1
REVIEW FOR TEST 2
L’Hopital Rule is a method for computing a limit of the indeterminate forms: 00
±∞
0 ∙ ∞ ∞ − ∞ 1 ∞ 0 0 ∞0
±∞
Let f and g be real functions, continuous on [a, b] and differentiable (a, b). Suppose that ∃c∈(a, b): f(c) = g(c) = 0 . Then:
𝑓′(𝑥)
𝑙𝑖𝑚 𝑓(𝑥)
= 𝑙𝑖𝑚
𝑥 → 𝑐 𝑔(𝑥)
𝑥 → 𝑐 𝑔′(𝑥)
provided that the second limit exists.
Suppose that instead of f(c) = g(c) = 0, we have that f(x)→ ∞ and g(x)→ ∞ as x→b- . Then:
𝑙𝑖𝑚 𝑓(𝑥)
𝑙𝑖𝑚 𝑓′(𝑥)
=
𝑥 → 𝑏 − 𝑔(𝑥) 𝑥 → 𝑏 − 𝑔′(𝑥)
Indeterminate forms:
0
0
and
±∞
±∞
provided that the second limit exists.
: use L’Hopital rule directly (use above theorem).
Indeterminate form: 0 ∙ ∞ use algebra to convert the expression to a fraction (0 ∙
Indeterminate form: ∞ − ∞ convert the expression into
0
0
or
1
0
=
0
0
), and then apply L'Hopital's Rule
±∞
±∞
Indeterminate forms: 1∞ 00 ∞0 can be handled by taking logs, computing the limit using the techniques above, and finally
exponentiating to undo the log. Remember that lim (log) = log (lim)
Parametric Equations and Curves
Explicit form of y = y(x) : y is expressed in terms of x.
Implicit form of y = y(x): parametric equations – both x and y are expressed in terms of parameter t
x = f(t) y = g(t)
Parametric curves have a direction of motion given by increasing t. Graphs include arrows that show the direction of motion.
Sketching a parametric curve:
1. method: Elimination of the parameter from the set of the equations and substitution to get explicit form y = y(x)
2. method: Using trigonometry to eliminate a parameter to get explicit form y = y(x)
3. method : there are some equations that you can’t get explicit form to get explicit form y = y(x), so use calculator
Polar Equations and Curves r = r ()
x = r cos 
𝑟 = √𝑥 2 + 𝑦 2
Different forms of symmetry
Symmetrical with respect to polar axis:
y = r sin 
 = arc tan
𝑦
𝑥
𝑦
( it is not always  = arc tan ; check picture always)
𝑥
r(−) = r()
Replace  with – . If an equivalent equation results, the graph has this type of symmetry.
Symmetrical with respect to the line  = /2 (y – axis): r( - ) = r()
Symmetrical with respect to the pole:
r( - ) = r()
Parametric form of polar equation: x = r cos  = f () cos 
y = r sin  = f () sin 
2
First and second derivative
Cartesian plane: y = y(x)
𝑑𝑦
1. derivative
𝑑2𝑦
2. derivative
𝑑𝑥
𝑑2𝑥
Parametric form: x = x(t), y = y(t)
𝑑𝑦
1. derivative
𝑑𝑥
𝑑2𝑦
2. derivative:
𝑑2𝑥
=
=
𝑑𝑦 𝑑𝑡
𝑑𝑡 𝑑𝑥
𝑑
𝑑𝑥
=
𝑑𝑦 1
=
𝑑𝑥
𝑑𝑡
𝑑𝑡
𝑑𝑦
𝑑
𝑑𝑥
𝑑𝑡
( )=
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
𝑑𝑦
( )
𝑑𝑡
𝑑𝑥 𝑑𝑥
y’ =
𝑑 𝑑𝑦
( )
𝑑𝑡 𝑑𝑥
𝑑𝑥
𝑑𝑡
=
=
𝑑𝑦′
𝑑𝑡
𝑑𝑥
𝑑𝑡
y’’ =
𝑦̇
is function of t
𝑥̇
𝑦′̇
𝑥̇
keep in mind that you already found y’ as
a function of t not x, so y’’ is a function of t.
