Section 10
10.3 Dot Product:
Dot product gives you a SCALAR
⟨𝑎1 , 𝑎2 ⟩ ∗ ⟨𝑏1 , 𝑏2 ⟩ = 𝑎1 𝑏1 + 𝑎2 𝑏2 = |𝑎||𝑏|cos⁡(𝜃)
*Dot product is zero when vectors are orthogonal!*
Projection:
𝑝𝑟𝑜𝑗𝑎 𝑏 =
𝑎∗𝑏
𝑎
𝑎∗𝑎
*this is a vector in the direction of a, we only changed the length of the vector not the direction*
10.4 Cross Product:
Cross product gives you a VECTOR; is orthogonal to the vectors that created it!
𝑖
𝑎 × 𝑏 = |𝑎1
𝑏1
𝑗
𝑎2
𝑏2
𝑘
𝑎3 | = |𝑎||𝑏|sin⁡(𝜃)
𝑏3
Parameterization:
For a Line:
For a Plane:
Ingredients:
Ingredients:
1. A point
2. A vector of the slopes ⟨𝑎, 𝑏, 𝑐⟩
1. A point on the plane⁡⁡𝑟 = (𝑥0 , 𝑦0 , 𝑧0 )
2. A normal vector
𝑥 = 𝑥0 + 𝑎𝑡
𝑛⃑ ∗ (𝑟 − 𝑟0 ) = 0
𝑦 = 𝑦0 + 𝑏𝑡
*bump into 4-D and use the gradient*
𝑧 = 𝑧0 + 𝑐𝑡
Linear Approximation:
𝑧 − 𝑧0 = 𝑓𝑥 (𝑥0 , 𝑦0 )(𝑥 − 𝑥0 ) + 𝑓𝑦 (𝑥0 , 𝑦0 )(𝑦 − 𝑦0 )
Chain Rule:
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
=
+
𝜕𝑡 𝜕𝑥 𝜕𝑡 𝜕𝑦 𝜕𝑡
MAX&MIN:
1.
2.
3.
4.
Take the derivative
Set equal to zero and solve for critical points
Take second derivative
2
𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − 𝑓𝑥𝑦
𝑖𝑓⁡𝐷 > 0, 𝑓𝑥𝑥 > 0⁡𝑚𝑖𝑛
𝑖𝑓⁡𝐷 < 0, 𝑠𝑎𝑑𝑑𝑙𝑒
𝑖𝑓⁡𝐷 > 0, 𝑓𝑥𝑥 < 0⁡𝑚𝑎𝑥
Absolute MAX&MIN:
1.
2.
3.
4.
5.
Take the derivative
Set equal to zero and solve for critical points
Check boundary conditions
Find corner points
The critical points from the boundary conditions, the corner points, and any critical points from
step 2 that are enclosed in the region need to be checked
6. No need to find D, just plug all the critical points found into the original function. The highest zvalue is your absolute max, the lowest z-value is your absolute min.
Section 12
12.1 Double Integrals:
For a curve y = f(x) going from x=a to b area under the curve and above the x axis is given by
𝑏
∫ 𝑓(𝑥)𝑑𝑥
𝑎
Similarly if z = f(x,y) represents an equation of a surface, then the volume under the surface is given by
𝑉 = ∬ 𝑓(𝑥, 𝑦)𝑑𝐴
𝑅
If f(x,y) is positive
According to FUBINI's THEOREM
𝑑⁡𝑏
𝑏⁡𝑑
∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ⁡ ∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦 = ⁡ ∬ 𝑓(𝑥, 𝑦)𝑑𝑥⁡𝑑𝑦⁡
𝑅
𝑐⁡𝑎
𝑎⁡𝑐
12.2
Consider two curves y = f(x) and y = g(x). The area between the two curves is given by
𝑏
𝐴 = ∫ (𝑓(𝑥) − 𝑔(𝑥))𝑑𝑥
𝑎
In this area
𝑏⁡𝑓(𝑥)
∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ⁡ ∬ 𝐹(𝑥, 𝑦)𝑑𝑦𝑑𝑥
𝑅
𝑎⁡𝑔(𝑥)
●If the x limits a and b have not been given, find them by doing f(x) = g(x)
12.3
To convert from Cartesian co-ordinates to polar co-ordinates put
𝑥 = 𝑟 cos 𝜃
y = r⁡sin 𝜃
2
r=√𝑥 2 + ⁡ 𝑦 2
𝛽⁡𝑏
●∬𝑅 𝑓(𝑥, 𝑦)𝑑𝐴 = ⁡ ∬𝛼⁡𝑎 𝑓(𝑟 cos 𝜃 , r⁡sin 𝜃)𝑟⁡𝑑𝑟𝑑𝜃⁡
12.5 TRIPLE INTEGRALS
𝑏⁡𝑓2 (𝑦)⁡𝑔2 (𝑥,𝑦)
𝑏𝑑⁡𝑓
∭ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑣 = ⁡ ∭ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑥⁡𝑑𝑦⁡𝑑𝑧 = ⁡
𝑣
∭
𝑎⁡𝑓1 (𝑦)⁡𝑔1(𝑥,𝑦)
𝑎⁡𝑐⁡𝑒
When f(x,y,z) = 1, above integral represents the volume.
