4.1 Related Rates
Ex.1. A small balloon is released at a point 150 ft. away from an observer, who is on level ground. If the balloon goes straight up, at a rate of 8 feet per second, how fast is the distance from the observer to the balloon increasing when the balloon is 50 ft. high
?
Figure.
Sol. We are given:
i):
dh
dt = 8;
ii): Distance between man and balloon is150 at t = 0. We are asked: dsdt :
) s2 = h2 + (150)2 ) 2sds
dt = 2hdh
dt ) ds
dt = h s
dh dt
= h
ph2 + (150)2 dh
dt:
Calculus 119 Now,
h = 50; dh
dt = 8;
i. e.
ds
dt = 50 50p
10 8 = 8 p10:
Ex.2 Water is pouring into a conical cistern at a rate of 8 cubic feet per minute. If the height of the cistern is 12 feet and the radius of its circular opening is 3 feet, how fast is the water level rising when the water is 4 feet deep?
Figure.
Sol. We are given:
i) Diameter of the opening of the cone: 6ft.;
ii) The height: 12 ft. i. e.
r
h = 3
12 = 1 4:
iii) pouring rate (8f t3= min) given. The volume is 13 r2h = V .
) V (h) = 1
3 r2h =
3(h
4)2h =
48h3:
Calculus 120 We want to know dhdt when h = 4.
* dV dt =
48 d
dt(h3) =
16h2dh
dt = 8f t3= min : ) dh
dt = 8 16 1
42 = 8 :
Ex.3 A plane ying north at 640 miles per hour over a certain town at noon, and a second plane going east at 600 miles per hour is directly over the same town 15 minutes later. If the planes are
ying at the same altitude, how fast will they be separating at 1:15PM?
Figure.
Sol. We are given:
i). y = 640t + 160;
ii). x = 600t, where t is in terms of hours.
We are asked:dsdt =? when t = 1 hr.?
Calculus 121
Ex.4 A woman standing on a cliff is watching a motor boat
Calculus 122 through a telescope as the boat approaches the shoreline directly below her. If the shoreline is 250 ft. above the water level and if the boat is approaching at 20 ft. per second, at what rate is the angle of telescope changing when the boat is 250 ft. from shore?
Figure.
Sol. We are given:
i) The height of the cliff =250f t;
ii).
dx
dt = 20 f t= sec We want to nd: ddt:
* tan = 250
x ; ) d
dt(tan ) = d
dt(250 x );
, sec2 d
dt = 250 x2
dx dt; , d
dt = cos2 250 x2
dx dt
= 250(cos
x )2dx dt:
Calculus 123 When x = 250,
d
dt = ( 250)
p1 2
2
250 250 ( 20) = 1 25:
Ex.5 Webster city monitors the height of water level in its cylindrical water tank with an automatic recording device. Water is constantly pumped into the tank at a rate of 2400 cubic feet per hour. During a certain 12 hr. period (beginning at midnight) , the water level rose and fell. If the radius of the tank is 20 ft., at what rate was water being used at 7:00AM ?
Figure.
Sol. We are given:
i) Water volume V = (20)2h;
ii) t (time) is in the unit of hour.
iii)dhdt is given by the graph (record,) d
dth(t = 7) :
= 3:
Calculus 124
* V = 400 h; dV
dt = 400 dh dt: ) dV
dt jt=7 :
= 400 ( 3) :
= 1200 :
Calculus 125 4.2 Differentials and Approximations
Def. Let y = f (x) be differentiable at x and suppose that dx, the differential of the independent variables x, denotes an arbitrary small increment of x. The corresponding differential dy of the dependent variable y is de ned by dy = f0(x)dx:
Ex.1 Find dy if
(a) y = x3; (b) y = p
x2 + 3x;
(c) y = sin (x4 3x2 + 11):
Sol. (a)
dy = 3x2dx:
(b)
dy = 2x + 3 2p
x2 + 3xdx:
(c)
dy = (4x3 6x) cos (x4 3x2 + 11)dx:
Note: (i)
f0(x) = lim
w!x
f (w) f (x)
w x
Set
y = f (w) f (x); x = w x;
Calculus 126
Hence we may interpret the derivative as the quotient of 2 differentials.
This is the formula of ”linear approximation.”
