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# Introduction To Calculus.pdf

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Mathematics

MAT1320

Kirill Zaynullin

Fall

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Calculus and Vectors – How to get an A+
1.1 Radical Expressions: Rationalizing Denominators
A Radicals Ex 1. Simplify:
a a =a a) 3 3 = 3
n 3 2
(n a) = a b) ( 5) = ( 5) 5 = 5 5
m
n m n m c) ( 7) = ( 7) ( 7) = 7( 7) = 7 7 3 2 = 7349
( a ) =an = ( a)
n
Note: If n is even, then ≥ 0 for a.
B Rationalizing Denominators (I) Ex 2. Rationalize:
a a c a c 2 2 5 2 5 2 5
= = = = =
b c b c c bc 3 5 3 5 5 3×5 15
C Conjugate Radicals Ex 3. For each expression, find the conjugate radical.
a + b ⇔ a − b a) 2 + 3 ⇒ 2 − 3
a + b ⇔ a − b b) 2 − 3 ⇒ 2 + 3
a +b c ⇔ a −b c c) 3 + 2 5 ⇒ 3 − 2 5
a b +c d ⇔ a b − c d d) 2 5 + 3 7 ⇒ 2 5 − 3 7
D Difference of squares identity Ex 4. Use the difference of squares identity to simplify:
(a + b)(a − b) = a − b a) (a + b)(a − b) = a − ( b) = a − b
b) ( a + b)( a − b) = ( a) − ( b) = a − b
c) ( a + b c)( a −b c) = ( a) − (b c) = a −b c 2
E Rationalizing Denominators (II) Ex 5. Rationalize the denominator:
Hint: Multiply and divide by the 3 3 1+ 2 3(1+ 2) 3(1+ 2)
conjugate radical e of the denominator.a) = = = = −3(1+ 2)
1− 2 1− 2 1+ 2 1 − ( 2) 2 1− 2
4 4 2− 3 5 4(2− 3 5) 4(2− 3 5) 4(2 −3 5)
b) = = 2 2 = = −
2+ 3 5 2 +3 5 2 − 3 5 2 − (3 5) 4− × 5 41
c)
2 2 3+ 6 2( 3 + 6) 2( 3 + 6) 2( 3 + 6)
= = = = −
3 − 6 3− 6 3+ 6 ( 3) 2− ( 6)2 3− 6 3
F Rationalizing Numerators Ex 6. Rationalize the numerator:
Hint: Multiply and divide by the 5 − 3 5 − 3 5 + 3 ( 5) − ( 3)2 2
conjugate radical of the numerator. = = =
2 − 1 2 −1 5 + 3 ( 2 −1)( 5 + 3) ( 2 −1)( 5 + 3)
G Equivalent Expressions Ex 7. Find equivalent expressions by rationalizing. State restrictions.
Hint: You may get equivalent x 1 x− 1 x+ 1 ( −1)( x +1)
expressions by rationalizing the a) = = = x +1, x≥ 0,x ≠ 1
numerator or denominator. x −1 x − 1 x+ 1 x −1
Note: State restrictions. x + 9 − = x + 9 − 3 x + 9 + = x = 1 ,
b) x x x + 9 + 3 x( x + 9 + 3) x + 9 + 3
x ≥ −9, x ≠ 0
1.1 Radical Expressions: Rationalizing Denominators
© 2010 Iulia & Teodoru Gugoiu - Page 1 of 2 Calculus and Vectors – How to get an A+
1 − 1 1 1
x + h x − 1
= x + h x = − ,
c) h ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
h⎜ + ⎟ x(x + h)⎜ + ⎟
⎝ x + h x ⎠ ⎝ x +h x ⎠
x+ h > 0,x > 0,h≠ 0
H More algebraic identities Ex 8. For each case, the numerator and denominator have a common
3 3 2 2 zero. Use algebraic identities to eliminate the common zero. State
a −b = (a −b)(a + ab + b ) restrictions.
3 3 2 2
a + b = (a + b)(a − ab + b ) x −1 ( x) −1 ( x −1)(( x) + x +1) 3 2 3
a −b = (a −b)(a + b)(a + b ) 2 a) 3 = 3 = 3 = x + x +1, x ≠1
x −1 x −1 x −1
4 2 2
b) x − 1 = (x− 1)(x+ 1)(x +1) = (x+1)( x +1) , x≠1
x 3− 1 (x− 1)(x2 + x 1) x2 + x1
Reading: Nelson Textbook, Pages 6-8
Homework: Nelson Textbook: Page 9, #1a, 2a, 3a, 4a, 5, 6a, 7ac
1.1 Radical Expressions: Rationalizing Denominators
© 2010 Iulia & Teodoru Gugoiu - Page 2 of 2 Calculus and Vectors – How to get an A+
1.2 The Slope of the Tangent
A Lines Ex 1. The equation of the lineL is:2x − 3y + 6 = .