Polar curve: r = r():
𝑑𝑥
x = r cos 
𝑑𝑦
1. derivative
2. derivative:
𝑑𝑥
𝑑2𝑦
𝑑2𝑥
=
=
𝑑𝜃
𝑑𝑦 𝑑𝜃
𝑑𝜃 𝑑𝑥
𝑑
𝑑𝑥
=
=
𝑑𝑟
𝑑𝜃
𝑑𝑦 1
𝑑𝜃 𝑑𝑥
𝑑
𝑑𝑥
𝑑𝜃
( ) =
=
𝑑𝜃
𝑑𝑦
y = r sin 
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
𝑑𝑦 𝑑𝜃
( )
𝑑𝑥 𝑑𝑥
𝑑𝑟
𝑑𝜃
𝑑𝑟
𝑑𝜃
=
=
𝑑𝑦
𝑑𝜃
=
𝑑𝑟
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑑𝜃
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
𝑑 𝑑𝑦
( )
𝑑𝜃 𝑑𝑥
𝑑𝑥
𝑑𝜃
=
1
𝑑
𝑑𝑥
𝑑𝜃
𝑑𝜃
(
𝑑𝑟
𝑑𝜃
𝑑𝑟
𝑑𝜃
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
and now good luck 
)
Note that rather than trying to remember this formula it would probably be easier to remember how we derived it .
Tangent line: y = m (x – x0) + y0
Cartesian plane: at point (x0, y0) of the curve y = y(x)
m=
Parametric form: at point t = t0 of the curve x = x(t), y = y(t)
x0 = x(t0), y0 = y(t0)
Polar curve: at point 0 of the r = r():
r0 = r(0), x0 = r0 cos 0
Parametric form: x = x(t), y = y(t)
Polar curve: at point 0 of the r = r():
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
y0 = r0 sin 0 m =
𝑑𝑦
𝑑𝑥
Parametric form: x = x(t), y = y(t)
Polar curve: at point 0 of the r = r():
Concavity:
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑥
=0 →
𝑑𝑦
𝑑𝑦
𝑑𝜃
= 0 → 0 →
r0 = r(0)
𝑑𝑥
𝑑𝑦
𝑑𝑥
=∞ →
𝑑𝑥
𝑑𝑡
𝑑𝑥
𝑑𝜃
at r0 = r(0)
𝑑𝑥
eq: y = y0
(𝑑𝑡 |𝑡=𝑡
y0 = r0 sin 0
eq: y = y0
(𝑑𝜃 |𝜃=𝜃
𝑑𝑦
𝑑𝑥
≠ 0)
0
𝑑𝑥
≠ 0)
0
=∞
= ∞ → x0
=∞ →
at t = t0
=0
= 0 → t0 → y0 = y(t0)
𝑑𝑡
𝑑𝑦
at (x0, y0)
eq: y = y0
Vertical tangent line: Vertical tangents will occur where the derivative is not defined:
Cartesian plane: y = y(x)
𝑑𝑦
= 0 → x0 (max or min) → y0 = y(x0)
=0 →
𝑑𝑥
m=
Horizontal tangent line: Horizontal tangent will occur where the derivative is zero:
Cartesian plane: y = y(x)
𝑑𝑦
eq: x = x0
𝑑𝑦
= 0 → t0 → x0 = x(t0)
= 0 → 0 →
r0 = r(0)
x0 = r0 cos 0
eq: x = x0
( 𝑑𝑡 |𝑡=𝑡
≠ 0)
eq: x = x0
(𝑑𝜃 |𝜃=𝜃
0
𝑑𝑦
0
≠ 0)
If second derivate is negative curve will be concave down, and for positive second derivate curve will be concave up.