Ex.2 Please approximate p
4:6 and p 8:2 Sol. Consider the function y = p
x: Then:
Calculus 127
ii) p
8:2 = f (x0 + x); x0 = 9; x = 0:8;
) f(x0 + x) ' f(x0) + f0(x0) x
= p
9 + 1 2p
9 ( 0:8) :
= 2:867 Def. (Big-O and Little-o)
i) f (x) = O(g(x)) iff.
xlim!c
f (x)
g(x) = L;
where c 2 R [ f+1g, and jLj is a nite number.
ii) f (x) = o(g(x)) iff.
xlim!c
f (x)
g(x) = 0;
where c 2 R [ f 1g:
Corollary. We know that f (x + x) ' f(x) + f0(x) x, when x is small. Set
"( x) = f (x + x) f (x) f0(x) x:
Calculus 128 Then "( x) = o( x), since
limx!0
"( x)
x = lim
x!0[f (x + x) f (x)
x f0(x)]
= lim
x!0
f (x + x) f (x)
x f0(x) = 0:
Therefore, we may always write
f (x + x) = f (x) + f0 (x) x + o ( x) :
Note that the result is consistent with Taylor expansion formula.
Ex.3 Use differentials to approximate the increment in the area of a soap bubble when its radius increases from 3 inches to 3.025 inches.
Figure.
For a sphere with radius r, the surface of the sphere is A (r) = 4 r2;
A = 4 [(r + r)2 r2]
= 4 [(r2 + 2r r + r2) r2]
= 4 r(2r + r) , where
Calculus 129
dA = 8 rdr; (Compare with A0 (r) r = 8 r r:) A = 8 r r + 4 ( r)2 = A0 (r) r + 4 ( r)2: Note that 4 ( r)2 = o ( r) ; since
limr!0
4 ( r)2
r = 0:
For small enough r, we have A ' 8 r r: Since r is very small,
A ' 8 r r = 8 3 (0:025) = 0:6 :
Summary: The difference between f (x + x) and f (x) can be approximated by the value of f0 (x) x; and is consistent with the form dy = f0 (x) dx: Of course, when x ! 0; i. e.
j xj becomes dx; the 2 quantities are the same. Hence we may consider the quantity dy = f0 (x) dx as the ”difference” of the in nitesimal case.
Calculus 130 4.3 Maxima and Minima
Def. Let S be the domain of f, containing c. We say that:
i). f (c) is the ”maximum value of f” on S if f (c) > f(x);
8x 2 S.
ii). f (c) is the ”minimum value of f” on S if f (c) 6 f(x);
8x 2 S.
iii). f (c) is an ”extreme value of f” on S if f (c) is either the maximum value or the minimum value.
Theorem: (Existence theorem for extreme value)
If f is continuous on a closed interval [a; b], then f attains both a maximum value and a minimum value.
Def. For an open interval (a; b), we say that a and b are the
”endpoints” of (a; b).
Def. If c is a point s.t. f0(c) = 0, then c is called a ”stationary point.”
Def. A point c is said to be an ”interior point” of an interval I if c is not an endpoint of I.
Def. If c is an interior point of I s.t. f0(c) does not exist, then c is said to be a ”singular point.”
Def. If c is a point s.t. c is an endpoint, a stationary point or a singular point, then c is called a ”critical point.”
Theorem: (critical point thm.) Let f be de ned on an interval I containing the point c. If f (c) is an extreme value, then c must
Calculus 131 be a critical point; that is, either c is:
(i) an endpoint of I;
(ii) a stationary point of f; or (iii) a singular point of f.
pf. of (ii): Suppose that f (c) is the maximum, then 8x 2 I;
f (x) f (c) 6 0 for
x > c; f (x) f (c)
x c 6 0 (1):
Also for
x < c; f (x) f (c)
x c > 0 (2):
If f0(c) exists, from (1), we have
xlim!c
f (x) f (c)
x c 6 0 (3):
Also from (2),
xlim!c
f (x) f (c)
x c > 0 (4):
Remark: The procedure of nding the extremum of a continuous function on a closed interval I is as the following:
1) Find the critical points,
2) Evaluate f at each of these critical points. The largest of these values is the maximum value; the smallest is the minimum value.
Calculus 132 Ex.1 Find the critical points of f (x) = 2x3 + 3x2 on [ 12; 2]:
Sol.
f0(x) = 6x2 + 6x = 6x( x + 1):
If
f0(x) = 0; x = 1 _ 0:
Then f has critical points: 12; 1; 0 & 2. Ex.2 Find the extremum of
f (x) = 2x3 + 3x2 on [ 12; 2]:
Sol. From previous example, we evaluate f at the following points:
f ( 1
2) = 2 ( 1
8) + 3 1
4 = 1;
f (0) = 0; f (1) = 1; f (2) = 4:
Hence the maximum is 1, minimum is 4.
Ex.3 F (x) = x23. Find its extremum on [ 1; 2]. Sol.
* F0(x) = 2
3x 13; ) F has critical points 1, 0, 2.
