MATH125: Calculus I
Spring 2014
Midterm Practice: February 12th
Problem 1 Let f (x) = | sin(cos(πx))|.
a. (3 pts) Find the domain of f (x).
Ans. f (x) is defined over R.
b. (1 pt) Write the definition of continuous function at x0 .
Ans. f (x) is continuous at x0 if
lim f (x) = f (x0 )
x→x0
c. (1 pt) Find where f (x) is continuous.
Ans. f (x) is a composition of functions which are continuous everywhere, therefore f (x) is continuous everywhere.
d. (1 pt) Write the definition of differentiable function at x0 .
Ans. A function f (x) is differentiable at x0 if f 0 (x0 ) is well-defined.
e. (4 pts) Find where f (x) is differentiable.
Ans. We first compute f 0 (x)
f 0 (x) =
sin(cos(πx))
cos(cos(πx))(− sin(πx))π
| sin(cos(πx))|
The domain of f 0 (x) is
1
R \ {x| sin(cos(πx)) = 0} = R \ {x| cos(πx) = 0} = R \ (Z + )
2
Problem 2 a. (3 pts) Write the precise definition of the statement limx→x+ f (x) = L.
0
Ans. We say limx→x+ f (x) = L if, for every ε > 0, there is a corresponding δ > 0 such that
0
x0 < x < x0 + δ ⇒ |f (x) − L| < ε
b. (6 pts) Use the above definition of limit to check the statement
lim x2 + x = 0
x→0+
Ans. Given ε > 0, we need δ > 0 such that
0 < x < δ ⇒ |x2 + x| < ε
The last inequality corresponds to (since x > 0)
2
x +x−ε<0⇔
−1 −
√
1 + 4ε
2
Then we need to choose δ as
0<δ≤
or, more specifically, δ =
−1 +
√
<x<
−1 +
√
1 + 4ε
2
1 + 4ε
2
√
−1+ 1+4ε
.
2
c. (1 pt) Use the answer above to find the right value of δ for ε = 10−6 (or analogous parameters).
1
Ans.
δ=
−1 +
√
1 + 4 · 10−6
2
Problem 3 Compute the following limit:
lim
x→0
cos(2x) − 1
x sin(2x)
Ans. We have
cos(2x) − 1
cos(2x) − 1
= lim
·
x→0 x sin(2x)
x→0 x sin(2x)
lim
= lim
x→0
cos(2x) + 1
cos(2x) + 1
− sin(2x) sin(2x)
=
x→0 x sin(2x)(cos(2x) + 1)
= lim
sin(2x)
−2
·
= −1
2x
cos(2x) + 1
Problem 4 a. (3 pts) Write the precise definition of the statement limx→x− f (x) = +∞.
0
Ans. We say limx→x− f (x) = +∞ if, for every M > 0, there is a corresponding δ > 0 such that
0
x0 − δ < x < x0 ⇒ f (x) > M
b. (6 pts) Use the above definition of limit to check the statement
lim
x→− 12 −
x−1
= +∞
2x + 1
Ans. For every M > 0 we need δ > 0 such that
−1/2 − δ < x < −1/2 ⇒
x−1
>M
2x + 1
Since 2x + 1 < 0 (x → −1/2− ) we have that the last inequality corresponds to
x − 1 < (1 + 2x)M ⇔ (1 − 2M )x < M + 1 ⇔ x >
M +1
1 − 2M
Therefore we need δ such that
M +1
1
M +1
1
−2M − 2 − 1 + 2M
3
≤− −δ ⇔δ ≤−
− ⇔δ≤
⇔δ≤
1 − 2M
2
1 − 2M
2
(1 − 2M )2
4M − 2
c. (1 pt) Use the answer above to find the right value of δ for M = 105 (or analogous parameters).
Ans.
δ=
3
4 · 105 − 2
Problem 5 Compute the following limits:
a. (3 pts)
x5 − x + 1
x→−∞ 2x5 + x4 − x2
lim
Ans.
(1 − 1/x4 + 1/x5 )
1
x5 (1 − 1/x4 + 1/x5 )
= lim
=
5
3
x→−∞ x (2 + 1/x − 1/x )
x→−∞ (2 + 1/x − 1/x3 )
2
lim
2
b. (3 pts)
lim x cos
x→−∞
1
−x
x
Ans.
lim x cos
x→−∞
1
1
cos(t) − 1
− sin(t)
sin(t)
− x = lim x(cos
− 1) = lim−
= lim−
·
=0
x→−∞
x
x
t
t
(cos(t) + 1)
t→0
t→0
c. (4 pts)
√
lim
x→1
√
x3 + 2x2 + 1 − x3 + x2 + 3x − 1
x2 − 1
Ans.
√
lim
x→1
√
x3 + 2x2 + 1 − x3 + x2 + 3x − 1
·
x2 − 1
!
√
√
x3 + 2x2 + 1 + x3 + x2 + 3x − 1
√
√
=
x3 + 2x2 + 1 + x3 + x2 + 3x − 1
x2 − 3x + 2
x3 + 2x2 + 1 − x3 − x2 − 3x + 1
√
√
√
√
= lim
=
x→1 (x2 − 1)( x3 + 2x2 + 1 +
x3 + x2 + 3x − 1) x→1 (x2 − 1)( x3 + 2x2 + 1 + x3 + x2 + 3x − 1)
= lim
= lim
x→1
(x2
− 1)(
√
x3
x−2
1
(x − 1)(x − 2)
√
√
√
= lim
=−
2
3
2
3
2
3
2
x→1
8
+ 2x + 1 + x + x + 3x − 1)
(x + 1)( x + 2x + 1 + x + x + 3x − 1)
Problem 6 (10 pts) Find A, B, C, D ∈ R such that the following function is continuous and differentiable over
R.
 2
x < −1
 x + Ax + B
Cx + D
−1 ≤ x ≤ 1
f (x) =

