Post on 18-Jan-2018
description
Working toward Rigor versus Bare-bones justification in Calculus
Todd Ericson
Background Info
► Fort Bend Clements HS ► 25 years at CHS after leaving University of Michigan► 4 years BC Calculus / Multivariable Calculus► 2014 School Statistics: 2650 Total Students 45 Multivariable Calculus Students 110 BC Calculus students 200 AB Calculus students ► 2013: 28 National Merit Finalists► BC Calculus AP Scores from 2011 – 2014 5’s : 316 4’s : 44 3’s : 11 2’s : 2 1’s : 0
Coached the 5A Texas State Championship for Men’s Soccer 2014.
Common Topics involving Justification
► Topics and Outline of Justifications:► Continuity at a point► Differentiability at a point► IVT and MVT (Applied to data sets)► Extrema (Both Relative and Absolute) and Critical values / 1st and 2nd Der. Tests► Concavity/Increasing decreasing Graph behavior including Points of Inflection► Justification of over or under estimates (First for Linear Approx, then Riemann Sums)► Behavior of particle motion (At rest , motion: up,down, left, right)► Error of an alternating Series► Lagrange Error for a Series► Convergence of a series► Justification of L’Hopital’s Rule
Both AB and BC topics are listed below.
References for problems
► Justification WS is 3 page document handed out as you entered.
► All documents will be uploaded to my wikispaces account. Feel free to use or edit as necessary.
► http://rangercalculus.wikispaces.com/
► As we work through problems, I will address certain points and thoughts given in document 2.
► Email for questions: todd.ericson@fortbendisd.com
See attached handout for justification outlines
Sample Problem 1
Continuity
Problem 1
1) Given this piecewise function, justify that the function is continuous at x = 2
2 , 2( )
4 4 , 2x x
f xx x
Continuity
Problem 1 Initial Solution (How can we create a more rigorous solution?
► 1) ) (2) 4) lim ( ) 4
2) (2) lim ( )
2( ) is continuous at x = 2
a fb f xx
c f f xx
f x
Continuity
Problem 1 Solution
► 1) ) (2) 4(2) 4 4 (Indicates we are using the appropriate piecewise function)) lim ( ) 4
2lim ( ) 4
2lim ( ) 4 (Indicates we check the limit from both the left and right side)
2) (2) lim ( )
2
a fb f x
xf x
xf x
xc f f x
x
( ) is continuous at x = 2f x
Sample Problem 2
Differentiability
Problem 2
► 2) Given this piecewise function, justify that the function is not differentiable at x = 2
Differentiability
Problem 2 Solution(How can we create a more rigorous solution)?
► 2)
► Or► The function is not continuous at x = 2 therefore it cannot be differentiable
at x = 2.
( ) (2)lim2
2
The function isn't differentiable at x = 2
f x f DNEx
x
Differentiability
Problem 2 Solution
► 2)
► Or
( ) (2)lim ( )2
2( ) (2)lim 4
22
The function f(x) isn't differentiable at x = 2
f x f DNE orx
xf x fx
x
Since lim ( ) lim ( )
2 2lim ( )
2Because of this, the function ( ) is not continuous at x=2 and
therefore cannot be differentiable at x = 2.
f x f x
x xf x DNE
xf x
Sample Problem 3
Extrema
Problem 3
► 3) Find the absolute maximum and minimum value of
the function in the interval from
Extrema
Problem 3 Solution(How can we create a more rigorous solution)?
► 3) x y
0 1
e
1
sin( )'( ) cos( ) xf x x e
Critical point at x = 2
The absolute maximum is ( , ).2
The absolute minimum is (0,1) and ( ,1).
e
Extrema
Problem 3 Solution
► 3)
x y
0 1
e
1
sin( )'( ) cos( ) xf x x e
'( ) 0 at x = 2
f x
The absolute maximum of f(x) is e and occurs at x = .2
The absolute minimum of f(x) is 1 and occurs at x = 0 and x = .
Sample Problem 4
IVT/MVT - Overestimate
Problem 44) Given this set of data is taken from a function v(t) and assuming it is continuous over the interval [0,10] and is twice differentiable over the interval (0,10)
T=0 hours T=1 hour T=2 hours T=4 hours T=6 hours T=10 hours
Vel=50mph Vel=60mph Vel=30mph Vel=38mph Vel=50mph Vel=70mph
a) Find where the acceleration must be equal to 4 mile per hour2 and justify.
b) Find the minimum number of times the velocity was equal to 35mph and justify.
c)Approximate the total distance travelled over the 6 hour time frame starting at t = 4 using a trapezoidal Riemann sum with 2 subintervals.
d) Assuming that the acceleration from 4 to 10 hours is strictly increasing. State whether the approximation is an over or under estimate and why.
IVT/MVT - Overestimate
Problem 4 Solution► a) Given that the function v(t) is continuous over the interval [0,10] and differentiable over the interval (0,10) and since and there must exist at least one c value between hours 2 and 4 such that by the Mean value theorem.
► b) Given the function v(t) is continuous over the interval [0,10] and since v(1)=60 and v(2) = 30 and since v(2)=30 and v(4) = 38 there must exist at least one value of c between hour 1 and hour 2 and at least one value between hour 2 and hour 4 so that v(c)=35 at least twice by the Intermediate Value theorem.
► c)
► d) Since the function v(t) is concave up and above the x-axis (because the derivative of velocity is increasing) . The top side of the trapezoid will lie above the curve and therefore the approximation will be an over estimate.
10
4
1 1( ) (2)(38 50) (4)(50 70) 328 miles2 2
v t dt
2(4) (2)( ) 44 2
v va c mph
2'( ) 4v c mph
( ) '( )a t v t
Sample Problem 5
Taylor Series
Problem 5
► 5) Given the functions
► a)Find the second degree Taylor Polynomial P2(x) centered at zero for
► b) Approximate the value of using a second degree Taylor Polynomial centered at 0.
► c) Find the maximum error of the approximation for if we used 2 terms of the Taylor series to approximate the value.
3
( ) sin
( )
g x x
h x x
1sin2e
( )( ) g xf x e
312e
Taylor Series
Problem 5 Solution
sin
2 sin sin
2
2
2
5 ) (0) 1
'( ) cos , '(0) 1
' '( ) cos sin , ' ' (0) 1
( ) 12
1 1 1 5) 12 2 8 8
) Since the Taylor Series is an alternating series, with terms decreasing in magnitude and w
x
x x
a f
f x x e f
f x x e x e f
xP x x
b P
c
6 9 3 13
hose values are approaching zero as x , the maximum error of a two term approximationis the absolute value of the third term.
( ): ( ) 1 ..... ...2! 3! ( 1)!
12
nx x xTaylor Series f x xn
Third term
6
1 1The two term Taylor series will have a maximum error of 2! 128 128
Additional Time - Additional Problem
Additional Problem
2014 Problem 3
Additional Problem
2014 Problem 3