Cartesian plane: y = y(x)
Parametric form: x = x(t), y = y(t)
Polar curve: at point 0 of the r = r():
𝑑2𝑦
𝑑2𝑥
𝑑2𝑦
𝑑2𝑥
𝑑2𝑦
𝑑2𝑥
at (x0, y0)
at t0
at (r0, θ0)
3
Arc length
The arc length of functions in Cartesian plane:
𝑏
𝑏
𝑏
𝑑𝑦
S = ∫𝑎 𝑑𝑠 = ∫𝑎 √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫𝑎 √1 + ( )2 𝑑𝑥
𝑑𝑥
dy =
𝑑𝑦
𝑑𝑥
𝑑𝑥
The arc length of parametric form is :
2
2
𝑏
𝑏
𝑏
𝑑𝑥
𝑑𝑦
S = ∫𝑎 𝑑𝑠 = ∫𝑎 √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫𝑎 √ ( 𝑑𝑡 ) + ( 𝑑𝑡 ) 𝑑𝑡
dx =
𝑑𝑥
𝑑𝑡
𝑑𝑡
dy =
𝑑𝑦
𝑑𝑡
𝑑𝑡
**The arc length of polar form is :
2
𝑏
𝑏
𝑏
𝑑𝑟
S = ∫𝑎 𝑑𝑠 = ∫𝑎 √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫𝑎 √ 𝑟 2 + (𝑑𝜃) 𝑑𝜃
2
2
𝑏
𝑏
𝑏
𝑑𝑥
𝑑𝑦
S = ∫𝑎 𝑑𝑠 = ∫𝑎 √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫𝑎 √ (𝑑𝜃) + (𝑑𝜃) 𝑑𝜃
x = r cos 
𝑑𝑥
𝑑𝜃
=
𝑑𝑟
𝑑𝜃
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
y = r sin 
dx =
𝑑𝑦
𝑑𝜃
=
𝑑𝑥
𝑑𝜃
𝑑𝑟
𝑑𝜃
𝑑𝜃
dy =
𝑑𝑦
𝑑𝜃
𝑑𝜃
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑑𝑥 2
𝑑𝑦 2
𝑑𝑟 2
𝑑𝑟
𝑑𝑟
( ) + ( ) = ( ) [𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃] + 𝑟 2 [ 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃] − 2𝑟
cos 𝜃 sin 𝜃 + 2𝑟
cos 𝜃 sin 𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
𝑑𝜃
Area enclosed by the curve and x axis
The area enclosed by function y = y(x) and x-axis in Cartesian plane:
𝑏
𝑏
A = ∫𝑎 𝑑𝐴 = ∫𝑎 𝑦(𝑥)𝑑𝑥
a ≤ x≤b
**The area enclosed by function y = y(x) and x-axis in parametric form:
𝛽
𝛽
A = ∫𝛼 𝑑𝐴 = ∫𝛼 𝑦(𝑡)
𝑑𝑥
𝑑𝑡
𝑑𝑡
α ≤ t ≤β
The area enclosed by a polar curve r = r():
𝜃2
1
1
2
A = ∫𝜃 𝑑𝐴 =
𝜃
2
∫𝜃 𝑟 2 𝑑𝜃
1
1 ≤  ≤ 2
dA = y(x) dx = y[x(t)]
𝑑𝑥
𝑑𝑡
𝑑𝑡
4
⃗ ≡ 𝒗 (bold letter)
𝒗
Vector’ algebra
Cartesian coordinate system: vector
𝐴 ≡ ⟨𝐴𝑥 , 𝐴𝑦 ⟩ is position vector of point (𝐴𝑥 , 𝐴𝑦 )
magnitude or absolute value of vector 𝐴 is nonnegative real number: |⟨𝐴𝑥 , 𝐴𝑦 ⟩| = √𝐴2𝑥 + 𝐴2𝑦
⃗ ≡ ⟨0,0⟩ is called the zero vector, has zero length and no direction.