Bx − x2
x>1
Ans.
• continuity at x = −1: 1 − A + B = −C + D
• differentiability at x = −1: −2 + A = C
• continuity at x = 1: C + D = B − 1
• differentiability at x = 1: C = B − 2
Then we get A = B (from 2 − 4), D = 1/2B (from 1-3), C = 1/2B − 1 (from 3), B = 2 (from 4), and therefore
(A, B, C, D) = (2, 2, 0, 1)
The corresponding graph is then
3
3
2
1
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
-2
-3
Problem 7 a. (3 pts) Write the definition of derivative of f (x) at a point a.
Ans. The derivative of f (x) is the function
f 0 (x) = lim
h→0
f (x + h) − f (x)
h
provided the limit exists.
b. (7 pts) Use the above definition to compute the derivative at a of
f (x) =
1−x
3x + 1
Ans. Using the definition
1
1−a−h
1−a
1 (1 − a − h)(3a + 1) − (1 − a)(3a + 3h + 1)
f 0 (a) = lim
−
= lim
=
h→0 h
h→0 h
3a + 3h + 1 3a + 1
(3a + 3h + 1)(3a + 1)
−h − 3h
1 3a + 1 − 3a2 − a − 3ah − h − 3a − 3h − 1 + 3a2 + 3ah + a
1
= lim
= lim
=
h→0 h
h→0 h
(3a + 3h + 1)(3a + 1)
(3a + 3h + 1)(3a + 1)
4
−4
=−
= lim
h→0 (3a + 3h + 1)(3a + 1)
(3a + 1)2
Problem 8 (10 pts) Compute the derivative of the following functions:
a. f (x) =
x2 −3x+1
x7 +3x4 −1
Ans.
f 0 (x) =
b. g(x) = (x4 +
(2x − 3)(x7 + 3x4 − 1) − (x2 − 3x + 1)(7x6 + 12x3 )
(x7 + 3x4 − 1)2
√
2x − 1)3 sin(3x2 + 1)
√
√
√
g 0 (x) = 3(x4 + 2x − 1)2 (4x3 + 2) sin(3x2 + 1) + (x4 + 2x − 1)3 cos(3x2 + 1) · 6x
c. h(x) = | cos(x2 )|
h0 (x) =
cos(x2 )
(− sin(x2 ))2x
| cos(x2 )|
d. s(x) = |x3 − sin(1/x)|−1
s0 (x) = (−1)|x3 − sin(1/x)|−2
√
e. t(x) = sin(cos( x))
x3 − sin(1/x)
1
1
(3x2 − cos( )(− 2 ))
|x3 − sin(1/x)|
x
x
√
√
1
t0 (x) = cos(cos( x)) · (− sin( x)) · √
2 x
4
Problem 9 (10 pts) Find the equation of the tangent line to the curve
y=
x2
3x + 1
√
− sin(π x)
at the point (1, 4).
Ans. The tangent has equation
y − 4 = f 0 (1)(x − 1)
then we have
where
f (x) =
3x + 1
√
x2 − sin(π x)
√
√
π
3(x2 − sin(π x)) − (3x + 1)(2x − cos(π x) 2√
)
x
√
f (x) =
(x2 − sin(π x))2
0
and
3(1) − (4)(2 + π2 )
= 3 − 8 − 2π = −5 − 2π
1
since sin(π) = 0 and cos(π) = −1. Therefore the tangent has equation
f 0 (1) =
y − 4 = (−5 − 2π)(x − 1)
Problem 10 A particle is moving along a line with motion law
s(t) = 2t3 − t + 2
a. (4 pts) Compute the velocity at t.
Ans.
v(t) = s0 (t) = 6t2 − 1
b. (2 pts) When is the particle at rest?
Ans. The particle is at rest when v(t) = 0, therefore
1
v(t) = 0 ⇔ 6t2 − 1 = 0 ⇔ t = √
6
1
t = −√
6
c. (2 pts) When is the particle moving backward? When is it moving forward?
Ans. The particle is moving backward when v(t) < 0 and it is moving forward when v(t) > 0. The velocity is
described by a convex downward parabola, therefore
1
1
v(t) < 0 ⇔ − √ < t < √
6
6
and
1
v(t) > 0 ⇔ t < − √
6
or
1
t> √
6
d. (2 pts) Compute the total distance in the interval of time [−1, 0].
Ans. The particle is moving forward in [−1, − √16 ] and backward in [− √16 , 0], therefore the total distance is
√
√
4
|s(−1/ 6) − s(−1)| + |s(0) − s(−1/ 6)| = √ + 1
3 6
with
s(−1) = 1
s(0) = 2
5
√
2
s(−1/ 6) = √ + 2
3 6
e. (4 pts) Find the acceleration at t.
Ans.
a(t) = s00 (t) = 12t
f. (2 pts) When is the particle speeding up? When is it slowing down?
Ans. The particle is speeding up when a(t) > 0 and speeding down when a(t) < 0, therefore
a(t) > 0 ⇔ t > 0
a(t) < 0 ⇔ t < 0
and
g. (4 pts) Sketch the graph of s(t) including all the information above.
Ans. We have the following information
√
√
• the tangent is horizontal at −1/ 6 and 1/ 6;
√
• the slope of the tangent is positive (s(t)
in (−∞, −1/sqrt6) ∪ (1/ 6, +∞) and the slope is
√ increasing)
√
negative (s(t) is decreasing) in (−1/ 6, 1/ 6);
• the slope is increasing when t > 0 (s(t) is convex downward) and decreasing when t < 0 (s(t) is convex
upward).
Therefore we get
5
2.5
-10
-7.5
-5
-2.5
0
-2.5
-5
6
2.5
5
7.5
10
MATH125: Calculus I
Spring 2014
Midterm Practice - Discussion Session
Problem 1 Let f (x) = | sin(πx)|.
a. (3 pts) Find the domain of f (x).
Ans. f (x) is defined over R.
b. (1 pt) Write the definition of continuous function at x0 .
Ans. f (x) is continuous at x0 if
lim f (x) = f (x0 )
x→x0
c. (1 pt) Find where f (x) is continuous.
Ans. f (x) is a composition of functions which are continuous everywhere, therefore f (x) is continuous everywhere.
d. (1 pt) Write the definition of differentiable function at x0 .
Ans. A function f (x) is differentiable at x0 if f 0 (x0 ) is well-defined.
e. (4 pts) Find where f (x) is differentiable.
Ans. We first compute f 0 (x).
f 0 (x) =
d
sin(πx)
| sin(πx)| =
cos(πx)π
dx
| sin(πx)|
From the expression above,
domain of f 0 (x) = R \ {x| sin(πx) = 0} = R \ Z
Problem 2 a. (3 pts) Write the precise definition of the statement limx→x+ f (x) = L.
0
Ans. We say limx→x+ f (x) = L if, for every ε > 0, there is a corresponding δ > 0 such that
0
x0 < x < x0 + δ ⇒ |f (x) − L| < ε
b. (6 pts) Use the above definition of limit to check the statement
lim+
x→0
1
=1
1+x
Ans. For every ε > 0, we need to find an appropriate δ > 0 such that
1
0 < x < δ ⇒ − 1 < ε
1+x
The last inequality is equivalent to
1 − 1 − x −x x
=
1 + x 1 + x = 1 + x < ε
where the last step follows from the fact that we are taking the limit for x →)+ , so that x > 0. Then we get
x < (1 + x)ε ⇔ (1 − ε)x < ε ⇔ x <
1
ε
1−ε
We are dealing with a limit, therefore we can assume ε to be rather small and therefore 1 − ε > 0 (as well as
1 + x > 0). Finally, we need to find δ > 0 such that
0<x<δ⇒0<x<
So we can take any 0 < δ ≤
ε
1−ε
ε
1−ε
and we can set in particular
δ=
ε
1−ε
c. (1 pt) Use the answer above to find the right value of δ for ε = 10−5 (or analogous parameters).
Ans. For ε = 10−5 , we have
δ=
1
ε
10−5
=
=
1−ε
1 − 10−5
9999
Problem 3 (10 pts) Compute the following limit:
lim
x→0
sin(3x)
sin(2x)
Ans. We want to use the special limit
lim
x→0
sin x
=1
x
Therefore
sin(3x)
6x sin(3x)
3 sin(3x) 2x
3
= lim
= lim
=
x→0 sin(2x)
x→0 6x sin(2x)
x→0 2
3x sin(2x)
2
lim
Problem 4 a. (3 pts) Write the precise definition of the statement limx→x− f (x) = +∞.
0
Ans. We say limx→x− f (x) = +∞ if, for every M > 0, there is a corresponding δ > 0 such that
0
x0 − δ < x < x0 ⇒ f (x) > M
b. (6 pts) Use the above definition of limit to check the statement
lim −
x→− 12
−1
= +∞
2x + 1
Ans. For every M > 0, we need to find δ > 0 such that
1
−1
1
>M
− −δ <x<− ⇒
2
2
2x + 1
This last condition corresponds to
−1
−M − 1
> M ⇔ −1 < (2x + 1)M ⇔
<x
2x + 1
2M
where the first step follows by observing that, since x → −1/2− , 2x + 1 < 0. Finally we get
1
1
− −
<x
2 2M
and therefore it is enough to set
0<δ≤
or, more specifically, δ = (2M )−1 .
2
1
2M
c. (1 pt) Use the answer above to find the right value of δ for M = 105 (or analogous parameters).
Ans. From the above formula we get
δ = (2M )−1 =
1
2 · 105
Problem 5 Compute the following limits:
a. (3 pts)
3x2 + 1
+ 3x − 2
lim
x→−∞ 2x2
Ans.
3
3x2 + 1
x2 (3 + 1/x2 )
3 + 1/x2
=
=
lim
=
lim
2
2
2
2
x→−∞ 2x + 3x − 2
x→−∞ x (2 + 3/x − 2/x )
x→−∞ 2 + 3/x − 2/x
2
lim
using limx→0 1/x = limx→0 1/x2 = 0.
b. (3 pts)
1
lim x sin
x→+∞
x
Ans.
1
1
= lim sin(t) = 1
lim x sin
x→+∞
x
t→0+ t
c. (4 pts)
lim
x→+∞
p
x2 + 1 −
p
x2 + x
Ans.
lim
x→+∞
p
x2 + 1 −
p
p
p
x2 + x = lim ( x2 + 1 − x2 + x) ·
x→+∞
!
√
√
x2 + 1 + x2 + x
√
√
=
x2 + 1 + x2 + x
1/x − 1
1
x2 + 1 − x2 − x
x(1/x − 1)
√
√ p
p
= lim p
=−
= lim √ p
2
2
2
2
2
2
x→+∞
x→+∞
x→+∞
2
x +1+ x +x
1 + 1/x + 1 + 1/x
x 1 + 1/x + x 1 + 1/x
√
using limx→0 1/x = limx→0 1/x2 = 0 and x2 = x for x > 0 (we are dealing with a limit for x → +∞).
= lim √
Problem 6 (10 pts) Find A, B, C, D ∈ R such that the following function is continuous and differentiable over
R.

x < −1
 Ax2 + x + B
Ax + C
−1 ≤ x ≤ 1
f (x) =

Dx − x2
x>1
Then sketch the graph of f (x).
Ans. f (x) is a piecewise defined polynomial function, so we obtain
• continuity at x = −1: A − 1 + B = −A + C
• differentiability at x = −1: −2A + 1 = A
• continuity at x = 1: A + C = D − 1
• differentiability at x = 1: A = D − 2
3
We have to solve the system


B


 A + B − 1 = −A + C



−2A + 1
=
A
1 − 3A
⇔
A
+
C
=
D
−
1
C






A
=
D−2
D
= −2A + C + 1
=
0
=
D−A−1
=
A+2
We find A with the second equation, then we progressively obtain D, C, B. The result is
7
1 4
(A, B, C, D) = ( , , 1, )
3 3
3
and
f (x) =
4
1 2
3x + x + 3
1
3x + 1
7
2
3x − x



