8c
s (t ) = 5t 2 + t + 1 = 5t 2 + 1 t +
8b
g (q ) = 1 q 2 + 2q + 1 ⇒ g '(q ) = 2 q + 2.
8d
N (t ) = 0, 01t − 0, 05t 2 + 0,1t ⇒ N '(t ) = 0, 03t − 0,1t + 0,1.
2
2
2
5
5
Voorkennis 9a
2 3
2
1 ⇒ s '(t ) = 10t + 1 . 2 2 2
4 Extremen berekenen (bladzijde 157)
f (x ) = 1 x 3 + 1 x 2 − 12x − 10 ⇒ f '(x ) = x 2 + x − 12. 3
2
f '(x ) = 0 ⇒ x 2 + x − 12 = 0 (x + 4)(x − 3) = 0 x = −4 ∨ x = 3. Maximum (zie een plot) f ( −4) = 24 2 en minimum (zie een plot) f (3) = −32 1 . 3
9b
3
2
2
2
g (x ) = −x + 9x − 50 ⇒ g '(x ) = −3x + 18x . g '(x ) = 0 ⇒ −3x 2 + 18x = 0 −3x (x − 6) = 0 x = 0 ∨ x = 6. Minimum (zie een plot) g (0) = −50 en maximum (zie een plot) g (6) = 58.
Voorkennis
5 Machten met negatieve en/of gebroken exponenten (bladzijden 158 en 159)
10a x −5 = 15 . x
1
3
10b x 3 = x 1 = 3 x .
10e
6x
10f
5x
1 3
3
= 6⋅ x1 = 6⋅3x .
− 41
= 5 ⋅ 11 = 45 . x
x4
10c x 10d x
3 21
−2 1
2
= 11 = x
2
1 2
x ⋅x
2
1 2
1
x 4 ⋅ x = x 4 ⋅x 2 = x
11b
x −2 = x −2−3 = x −5 . x3
11d
1
1
=
11a
11c
2
x ⋅ x
4 21
1
1
.
10g
2 31 3 2 x = x 3 = x .
10h
(x )
11e
1 1 −2 − 2 2 x⋅ x = x ⋅ x2 = x 2 = x 2 = x 2. x2 x x
3
11g 2
−2 −1 3 x = x 2 = x 3 = x 3. 2 x x
1
−5
( )
= x −1
−5
= x 5.
1
.
11f
x = x1 = x1−4 = x 4. 1 4 x x4 3
2
1
= x3 ⋅x 2 = x3 ⋅ x .
11h
x4 x⋅ x
x 1⋅x 2
)
1
1
1
x4
=
11
1 2
1
4 4 −1 2 2 = x 2. = x 1 =x
x
1
2
2
2
1 11 = x 1 ⋅ x 2 = x 2 = x 3.
(
x⋅ x
(
2 2 1 2 11 22 x ⋅ 3 x = x 1 ⋅ x 3 = x 3 = x 3 .
)
