Download Math 111 – Calculus I

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Math 111 – Calculus I.
Week Number Nine Notes
Fall 2003
I.
Related Rates Problems
Example 9.1: (section 4.1, p. 269, problem 12): A baseball diamond is square with side
length 90 feet. A batter hits the ball and runs toward first base at a speed of 24 ft/s.
(a) At what rate is the distance from second base decreasing when he is
halfway to first base?
(b) At what rate is his distance to third base increasing at the same moment?
Example 9.2(Section 4.1, p. 270, problem 27): If two resistors with resistances R1 and R2
are connected in parallel, then the total resistance R, measure in ohms is given by the
following equation.
1
1
1


R R1 R 2
If R1 is increasing at a rate of 0.3 ohms/sec and R2 is increasing at a rate of 0.2 ohms/sec,
how fast is R changing when R1 = 80 ohms and R2 = 100 ohms, respectively.
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Example 9.3(section 4.2, p. 271, problem 33): A runner sprints around a circular track of
radius 100 m. at a constant speed of 7 m/s. The runner’s friend is standing at a distance
200 m. from the center of the track. How fast is the distance between the friends
changing when the distance between them is 200 meters?
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II.
Relative vs Absolute Extreme Values of Functions
First, we’ll start with some terminology.
Definition 9.4: A function f (with domain D) has an absolute maximum/minimum at the
real number c if f(c) > (respectively <) f(x) for all x in the domain D. The number f(c) is
called the maximum (respectively minimum) value of f on D. These values are also
called extreme values of f.
The following is a fundamental result involving absolute maximums and minimums of
continuous functions. It is called the Extreme Value Theorem. The formal statement of
the theorem follows.
Theorem 9.5(the Extreme Value Theorem): Assume f is a continuous function on a
closed interval [a,b]. Then, f attains an absolute maximum value f(c) for some c on [a,b]
and attains an absolute minimum value f(d) for some d on [a,b].
To find global maximums of continuous functions on [a,b], we will utilize the following
“search technique”.
Method for finding global extrema on [a,b]
(i)
Find the values of f at a and b respectively (the interval endpoints).
(ii)
Locate any local extreme values of f by identifying f’s critical points (i.e.
values in the domain of f such that the derivative of f is either 0 or DNE).
(iii)
The maximum of the values generated in (i) and (ii) is the absolute maximum,
the smallest is the absolute minimum of f on D.
Let us now define the term “local extreme value” and discover how to find them.
Definition 9.6: A function f has a local (or relative) maximum/minimum at a real number
c if f(c) > (respectively < ) f(x) for all x in some open interval about c. These values are
called local extreme values of f.
Theorem 9.7 (Fermat’s Theorem): If f has a local maximum or minimum at c and if f is
differentiable at c, then the derivative of f at c is 0.
Hence, we define a critical number of a function f as a number in the domain of f where
either (i) the derivative of f at c is 0 or (ii) the derivative of f at c DNE.
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To finish up this section, we’ll do two examples that will illustrate how to identify both
relative and absolute extreme values of a function f.
Example 9.8(section 4.2, p. 277, problem 27): Find the critical numbers of the following
function.
f(x) 
x
x 1
2
Identify which, if any, of these critical values are local extreme values (i.e. local
maximums or local minimums).
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Example 9.9(section 4.2, p. 476 of textbook): The Hubble Space Telescope was
deployed on April 24, 1990 by the space shuttle Discovery. A model for the velocity of
the shuttle during this mission from liftoff at t = 0 until the solid rocket boosters were
jettisoned at t = 126 seconds is given by the following equation.
v(t)  0.001302t 3  0.09029t 2  23.61t - 3.083 ft/sec
Estimate the absolute maximum and minimum values of the acceleration of the shuttle
between liftoff and booster jettison.
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Non Hand-In Homework Problems Associated with Week #9 Notes
Sections 4.1 and 4.2
Section 4.1: 1,2,3,6,11,15,16,17,21,25,26,34
Section 4.2: 3,5,7,10,15 – 43 odd, 49, 51, 53
Read Section 4.3-4.5 of the textbook