x < −1
−1 ≤ x ≤ 1
x>1
2.4
1.6
0.8
-4.8
-4
-3.2
-2.4
-1.6
-0.8
0
0.8
1.6
2.4
3.2
4
4.8
-0.8
-1.6
-2.4
Problem 7 a. (3 pts) Write the definition of derivative of f (x) at a point a.
Ans. The derivative of f (x) is the function
f 0 (x) = lim
h→0
f (x + h) − f (x)
h
provided the limit exists.
b. (7 pts) Use the above definition to compute the derivative at a of
f (x) =
x
1−x
Ans. We have
1
f (a) = lim
h→0 h
0
a+h
a
(a + h)(1 − a) − a(1 − a − h)
1 a − a2 + h − ah − a + a2 + ah
−
= lim
= lim
=
h→0
h=0 h
1−a−h 1−a
(1 − a − h)(1 − a)
(1 − a − h)(1 − a)
1
1
h
1
lim
= lim
=
h=0 h
h=0 (1 − a − h)(1 − a)
(1 − a − h)(1 − a)
(1 − a)2
Problem 8 (10 pts) Compute the derivative of the following functions:
a. f (x) =
3x2 −x+2
4x+1
Ans. Quotient rule:
f 0 (x) =
(6x − 1)(4x + 1) − (3x2 − x + 2)4
(4x + 1)2
4
√
b. g(x) = ( 2x − 1)3 cos(x2 )
Ans. Product rule and chain rule:
√
√
√
g 0 (x) = 3( 2x − 1)2 2 cos(x2 ) + ( 2x − 1)3 (− sin(x2 ))2x
c. h(x) = | cos(πx)|
Ans. Derivation of absolute value and chain rule (twice):
h0 (x) =
cos(πx)
(sin(πx))π
| cos(πx)|
d. s(x) = |x2 − 1|−1
Ans. Power rule, absolute value and chain rule (twice):
s0 (x) = (−1)|x2 − 1|−2 ·
x2 − 1
· 2x
|x2 − 1|
e. t(x) = sin(cos(x))
Ans. Chain rule:
t0 (x) = cos(cos(x)) · (− sin(x))
Problem 9 (10 pts) Find the equation of the tangent line to the curve
y=
x+1
cos(πx)
at the point (1, −2).
Ans. The tangent line at (1, −2) has equation
y + 2 = f 0 (1)(x − 1)
We have
f 0 (x) =
f (x) =
x+1
cos(πx)
cos(πx) − (x + 1)(− sin(πx))π
cos2 (πx)
and f 0 (1) = −1 (since cos π = −1 and sin π = 0). Therefore the tangent has equation
y = −1 − x
Problem 10 A particle is moving along a line with motion law
s(t) = t3 − t + 1
a. (4 pts) Compute the velocity at t.
Ans. The velocity is given by the first derivative
v(t) = s0 (t) = 3t2 − 1
b. (2 pts) When is the particle at rest?
Ans. The particle is at rest when v(t) = 0. Namely,
1
3t2 − 1 = 0 ⇔ t = ± √
3
5
c. (2 pts) When is the particle moving backward? When is it moving forward?
Ans. The particle is moving forward when v(t) > 0, namely
1
3t2 − 1 > 0 ⇔ t > √
3
1
t < −√
3
or
and it is moving backward when v(t) < 0, namely
1
1
3t2 − 1 < 0 ⇔ − √ < t < √
3
3
d. (2 pts) Compute the total distance in the interval of time [0, 1].
√
√
Ans. The particle is moving backward in [0, 1/ 3] and forward in [1/ 3, 1]. The total distance then is
√
√
2
2
4
|s(1/ 3) − s(0)| + |s(1) − s(1/ 3)| = | √ | + | √ | = √
3 3
3 3
3 3
e. (4 pts) Find the acceleration at t.
Ans. The acceleration is the second derivative of s(t):
a(t) = s00 (t) = 6t
f. (2 pts) When is the particle speeding up? When is it slowing down?
Ans. The particle is speeding up when a(t) > 0, that is t > 0, and it is slowing down when a(t) < 0, that is t < 0.
g. (4 pts) Sketch the graph of s(t) including all the information above.
Ans. We have the following information
√
√
b. the tangent is horizontal at (− √13 , 2+3 3 ) and ( √13 , 4+3 3 )
√
√
c. s(t) is increasing (tangent with positive slope) in the
√ interval
√ (−∞, −1/ 3)∪(1/ 3, +∞) and decreasing
(tangent with negative slope) in the interval (−1/ 3, 1/ 3).
d. s(t) is convex upward (tangent with decreasing slope) in (−∞, 0) and is convex downward (tangent with
increasing slope) in (0, +∞).
By simple evaluation, we can guess that s(t) has a zero in [−2, −1] (s(−2) = −5, s(−1) = 1) by the intermediate value theorem. The result is the graph
2
1.5
1
0.5
-2.5
-2
-1.5
-1
-0.5
0
-0.5
6
0.5
1
1.5
2
MATH125: Calculus I
Spring 2014
Midterm Practice: April 2nd - SOLUTION
Problem 1 Find the equation of the tangent at (2, 2) to the elliptic curve
y 2 = x3 − 2x
Ans: We first check that (2, 2) is a point on the curve: 4 = 8 − 2 · 2. By implicit differentiation we get
2yy 0 = 3x2 − 2 ⇒ y 0 =
3x2 − 2
2y
The tangent at (2, 2) is given by
y−2=
dy
10
10
|(2,2) (x − 2) ⇒ y − 2 =
(x − 2) ⇒ y =
x−3
dx
4
4
Problem 2 The volume of a right circular cylinder is V = πr2 h, where r is the radius of the base and h is the
height.
(a) Find the rate of change of the volume with respect to the height if the radius is constant.
(b) Find the rate of change of the volume with respect to the radius if the height is constant.
And: We have V = πr2 h. We get dV /dh = πr2 , if r is constant, and dV /dr = 2πrh if h is constant.
√
Problem 3 Use linearization process to approximate 1.43.
√
√
And: We want to approximate 1.43 using the linearization of the function f (x) = x at x0 = 1.44.
f (1.44) = 1.2
1
f 0 (x) = √
2 x
f 0 (1.44) =
1
5
=
2.4
12
We use the formula
f (x) ∼ f (x0 ) + f 0 (x0 )(x − x0 )
for x near x0 . We obtain
√
1.43 ∼
12
5
6
5 1
6 · 48 − 1
287
+ (1.43 − 1.44) = −
=
=
= 1.19583̄
10 12
5 12 100
240
240
while 1.43 = 1.195826074....
Problem 4 Show that the equation y = cos(x) + x3 + x has only one real root.
Ans: We first observe that for f (x) = cos(x) + x3 + x we have
f (0) = 1
and
π
π3
π
f (− ) = −
− <0
2
8
2
Since f is continuous over R by the intermediate value theorem there exists a ∈ [−π/2, 0] such that f (a) = 0.
Now assume there are two such points a, b ∈ R, a 6= b, f (a) = f (b) = 0. Then by mean value theorem, there
exists a point c ∈ (a, b) such that f 0 (c) = 0. Since f 0 (x) = − sin(x) + 3x2 + 1 ≥ 3x2 ≥ 0 for all value x ∈ R. In
particular, for x = 0, f 0 (x) > 0, therefore f 0 (x) > 0 for all x ∈ R and we obtain a contradiction. We conclude
that there exists only one real solution to the equation f (x) = 0.
Problem 5 A metal storage tank with volume V is to be constructed in the shape of a right circular cylinder
surmounted by a hemisphere. What dimensions will require the least amount of metal?
1
And: By construction, the volume and the area of the surface of the tank are given by the formulae
2
V = πr2 h + πr3
3
S = 2πrh + πr2 + 2πr2 = 3πr2 + 2πrh
We want to minimize S assuming V is fixed. We then recover the relation
2
1
h = 2 V − πr3
πr
3
and we get
1
S = 3πr + 2πr 2
πr
2
2
V − πr3
3
2
= 3πr +
r
2
2
V − πr3
3
= 3πr2 +
2V
4
− πr2
r
3
Since h ≥ 0, we get
2
V − πr3 ≥ 0 ⇔ 0 < r ≤
3
r
3
3V
2π
p
Now we are reduced to find the minimum of a continuous function in the interval (0, 3 3V /2π]. We should
remark that since the interval is not closed we do not know for sure that the minimum exists. The derivative of
S is
dS
2V
8
18πr3 − 6V − 8πr3
10πr3 − 6V
= 6πr − 2 − πr =
=
dr
r
3
3r2
3r2
and we get
r
r
dS
3 6V
3 3V
=0⇔r=
=
dr
10π
5π
We observe that (for r > 0)
r
r<
3
3V
dS
⇒
<0
5π
dr
r
and
r>
3
3V
dS
⇒
>0
5π
dr
q
p
3
3V /2π] (please notice that in the
Therefore r = 3 3V
5π corresponds to the absolute minimum in the interval (0,
same interval we have no absolute max).
The minimum amount of material required is obtained with dimensions
r
r
5π
1
2 3V
3 3V
r=
and
h= q
V − π
=
5π
3
5π
3V
3
3V 2
π ( )
5π
Problem 6 Sketch the graph of y = x3 − x2 − x + 1.
And: We first observe that f (x) = x3 − x2 − x + 1 = (x − 1)2 (x + 1). Therefore
f (x) ≥ 0
in
[−1, +∞]
and
f (x) ≤ 0
in
[−∞, −1]
Moreover, f (x) = 0 at x = ±1.
There are no vertical neither horizontal asymptote, and clearly
lim f (x) = +∞
x→+∞
and
lim f (x) = −∞
x→−∞
The derivatives are
1
f 0 (x) = 3x2 − 2x − 1 = 3(x − 1)(x + )
3
2
and
f 00 (x) = 6x − 2
Therefore
is increasing in(−∞, −1/3) ∪ (1, +∞)
f (x)
and
f (x)
is decreasing in
(−1/3, 1)
and
f (x)
is concave downward in