Calculus 133
F ( 1) = 1; F (0) = 0; F (2) = p3 4:
Hence the maximum: p3
4; minimum: 0:
Ex.4 Given f (x) = x1; 8x 2 (0; 1] : What are the maximum and minimum values?
Sol. Since lim
x!0+ f (x) = 1; we cannot pinpoint the
happening location of the maximum value (c) as well as its exact value (f (c)) inside of the considered domain. Therefore there is no maximum value. The minimum value, however, can be obtained as 1:
Ex.5 Again, given f (x) = 1x; 8x 2 (0; 1): What are the maximum and minimum values?
Sol. In this case, both maximum and minimum values do not exist.
Ex.6 A rectangle box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical
squares from the four corners and turning up the sides. Find the dimensions of the box of maximum volume. What is the volume?
Sol. Figure.
Calculus 134 The volume is
(24 2x)(9 2x)x;
for x > 0; x < 92:
) V (x) = 216x 66x2 + 4x3; and consider0 6 x 6 92 = 4:5: Then
dV
dx = 216 132x + 12x2 = 12(9 x)(2 x) = 0;
) x = 9 _ 2:
i. e. The critical points for V (x) on [0; 92] is 0; 2 & 92: V (0) = 0; V (2) = 200; V (9
2) = 0 ) Maximum volume 200 at x = 2.
Ex.7 Farmer Brown has 100 meters of wire fence with which he plans to build two identical adjacent pens, as shown in Figure.
below. What are the dimensions of the total enclosure for which its area is a maximum?
Figure.
Sol. 3x + 2y = 100 or y = 50 32x: The enclosed area is
Calculus 135
A(x) = xy = 50x 3
2x2; 0 6 x 6 100 3 : ) A0(x) = 50 3x = 0 iff. x = 50
3 : ) A(x) has 3 critical points 0; 503 ; 1003 on [0; 1003 ];
A(0) = 0; A(50
3 ) = 4162
3; A(100
3 ) = 0:
Hence the maximum is 41623 when x = 503 ( and y = 25.)
Ex.8 The operating cost for a certain truck is estimated to be (30 + v2) cents per mile when it is driven at a speed of v miles per hour. The driver is paid $14 per hour. What speed will minimize the cost of making a delivery to city k miles away? Assume that the law restricts the speed to 40 6 v 6 60:
Sol. The total operating cost
c = driver cost + operating cost
= kv 1400 + k (30 + v2);
dc
dv = 1400kv 2 + k 2: If
dc
dv = 0 ) 1400
v2 = 1 2; i. e. v = 20p
7 53; therefore, we nd that
c(40) = 85k; c(53) = 82:9k; c(60) = 83:3k:
Calculus 136 Hence v 53 miles per hour.
?Ex 9. (Snell's Law) Figure.
v1
v2 = sin 1 sin 2:
Since light always ”chooses” to travel in a fastest way, we can obtain the above result. i. e. from point A to point B. light
”chooses” to pass point C so that the time it uses to travel from A to B is the shortest, i. e.
T (x) =
pL2 + x2 v1 +
pL2 + (a x)2 v2
reaches minimum, x 2 [0; a]. Note that
sin 1 = x
pL2 + x2; and
sin 2 = a x
pL2 + (a x)2: #
Calculus 137 4.4 Monotonicity, concavity, and the construction of the
graph of a function
Def. Let f be de ned on an interval I ( open, closed, or neither.) We say that:
i). f is ”increasing” on I if 8x1; x2 2 I; x1 < x2 ) f (x1) < f (x2):
ii). f is ”decreasing” on I if 8x1; x2 2 I; x1 < x2 ) f (x1) > f (x2):
iii). f is ”strictly monotonic” if it is either increasing or decreasing on I.
Theorem: (Monotonicity Theorem)
Let f be continuous on I (i. e. f 2 C(I)) and differentiable at all interior points on I:
i). If f0(x) > 0 for all x interior to I, then f is increasing on I.
ii). If f0(x) < 0 for all x interior to I, then f is decreasing on I.
Figure.
Def. f is continuous on an open interval I. We say that f (as well as its graph) is ”concave up” (or upward) on I if f0
Calculus 138 is increasing on I and we say that f is ”concave down” ( or downward) on I if f0 is decreasing on I .
Figure.
Theorem (Concavity Theorem):
Let f be twice differentiable on the open interval I. i). If f00(x) > 0; 8x 2 I, then f is concave up.
ii). If f00(x) < 0; 8x 2 I, then f is concave down.
