MATH 100, Section 110 (CSP)
Week 5: Marked Homework Solutions
2010 Oct 14
1. [3] Check that x = −1, y = −1 in fact satisfies the equation. Implicitly differentiating
the equation gives
d
d −1
(y x) + 3(x−1 y)2 · (x−1 y) = 0,
dx
dx
and then
−y −2 y 0 x + y −1 + 3(x−1 y)2 (−x−2 y + x−1 y 0 ) = 0,
or
1 3y 3
x
3y 2
− 4 + − 2 + 3 y 0 = 0.
y
x
y
x
Substituting (x, y) = (−1, −1) we get 2 − 2y 0 = 0, so y 0 = 1 at the point (−1, −1), and an
equation of the tangent line is
y = −1 + 1 · (x + 1),
or more simply
y = x.
2. [3] Check that x = 1, y = 0 satisfies the equation. Implicit diffferentiation of the equation
gives
y 0 ex−1 + yex−1 + ey + xey y 0 = 0.
Substituting (x, y) = (1, 0) gives 1 + 2y 0 = 0, so
y0 = −
1
2
is the slope of the tangent line at (1, 0).
3. (a) [2]
dx
dt
= −√
1
1−(sin−1 t)2
·
d
dt
sin−1 t
1
= −p
.
√
1 − (sin−1 t)2 1 − t2
d
ln[x2 + (x3 + x)1/2 ] = x2 +(x31+x)1/2 · 2x + 12 (x3 + x)−1/2 (3x2 + 1)
(b) [2] dx
1
3x2 + 1
√
=
· 2x + √
.
x2 + x 3 + x
2 x3 + x
(c) [4] Use logarithmic
differentiation:
Let y = f (x), then ln y = ln (xarctan x ) = arctan x ln x,
dy
dy
1
1
= 1+x
(ln x) + (arctan x) x1 , solving for dx
= f 0 (x) and using y = xarctan x we get
2
y dx
0
f (x) = x
arctan x
arctan x
ln x
+
.
1 + x2
x
1
(d) [4] Use logarithmic differentiation: ln y = ln (cos x)sin x = sin x ln(cos x),
dy
sin x cos1 x (− sin x), solving for dx
and using y = (cos x)sin x we get
1 dy
y dx
= cos x ln(cos x)+
dy
= (cos x)sin x [cos x ln(cos x) − sin x tan x].
dx
4. [3] f (0) (x) = f (x) = ln(x/2) = ln x − ln 2
f (1) (x) = f 0 (x) = x1 = x−1 since ln 2 is a constant (or use the Chain Rule directly on
ln(x/2))
f (2) (x) = f 00 (x) = (−1)x−2
f (3) (x) = f 000 (x) = (−1)(−2)x−3
f (4) (x) = f 0000 (x) = (−1)(−2)(−3)x−4
1
Following the pattern, f (101) (x) = (−1)(−2)(−3) · · · (−100)x−101 = 100! x101
, and therefore
1
(101)
f
(2) = 100! 2101 .
5. (a) [3] Differentiating the suggested function ψ(x) = C sin 2π
x , we get (using the Chain
λ
2
Rule) ψ 0 (x) = 2π
C cos 2π
x , and differentiating again we get ψ 00 (x) = − 4π
C sin 2π
x .
λ
λ
λ2
λ
Therefore
4π 2
4π 2
2π
00
ψ (x) = − 2 C sin
x = − 2 ψ(x),
λ
λ
λ
which verifies that the suggested function is indeed a solution of the differential equation.
Substituting x = 0 into the function ψ(x) gives
ψ(0) = C sin 0 = 0,
which verifies that the function also satisfies the first condition for any value of C.
(b) [4] Substituting x = 1 into the function ψ(x) and its derivative ψ 0 (x) (calculated above)
gives
2π
2π
2π
0
+ C cos
= C (sin X + X cos X),
ψ(1) + ψ (1) = C sin
λ
λ
λ
where
2π
λ
0
is positive. Therefore ψ(1) + ψ (1) = 0 for any value of C if and only if sin X + X cos X = 0
if and only if
tan X = −X
X=
for X > 0. But by Question 2 of the Week 2: Marked Homework Assignment (see the graph
on p. 2 of the Solutions), there are infinitely many positive values of X which satisfy this
last equation,
and therefore there are infinitely many values of λ = 2π
such that ψ(x) =
X
2π
0
C sin λ x satisfies the second condition ψ(1) + ψ (1) = 0 for any value of C. (Note that
these values of λ are discrete, as in the “particle in a box” example, but in this case (due
to the condition ψ(1) + ψ 0 (1) = 0), simple expressions for the exact values of λ cannot be
found.)
2