(−∞, 1/3)
and
f (x)
is concave upward in
(1/3, +∞)
We obtain a local maximum at x = −1/3, a local minimum at x = 1, and an inflection point at x = 1/3.
2
1.6
1.2
0.8
0.4
-2.4
-2
-1.6
-1.2
-0.8
-0.4
0
0.4
0.8
1.2
1.6
2
2.4
-0.4
-0.8
Problem 7 Write the definition of indefinite integral of a function f and of definite integral of a function f
from a to b.
And: Let f be a function. The indefinite integral of f is the family of functions
Z
dF
f (x)dx := F (x) + C
s.t.
= f, C ∈ R
dx
The definite integral of f on [a, b] is
Z
b
f (x)dx :=
a
lim
max∆xi →0
n
X
f (x∗i )∆xi
i=1
where the limit is taken over all the possible choices of partitions of [a, b] and sample points {x∗i }.
Problem 8 Use the definition of definite integral to evaluate
Z 3
(x2 − x + 3)dx
0
2
And: Since f (x) = x − x + 3 is continuous in [0, 3] the integral is well defined and we can choose the most
convenient partition to compute it (e.g. right endpoint partition). Moreover by linearity
Z 3
Z 3
Z 3
Z 3
(x2 − x + 3)dx =
x2 dx −
xdx + 3
dx
0
0
0
0
We get
3
Z
dx
=
xdx
=
x2 dx
=
0
Z
3
0
Z
3
0
n
X
3
=3
n→∞
n
i=1
n n
3 X 3i
9 X
9
9n(n + 1)
lim
= lim 2
i = lim
=
2
n→∞ n
n→∞
n→∞ n
n
2n
2
i=1
i=1
n
n
2
3 X 3i
27 X 2
27n(n + 1)(2n + 1)
54
lim
= lim 3
i = lim
=
=9
3
n→∞ n
n→∞ n
n→∞
n
6n
6
i=1
i=1
lim
Therefore we have
Z
3
2
Z
(x − x + 3)dx =
0
3
2
3
Z
x dx −
0
Z
0
3
3
dx = 9 −
xdx + 3
0
9
27
+9=
2
2
Problem 9 Evaluate the following definite integrals:
R3
• 1 (x − sin(x))dx
Ans:
3
Z
Z
3
(x − sin(x))dx =
R5
0
3
(− sin(x))dx =
1
1
1
•
Z
xdx +
3
x2 8
3
+ cos(x)|1 = + cos(3) − cos(1)
2 1
2
dx
Ans:
5
Z
dx = x|50 = 5
0
•
R0 √
(3 x − x5 )dx
1
Ans:
0
Z
√
Z
5
(3 x − x )dx = 3
1
•
R1
−1
0
√
0
Z
5
xdx −
x dx = (2x
1
3/2
1
0
0
1
11
x6 = −2 + = −
) −
6 1
6
6
1
sin3 (x)dx
Ans:
1
Z
sin3 (x)dx = 0
−1
because the function sin3 (x) is odd and the interval is symmetric with respect to zero. More precisely, using
sin(−x) = − sin(x) we get
Z
1
Z
sin3 (x)dx =
Rb
a
−1
f (−x)dx =
1
Z
sin(x)dx +
−1
and since
0
R −a
Z
−1
0
1
Z
3
sin (x)dx = −
−1
R π/4
0
Z
(− sin(−x))dx +
1
sin(x)dx
0
f (x)dx, we get
−b
Z
•
0
sin(x)dx =
1
Z
1
sin(x)dx +
0
sin(x)dx = 0
0
1+cos2 θ
cos2 θ dθ
Ans:
Z
0
π/4
1 + cos2 θ
dθ =
cos2 θ
Z
π/4
sec2 (x)dx +
0
Z
0
π/4
π/4
dx = tan(x)|0
Problem 10 Compute the area of the region
S = {(x, y) | − π ≤ x ≤ π, sin(x) ≤ y ≤
And: We need to compute the area
4
p
π 2 − x2 }
+
π
π
=1+
4
4
7.5
5
2.5
-10
-7.5
-5
-2.5
0
2.5
5
7.5
-2.5
We can immediately notice that the two areas (respectively removed and added to the hemisphere) are equal.
Therefore the total area is π 3 /2. More precisely
Z
π
A=
p
π 2 − x2 dx −
−π
π
Z
Z
0
| sin(x)|
sin(x)dx +
−π
0
where we add the absolute value to the last summand because we are ”adding an area with negative sign”. We
get
Z π
Z 0
Z π p
2
2
π − x dx −
sin(x)dx −
sin(x) =
A=
−π
Z
π
=
−π
−π
0
Z
p
2
2
π − x dx −
π
Z
sin(x)dx +
0
π
p
π3
π 2 − x2 dx =
2
−π
Z
sin(x) =
0
5
π
MATH125: Calculus I
Spring 2014
Midterm Practice - Discussion Session
Problem 1
(a) Compute dy/dx using implicit differentiation for
x
tan( ) = x + y
y
Ans: By implicit differentiation we get
x y − xy 0
= 1 + y0
sec2 ( ) ·
y
y2
therefore
y sec2 ( xy ) − y 2
dy
=
dx
x sec2 ( xy ) + y 2
(b) Find the equation of the tangent at (0, 0) to the curve
xy = sin(x + y)
And: First we observe that (0, 0) belongs to the curve. Then, the tangent at (0, 0) has equation y = mx with
m = dy/dx|(0,0) , that is
y + xy 0 = cos(x + y)(1 + y 0 ) ⇒ y − cos(x + y) = (cos(x + y) − x)y 0 ⇒ y 0 =
y − cos(x + y)
cos(x + y) − x
therefore at (0, 0) the tangent has equation
y = −x
Problem 2 Apply linear approximation to the function
f (x) =
to approximate the number
(1 − x)2
1 + (1 + x)2
at
x=0
(0.99)2
1+(1.01)2 .
And: By linear approximation, if x is near 0, we get
f (x) ∼ f (x0 ) + f 0 (x0 )(x − x0 )
In this case we have
x0 = 0
Then
f 0 (x) =
and
f (0) =
1
2
−2(1 − x)(1 + (1 + x)2 ) − 2(1 − x)2 (1 + x)
(1 + (1 + x)2 )2
and
f 0 (0) = −
3
2
and we obtain
1 3 1
97
f (0.01) ∼ − − ·
=
= 0.485
2 2 100
200
while
f (0.01) =
(0.99)2
= 0.485174
1 + (1.01)2
Problem 3 Show that the equation sin(x) − cos(x) − 3x = 0 has only one real root.
1
Ans: Let f (x) = sin(x) − cos(x) − 3x. We have
f (0) = −1 < 0
and
f (−π) = 1 + 3π > 0
therefore, by intermediate value theorem, there is at least one root in the interval (−π, 0). Assume now that
there are two distinct roots, a, b. By mean value theorem (or rather Rolle’s theorem), there is at least one point
c ∈ (a, b) such that f 0 (c) = 0. Since
f 0 (x) = cos(x) + sin(x) − 3 < 0
x∈R
for all
we get a contradiction and therefore there is only one real root. Alternatively, we can simply remark that f 0 < 0
implies that f is decreasing.
Problem 4 Assuming that the volume of water in a conic tank (with radius 2m and height 6πm) is increasing
at a rate of 1m3 /h, how fast is the water level raising when the water is 2πm deep?
And: The quantities involved are V = V (t), r = r(t), h = h(t). We know that
V =
1 2
πr h
3
dV
=1
dt
h
3π
We want to find dh/dt at h = 2π. We substitute r =
V =
and
r
2
=
h
6π
and we obtain
h3
dV
h2 dh
dh
9π
⇒
=
⇒
= 2
27π
dt
9π dt
dt
h
Therefore, at h = 2π, we get
dh 9
=
dt h=2π
4π
Problem 5 Study the function f (x) =
x
x2 −4
Ans: The domain of the function is R \ 2, −2. We have
f (x) > 0
in
(−2, 0) ∪ (2, ∞)
and
f (x) < 0
(−∞, −2) ∪ (0, 2)
in
We have vertical asymptotes at x = 2 and x = −2, in particular
lim f (x) = −∞
lim f (x) = +∞
x→2−
x→2+
lim f (x) = −∞
lim f (x) = +∞
x→−2−
x→−2+
We have horizontal asymptotes
lim f (x) = 0−
lim f (x) = 0+
x→+∞
The first derivative is
f 0 (x) =
x→−∞
x2 + 4
(x2 − 4) − 2x2
=
−
<0
(x2 − 4)2
(x2 − 4)2
This implies that the function f (x) is always decreasing (but remember we have discontinuities at x = ±2).
The second derivative is
f 00 (x) =
−2x(x2 − 4)2 + (x4 )2(x2 − 4)2x
−2x3 + 8x + 4x3 + 16x
2x3 + 24x
2x(x2 + 12)
=
=
=
2
4
2
3
2
3
(x − 4)
(x − 4)
(x − 4)
(x2 − 4)3
Therefore we have
f 00 (x) < 0
in
(−∞, −2) ∪ (0, 2)
and
f 00 (x) > 0
in
(−2, 0) ∪ (2, +∞)
We have an inflection point at x = 0. The function has no absolute maximum nor absolute minimum in the
connected components of the domain. We obtain the following graph:
2
3
2
1
-5
-4
-3
-2
-1
0
1
2
3
4
-1
-2
-3
Problem 6
(a) Give an example of a continuous function with no absolute maximum on (0, 1].
And: For example, f (x) = 1 − x has no absolute maximum in (0, 1]. The maximum should be attained at x = 0,
but the point is not part of the interval.
(b) Give an example of a continuous function with no absolute minimum on [0, 1].
And: There is no such function. Every continuous function admits maximum and minimum on a closed interval.
Problem 7 Use the definition of definite integral to compute
Z 1
4x3 dx
−1/2
And: Since f (x) = 4x3 is continuous on the interval [−1/2, 1], f (x) is integrable and we can compute the
integral using the standard right-endpoint partition.
Z
1
3
3
n
n
n
3 X
−n + 3i
1
3i
3 X
3 X
4 − +
4
(27i3 −27i2 n+9in2 −n3 ) =
= lim
= lim 4
n→∞ 2n
n→∞ 2n
n→∞ (2n)4
2
2n
2n
i=1
i=1
i=1
4x3 dx = lim
−1/2
!
!
!
n
n
n
X
X
X
3
= lim 4
(27
i3 − 27n
i2 + 9n2
i − n4 ) =
n→∞ (2n)4
i=1
i=1
i=1
4
3
3
n + 2n + n
n(n + 1)(2n + 1)
n(n + 1)
2
= lim 4
(27
−
27n
+
9n
− n4 ) =
n→∞ (2n)4
4
6
2
12 81 − 108 + 54 − 12
12 27 27 · 2 9
12 15
15
−