Note: For the proof of the theorem, we may use ”mean value theorem” or simply consider this:
f0(x) f0(c) > 0 if x > c; i. e. f0 : increasing on (c; x) ) f0(x) fx c0(c) > 0
) lim
x!c
f0(x) f0(c)
x c > 0: i. e. f00(c) > 0:
Similarly, if f0 is decreasing, we may conclude thatf00(c) <
0:
Ex.1 f (x) = 2x3 3x2 12x + 7. Find where f is increasing and where f is decreasing.
Sol. f0(x) = 6x2 6x 12 = 6(x2 x 2) = 6(x+1)(x 2);
f0(x) > 0 ) x < 1 _ x > 2;
f0(x) < 0 ) x > 1 ^ x < 2( 1 < x < 2):
Calculus 139 i. e. f increasing on x < 1 _ x > 2; and f decreasing on 1 < x < 2:
Ex.2 f (x) = 13x3 x2 3x + 4. Find where f is concave up or down.
Sol. f0(x) = x2 2x 3; f00(x) = 2x 2 ) f00(x) = 2(x 1) > 0 ) x > 1
2(x 1) < 0 ) x < 1
i. e. when x > 1, f is concave up, and when x < 1, f is concave down.
Ex.3 g(x) = 1+xx 2. Find where g is increasing, decreasing, concave up or down.
Sol. g(x) = 1+xx 2;
) g0(x) = (x(1 + x2) 1)0
= (1 + x2) 1 x(1 + x2) 2(2x)
= 1 + x2 2x2
(1 + x2)2 = 1 x2 (1 + x2)2:
Calculus 140
) g00(x) = [(1 x2)(1 + x2) 2]0
= ( 2x)(1 + x2) 2 2(1 x2)(1 + x2) 3(2x)
= (1 + x2) 3( 2x 2x3 4x + 4x3)
= 2x(x2 3) (1 + x2)3 :
) g0(x) > 0 ) 1 < x < 1;
g0(x) < 0 ) x > 1 _ x < 1;
g00(x) > 0 ) x > p
3 _ p
3 < x < 0;
g00(x) < 0 ) x < p
3 _ 0 < x < p 3:
Hence g: increasing:
1 < x < 1;
decreasing:
x > 1 _ x < 1;
concave up:
x > p
3 _ p
3 < x < 0;
concave down:
x < p
3 _ 0 < x < p 3:
Def. Let f be continuous at c. We call (c; f (c)) an ”in ection
Calculus 141 point” of (the graph of) f is concave up on one side & concave down on the other side.
Ex.4 Find all in ection points of the graph of f (x) = x63 2x:
Sol. f0(x) = x22 2; f00(x) = x;
) f00(x) > 0 ) x > 0; f00(x) < 0 ) x < 0:
Then y = 0
x = 0 is the in ection point.
Ex.5 Find all the points of in ection for F (x) = x13 + 2:
Sol. F0(x) = 13x 23; F00(x) = 29x 53: Then F00(x) > 0 ) x 53 < 0 ) x < 0;
F00(x) < 0 ) x 53 > 0 ) x > 0:
Since F (x) is continuous at x = 0, hence (0; 2) is the point of in ection.
Figure.
Calculus 142 4.5 Local Maxima and Minima
Def. Let S, the domain of f, contain the point c. We say that:
(i). f (c) is a ”local maximum” (maximum value) of f if there is an interval (a; b) containing c s.t. f (c) is the maximum value of f on (a; b) \ S.(local maximum are also referred as ”relative maximum.”)
(ii). f (c) is called the ”global (or absolute) maximum” if f (c) is the maximum on S.
(iii). f (c) is a ”local (or relative) minimum (or minimum value)” of f if there is an interval (a; b) containing c s.t. f (c) is the minimum of f on (a; b) \ S:
(iv). f (c) is called the ”global(or absolute) minimum (or minimum value)” of f if f (c) is the minimum of f on S.
(v). f (c) is a ”local (or relative) extremum (or extremum value)” of f is f (c) is either a local maximum or local minimum.
(vi). f (c) is a ”global (or absolute) extremum (or extremum value)” if f (c) is a global maximum or a global minimum.
Figure.
Def. is said to be the ”supremum (sup)” or ”least upper
Calculus 143 bound (l:u:b:)” of a function f on a domain I if
(i). 8x 2 I; f (x) ; and
(ii). 8 > 0; 9x0 2 I 3 f (x0) > : We usually write = sup
x2I
f (x) = l:u:b:
x2I f (x):
Def. is said to be the ”infremum (inf)” or ”greatest lower bound (g:l:b:)” of a function f on a domain I if
(i). 8x 2 I; f (x) ; and
(ii). 8 > 0; 9x0 2 I 3 f(x0) < + : We usually write = inf
x2If (x) = g:l:b:
x2I
f (x):
Ex. Given f (x) = x; x 2 (0; 1): Then there is no maximum nor minimum values in (0; 1): However, the infremum 0 as well as supremum 1 both exist.