1
a) Find the slope of the linL1.
y = 2 x +2 ⇒ ∴m = 2
3 1 3
b) Find the equation of the linL 2, that is parallel to the
lineL and passes through the point P(2,−1).
1
2 2 4
y −( 1) = (x −2) ⇒ y = x −1− ⇒
3 3 3
2 7
rise ∆y y2 − y1 ∴L 2 :y = x −
The slope of a line:m = = = 3 3
run ∆x x2 − 1 c) Find the equation of the linL 3, that is perpendicular
Slope y-intercept equation of a line:y = mx + b to the lineL1and passes through the point Q(4,2).
Slope-point equation of a line:y − y = m(x − x )
1 1 m = − 1 = −3 ⇒ y−2 = − 3(x− 4) ⇒
For parallel lines, slopes are equalm 1 m 2 3 m1 2 2
For perpendicular lines, slopes are negative 3
reciprocal: ∴ L3: y= − 2x +8
1
m2= − or m1m 2 −1
m1
B Function Notation
Ex 2. Let f (x) = x −1 . Find:
y = f (x)
where: a) f (1)
x is the argument, input or independent variable f (1) = 1 −1 = 0 ⇒ ∴ f (1) = 0
y is the value, output or dependant variable
b) f (a)
f is the name of the function. 2
∴ f (a) = a −1
c) f (a +1)
f (a +1) = (a +1) −1 = a + 2a +1−1 = a + 2a
∴ f (a +1) = a + 2a
C Secant Line 2
Let y = f (x)be a function and P(x , y )and Q(x , y ) Ex 3. Consider f( ) = . Find the equation of the
1 1 2 2 x +1
two points on its graph. secant line that passes through the points A(0,2) and
The slope of the secant line that passes through the
points P and Q is given by: B(−3,−1) .
y − y (−1) −2)
rise ∆y y − y f(x )− f(x ) m = 2 1= =1
m = = = 2 1= 2 1 x2 − x1 (−3)−(0)
run ∆x x2 − 1 x2− x1
y −2 =1( −0)
∴ = x + 2
1.2 The Slope of the Tangent
©2010 Iulia & Teodoru Gugoiu - Page 1 of 3 Calculus and Vectors – How to get an A+
IfP(a, f (a))and Q(a + h, f (a + h)then the slope of
the secant line is given by:
m = f (a + h) − f (a)
h
D Tangent Line
As the point Q approaches the point P , the secant
line approaches the tangent line at P . See the
diagram on the right side.
The slope of the tangent line at P(a, f (a)is:
m =lim f(a+ h) − f( ) (1)
h→0 h
E Graphical Computation Ex 4. Find the slope of the tangent line to the curve
1. Draw the tangent line using a ruler
y = 2 x −1 at the point P(5,4).
2. Choose two points on the tangent line A(x1, 1 )and
B(x , y )
2 2
3. Use the formula:
y −y
m ≅ 2 1
x2 −x1
to estimate the slope of the tangent line.
Note. The slope of the tangent line can be calculated
using:
m = tanα
where α is the angle between the tangent line and the
positive direction of the x-axis.
A (1,2), B(9,6)
x =1,y = 2,x = 9,y = 6
1 1 2 2
y2− y1 6 2 1 1
m = = = ∴ m =
x2− x1 9 1 2 2
F Numerical Computation Ex 5. Consider y = f (x) = x. Estimate numerically the
f (a + h) − f (a)
1. Use the formula m ≅ (2) slope of the tangent line atP(1,1)using
h −20
2. Choose a sequence h , 1 , 2 ,3..→0 . h1= 0.1, 2 = 0.0001,h3=10 .
a = 1, (1) 1
3. Compute m , 1 ,2m ,3..
4. Observe the pattern and conclude. (1 0.1) 3−1
m 1 ≅ 3.31
5. Be careful at “difference catastrophe”. 0.1
3
Note. The difference catastrophe is related to the m 2 (1+0.0001) −1 ≅ 3.0003
0.0001
limited capacity of memorizing numbers by any −20 3
technologic device (scientific calculator, computer, m ≅ (1+10 ) −1 ≅ 0 (difference catastrophe)
etc). If two numbers are very close, then they have the 3 10 −20
same internal representation in the memory.
∴m ≅ 3
1.2 The Slope of the Tangent
©2010 Iulia & Teodoru Gugoiu - Page 2 of 3 Calculus and Vectors – How to get an A+
G Algebraic Computation Ex 6. Find the slope of the tangent line to the graph of
2
1. Use the formula m = lim f a+ h− (f a . y= f(x) =x − 3x at the point (1,−2).
h→0 h a =1, f(1)= −2
2. Do not substitute h by0 because you will get the
f( + ) = f(1+ ) = (1+ ) − 3(1+ ) =
indeterminate case 0 .
