MATH 100, Section 110 (CSP)
Week 4: Marked Homework Solutions
2010 Oct 07
1. [3] Using the Product Rule and the Chain Rule, differentiating g gives
d
d
g 0 (x) = 3x2 h(cos πx)+x3 dx
h(cos πx) = 3x2 h(cos πx)+x3 h0 (cos πx) dx
(cos πx) = 3x2 h(cos πx)+
d
2
3 0
3 0
x h (cos πx) (− sin πx) dx (πx) = 3x h(cos πx) + x h (cos πx) (− sin πx)(π).
When x = 2 we have
g 0 (2) = 3 · 22 · h(cos 2π) − 23 h0 (cos 2π) · π sin 2π = 12 · h(1) − 8 · h0 (1) · 0 = 24.
2. [3 each part] (a)
=
2x2
(x3 +1)2/3 (x3 −1)4/3
x3 −1
x3 +1
= √
3
1/3
=
1
3
x3 −1
x3 +1
−2/3
2
2x√
(x3 −1)2 3 (x3 +1)4
(3x2 )(x3 +1)−(x3 −1)(3x2 )
(x3 +1)2
=
6x2
1 (x3 +1)2/3
3 (x3 −1)2/3 (x3 +1)2
= 2t cos2 t + (1 + t2 ) dtd (cos t)2 = 2t cos2 t + (1 + t2 )2(cos t)(− sin t)
= 2t cos2 t − 2(1 + t2 ) cos t sin t
d
2
(c) dx
=
2t
cos
2t
+
(1
+
t
)
cos(2t)
= 2t cos 2t + (1 + t2 )(− sin 2t)(2)
dt
dt
2
= 2t cos 2t − 2(1 + t ) sin 2t
2z
d
(d) dz
(1 + tan z)1/2 = 21 (1 + tan z)−1/2 sec2 z = 2√sec
1+tan z
(b)
dx
dt
d
dx
= −6 sin(3t + π/3), so when t = 2π/3, we
3. [6] (a) At any time t, the velocity is v(t) = dy
dt
√
have v(2π/3) = −6 sin(π/3) = −3 3 cm/s.
√
√
(b) The speed is |v| = | − 3 3| = 3 3 cm/s.
(c) The particle is moving in the negative y-direction (i.e. downwards, if y is assumed to be
measured vertically upwards from some reference point y = 0).
2
(d) At any time t, the acceleration is a(t) = ddt2y = dv
= −18 cos(3t + π/3), so when t = 2π/3,
dt
we have a(2π/3) = −18 cos(π/3) = −9.
The acceleration is in the negative y-direction, in the same direction as the velocity, so the
speed is increasing.
(The velocity is getting more negative, but it is already negative, so its absolute value is
increasing.)
4. [2] (a) sin (arcsin(1/2)) = 1/2 (since 1/2 is in the domain [−1, 1] of arcsin = sin −1 ).
[2] (b) sin−1 (sin(5π/6)) = sin−1 (1/2) = π/6 (note that 5π/6 is not in [−π/2, π/2]).
[3] (c) sec2 (arctan 10) = 1 + tan2 (arctan 10) = 1 + [tan(tan−1 10)]2 = 1 + 102 = 101.
√
√
√
q
q
2
−1
2
−1
[3] (d) cos sin ( 5/4) = 1 − sin sin ( 5/4) = 1 − sin sin−1 ( 5/4)
q
√ 2 √
5/4 = 11/4. Note that we take the positive square root because the angle
= 1−
√
−1
y = sin ( 5/4) is in the first quadrant, and therefore cos y is positive.
5. [7] f (x) = x2 − 2x, −∞ < x ≤ 1.
(a) The graph of y = x2 − 2x is a parabola opening upwards, with vertex and minimum at
the point (1, −1). With the specified domain, the graph of y = f (x) is the left-hand side of
the parabola, including the vertex. The function passes the Horizontal Line Test (p. 64), so
it is one-to-one. See the figure below in part (c).
1
(b) The given function f has Domain(f ) = (−∞, 1] and Range(f ) = [−1, ∞), so its inverse
function f −1 has Domain(f −1 ) = [−1, ∞) and Range(f −1 ) = (−∞, 1].
(c) The graph of y = f −1 (x) is the reflection of the graph of y = f (x) about the diagonal
line y = x. y = f −1 (x) is the bottom half of a parabola opening to the right, with vertex at
the point (−1, 1).
2
(d) Put y = x2 − 2x. To solve for √
x, write the equation
√ as x − 2x − y = 0 and use the
quadratic formula to get x = (2 ± 4 + 4y)/2 = 1 ± √1 + y. But −∞ < x ≤ 1, so we
must take
x = 1 − 1 + y. Exchanging x and y gives
√ the negative square root, and thus −1
y = 1 − 1 + x, and the explicit formula for f (x) is
√
−1 ≤ x < ∞.
f −1 (x) = 1 − 1 + x,
6. [4] f (x) = sec x, 0 ≤ x < π/2 or π ≤ x < 3π/2.
(a) The graph of y = f (x) consists of the two portions of the graph of y = sec x that are
above or below the two given intervals on the x-axis. The function passes the Horizontal
Line Test, so it is one-to-one. See the figure below in part (c), and see also Figure 25 (p. 74;
the red part of the graph).
(b) The given function f has Domain(f ) = [0, π/2) ∪ [π, 3π/2) and Range(f ) = (−∞, −1] ∪
[1, ∞), so its inverse function f −1 has Domain(f −1 ) = (−∞, −1] ∪ [1, ∞) and Range(f −1 ) =
[0, π/2) ∪ [π, 3π/2).
(c) The graph of y = f −1 (x) is the reflection of the graph of y = f (x) about the diagonal
line y = x.
2