Ex. Given f (x) = x; x 2 [0; 1] : The maximum value of f (x) in [0; 1] is 1; and the supremum value is also 1: The minimum value and the infremum value of f (x) are both 0:
Theorem (First derivative test for local extrema):
Given f 2 C (I) and f : differentiable on interior points of I\ neighborhood of cn fcg : Then we have
i). f (c) : local maximum if
f0 c > 0 and f0 c+ < 0;
Calculus 144 ii). f (c) : local minimum if
f0 c < 0 and f0 c+ > 0;
iii). c : left endpoint of I;
a). f (c) : local maximum if f0 (c+) < 0;
b). f (c) : local minimum if f0 (c+) > 0;
iv). c : right endpoint of I;
a). f (c) : local maximum if f0 (c ) > 0;
b). f (c) : local minimum if f0 (c ) < 0:
Theorem (Second derivative test for local extrema):
i). If f0 (c) = 0; and f00 (c) > 0; then f (c) is a local minimum.
ii). If f0 (c) = 0; and f00 (c) < 0; then f (c) is a local maximum.
Ex. Find the local extrema of y = x 4x2 : Sol. Since y0 = x(x 8)
(x 4)2; we nd i). y0 > 0 if x < 0 or x > 8; and
ii). y0 < 0 if 0 < x < 8: Therefore, it is obvious that y (0) is the local maximum, and y (8) is the local minimum by the 1st derivative test.
Ex. What are the local extrema of h (x) = x23? Sol. h0 (x) = 3p23
x: Then we nd that
Calculus 145 i). h0 (x) > 0 if x > 0; and
ii). h0 (x) < 0 if x < 0: So h(0) = 0 is the local minimum by the 1st derivative test. (Note that h0 (0) doesn't exist.)
Figure.
Ex. Determine the local extrema of the function f (x) = 5x3 2x2 3x + 9:
Sol.
f0(x) = 15x2 4x 3 = (3x + 1) (5x 3);
where f0(x) = 0 ) x = 13 or x = 35: Also f00(x) = 30x 4;
and we nd that f00( 13) = 14 < 0; f00(35) = 14 > 0: i. e.
f 13 is a local maximum, and f 35 is a local minimum by the 2nd derivative test.
Calculus 146 4.6 Drawing the graph of a function
There are 4 major components of drawing graphs of functions:
i). Monotonicity, ii). Concavity,
iii). The coordinates of key points, and iv). Asymptotes.
Remark: There are also 3 types of asymptotes:
i). Horizontal, ii). vertical, and iii). slant.
Def. We say that y = ax + b is the ”slant asymptote” of a function f if
x!+1lim [f (x) (ax + b)] = 0:
Remark:
i). If a function with the form h(x)g(x); and deg g (x) = deg h (x) + 1; then there is a slant asymptote.
ii). More simply speaking, if a function f (x) with deg f (x) = 1; then there is a slant asymptote.
iii). a can be determined as the quotient of the leading coef cient of g(x) over the leading coef cient of h(x):
iv). b can be determined through estimate, or simply compute the limit
x!+1lim [f (x) ax] = b:
Calculus 147 Ex. f (x) = x 4x2 : Draw the graph.
Sol. f0 (x) = x(x 8)
(x 4)2; and f00(x) = 32
(x 4)3: Therefore, f0 (x) > 0 ) x > 8 or x < 0;
f0 (x) < 0 ) 0 < x < 8; and f00 (x) > 0 ) x > 4;
f00 (x) < 0 ) x < 4: Also we nd that
xlim!+
x2
x 4 (x + 4) = 0;
i. e. y = x + 4 is a slant asymptote, and x = 4 is a vertical asymptote.
Figure.
Ex. Draw the graph of the function h (x) = x2x+13 : Sol. f0 (x) = x
2(x2+3)
(x2+1)2 ; and f00 (x) = 2x(3 x2)
(x2+1)3 : Therefore, f0 (x) > 0; 8x; and
f00 (x) > 0 ) x < p
3 or 0 < x < p 3;
f00 (x) < 0 ) p
3 < x < 0 or x > p
3: Also we nd that
xlim!+
x3
x2 + 1 x = 0;
Calculus 148 i. e. y = x is a slant asymptote.
Figure.
Calculus 149 4.7 Mean value theorem
Theorem (Rolle's theorem):
f 2 C [a; b] ; f : differentiable on (a; b) ; and f (a) = f (b) : Then 9c 2 (a; b) 3 f0 (c) = 0:
pf. Since f 2 C [a; b] ; therefore maximum value and minimum value of f both exist in [a; b] : Then consider all of the critical points of f :
i). endpoints are a and b:
ii). there will be no singular point (* f is differentiable on (a; b) :)
iii). there may be some c 2 (a; b) such that f0 (c) = 0 (i. e.
stationary point.)