+ −1 =
= lim
=
=
n→∞ 16
4
6
2
16
12
16 12
16
We can check the answer by
Z
1
1
15
1
4x3 dx = x4 −1/2 = 1 −
=
16
16
−1/2
Problem 8 Evaluate the following integrals:
(a)
Z
1
√
(x( x − 1))dx =
0
Z
1
x3/2 dx −
0
Z
1
xdx =
0
2 1
1
2x5/2 1 x2 1
| − |0 = − = −
5 0
2
5 2
10
(b)
Z
5
Z
1
|x − 1|dx =
−1
Z
(1 − x)dx +
−1
=1−
5
(x − 1)dx =
1
1
2
5
x2 x
x−
+
− x =
2 −1
2
1
1
1 25
1
+1+ +
− 5 − + 1 = 12 − 2 = 10
2
2
2
2
3
Problem 9 Compute the area of the region
x
2 − 2| sin( )| ≤ y ≤ cos(x) + 1}
2
S = {(x, y) | − π ≤ x ≤ π,
And: We need to compute the area of the region
3
2
1
-5
-4
-3
-2
-1
0
1
2
3
4
-1
-2
-3
It corresponds to the difference of the integrals
Z π
Z
A(S) =
(cos(x) + 1)dx −
−π
We have
Z
π
(2 − 2| sin(x)|)dx
−π
π
(cos(x) + 1)dx = sin(x) + x|π−π = 2π
−π
Z
π
Z
0
(2 − 2| sin(x/2)|)dx =
−π
Z
−π
π
(2 − 2 sin(x/2))dx =
(2 + 2 sin(x/2))dx +
0
0
π
= (2x − 4 cos(x/2))|−π + (2x + 4 cos(x/2))|0 = 2π − 4 + 2π − 4 = 4π − 8
Therefore
A(S) = 8 − 2π
4
Final - Practice
List of exercises from previous finals.
The final exam will have nine exercises, following the description below.
Problem 1 Find the limit if it exists. If the limit does not exist, explain why.
1.
lim
p
x→∞
x4 − 2x2 − x2
Ans: We have
lim
p
x→∞
x4 − 2x2 − x2 = lim x2
p
x→∞
p
1 − 2/x2 + 1
1 − 2/x2 − 1 p
1 − 2/x2 + 1
=
−2
1 − 2/x2 − 1
= lim p
= −1
= lim x2 p
2
x→∞
x→∞
1 − 2/x + 1
1 − 2/x2 + 1
2.
π i
+
1
|x|
x→0
x2
i
h π + 1 |x| ≤ 2|x|
0 ≤ sin 2
x
lim
Ans: We have
h
sin
by the squeeze theorem
h π h π i
i
0 ≤ lim sin 2
+ 1 |x| ≤ lim 2|x| = 0 ⇒ lim sin 2
+ 1 |x| = 0
x→0
x→0
x→0
x
x
3.
sin(x)3
x→0 x2 − x
lim
Ans:
sin(x)3
x3 sin(x)3
x3
x2
x
=
lim
=
lim
=
lim
=0
x→0 x2 − x
x→0 x3 x2 − x
x→0 x2 − x
x→0 x2 1 − 1/x
lim
4.
lim
sin(x) + cos
x→0
Ans:We have
−2x2 ≤
sin(x) + cos
1
x2
x
1
x2 ≤ 2x2
x
therefore by squeeze theorem
1
1
2
2
2
0 = lim 2x ≤ lim
sin(x) + cos
x ≤ lim 2x = 0 ⇒ lim
sin(x) + cos
x2 = 0
x→0
x→0
x→0
x→0
x
x
5.
x2 − 1
x→1 |x − 1|
lim
Ans: We have
x2 − 1
x−1
= lim
(x + 1)
x→1 |x − 1|
x→1 |x − 1|
lim
and clearly
lim
x→1+
x−1
(x + 1) = lim (x + 1) = 2
|x − 1|
x→1+
and
Therefore the limit does not exist.
1
lim
x→1−
x−1
(x + 1) = lim −(x + 1) = −2
|x − 1|
x→1−
6.
lim
x→−∞
p
x2 + 2x + 3
x+
Ans:
x − √x2 + 2x + 3
p
p
2
2
√
lim x + x + 2x + 3 = lim x + x + 2x + 3
=
x→−∞
x→−∞
x − x2 + 2x + 3
x
x2 − x2 − 2x − 3
−2 − 3/x
p
p
= lim
= −1
= lim
·
x→−∞ x 1 +
x→−∞ x − |x| 1 + 2/x + 3/x2
1 + 2/x + 3/x2
7.
x2 − 6x + 5
x→5
x2 − 5x
lim
Ans:
x2 − 6x + 5
x−5 x−1
4
= lim
·
=
x→5
x→5 x − 5
x2 − 5x
x
5
lim
8.
lim
x→+∞
p
x2 + x + 1 − x
Ans:
lim
p
x→+∞
x2
+ x + 1 − x = lim x
p
x→+∞
1 + 1/x +
1/x2
p1 + 1/x + 1/x2 + 1
−1 p
=
1 + 1/x + 1/x2 + 1
1 + 1/x + 1/x2 − 1
1 + 1/x
1
= lim x p
= lim p
=
2
2
x→+∞
x→+∞
2
1 + 1/x + 1/x + 1
1 + 1/x + 1/x + 1
9.
lim+
x→1
x2 − 9
x2 + 2x − 3
Ans:
lim+
x→1
x−3 x+3
x−3
x2 − 9
= lim
·
= lim
x2 + 2x − 3 x→1+ x − 1 x + 3 x→1+ x − 1
Since x − 3 < 0 near 1 and x − 1 → 0+ for x → 1+ , we have
lim+
x→1
10.
x2
x2 − 9
= −∞
+ 2x − 3
t3
t→0 tan3 (2t)
lim
Ans:
t3
lim
= lim
t→0 tan3 (2t)
t→0
11.
t3
sin3 (2t)
3
cos (2t) = lim
t→0
2t
sin(2t)
3
·
1
1
cos3 (2t) =
8
8
ex − 1
x→0
x
lim
Ans:
ex − 1
eh+0 − e0
d t lim
= lim
=
e
= e0 = 1
x→0
h→0
x
h
dt t=0
2
12.
n
2X
lim
n→∞ n
i=1
s
4−
2
2
i+1 −1
n
Ans:
n
2X
lim
n→∞ n
i=1
s
4−
2 Z 3 p
2
4 − (x − 1)2 dx
i+1 −1 =
n
1
p
The function 4 − (x − 1)2 represents the semicircle of radius 2 centered in (1, 0). We can interpret the
integral as an area and we obtain
s
2 Z 3 n
p
4π
2X
2
lim
4−
i+1 −1 =
4 − (x − 1)2 dx =
=π
n→∞ n
n
4
1
i=1
3
Problem 2 Differentiate the following functions.
1.
y = tan
Ans:
p
1 + x2
p
1
y 0 = sec2 ( 1 + x2 ) · (1 + x2 )−1/2 · 2x
2
2.
y = (x4 + 4)x
Ans:
y = eln(x
4
+4)x
⇒ y 0 = eln(x
4
+4)x
3.
·
2
y = ex(x
Ans:
3
y 0 = ex
+x
4.
x
4x3 + ln(x4 + 4)
4
x +4
+1)
· (x2 + 1)
x
y = e x+1
Ans:
x
y 0 = e x+1 ·
1
(x + 1)2
5.
y = (2 + cos(x))x
Ans:
y=e
ln(2+cos(x))·x
0
⇒y =e
ln(2+cos(x))·x
x sin(x)
+ ln(2 + cos(x))
· −
2 + cos(x)
6.
y = ln (x2 + a2 )(x + b)6 (x + c)7
Ans:
y = ln(x2 + a2 ) + 6 ln(x + b) + 7 ln(x + c) ⇒ y 0 =
2x
6
7
+
+
x2 + a2
x+b x+c
7.
Z
x2
2
e−t dt
y=
x
Ans:
Z
y=
0
x2
2
e−t dt −
Z
x
2
2
e−t dt ⇒ y 0 = 2xe−x − e−x
0
8.
y = tan2 (x3 + 1)
Ans:
y 0 = 2 tan(x3 + 1) · sec2 (x3 + 1) · 3x2
9.
y = xx
Ans:
y = eln(x)x ⇒ y 0 = eln(x)x (1 + ln(x))
4
2
10.
x2
Z
et sin(t)dt
y=
1
Ans:
2
y 0 = ex sin(x2 ) · 2x
11.
Z
2
p
1 + t4 dt
y=
cos(x)
Ans:
Z
y=−
cos(x)
p
1 + t4 dt ⇒ y 0 = −
p
1 + (cos(x))4 · (− sin(x))
2
12.
y = cos(x)x
Ans:
x sin(x)
+ ln(cos(x))
y = eln(cos(x))x ⇒ y 0 = eln(cos(x))x · −
cos(x)
13.
3t − 2
y=√
2t + 1
Ans: We use the logarithmic derivative:
ln(y) = ln(3t − 2) −
Therefore
1
y0
3
1 2
ln(2t + 1) ⇒
=
−
2
y
3t − 2 2 2t + 1
3t − 2
y0 = √
·
2t + 1
5
3
1 2
−
3t − 2 2 2t + 1
Problem 3 Evaluate the following integrals.
1.
1
Z
√
x x − 1dx
0
Ans: The function is not defined in the interval [0, 1].
2.
ln(x2 )
dx
x
Z
Ans: Apply the substitution
du
2x
dx
u = ln(x2 )
du = 2 dx ⇒
=
x
2
x
Z
Z
ln(x2 )
u2
1
2 √
udu =
dx =
+ C = ln ( x) + C
x
2
4
3.
3
Z
√
cos3 ( s)dx
1
Ans:
3
Z
√
√
√
cos3 ( s)dx = cos3 ( s)(3 − 1) = 2 cos3 ( s)
1
4.
5
Z
|x − 1|dx
−1
Ans:
Z
5
Z
1
|x − 1|dx =
−1
5
Z
(x − 1)dx = (x − x2 /2)|1−1 + (x2 /2 − x)|51 =
(1 − x)dx +
−1
1
= 1 − 1/2 + 1 + 1/2 + 25/2 − 5 − 1/2 − 1 = 2 + 16/2 = 10
5.
Z
sin(x)
dx
1 + cos(x)
Ans: We apply the substitution
u = cos(x) + 1
du = − sin(x)dx
Z
sin(x)
du
1
dx = −
= − ln(u) + C = ln
+C
1 + cos(x)
u
1 + cos(x)
Z
6.
Z
1
−1
sin θ + sin θ tan2 θ
dθ
sec2 θ
Ans:
Z
1
−1
sin θ + sin θ tan2 θ
dθ =
sec2 θ
1
1 + tan2 θ
sin θ
dθ =
sec2 θ
−1
Z
6
Z
1
sin θdθ = 0
−1
7.
1/4
Z
tan3 (x + sin(x))dx
−1/4
Ans: The integrand function is odd and the interval is symmetric. Indeed
tan3 (−x + sin(−x)) = − tan3 (x + sin(x))
therefore
Z
1/4
tan3 (x + sin(x))dx = 0
−1/4
8.
Z
Ans:
Z
x3 (1 +
9.
√
4
x3 (1 +
√
4
x3 )dx
x4
4x15/4
+
+C
4
15
x3 )dx =
π
Z
x sin(x2 + π)dx
0
Ans: Apply the substitution
u = x2 + π
Z
π
sin(u)du = − cos(u)|ππ
x sin(x + π)dx =
0
u(π) = π 2 + π
u(0) = π
π 2 +π
Z
2
du = 2xdx
2
+π
= − cos(π 2 + π) + cos(π) = cos(π 2 ) − 1
π
10.
1
Z
0
ex − e−x
dx
(ex + e−x )2
Ans: Apply the substitution
u = ex + e−x
Z
0
1
du = (ex − e−x )dx
ex − e−x
dx =
(ex + e−x )2
11.
Z
e+e−1
Z
2
u(0) = 2
u(1) = e + e−1
e+e−1
du
1 1
1
=− = −
u2
u 2
2 e + e−1
−π/4
(cos θ + sec2 θ)dθ
π/4
Ans:
Z
√
√
2
2
−
−1−1=− 2−2
=−
2
2
√
−π/4
2
(cos θ + sec θ)dθ =
π/4
−π/4
sin θ|π/4
12.
Z
0
1
+
−π/4
tan θ|π/4
x+1
dx
x2 + 2x + 1
Ans: Apply the substitution
u = x2 + 2x + 1
Z
0
1
du = 2(x + 1)dx
x+1
1
dx =
2
x + 2x + 1
2
7
Z
1
4
u(0) = 1
u(1) = 4
√
du
= ln( u)|41 = ln(2)
u
13.
Z
sin(t)
dt
cos(t) + 2
Ans: Apply the substitution
Z
u = cos(t) + 2
du = − sin(t)dt
Z
sin(t)
1
du
dt = −
= ln(1/u) + C = ln
+C
cos(t) + 2
u
cos(x) + 2
14.
e2
Z
1
√
ln x
dx
x
Ans: Apply the substitution
u = ln(x)
Z
1
e2
du =
√
dx
x
ln x
dx =
x
u(1) = 0
Z
2
u1/2 du =
0
8
u(e2 ) = ln(e2 ) = 2
√
2
4 2
2u3/2 =
3 0
3
Problem 4 Study the following functions:
• find the domain
• find where f is positive, negative or zero (if doable)
• find horizontal and vertical asymptotes (if any)
• find f 0
• find where f is increasing or decreasing
• find maxima and minima
• find f 00
• find where f is concave up or concave down
• find inflection points
• sketch the graph
1.
f (x) =
x2
x
−4
Ans:
• Domain: the domain of the function is R \ {±2}
•

