Filetype PDF | Posted on 25 Jan 2023 | 3 years ago
Authentication
digital resources
339x Filetype PDF File size 0.49 MB Source: ocw.mit.edu
SOLUTIONS TO 18.01 EXERCISES
2. Applications of Differentiation
2A. Approximation
d √ b √ b
2A-1 dx a + bx = √ ⇒ f(x) ≈ a + √ x by formula.
2 a + bx 2 a
�
√ √ bx √ bx
By algebra: a + bx = a 1+ ≈ a(1 + ), same as above.
a 2a
2A-2 D( 1 ) = −b ⇒ f(x) ≈ 1 − b x; OR: 1 = 1/a ≈
2 2
a + bx (a + bx) a a a + bx 1+ b/ax
1 b
(1 − x).
a a
3/2 3 3/2 3/2
(1 + x) (1+2x) · 2 · (1 + x) − (1 + x) · 2 � 1
2A-3 D( 1+2x )= (1 + 2x)2 ⇒ f (0) = −
2
3/2
1 (1 + x) 3 1
⇒ f(x) ≈ 1− x; OR, by algebra, ≈ (1+ x)(1−2x) ≈ 1− x.
2 1 + 2x 2 2
h g 2 2h
2A-4 Put R = �; then w = (1+ �)2 ≈ g(1 − �) ≈ g(1 − 2�) = g(1 − R ).
2A-5 A reasonable assumption is that w is propotional to volume v, which is in
turn proportional to the cube of a linear dimension, i.e., a given person remains
3; since 5 feet = 60
similar to him/herself, for small weight changes.) Thus w = Ch
inches, we get
w(60 + �) C(60 + �)3 � 3 3� 1
= 3 = (1+ ) ⇒ w(60+�) ≈ w(60)·(1+ ) ≈ 120·(1+ ) ≈ 126.
w(60) C(60) 60 60 20
[Or you can calculate the linearization of w(h) arround h = 60 using derivatives,
and using the value w(60) to determine C. getting w(h) ≈ 120 + 6(h − 60)
COPYRIGHT DAVID JERISON AND MIT 1996, 2003
1
E. Solutions to 18.01 Exercises 2. Applications of Differentiation
2A-6 tan θ = sin θ ≈ θ ≈ θ(1 + θ2//2) ≈ θ
cos θ 1 − θ2/2
sec x 1 1 1 2
2A-7 √ 2 = √ 2 ≈ 1 2 1 2 ≈ 1 − x2 ≈ 1+ x
1 − x cos x 1 − x (1 − x )(1 − x )
2 2
2A-8 1 = 1 = 1 = 2
1 − x 1 1 1 − 2Δx
1 − ( +Δx) − Δx
2 2
2 1 1 2
≈ 2(1 + 2Δx + 4(Δx) ) ≈ 2 + 4(x − ) + 8(x − )
2 2
r � r−1 �� r−2
2A-10 y = (1+ x) ,y = r(1 + x) , y = r(r − 1)(1 + x)
� �� r
Therefore y(0) = 1,y (0) = r,y (0) = r(r − 1), giving (1 + x) ≈ 1 + rx +
r(r − 1) 2
2 x .