Assume that there is no stationary point. Then there are some possibilities of it:
a). f (a) = f (b) is the maximum and the minimum. i. e.
f (x) is a constant, therefore f0 (c) = 0; 8c 2 [a; b] :
b). f (a) = f (b) is the maximum but not the minimum.
Then there must be a c 2 (a; b) such that f (c) is the minimum value in [a; b] : Then c must be a stationary point (i. e. f0 (c) = 0.)
c). f (a) = f (b) is the minimum but not the maximum.
Similarly as in case b)., 9c 2 (a; b) such that f (c) is the maximum value and therefore, f0 (c) = 0.
Theorem (Mean value theorem):
Calculus 150 f 2 C [a; b] and f : differentiable on (a; b) : Then
9c 2 (a; b) 3 f0 (c) = f (b) f (a)
b a :
pf. Set
h(x) = [f (x) f (a)] f (b) f (a)
b a (x a) ;
then we nd that h (x) 2 C [a; b] ; h (x) is differentiable on (a; b) ; and h (a) = h (b) = 0: Therefore, by applying Rolle's theorem, we nd that 9c 2 (a; b) such that h0(c) = 0; i. e.
f0 (c) = f (b) f (a)
b a for some c 2 (a; b) : Theorem (Cauchy's mean value theorem):
f and g 2 C [a; b], f and g : both differentiable on (a; b) ; g (b) 6= g (a) and g0(x) 6= 0 on (a; b) : Then
9c 2 (a; b) 3 f0 (c)
g0(c) = f (b) f (a) g (b) g (a): pf. Set
h (x) = [f (x) f (a)] f (b) f (a)
g (b) g (a) [g (x) g (a)] ; then again, we nd that h (x) 2 C [a; b] ; h (x) is differentiable on (a; b) ; and h (a) = h (b) = 0: Therefore, by applying Rolle's
Calculus 151 theorem, we nd that 9c 2 (a; b) such that h0(c) = 0; i. e.
f0 (c)
g0 (c) = f (b) f (a)
g (b) g (a) for some c 2 (a; b) :
Ex. 1 Given f 2 C [a; b] and f : differentiable on (a; b) ; prove that f0(x) > 0; 8x 2 (a; b) implies f is increasing on [a; b]
by using Mean Value Theorem.
pf. 8x1; x2 2 [a; b] ; assume that x1 < x2: Then 9c 2 (x1; x2) s. t.
f0(c) = f (x2) f (x1)
x2 x1 > 0:
We know that x1 < x2; therefore f (x1) < f (x2) ; i. e. f is increasing–since x1 and x2 are arbitrary.
Ex. 2 (Corollary) If f0 (x) = 0; 8x 2 I: Then f (x) is a constant function.
pf. 8x1; x2 2 I; x1 6= x2: Then obviously f 2 C [x1; x2] and f : differentiable on (x1; x2) ; therefore we have
f0 (c) = f (x2) f (x1)
x2 x1 = 0 for some c in (x1; x2) : Since x1 6= x2; we can only nd that f (x2) = f (x1); for all arbitrary x1; x2 2 I: i. e. f is a constant.
Ex. 3 (Corollary) If f0 = g0 on certain interval I; then we must have f (x) = g (x) + c; where c is a certain constant.
pf. Assume that h (x) = f (x) g (x) ; then it is clear that
Calculus 152 h is continuous on I and differentiable on I: Then we may apply MVT as well as Ex. 2, and nd that
h (x) = c for some constant c:
Hence the corollary is proved.
Remark. f0 = g0 , f (x) = g (x) + c; where c is a certain constant, 8x 2 I:
Calculus 153 4.8 Economic Applications
Def. If a company sells x units of speci c product in a xed period of time, we de ne p(x) as the ”price” that it charges for each unit.
Def. ”Total revenue” of a company's sale over a speci c product is de ned as R(x) = x p(x), where x and p(x) are de ned as the above.
Def. C(x) is de ned as the ”total cost” in order to produce and market x units of a speci c product. Usually C(x) contains 2 parts of cost: Fixed cost (of ce utilities, real estate taxes, and so on) and variable cost (depends directly on the no. of units produced—x:)
Def. ”Total pro t” is de ned as R(x) C(x) P (x). Note: In real world, most problem in social sciences are properly viewed as discrete in nature. But calculus mainly consider the continuous cases.
Figure.