 f (x) > 0 if x ∈ (−2, 0) ∪ (2, +∞)
f (x) = 0 if x = 0

f (x) < 0 if x ∈ (−∞, −2) ∪ (0, 2)
• We have

limx→+∞ f (x) = 0+




limx→−∞ f (x) = 0−



limx→+2+ f (x) = +∞
lim

x→+2− f (x) = −∞



lim

x→−2+ f (x) = −∞


limx→−2− f (x) = +∞
• The first derivative is
f 0 (x) =
2
1
x2 + 4
2
x
−
4
−
2x
=
−
(x2 − 4)2
(x2 − 4)2
It follows that f 0 (x) < 0 on its domain.
• The second derivative is
f 00 (x) =
thus
2x3 − 8x − 4x3 − 16x
1
x2 + 12
2
2
2x(x
−
4)
−
4x(x
+
4)
=
=
−2x
(x2 − 4)3
(x2 − 4)3
(x2 − 4)3
 0
 f (x) > 0 if x ∈ (−2, 0) ∪ (2, +∞)
f 0 (x) = 0 if x = 0
 0
f (x) < 0 if x ∈ (−∞, −2) ∪ (0, 2)
We have inflection points at x = 0 and a discontinuity at ±2.
• The graph is
9
5
2.5
-7.5
-5
-2.5
0
2.5
5
7.5
10
-2.5
-5
2.
f (x) =
x−1
x2
Ans:
• Domain: the domain of the function is R \ {0}
• Since x2 > 0, we have