k −k −k −k Δv −k
2A-11 pv = c ⇒ p = cv = c((v +Δv) = cv (1 + )
0 0 v
0
c Δv k(k + 1) Δv 2
≈ v k (1 − k v + 2 ( v ) )
0 0 0
ex x2 5
2 2
2A-12 a) ≈ (1 + x + )(1 + x + x ) ≈ 1 +2x + x
1 − x 2 2
b) ln(1 + x) ≈ x ≈ 1 − x
x
xe x(1 + x)
2
−x 2 x
c) e ≈ 1 − x [Substitute into e ≈ 1+ x]
x2 x2
d) ln(cos x) ≈ ln(1 − 2 ) ≈ − 2 [since ln(1 + h) ≈ h]
2 2
e) x ln x = (1 + h) ln(1 + h) ≈ (1 + h)(h − h ) ≈ h + h ⇒ x ln x ≈
2 2 2
(x − 1) + (x − 1)
2
2A-13 Finding the linear and quadratic approximation
2
2. Applications of Differentiation E. Solutions to 18.01 Exercises
a) 2x (both linear and quadratic)
b) 1, 1 − 2x2
2 −1 2
c) 1, 1 + x /2 (Use (1 + u) ≈ 1 − u with u = x /2:
sec x = 1/ cos x 2 2 −1 2
≈ 1/(1 − x /2) = (1 − x /2) ≈ 1+ x /2
d) 1, 1 + x2
−1 2
e) Use (1 + u) ≈ 1 − u + u :
−1 −1 −1 −1 2
(a + bx) = a (1 + (bx/a)) ≈ a (1 − bx/a +(bx/a) )
Linear approximation: (1/a) 2
− (b/a )x
2 2 3 2
Quadratic approximation: (1/a) − (b/a )x +(b /a )x
� −2 �� 2 −3
f) f(x) = 1/(a + bx) so that f (1) = −b(a + b) and f (1) = 2b /(a + b) .
We need to assume that these numbers are defined, in other words that a + b �= 0.
Then the linear approximation is
1/(a + b) 2
− (b/(a + b) )(x − 1)
and the quadratic approximation is
1/(a + b) 2 3 2
− (b/(a + b) )(x − 1) + (b/(a + b) )(x − 1)
Method 2: Write
1/(a + bx) = 1/(a + b + b(x
− 1))
Then use the expansion of problem (e) with a+b in place of a and b in place of b and
− 1) in place of x. The requirement a �= 0 in (e) corresponds to the restriction
(x
a + b �= 0 in (f).
2A-15 f(x) = cos(3x), f�(x) = −3sin(3x), f��(x) = −9cos(3x). Thus,
(1) f(0) = 1, f(π/6) = cos(π/2) = 0, f(π/3) = cos π = −1
(2) f�(0) = −3sin0 = 0, f�(π/6) = −3sin(π/2) = −3, f�(π/3) = −3sin π = 0
(3) f��(0) = −9, f��(π/6) = 0, f��(π/3) = 9
Using these values, the linear and quadratic approximations are respectively:
2
(4) for x ≈ 0 : f(x) ≈ 1 and f(x) ≈ 1 − (9/2)x
(5) for x ≈ π/6 : both are f(x) ≈ −3(x − π/6)
(6) for x ≈ π/3 : f(x) ≈ −1 and f(x) ≈ −1 + (9/2)(x − π/3)2
3
E. Solutions to 18.01 Exercises 2. Applications of Differentiation
2A-16 a) The law of cosines says that for a triangle with sides a, b, and c, with
θ opposite the side of length c,
2 2 2
c = a + b − 2ab cos θ
Apply it to one of the n triangles with vertex at the origin: a = b = 1 and θ = 2π/n.
So the formula is �
c = 2 − 2cos(2π/n)
�
b) The perimeter is n 2 − 2cos(2π/n). The quadratic approximation to cos θ
near 0 is
cos θ ≈ 1 − θ2/2
Therefore, as n → ∞ and θ = 2π/n → 0,
� � 2 � 2
n 2 − 2cos(2π/n) ≈ n 2 − 2(1 − (1/2)(2π/n) ) = n (2π/n) = n(2π/n) = 2π
In other words, �
lim n 2 − 2cos(2π/n) = 2π,
n→∞
the circumference of the circle of radius 1.
2B. Curve Sketching
3 � 2 �
2B-1 a) y = x − 3x + 1, y = 3x − 3 = 3(x − 1)(x + 1). y = 0 =⇒ x = ±1.
Endpoint values: y → −∞ as x → −∞, and y → ∞ as x → ∞.
Critical values: y(−1) = 3, y(1) = −1.
Increasing on: −∞ < x < −1, 1 < x < ∞.
Decreasing on: −1 < x < 1.
Graph: (−∞, −∞) � (−1, 3) � (1, −1) � (∞, ∞), crossing the xaxis three
times.
(-1, 3)
1a (1,-1)
4 � 3 � 1/3
b) y = x − 4x + 1, y = x − 4. y = 0 =⇒ x = 4 .
4
no reviews yet
Please Login to review.