Remark: Mathematics involving studying discrete models and problems are ”discrete mathematics” and algebra. Yet calculus remains the major tool in studying social science (and other sciences) because of its power.
Calculus 154 Ex.1 C(x) = 10; 000 + 50x where 10,000 is the ” xed cost”
and 50x is the ”variable cost”.
Ex.2 C(x) = 10; 000 + 45x + 100p
x, where 10,000 is again the xed cost, and 45x + 100p
x is the variable cost.
Note: In this example, 45x+100x px = 45 + 100p
x is called the
”average variable cost”. per item.
Remark: The determination of price function or cost function is a nontrivial task.
Def. ”Marginal cost at x” is de ned as the cost of producing one additional unit, i. e. C(x + 1) C(x).
Remark: In mathematical model, since C(x + 1) C(x) = C0(x) 1 + Q(1) , if x is large enough, we may understand that C(x + 1) C(x) C0(x): (Note that Q(1) is the error term.) Hence we will use C0(x) as ”marginal cost at x ”. #
Def. ”Marginal price” is de ned as dxdp, and ”marginal revenue” is de ned as dRdx, and”marginal pro t” is de ned as dPdx. (The reason is the same as the remark above).
Ex. C(x) = 8300 + 3:25X + 40 p
3x dollars. Find the average cost per unit and the marginal cost and evaluate them when x = 103.
Sol. i) C(x)x = 8;300x + 3:25 + p40
3x2
ii) C0(x) = 3:25 + 403 p1
3x2
iii) C(x)x jx=103 = 8:3 + 3:25 + 0:4 = 11:95;
Calculus 155 C0(1; 000) = 3:25 + 13:33 0:01 = 3:37:
(Note: C(x + 1) C(x) = 3:25 + 40(p
3x + 1 p
3x); C(1001) C(1000) :
= 3:25 + 0:13 = 3:37.)
Ex. A company estimates that it can sell 1,000 units per week if it sets the unit price at $3:00, but that its weekly sales will rise by 100 unit for each 10 cent decrease in price. If x is the number of units sold each week (x 1; 000), nd
a) the price function p(x);
b) the # of units and corresponding price that will maximize weekly revenue;
c) the maximum weekly revenue.
Sol. a) * x = 1; 000 + 3 p(x)0:1 100
) p(x) = 3 x 1;000100 0:1 = 4 0:001x.
b)R(x) = xp(x) = 4x 0:001x2; R0(x)(= dRdx) = 4 0:002x;
R0(x) = 0 ) x = 2; 000.
By the idea of 1st derivative test, R0(x) > 0 ) x 2 [1; 000; 2; 000), R0(x) < 0 ) x 2 (2; 000; 1),
) R(2; 000) is the maximum, (or, R00(x) = 0:002 , R00(2; 000) < 0: By 2nd derivative test, R reaches maximum when x = 2; 000.)
c) R(2; 000) = $ 4; 000:#
Calculus 156 4.9 L'Hôpital Rule and its proof
Theorem: (L'Hôpital Rule)
f, g: differentiable on (a; b) and g0(x) 6= 0 on (a; b), where 1 a < b 1. Either
(1): lim
x!a+f (x) = lim
x!a+g(x) = 0;
or
(2): lim
x!a+f (x) = 1; and lim
x!a+g(x) = 1:
) lim
x!a+
f (x)
g(x) = L, if lim
x!a+
f0(x)
g0(x) = L pf. Under condition (1):
Case a)
Consider a : nite, 1 L 1.
De ne f (a) = 0 = g(a). Then f and g become continuous on [a; b). 8x > a; and then by Cauchy's MVT, we nd .
f (x) f (a)
g(x) g(a) = f0(c) g0(c); where c 2 (a; x). i. e.
f (x)
g(x) = f0(c) g0(c);
Calculus 157 nite, the proof are the same as the above, except for some minor changes in the application of de nitions.
Under condition (2):
Calculus 158 Case a)
a = 1, L: nite
* lim
x!a+
f0(x)
g0(x) = L;
) 8" > 0; 9x" : a < x < x" ) jf0(x)
g0(x) Lj < ":
Choose y so that a < y < x" and keep y xed. De ne r(x) = f (x) f (y)
g(x) g(y); a < x < y:
) 9 2 (x; y) 3 f (x) f (y)
g(x) g(y) = f0( ) g0( ); i. e.