 f (x) > 0 if
f (x) = 0 if

f (x) < 0 if
• We have

limx→+∞ f (x) =



limx→−∞ f (x) =
 limx→0+ f (x) =


limx→0− f (x) =
• The first derivative is
f 0 (x) =
It follows
x ∈ (1, +∞)
x=1
x ∈ (−∞, 1)
d
dx
1
1
− 2
x x
=−
0+
0−
−∞
−∞
1
1
1−x
+ 3 =
2
x
x
x3
 0
 f (x) > 0 if x ∈ (0, 1)
f 0 (x) = 0 if x = 1
 0
f (x) < 0 if x ∈ (−∞, 0) ∪ (1, +∞)
We have a local max at x = 1.
• The second derivative is
f 00 (x) =
thus
d
dx
1
1
− 2
x3
x
 0
 f (x) > 0 if
f 0 (x) = 0 if
 0
f (x) < 0 if
=−
1
1
x−1
+ 3 =
x4
x
x4
x ∈ (1, +∞)
x=1
x ∈ (−∞, 1) \ {0}
We have inflection points at x = 1 and a discontinuity at 0.
• The graph is
2.5
-7.5
-5
-2.5
0
-2.5
-5
-7.5
10
2.5
5
7.5
10
3.
f (x) =
x
x+1
2
Ans:
• Domain: the domain of the function is R \ {−1}
• f (x) > 0 for x 6= 0, −1, and f (x) = 0 at x = 0.
• We have
f (x) =
thus
x
x+1
2
=
1
1 + 1/x
2

limx→+∞ f (x) = 1−



limx→−∞ f (x) = 1+
 limx→−1+ f (x) = +∞


limx→−1− f (x) = −∞
• The first derivative is
f 0 (x) = 2
It follows
x
x+1
1
2x
=
2
(x + 1)
(x + 1)3
 0
 f (x) > 0 if x ∈ (−∞, −1) ∪ (0, +∞)
f 0 (x) = 0 if x = 0
 0
f (x) < 0 if x ∈ (−1, 0)
We have local min at x = 0 and a critical point at x = −1 where the function is not differentiable.
• The second derivative is
f 00 (x) =
thus
2
2
2(x + 1)3 − 6x(x + 1)2 =
[2 − 4x]
(x + 1)6
(x + 1)4
 0
 f (x) > 0 if x ∈ (−∞, +1/2) \ {−1}
f 0 (x) = 0 if x = 1/2
 0
f (x) < 0 if x ∈ (1/2, +∞)
We have inflection points at x = 1/2 and a discontinuity at −1.
• The graph is
10
7.5
5
2.5
-10
-7.5
-5
-2.5
0
-2.5
4.
f (x) = x2/3 (x − 1)
Ans:
• Domain: the domain of the function is R
11
2.5
5
7.5
• Since x2/3 ≥ 0, we have

 f (x) > 0 if
f (x) = 0 if

f (x) < 0 if
• We have
x ∈ (1, +∞)
x=1
x ∈ (−∞, 1)
limx→+∞ f (x)
limx→−∞ f (x)
=
=
+∞
−∞
• The first derivative is
f 0 (x) =
It follows
d 5/3
5
2
5x − 2
(x − x2/3 ) = x2/3 − x−1/3 = √
dx
3
3
33x
 0
 f (x) > 0 if x ∈ (−∞, 0) ∪ (2/5, +∞)
f 0 (x) = 0 if x = 2/5
 0
f (x) < 0 if x ∈ (0, 2/5)
We have local min at x = 2/5 and a critical point at x = 0 where the function is not differentiable.
• The second derivative is
f 00 (x) =
thus
10x + 2
10 −1/3 2 −4/3
x
+ x
= √
3
9
9
9 x4
 0
 f (x) > 0 if x ∈ (−1/5, +∞)
f 0 (x) = 0 if x = −1/5
 0
f (x) < 0 if x ∈ (−∞, −1/5)
We have inflection points at x = −1/5 and a discontinuity at zero.
• The graph is
3
2
1
-3
-2
-1
0
1
2
3
4
5
6
-1
-2
5.
f (x) =
1
1
−
x x+1
Ans:
• Domain: the domain of the function is R \ {0, −1}
• We have
f (x) =
thus
x+1−x
1
=
x(x + 1)
x(x + 1)
f (x) > 0 if x ∈ (−∞, −1) ∪ (0, +∞)
f (x) < 0 if x ∈ (−1, 0)
Moreover, f (x) 6= 0 on its domain.
12
• We have

limx→+∞ f (x)




lim

x→−∞ f (x)


limx→0+ f (x)
limx→0− f (x)




lim

x→−1+ f (x)


limx→−1− f (x)
• The first derivative is
f 0 (x) = −
It follows
=
=
=
=
=
=
0+
0−
+∞
−∞
−∞
+∞
1
2x + 1
1
+
=− 2
x2
(x + 1)2
x (x + 1)2
 0
 f (x) > 0 if x ∈ (−∞, −1/2) \ {−1}
f 0 (x) = 0 if x = −1/2
 0
f (x) < 0 if x ∈ (−1/2, +∞) \ {0}
We have local max at x = −1/2.
• The second derivative is
2
1
2x (x + 1)2 − (2x + 1)2x(x + 1)(2x + 1) =
4
+ 1)
2
1
1
[2x(x + 1) − 2(2x + 1)(2x + 1)] = − 3
2x + 2x − 8x2 − 4x − 2 =
=− 3
x (x + 1)3
x (x + 1)3
2
2
= 3
3x + x + 1
3
x (x + 1)
f 00 (x) = −
x4 (x
We observe that 3x2 + x + 1 > 0, therefore
00
f (x) > 0 if x ∈ (−∞, −1) ∪ (0, +∞)
f 00 (x) < 0 if x ∈ (−1, 0)
• The graph is
5
2.5
-7.5
-5
-2.5
0
2.5
5
-2.5
-5
6.
f (x) = esin(πx)
x ∈ [0, 2]
Ans:
• Domain: the domain of the function is [0, 2].
• f (x) > 0 on its domain.
• The function is periodic over R and continuous. Therefore
limx→0+ f (x) = 1+
limx→2− f (x) = 1−
13
7.5
• The first derivative is
f 0 (x) = π cos(πx)esin(πx)
It follows
 0
 f (x) > 0 if x ∈ (0, 1/2) ∪ (3/2, 2)
f 0 (x) = 0 if x = 1/2, 3/2
 0
f (x) < 0 if x(1/2, 3/2) ∈
We have local max at x = 1/2 and local min at x = 3/2.
• The second derivative is
f 00 (x) = −π 2 sin(πx)esin(πx) + π 2 cos2 (πx)esin(πx) = π 2 esin(πx) (− sin2 (πx) − sin(πx) + 1)
and it is equal to zero for
√
1± 5
sin(πx) =
2
• The graph is
2.8
2.4
2
1.6
1.2
0.8
0.4
-1.2
-0.8
-0.4
0
0.4
0.8
1.2
1.6
2
2.4
2.8
3.2
3.6
7.
f (x) = ex+ln(x
2
)
Ans:
• Domain: the domain of the function is R \ {0}
• f (x) > 0 on its domain.
• We have for x 6= 0
2
f (x) = ex+ln(x
thus

limx→+∞ f (x)



limx→−∞ f (x)
limx→0− f (x)