0 =1+ 2 + h 2− 3− 3 = −2 − h +h 2
3. Compute algebraic the difference quotient
DQ = f(a+ h)− f( ) = f(1+ h)− f (1=
DQ = f a h)− (f a until you succeed to cancel out h h
h 2 2
= (−2 − h+ h ) − (−2)= h −h = h −1
the factor h . h h
4. Substitute in the remaining expressionhby 0.
m = h→0(h −1) = 0 −1= −1 ⇒ ∴m = −1
2
Ex 7. Find the equation of the tangent line to the grapEx 8. Consider y = f (x) = x − 2.
1
of y = f (x) = at the pointP(2,1).
x −1 a) Find the slope of the tangent line at the generic point
a = 2, f( )= f (2) 1 P(a, f (a).
2
f(a+ h)= f (2+h ) 1 = 1 f a) =a −2a
2 + − 1 1 +h 2 2 2
f a h) = (a h) −2( +a h= a +2 ah h −2 a2 h
1 −1 1 −1 1 +1 f a h)− (f a (a2 +2 ah h 2−2 a2 )h( a2 −2 a
f(a+ h)− f( ) 1+ h 1+ h 1 +h DQ = =
DQ = = = h h
h h h 1 +1 2
1 +h = 2ah+h −2h = 2a+h−2
h
1 −1 m = lim(2a+h−2) = 2a−2 ⇒ ∴m = 2a−2
1+ h − h −1 x→0
= ⎛ ⎞ = ⎛ ⎞ = ⎛ ⎞
h⎜ 1 +1⎟ h(1+h ) 1 +1⎟ (1+h)⎜ 1 + 1⎟
⎝ 1+ h ⎠ ⎝ 1+ h ⎠ ⎝ 1+ h ⎠ b) Find the point where the tangent line is horizontal.
⎧m = 0
m = lim −1 = − 1 = − 1 ⎨ ⇒ 0 = 2 − 2⇒ =1, (1) = −1⇒ ∴ (1,−1)
x→0 ⎛ 1 ⎞ ⎛ 1 ⎞ 2 ⎩m = 2 − 2
(1+h)⎜ +1⎟ (1 0) ⎜ +1 ⎟
⎝ 1+ h ⎠ ⎝ 1 0 ⎠
c) Find the pointP such that m P 2.
⎧ m = −1 2a − 2 = 2 ⇒ a = 2, f (2) = 0 ⇒ ∴P(2,0)
⎨ 2 ⇒ y −1= − 1 (x −2)⇒∴y = − 1 x + 2
⎪ 2 2
⎩ P(2,1) d) Find the pointP such that the tangent line at P is
perpendicular to the line L : x −3y =3.
2
1 1 1
L 2:y = x − ⇒ m 2= ⇒ m = − = −3
3 3 m2
⎧m = −3
⎨ ⇒ − = 2 a − ⇒ = − 1,f ⎜−1 ⎟= 1 + = 5
⎩m =2a − 2 2 ⎝ 2 ⎠ 4 4
⎛ 1 5 ⎞
∴ P ⎜− 2 4⎟
⎝ ⎠
Reading: Nelson Textbook, Pages 10-18
Homework: Nelson Textbook: Page 18, #1a, 2a, 3ac, 4a, 5a, 6a, 8a, 9a, 10a, 11a, 15, 21, 25
1.2 The Slope of the Tangent
©2010 Iulia & Teodoru Gugoiu - Page 3 of 3 Calculus and Vectors – How to get an A+
1.3 Rate of Change
A Average Rate of Change 2
Ex 1. Consider y = f (x) = (x +1. Find the rate of change in
y = f (x), y1= f (x1), y 2 f (x 2 the y variable over the interval[−1,2].
∆x = x − x ( change in variable x )
2 1 x = −1, y = f (x ) = f (−1) = (−1+1) = 0
∆ y = y2− y1( change in variable y ) 1 1 1
x = 2, y = f (x ) = f (2) = (2+1) = 9
The Average Rate of Change (ARC) in y 2 2 2
variable over the interval [x ,x ] is given ∆x = x2− x 1 2−(−1) = 3, ∆y = y 2 y 1 9−0 = 9
1 2
by: ARC = ∆y = 9 = 3 ∴ARC = 3
∆x 3
ARC = rise = ∆y = y 2− y1 = f(x 2)− f( 1
run ∆x x 2− x1 x2 − x1
Note: The Average Rate of Change is the
same as the slope of the secant line
passing through the points P(x , y ) and
1 1
Q (x2, y2).