r(x) = f0( )
g0( ); and jr(x) Lj < ":
Since
f (x) = f (y) + r(x) [g(x) g(y)] ; we nd
f (x)
g(x) = f (y)
g(x) + r(x) 1 g(y) g(x) ) f (x)
g(x) L = f (y)
g(x) + r(x) L r(x) g(y) g(x):
Calculus 159
Calculus 160 , 8A > 0; 9xA > 0 : a < x < xA
) f0(x)
g0(x) > A:
Choose y so that a < y < xA and keep y xed. De ne r(x) = f (x) f (y)
g(x) g(y); a < x < y < xA.(By MVT,) 9 2 (x; y) 3
f (x) f (y)
g(x) g(y) = f0( ) g0( ) i: e:
r(x) = f0( ) g0( ); and r(x) > A. Since
f (x) = f (y) + r(x) (g(x) g(y)) ; we obtain that
f (x)
g(x) = f (y)
g(x) + r(x) 1 g(y) g(x) :
* lim
x!a+g(x) = 1;
) lim
x!a+
f (y)
g(x) = 0; lim
x!a+
g(y)
g(x) = 0:
Calculus 161 or, 9XA0 < y : a < x < x0A;
) jf (y)
g(x)j < A
4 ; jg(y)
g(x)j < 1 2 ) f (x)
g(x) > A
4 + A 1
2 = A 4 ) lim
x!a+
f (x)
g(x) = 1: # Note: In case
xlim!a+
f0(x)
g0(x) = L; L = 1:
Then consider h(x) = g(x), we nd
xlim!a+
f0(x)
h0(x) = 1 ) lim
x!a+
f (x)
h(x) = 1 ) lim
x!a+
f (x)
g(x) = 1: #
Calculus 162 4.10 The indeterminate forms and variations of L'Hôpital rule ) By using L'Hôpital Rule,
xlim!0
) By using L'Hôpital Rule,
xlim!0
Calculus 163 Ex.3
xlim!0
sin x cos x =?
Sol. In fact,
xlim!0
sin x
cos x = 0:
However, if we accidentally use L'Hôpital Rule, we nd
xlim!0
cos x
sin x ) doesn't exist!
The result is wrong, of course.
Ex.4
xlim!0
sin x x x3 =?
Sol.(This is an important example).
xlim!0
Calculus 164
(Note: in this case, we may only use L'Hôpital Rule once.) Ex.6
Calculus 165
(This problem has nothing to do with L'Hôpital Rule.
However, to combined both undetermined factor is the proper method to do it.)
Note: (Review of a theorem) f: continuous at L, and
xlim!ag(x) = L exists.
) f(lim
x!ag(x)) = lim
x!af g(x): #
Calculus 166 Ex.11
xlim!0(x + 1)cot(x) =?
Sol.
* (x + 1)cot(x) = eln(x+1)cot(x) = e(cot(x))ln(x+1); and
* lim
x!0(cot(x))ln(x + 1)
= lim
x!0
ln(x + 1) tan(x)
L0:
= lim
x!0
1 x+1
sec2(x) = 1:
Since ex is a continuous function we nd (By theorem,)
ex!0lim(cot(x))ln(x+1)
= lim
x!0e(cot(x))ln(x+1)
= lim
x!0(x + 1)cot(x) = e1 = e: # Ex.12
xlim!0+(sin x)cot(x) =?
Sol. In fact, this problem doesn't require L'Hôpital Rule to solve it. However, we may see something from its complicated form.
(sin x)cot x = e(cot x)ln(sin x):
Calculus 167
Sol. Similarly as before, 1 + 1
x
x
= eln(1+1x)x = ex ln(1+x1):
Let's look at the limit
xlim!1x ln 1 + 1
which exists. By the theorem mentioned above, we nd that
xlim!1 1 + 1 x
x
= ex!1lim x ln(1+x1)
= e1 = e:
Calculus 168
5 Integrals
5.1 Antiderivative (inde nite integrals)
Def. We call F an ”antiderivative” of f on the interval I if DxF = f (x) on I, i. e. if
F0(x) = f (x); 8x 2 I.
(note: if x is an endpoint of I, F0(x) needs to be a one-sided derivative.)
Ex.1 Find an antiderivative of the function f (x) = 4x3 on ( 1; 1).
Sol. Let F (x) = x4, then F0(x) = 4x3. i. e. x4 is an antiderivative of 4x3: #
Note: Let F and G are antiderivatives of f on I. Then F (x) = G(x) + c; 8x 2 I: #
Also please notice that c is an arbitrary constant.
Def. Consider an antiderivative F of a function f on I. We say that F (x) + c is a ”general” antiderivative of f (x) on I; where c is an arbitrary constant. Note that the general antiderivative is usually called the ”inde nite integral”.
Ex.2 Find the general antiderivative of f (x) = x2 on ( 1; 1).
Sol. * F(x) = 13x3 ) F0(x) = x2.
) The general antiderivative of f (x) is 13x3 + c, where c is an