limx→0+ f (x)
)
= x2 ex
=
=
=
=
+∞
0+
0+
0+
• The first derivative is
f 0 (x) = 2xex + x2 ex = xex (2 + x)
It follows
 0
 f (x) > 0 if x ∈ (−∞, −2) ∪ (0, +∞)
f 0 (x) = 0 if x = 0, −2
 0
f (x) < 0 if x ∈ (−2, 0)
We have local max at x = −1 and local min at x = 0 (technically, x = 0 is not part of the domain,
but the discontinuity is removable).
14
• The second derivative is
f 00 (x) = 2ex + 4xex + x2 ex = (x2 + 4x + 2)ex
and its sign depends on the roots
x1,2 =
−4 ±
√
16 − 8
2
= −2 ±
√
2
thus
√
√
 0
−
2)
∪
(−2
+
2, +∞)
 f (x) > 0 if x ∈ (−∞, −2
√
0
f (x) = 0 if x = −2 ± √2
√
 0
f (x) < 0 if x ∈ (−2 − 2, −2 + 2)
√
We have inflection points at x = −2 ± 2.
• The graph is
3
2
1
-5
-4
-3
-2
-1
0
-1
-2
-3
15
1
2
3
4
5
Problem 5
1. Show that sin x − cos x + 3x = 0 has only one real solution.
Ans: Let f (x) = sin x − cos x + 3x. By intermediate value theorem, since f (0) = −1 < 0 and
f (π) = 1 + 3π > 0, there is a solution in the interval (0, π).
Moreover, f 0 (x) = 3 + cos(x) + sin(x) > 0, therefore, f (x) is increasing and the solution is unique.
Alternatively, given two roots a 6= b such that f (a) = f (b), by Rolle’s theorem there exists a point
c ∈ (a, b) such that f 0 (c) = 0. This is a contradiction, since f 0 (x) > 0, and a = b.
2. Show that x4 + 4x − 3 = 0 has exactly two real solutions.
Ans: Let f (x) = x4 + 4x − 3. Since f 00 (x) = 12x2 ≥ 0, the function does not change concavity, therefore
there are at most two solution.
Moreover,
f (−2) = 5
f (0) = −3
f (1) = 2
therefore we can apply the intermediate value theorem on (−2, 0) and (0, 1) to prove that the are at least
two roots.
3. Show that ex −
x2
2
− 2 = 0 has exactly one solution.
2
Ans: Let f (x) = ex − x2 − 2. By intermediate value theorem, since f (0) = −1 < 0 and f (ln(4)) =
2 − ln2 (2) > 0, there is a solution in the interval (0, 2 ln(2)).
Moreover, f 0 (x) = ex − x. Therefore f 0 (x) > 0 and f (x) is increasing. Thus, the solution is unique.
Alternatively, given two roots a 6= b such that f (a) = f (b), by Rolle’s theorem there exists a point
c ∈ (a, b) such that f 0 (c) = 0. This is a contradiction, since f 0 (x) > 0, and a = b.
4. Show that x5 − 2x + 1 = 0 has exactly three real solutions.
Ans: Let f (x) = x5 − 2x + 1. Since f 00 (x) = 20x3 , the concavity changes only one time, therefore there
are at most three solutions.
p
Clearly, f (1) = 0. Then f 0 (x) = 5x4 − 2 and we have critical points at x = ± 4 2/5 and
p
p
p
8p
f ( 4 2/5) = 2/5 4 2/5 − 2 4 2/5 + 1 = − 4 2/5 + 1 < 0
5
p
p
p
8p
f (− 4 2/5) = −2/5 4 2/5 + 2 4 2/5 + 1 = 4 2/5 + 1 > 0
5
p
p
4
4
therefore we have another
root in (− 2/5, 2/5). Finally, f (−2) < 0, therefore there is a third root in
p
4
the interval (−2, − 2/5).
5. Show that x4 + 2x2 − 1 = 0 has exactly two real solutions.
Ans: Let f (x) = x4 + 2x2 − 1. Then f 00 (x) = 12x2 + 4 > 0, therefore the concavity does not change and
the equation has at most two solution. Since limx→±∞ f (x) = +∞ and f (0) = −1, by intermediate value
theorem there are two roots, one in (−∞, 0) and one in (0, +∞).
16
Problem 6 Find the equation of the tangent to the curve at the given point.
1. 2(x2 + y 2 )2 = 25xy at (1, 2).
Ans: By implicit differentiation
4(x2 + y 2 ) · (2x + 2yy 0 ) = 25y + 25xy 0 ⇒ y 0 =
Therefore
8x(x2 + y 2 ) − 25y
25x − 8y(x2 + y 2 )
dy
40 − 50
10
2
|(1,2) =
=
=
dx
25 − 80
55
11
The equation of the tangent at (1, 2) is
y−2=
2
(x − 1)
11
3
2
1
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
-2
-3
2. ln y = y 2 ln x at (1, 1).
Ans: By implicit differentiation
y2
y3
y0
= 2yy 0 ln(x) +
⇒ y0 =
y
x
x(1 + 2y 2 ln(x))
Therefore
dy
|(1,1) = 1
dx
and the equation of the tangent at (1, 1) is
y−1=x−1⇒y =x
5
4
3
2
1
0
1
2
3
4
5
6
7
8
9
3. sin(πy) − cos(πx) = 1 + ln(xy) at (1, 1)
Ans: By implicit differentiation
cos(πy)πy 0 + sin(πx)π =
sin(πx)πy − xy
1 y0
+
⇒ y0 =
x
y
1 − cos(πy)πy
17
Therefore
dy
1
|(1,1) = −
dx
1+π
and the equation of the tangent at (1, 1) is
y−1=−
1
(x − 1)
π+1
4
3
2
1
-3
-2
-1
0
1
2
3
4
5
6
-1
-2
4. ex + ln y = x2 − y 2 + 2 at (0, 1).
Ans: By implicit differentiation,
ex +
y0
2xy − ex y
= 2x − 2yy 0 ⇒ y 0 =
y
1 + 2y 2
therefore
1
dy
|(0,1) = −
dx
3
and the equation of the tangent at (0, 1) is
1
y−1=− x
3
2.8
2.4
2
1.6
1.2
0.8
0.4
-2.8
-2.4
-2
-1.6
-1.2
-0.8
-0.4
0
0.4
0.8
1.2
1.6
2
√
√
5. (x2 + y 2 )3/2 = 2xy at (1/ 2, 1/ 2).
Ans: By implicit differentiation
3x(x2 + y 2 )1/2 − 2y
3 2
(x + y 2 )1/2 (2x + 2yy 0 ) = 2y + 2xy 0 ⇒ y 0 =
2
2x − 3y(x2 + y 2 )1/2
therefore
1
√
dy
2
|(1/√2,1/√2) =
= −1
dx
− √12
√
√
and the equation of the tangent at (1/ 2, 1/ 2) is
1
1
y − √ = −(x − √ )
2
2
18
1.6
1.2
0.8
0.4
-2.4
-2
-1.6
-1.2
-0.8
-0.4
0
0.4
0.8
1.2
1.6
2
2.4
-0.4
-0.8
-1.2
-1.6
6. y = exy at (−e, 1/e).
Ans: By implicit differentiation
y 0 = exy (y + xy 0 ) ⇒ y 0 =
yexy
1 − xexy
therefore
dy
1
|(−e,1/e) = 2
dx
2e
and the equation of the tangent at (−e, 1/e) is
y−
1
1
= 2 (x + e)
e
2e
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
-1
0
-1
-2
19
1
Problem 7
1. Find the biggest right cone inside a sphere of radius r.
Ans: Consider a cone inscribed in a sphere of radius r. Let’s denote by rC = y the radius of the base of
the cone and the height of the cone by hC = x + r. It is easy to check that
x2 + y 2 = r 2
It follows that the volume of the cone
V =
π 2
π
π
π
rC hC = y 2 (x + r) = (r2 − x2 )(x + r) = (r3 + xr2 − x2 r − x3 )
3
3
3
3
It follows that
√
dV
π 2
r
2r ± 16r2
2
= (r − 2xr − 3x ) = 0 ⇔ x =
=
dx
3
−6
3
We obviously eliminate the value −r and we notice that
dV
<0
dx
if x >
r
3
and
dV
>0
dx
if
−r
or
0<x<
r
3
Therefore the absolute maximum is attained at x = r/3.
2. Find the smallest right cone around a sphere of radius r.
Ans: Consider a cone circumscribed to a sphere of radius r. Let’s denote the radius of the base of the cone
by rC = y and the height of the cone by hC = x + 2r. Comparing similar triangles (or simply studying
the tangent of the top angle, it follows that
p
(x + 2r)2 + y 2
x+r
(x + r)2
(x + 2r)2 + y 2
=
⇒
=
2
r
y
r
y2
Expanding the expression above we get
y2 = r2
(x + 2r)2
(x + r)2 − r2
The volume of the cone is given by the expression
V =
π
π
(x + 2r)2
π 2
rC hC = y 2 (x + 2r) = r2
(x + 2r)
3
3
3 (x + r)2 − r2
Since the volume of the cone is never going to be zero, we apply the logarithm to compute the derivative
and we obtain
dV
3
2(x + r)
x2 − 4r2
=V
−
=
V
dx
x + 2r (x + r)2 − r2
(x + r)((x + r)2 − r2 )
It follows that dV /dx = 0 if x = 2r. Moreover, we notice that dV /dx < 0 if x < 2r and dV /dx > 0 if
x > 2r, therefore we obtain the minimum volume at x = 2r.
3. Find the biggest isosceles triangle inside a circle of radius r.
Ans: Consider a triangle inscribed in a circle of radius r. Let’s denote by bT = 2y the base of the triangle
and the height of the triangle by hT = x + r. It is easy to check that
x2 + y 2 = r 2
It follows that the area of the triangle
A=
p
bT hT
= y(x + r) =
r2 − x2 (x + r)
2
20
It follows that
−2x(x + r) + 2(r2 − x2 )
p
dA
2x2 + xr − r2
2x(x + r)
√
+
r 2 − x2 =
=− √
=− √
dx
2 r 2 − x2
2 r 2 − x2
r 2 − x2
We have dA/dx = 0 at
√
−r ± 9r2
r
x=
=
4
2
We obviously eliminate the value −r and we notice that
dA
<0
dx
if x >
r
2
−r
or
dA
>0
dx
and
if
0<x<
r
2
Therefore the absolute maximum is attained at x = r/2.
4. Find the smallest isosceles triangle around a circle of radius r.
Ans: Consider a triangle circumscribed to a circle of radius r. Let’s denote the base of the triangle by
bT = 2y and the height of the triangle by hC = x + 2r. Comparing similar triangles (or simply studying
the tangent of the top angle, it follows that
p
(x + 2r)2 + y 2
(x + r)2
(x + 2r)2 + y 2
x+r
=
⇒
=
r
y
r2
y2
Expanding the expression above we get
y2 = r2
(x + 2r)2
r(x + 2r)
⇒y= √
2
2
(x + r) − r
x2 + 2rx
The area of the triangle is given by the expression
A=
bT hT
r(x + 2r)2
= y(x + 2r) = √
2
x2 + 2rx
Since the area of the triangle is never going to be zero, we apply the logarithm to compute the derivative
and we obtain
dA
2
x+r
x−r
=A
−
=A
dx
x + 2r x2 + 2rx
x(x + 2r)
It follows that dA/dx = 0 if x = r. Moreover, we notice that dA/dx < 0 if x < r and dA/dx > 0 if x > r,
therefore we obtain the minimum area at x = r.
2
5. Find the largest rectangle which has its base on the x-axis, and its vertices on the curve y = e−x .
2
Ans: The function e−x is even and a rectangle is uniquely determined by a positive coordinate x > 0. It
follows that the area of the rectangle is given by the formula
2
2
2
2
A(x) = 2xe−x ⇒ A0 (x) = 2e−x − 4x2 e−x = 2e−x (1 − 2x2 )
√
√
√
It follows that A0 (x) = 0 at x = 1/ 2. Moreover,
since A0 > 0 if x < 1/ 2 and A0 < 0 if x > 1/ 2, it
√
follows that the area is maximal at x = 1/ 2.
6. Find the point on the curve y = e−x − ln x which is the closest to (0, 0).
Ans: The general point on the curve has the form (x, e−x − ln(x)). Its distance from (0, 0) is given by
D(x) =
p
x2 + (e−x − ln(x))2 ⇒ D0 (x) =
2x + 2(e−x − ln(x))(e−x − x1 )
2D(x)
Therefore D0 (x) = 0 if
x + (e−x − ln(x))(e−x −
1
1
ln(x)
) = 0 ⇔ e−2x − (ln(x) + )e−x + (
+ x) = 0
x
x
x
It is easy to find that the solution exists, it’s unique, and lies in the interval (0, 1).
21
√
1. Use linearization process to approximate 35.9.
√
Ans: Consider f (x) = x. We want to apply linearization at x0 = 36. In this case we have
Problem 8
1
f 0 (x) = √
2 x
f (36) = 6
The equation of the tangent at x0 is
f 0 (36) =
1
12
1
(x − 36)
12
y =6+
It follows that
1
1 1
720 − 1
719
(35.9 − 36) = 6 −
=
=
= 5.9916̄
12
12 10
120
120
√
On the other hand, using a calculator 35.9 = 5.991660 . . . .
f (35.9) ' 6 +
2. Consider the curve that is given by x3 + xy 3 = 9(y − 1). Notice that the point (1, 2) lies on this curve.
Approximate the y–coordinate of the point of the curve with x–coordinate 0.9.
Ans: By implicit differentiation,
3x2 + y 3 + 3xy 2 y 0 = 9y 0 ⇒ y 0 =
Therefore
3x2 + y 3
9 − 3xy 2
11
dy
|(1,2) = −
dx
3
and the equation of the tangent is
y−2=−
11
(x − 1)
3
2.8
2.4
2
1.6
1.2
0.8
0.4
-2
-1.6
-1.2
-0.8
-0.4
0
0.4
0.8
1.2
1.6
2
2.4
Therefore, if we want to approximate the y–coordinate corresponding to x = 0.9 it is enough to compute
the corresponding value on the tangent line
y|x=0.9 ' 2 −
11
11
(0.9 − 1) = 2 +
= 2.36̄
3
30
3. Consider the curve given by the equation x ln y + y = 1 − ln x. Notice that the point (1, 1) lies on this curve.
Approximate the y–coordinate of the point of the curve with x–coordinate 0.9.
Ans: By implicit differentiation,
ln(y) +
Therefore
ln(y −1 ) −
1
xy 0
+ y0 = − ⇒ y0 = y0 =
x
y
x
y +1
dy
1
|(1,1) = −
dx
2
and the equation of the tangent is
1
y − 1 = − (x − 1)
2
22
1
x
2.4
2
1.6
1.2
0.8
0.4
-0.8
-0.4
0
0.4
0.8
1.2
1.6
2
2.4
2.8
3.2
3.6
-0.4
Therefore, if we want to approximate the y–coordinate corresponding to x = 0.9 it is enough to compute
the corresponding value on the tangent line
1
1
= 1.05
y|x=0.9 ' 1 − (0.9 − 1) = 1 +
2
20
4. Use linearization process to approximate ln(1.1).
Ans: Consider f (x) = ln x. We want to apply linearization at x0 = 1. In this case we have
f 0 (x) =
f (1) = 0
1
x
f 0 (1) = 1
The equation of the tangent at x0 is
y =x−1
It follows that
f (1.1) ' 0.1
On the other hand, using a calculator ln(1.1) = 0.095 . . . .
5. Use linearization process to approximate e−0.1 .
Ans: Consider f (x) = ex . We want to apply linearization at x0 = 0. In this case we have
f (0) = 1
f 0 (x) = ex
f 0 (0) = 1
The equation of the tangent at x0 is
y =1+x
It follows that
f (−0.1) ' 1 − 0.1 = 0.9
On the other hand, using a calculator
√
35.9 = 0.904 . . . .
23
Problem 9
1. The population of the city of Metropolis grows exponentially. In 2000, the population was
2 million. In 2010, the population was 3 million. Let P denote the population (measured in millions of
people) and let t denote the number of years that have passed since the year 2000.
a) Find an expression for P as a function of t.
b) In what year did the population reach 2.5 million?
Ans: First we have
P (t) = #population in 2000+t years = P (0)ekt
with P (0) = 2 · 106 and P (10) = 3 · 106 . It follows
3 · 106 = 2 · 106 e10k ⇒ ln(3/2) = 10k ⇒ k = ln
It follows
p
10
3/2
t
P (t) = 2 · 106 e 10 ln(3/2)
To answer (b) we need to solve
t
5
ln(5/4)
= e 10 ln(3/2) ⇒ t = 10
4
ln(3/2)
t
2.5 · 106 = 2 · 106 e 10 ln(3/2) ⇒
2. Suppose that the acceleration of a particle (in m/sec2 ) at time t (in sec) is given by the function
a(t) = 6t − 6
a) Find v(t) if v(0) = −9
t≥0
m/sec.
b) Find total displacement in [0, 10].
c) Find total distance in [0, 10].
Ans: First of all, we get
Z
a(t)dt = 3t2 − 6t + C ⇒ v(t) = 3t2 − 6t − 9
v(t) =
since v(0) = −9. It follows
Z
s(t) =
The total displacement is
Z
v(t)dt = t3 − 3t2 − 9t + C
10
v(t)dt = (t3 − 3t2 − 9t)|10
0 = 1000 − 300 − 90 = 610
0
The total distance is instead
10
Z
|v(t)|dt
0
In this regard, we notice that
v(t) = 0 ⇔ t =
6±
√
6
144
⇔ t = 3, −1
It follows that v(t) ≤ 0 if 0 ≤ t ≤ 3 and v(t) ≥ 0 if 10 ≥ t ≥ 3 and
Z
10
Z
|v(t)|dt = −
0
3
3
Z
v(t)dt +
0
10
v(t)dt =
3
= −(t − 3t2 − 9t)|30 + (t3 − 3t2 − 9t)|10
3 = 664
24
3. The perimeter of a rectangular copper sheet is 12 inches. The two opposite edges are welded together to
form a cylindrical pipe. What dimensions of the rectangle will result in a cylinder of maximum volume?
Ans: Denote by x, y the width and the length of the rectangular sheet
x + y = 12
The cylinder has radius 2πr = y and height h = x. It has volume
V =
1
(12 − y)y 2
4π
We have to maximize V . We get
1
3
dV
=
(24y − 3y 2 ) =
y(8 − y)
dy
4π
4π
The critical points are at y = 0 and y = 8. In particular, since dV /dy < 0 if y > 8 and dV /dy > 0 if y < 8,
the function attains its maximum at y = 8. We get
x = 4, y = 8, V =
64
π
4. Santa Claus needs to budget his time in order to be able to deliver presents to all the world’s children. In
order to deliver presents to Billy’s city he needs
T (p) =
3p3/2
p+4
sec
where p is the population of the city measured in hundreds of thousands of people. The city presently has
4 · 105 people in it (i.e. p = 4).
a) How many seconds does it now take for Santa to deliver presents to Billy’s city?
b) Billy’s city is presently growing at the rate of 8,000 (i.e. 0.08 hundred thousand) people per year.
How fast is the amount of time that Santa needs to deliver presents to Billy’s city increasing when
p = 4.5? Remember to include units in your answer. (Hint: use related rates.)
Ans: At the present time, T (4) = 3. To compute dT /dt at p = 4.5, we first notice that
8
dp
=
dt
100
Since
1/2
dT
1
p
3/2
=
(p + 4) − 3p
dp
(p + 4)2
2
we get
dT
8
=
dt
200
√
and
dT
8
|p=4.5 =
dt
200
where T (4.5) =
dT
dT dp
=
dt
dp dt
and
p − 2T (p)
(p + 4)
√
4.5 − 2T (4.5)
8.5
108
√ .
17 2
5. Water is poured into a tank shaped as an inverted cone of radius 15 m and of height 10 m. The water
flows at the rate of 2 m3 per minute when the water is 6 m deep. How fast is the water level rising at that
time?
25
Ans: Let us denote by r = r(t), h = h(t) the radius and the height of the cone of water at time t. We
have
r
15
π
3π 3
=
and
V = r2 h ⇒ V =
h
h
10
3
4
It follows that
dV
9π 2 dh
=
h
dt
4
dt
and at h = 6, we get dV /dt = 2 and therefore
2 = 81π
dh
dh
2
⇒
=
dt
dt
81π
6. The body of a murder victim was found at noon in a room that is kept at 20 degrees. From Newton?s law
of cooling, it is known that the temperature H of the body varies as H = A + Be−kt , where t is the time
elapsed since the body was found and where A, B and k are positive constants.
a) Find the value of A. (Hint: In the long run, the body will reach room temperature.)
b) When the body was found, its temperature was 35 degrees. One hour later, its temperature was 31
degrees. Use this information to find B and k.
c) Assuming that, at the time of the murder, the victim?s body had the normal body temperature of 37
degrees, when did the murder occur? You do not need to simplify your answer.
Ans: First of all we notice that
lim H = 20 ⇒ lim A + Be−kt = A ⇒ A = 20
t→∞
t→∞
We know that H(0) = 35, therefore 35 = 20 + B and B = 15. On the other hand, using H(1) = 31, we get
15
−k
31 = 20 + 15e ⇒ k = ln
11
then we obtain
H(t) = 20 + 15eln(11/15)t
Finally, if we want to find the time of the crime we need to solve
H(t) = 37 ⇔
17
ln(17/15)
= eln(11/15)t ⇔ t =
15
ln(11/15)
7. The population of a certain bacteria grows with time t as
p(t) =
1
1 + Ae−kt
where A and k are positive constants.
(a) Find limt→∞ p(t).
(b) Find the rate of population growth and evaluate it at t = 0.
Ans: Clearly,
lim p(t) = lim
t→+∞
Then we have
t→∞
1
=1
1 + Ae−kt
dp
kAe−kt
dp
kA
=
⇒
|t=0 =
dt
(1 + Ae−kt )2
dt
(1 + A)2
8. Suppose the number of bacteria in a culture at time t is given by N = c(25 + te−t/20 ) for some positive
constant c.
26
(a) Find the critical values of N.
(b) Find the largest and smallest number of bacteria in the culture during the time interval 0 ≤ t ≤ 100.
Ans: The critical values of N are given by
dN
t2 −t/20
t
−t/20
−t/20
= ce
−c e
= 0 ⇔ ce
1−
=0
dt
20
20
that is t = 20. Moreover, dN/dt > 0 if t < 20 and dN/dt < 0 if t > 20. Therefore there is a maximum at
t = 20, corresponding to the value N (20) = c(25 + 20e−1 ). The minimum number of bacteria is instead
min{N (0), N (100)} which is N (0) = c25.
27