Download Problem Solving Presentation 1

Beth Van Schoyck, Allen Dickerson, Kevin Privott
October 4, 2011
Math 280: Problem Solving
Presentation 1: Group D
Order of Presentation:
1.
Draw graphs of trig. functions (and leave on the board the entire presentation for
reference).
2.
Explain the definition of a limit.
Give an example of a limit.
3.
Explain the definition of a derivative and how the derivative gives the slope of the
tangent line.
Give an example of a derivative.
4.
Show the correlation (or relationship) between the slope of the graph (or derivative) of
sin(x) and the values on the graph of cos(x).
5.
Write on the board the limits of trigonometric functions.
6.
Write on the board the derivatives of the trigonometric functions.
7.
Prove that the limit of (sin x)/x is equal to 1.
8.
Prove that the derivative of sin(x) is equal to cos(x).
1. Graphs of trigonometric functions:
y = sin x
y = cos x
y = csc x
y = sec x
y = tan x
y = cot x
2. Definition of a Limit:
lim f(x) = L
x→ a
*** This statement is read as “the limit of f(x), as x approaches a,
is equal to L.”
•
The closer x gets to a (from the left or right), the closer f(x) gets to L.
•
One of the reasons we use limits is in case x = a is undefined. This way we can still get
sufficiently close to our value L.
lim f(x) = lim tan(x) = ∞
x→a
x→a
When a = п/2, the value f(x) is undefined, but if we take the limit as x approaches
a, then we can see that f(x) approaches infinity.
Example:
3. Definition of a Derivative:
•
The derivative of a function can be interpreted geometrically as the slope of the tangent
line to the graph of f(x) at the point (x, f(x)).
•
The ∆x can also be portrayed as a value, such as the variable h, when the derivative is
being used in proofs or problem solving.
4. Correlation (or relationship) between the slope of the tangent line of sin(x)
and the values of cos(x):
**The slope of any point on the graph of sin(x) is equal to the value of that same point on the
graph of cos(x). The slope (or derivative) of sin(x) is always equal to cos(x).
•
•
Using the graphs of y = sin(x) and y = cos(x):
At the point x = 0 we can see that the slope of the tangent line of sin(x) is equal to 1.
At this same point, cos(x) = 1. This illustrates that at x = 0, the tangent line of
sin(x) is equal to
cos(x) or:
d/dx (sin x) = cos x
We can see this pattern at x = п as well:
d/dx (sin п) = - 1
cos п = -1
d/dx (sin x) = cos x
5. The limits of trigonometric functions:
For values of c (where f(c) exists:
(1) lim sin (x) = sin (c)
x→c
(3) lim [sin(∆x)] / (∆x) = 1
∆x→0
(2) lim cos (x) = cos (c)
x→c
(4) lim [1 – cos(∆x)] / (∆x) = 0
∆x→0
6. The derivatives of trigonometric functions:
(1) (d / dx) sin(x) = cos(x)
(2) (d / dx) cos(x) = - sin(x)
(3) (d / dx) csc(x) = - csc(x) cot(x)
(4) (d / dx) sec(x) = sec(x) tan(x)
(5) (d / dx) tan(x) = sec 2 (x)
(6) (d / dx) cot(x) = - csc 2 (x)
7. Proof that the limit, as x approaches 0, of (sin x)/x equals 1.
Area of ∆ABC = (½) (1) (sin x) = ½ sin x
Area of Arc ABC = (x / 2л) (л) (12) = x / 2
Area of ∆ADC = (½) (1) (tan x) = ½ tan x
As seen in the model:
(1) < (2) < (3)
or
Area of ∆ABC < Area of Arc ABC < Area of ∆ADC
Therefore,
½ sin x
<
x/2
<
½ tan x
Taking the absolute value
(since sin and tan are negative in 4th quadrant):
½ │sin x│ < │x│/2 < ½ │tan x│
Multiplying equation by 2:
│sin x│ < │x│
Divide equation by │sin x│:
<
│tan x│
1 < │x│/ │sin x│
<
Take inverse of everything (which flips inequality signs): 1
1/ │cos x│
>
│sin x│/│x│
>
│cos x│
Absolute values are no longer needed since sin x / x and cos x are positive in 1st and 4th quadrant:
1 > (sin x) / x > cos x
Using Squeeze Theorem:
lim 1 = 1
lim cos x = 1
x→0
x→0
The limits of 1 and cos x both approach 1 as x approaches 0. Therefore, using the Squeeze
Theorem, the limit of (sin x)/x, as x approaches 0, must be 1.
8. Proof that derivative of sin(x) equals cos(x).
This proof uses L'Hopital's rule which states:
For functions f and g:
If the
lim f(x) = lim g(x) = 0 or ∞ and lim f '(x) / g '(x)
x→c
x→c
x→c
Then
exists,
lim f(x) / g(x) = lim f '(x) / g '(x)
x→c
x→c
(d / dx) [sin (x)] = lim [sin (x + ∆x) – sin (x)] / (∆x)
∆x→0
= lim [ sin(x) cos(∆x) + cos(x) sin(∆x) – sin(x)] / (∆x) ]
∆x→0
= lim [ cos(x) ( sin(x) / (∆x)) - sin(x) ( (1 – cos(∆x)) / (∆x) )]
∆x→0
= (cos(x)) [ lim (sin(x) / (∆x)) ] - ( sin(x)) [ lim ((1 – cos(∆x)) / (∆x)) ]
∆x→0
∆x→0
= ( cos(x)) (1) – ( sin(x)) (0) = cos (x)
Therefore,
(d / dx) ( sin(x)) = cos(x)