vector 0
If initial point (tail) is Q (Qx, Qy) and terminal point is R (Rx, Ry) (head): vector ⃗⃗⃗⃗⃗
𝑄𝑅 = ⟨𝑅𝑥 − 𝑄𝑥 , 𝑅𝑦 − 𝑄𝑦 ⟩
Unit vector
⃗⃗⃗
𝐴
|⃗⃗⃗
𝐴|
is a vector of magnitude 1 in the direction of 𝐴 . Sometimes called direction vector:
⃗⃗⃗
𝐴
|⃗⃗⃗
𝐴|
= ⟨cos 𝜃, 𝑠𝑖𝑛𝜃⟩
Components of a Vector
if (A,θ) 
known
A x = Acosθ
A = |𝐴|
A y = Asinθ
𝐶𝑥 = 𝐴𝑥 + 𝐵𝑥 = 𝐴 cos 𝜃𝐴 + 𝐵𝑐𝑜𝑠 𝜃𝐵
𝐶𝑦 = 𝐴𝑦 + 𝐵𝑦 = 𝐴 𝑠𝑖𝑛 𝜃𝐴 + 𝐵𝑠𝑖𝑛 𝜃𝐵
⃗
𝐶 = 𝐴+ 𝐵
Addition of vectors:
𝐴 = ⟨𝐴 cos 𝜃 , 𝐴 sin 𝜃⟩
𝐶 = √𝐶𝑥2 + 𝐶𝑦2 ;  from the picture
Velocity, Speed, Acceleration, and Direction of Motion
Suppose a particle moves along a smooth curve in the plane so that its position at any time t is ⟨𝑥(𝑡), 𝑦(𝑡)⟩ , where x and y are
differentiable functions of t.
1. The particle’s position vector is 𝑟 (𝑡) = ⟨𝑥(𝑡), 𝑦(𝑡)⟩ .
𝑑𝑥 𝑑𝑦
2. The particle’s velocity vector is 𝑣(𝑡) = ⟨ , ⟩.
𝑑𝑡 𝑑𝑡
3. The particle speed is the magnitude of 𝑣 , denoted by |𝑣| . Speed is a scalar.
4. The particle’s acceleration vector is 𝑎(𝑡) = ⟨
𝑑2𝑥
𝑑𝑡 2
,
𝑑2𝑦
𝑑𝑡 2
⟩.
⃗𝒗
5. The particle’s direction of motion is the direction vector |𝑣⃗| .
Displacement and Distance Traveled
𝑏
𝑏
1– D motion: displacement : 𝑥 = ∫𝑎 𝑣(𝑡)𝑑𝑡 distance traveled: 𝑠 = ∫𝑎 |𝑣(𝑡)|𝑑𝑡
2 – D motion: Suppose a particle moves along a path in the plane at velocity 𝑣(𝑡) = ⟨𝑣𝑥 (𝑡), 𝑣𝑦 (𝑡)⟩, where 𝑣𝑥 and 𝑣𝑦
are integrable functions of t. The displacement from t = a to t = b is given by the vector:
𝑏
𝑏
⃗ + ⃗𝑟
𝑛𝑒𝑤 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛: ⃗𝑏 = 𝑎
𝑟 = ⟨𝑥, 𝑦⟩ = ⟨∫𝑎 𝑣𝑥 (𝑡) 𝑑𝑡 , ∫𝑎 𝑣𝑦 (𝑡) 𝑑𝑡 ⟩