If 1 = a and x 2 =a + h then:
ARC = f (a + h)− f (a)
h
B Average Velocity Ex 2. A rock is launched vertically upward. The height of the rock is
2
Let s = s(t)be the position function, where given by s(t) =100t −10t . Find the average velocity over the time
s is position in meters and tis the time in
interval[1,2].
seconds. 2
s = s(t), s1= s(1 ), s2= s(t2) t1=1, s1= ( 1 = (1) =100(1)−10(1) = 90 m
2
∆t = t2−t1( time duration) t2 = 2, s2= ( )2= (2) =100(2)−10(2) =160 m
∆ t t − t = 2−1=1 ,s ∆s= s − s =160−90 = 70 m
∆s = 2 − s1( displacement) 2 1 2 1
The Average Velocity (AV) over the time ∆ s 70
AV = = = 70 ∴AV = 70m/s
interval [1 2t is given by: ∆ t 1
rise ∆s s −s s( ) − s( )
AV = = = 2 1 = 2 1
run ∆t t2 −t1 2 −t1
Note: The unit of measurement for velocity
is m/s .
C Instantaneous Rate of Change Ex 3. Consider the following position function: s(t) = t −4t.
As h → 0 the Average Rate of Change a) Find the instantaneous velocity at t = 3s.
approaches to the Instantaneous Rate of 2
Change (IRC): a = 3, s a) =s(3) =3 −4(3) = −3
2 2
IRC = RC = lim f a+ )h ( f a s a h) = (s+ ) h (3+ ) h4(3+ ) = 9h6 + h h −12−4 h
h→0 h 2
Note: The Instantaneous Rate of Change s a h)− (s a (9+6 +h h −12−4 )h(−3)
v = h→0 h = h→0 h =
(IRC) is the same as the slope of the
tangent line at the point P(a, f (a). h(2+ h
= h→0 h = h→0(2+h) = 2+0 = 2 ∴v = 2m/s
Similarly, the Average Velocity (AV) b) Find the instantaneous velocity at the generic moment t = a
approaches Instantaneous Velocity (IV): 2
s a) = a −4a
IV = v = lims a+ )h ( s a 2 2 2
h→0 h s a h) = ( a h −4( + )a h a +2 ah h −4 a4 h
2 2 2
v = lim s a h)− (s a = lim(a +2 ah h −4a−4 )h( a −4 a =
h→0 h h→0 h
= lim h(2a+h−4) = lim(2a+h−4) = 2a+0−4 = 2a−4
h→0 h h→0
∴v = 2a−4
1.3 Rate of Change
© 2010 Iulia & Teodoru Gugoiu - Page 1 of 2 Calculus and Vectors – How to get an A+
c) Use the formula at part b) to compute the velocity at time5s .
⎧v= 2a−4
⎨a= 5s ⇒ =v 2(5) 4 6 ∴ =v 6m s
⎩
d) Find the moment(s) of time at which the velocity is zero.
⎧v =2 a− 4
⎨ ⇒ =0 2 a− 4 = a 2 ∴ a 2s
⎩v = 0
Ex 4. A spherical balloon is inflated. Find the instantaneous rate of
change in volume of the balloon with respect to its radius when the
radius is10m.
4π 3
V r )= r , r =a= 10m
3
4π 3 4π 3
V a )= V(10)= (10) , V a h ) =V (10+h) = (10+ h)
3 3
4π 3 4π 3
∆V V a h )−V a ) 3 (10+ h) − 3 (10)
IRC = lim = lim = lim =
∆r→ 0∆r h→0 h h→0 h
4π (10+h ) −(10)3 4π h[(1+ h)2+ (10+h)(10)+10 ]
= lim = lim =
3 h→ 0 h 3 h→0 h
4π 2 2 4π 2 2
= 3 h→i0[(1+ h) + (10+h)(10)+10 ]= 3 [(10+0) +(10+0)(10)+10 ]
∴ IRC = 400π m 2
Reading: Nelson Textbook, Pages 22-28
Homework: Nelson Textbook: Page 28 #2a, 7, 12, 14, 15b, 20, 22
1.3 Rate of Change
© 2010 Iulia & Teodoru Gugoiu - Page 2 of 2 Calculus and Vectors – How to get an A+
1.4 Limit of a Function
A Left-Hand Limit Ex 1. Use the function y = f (x)defined by the following
If the values of y = f (x)can be made arbitrarily graph to find each limit.
close to L by taking x sufficiently close to a with
x < , then:
lim f (x) = L
x→a −
Read: The limit of the function f (x) as x
approaches a from the left is L .
a) lim−f (x) = DNE
Notes: x→−4
1. The function may be or not defined at a . b) lim f (x) = 2
x→−2−
2. DNE stands for Does Not Exist. c) lim f (x) = 2
3. L must be a number. x→−1−
4. ∞ is not a number.
d) lim−f (x) =1
x→3
B Right-Hand Limit Ex

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