The distance traveled from t = a to t = b is:
𝑏
𝑏 𝑑𝑟
𝑏
𝑏
s = ∫𝑎 |𝑑𝑟 | = ∫𝑎 | | 𝑑𝑡 = ∫𝑎 |𝑣(𝑡)|𝑑𝑡 = ∫𝑎 √(𝑣𝑥 )2 + (𝑣𝑦 )2 𝑑𝑡
𝑑𝑡
𝑏
𝑏
𝑏
𝑑𝑥
𝑑𝑦
s = ∫𝑎 |𝑑𝑟 | = ∫𝑎 𝑑𝑠 = ∫𝑎 √( )2 + ( )2 𝑑𝑡
𝑑𝑡
𝑑𝑡
5
Improper Integral
An improper integral is a definite integral that has
1. one or both limits infinite
if f(x) is continuous on [a, )
∞
𝑏
then ∫𝑎 𝑓(𝑥) 𝑑𝑥 = lim ∫𝑎 𝑓(𝑥)𝑑𝑥
𝑏→∞
𝑏
𝑏
lim ∫ 𝑓(𝑥)𝑑𝑥
𝑎→−∞ 𝑎
∞
𝑐
lim ∫ 𝑓(𝑥)𝑑𝑥
𝑎→−∞ 𝑎
if f(x) is continuous on (-, b] then ∫−∞ 𝑓(𝑥) 𝑑𝑥 =
if f(x) is continuous on (-, ) then ∫−∞ 𝑓(𝑥) 𝑑𝑥 =
𝑏
c ∈𝑅
+ lim ∫𝑐 𝑓(𝑥)𝑑𝑥
𝑏→∞
No limit → improper integral diverges
∞
∫1 𝑥 −2 𝑓(𝑥) 𝑑𝑥 = 1
Improper Integral with Infinite Discontinuity
Integral of a function that becomes infinite at a point within the interval of integration.
𝑏
if f(x) is continuous on (a, b]
if f(x) is continuous on [a, b)
𝑏
then ∫𝑎 𝑓(𝑥) 𝑑𝑥 = lim+ ∫𝑐 𝑓(𝑥)𝑑𝑥
𝑐→𝑎
𝑏
𝑐
then ∫𝑎 𝑓(𝑥) 𝑑𝑥 = lim− ∫𝑎 𝑓(𝑥)𝑑𝑥
𝑐→𝑏
𝑏
𝑐
𝑏
if f(x) is continuous on [a,c) U (c,b] then ∫𝑎 𝑓(𝑥) 𝑑𝑥 = ∫𝑎 𝑓(𝑥) 𝑑𝑥 + ∫𝑐 𝑓(𝑥) 𝑑𝑥
Comparison Test
Show that
∞
∫0
x < 1 → x4 < x
𝑑𝑥
√𝑥+ 𝑥 4
so you can neglect x4 in 𝑥 + 𝑥 4 to see its behavior for small x, or in other words:
𝑥 + 𝑥4 ≈ x
x > 1 → x4 > x
converges.
→
1
𝑥+ 𝑥 4
≈
1
correct
𝑥
1
𝑥+ 𝑥 4
≤
1
1
or
𝑥
√𝑥+ 𝑥 4
≤
→
1
𝑥+ 𝑥 4
≈
1
𝑥4
correct
1
𝑥+ 𝑥 4
≤
1
𝑥4
or
1
√𝑥+ 𝑥 4
conclusion:
∞
∞
∫0
𝑑𝑥
√𝑥+ 𝑥 4
𝑑𝑥
√𝑥+ 𝑥 4
1 𝑑𝑥
≤ ∫0
≤
√𝑥
so you can neglect x in 𝑥 + 𝑥 4 to see its behavior for big x, or in other words:
𝑥 + 𝑥 4 ≈ x4
∫0
1
3
√𝑥
∞ 𝑑𝑥
+ ∫1
√ 𝑥4
1
𝑏 𝑑𝑥
1 𝑏
= (2√𝑥)0 + lim ∫1 2 = 2 + lim (− 𝑥) = 3
𝑥
𝑏→∞
𝑏→∞
1
it is converging. upper limit is 3.
≤
1
√𝑥 4