12.6
To convert to cylindrical co-ordinates
𝑥 = 𝑟 cos 𝜃, y = r⁡sin 𝜃
2
𝑦
r=√𝑥 2 + ⁡ 𝑦 2 , tan 𝜃 = ⁡ 𝑥
𝑑𝑥⁡𝑑𝑦⁡𝑑𝑧 = 𝑟⁡𝑑𝑟⁡𝑑𝜃⁡𝑑𝑧
𝑓(𝑥, 𝑦, 𝑧)𝑑𝑧⁡𝑑𝑦⁡𝑑𝑥
12.7
To convert to spherical system, put
𝑥 = ⁡𝜌⁡ sin 𝜑 cos 𝜃 , 𝑦 = ⁡𝜌⁡ sin 𝜑⁡ sin 𝜃, 𝑧 = 𝜌⁡ cos 𝜑
𝜌 = ⁡ √𝑥 2 + 𝑦 2⁡ + 𝑧 2
𝑑𝑥⁡𝑑𝑦⁡𝑑𝑧 = ⁡ 𝜌2⁡⁡ sin 𝜑⁡𝑑𝜌⁡𝑑𝜃⁡𝑑𝜑⁡
SECTION 13: VECTOR CALCULUS:
13.1
A vector field F is called conservative vector field if there exists a function f such that
F=
f
For finding vector fields corresponding to a particular set of equations, consider sample points in the
field (ex. =0, y=0 etc.) and see which equations satisfy a given vector field.
13.2
For a curve C defined by 𝑥(𝑡)𝑖 + 𝑦(𝑡)𝑗 , line integral
𝑏
2
𝜕𝑥 2
𝜕𝑦 2
∫ 𝑓(𝑥, 𝑦) 𝑑𝑠 = ∫ 𝑓(𝑥(𝑡), 𝑦(𝑡)) √( ) + ⁡ ( ) ⁡⁡ 𝑑𝑡
𝜕𝑡
𝜕𝑡
𝑎
𝑐
●To parameterize a line segment starting at r0 and ending at r1, use r(t) = (1-t)* r0 + t*r1
For example to parametrize a line joining (-5,3) to (0,2) put
x = (1-t)(-5) + t(0) = 5*t -5
y = (1-t)(-3) + t(2) = 5*t -3
● Work done by a force F is given by
∫ 𝑓(𝑥, 𝑦) 𝑑𝑟
𝑐
13.3 FUNDAMENTAL THEOREM FOR LINE INTEGRAL
0 <= t <=1
Hence, if F =
f (i.e. F is conservative) then the integral can be evaluated by using the RHS above
●If a vector field is conservative, then
∫ 𝐹 𝑑𝑟
is path independent. Hence, the integral equates to the difference in between the two points.
●
∫ 𝐹 𝑑𝑟
For a closed path is equal to zero.
● If
𝐹(𝑥, 𝑦) = ⁡𝑃(𝑥, 𝑦)⁡𝑖 + 𝑄(𝑥, 𝑦)⁡𝑗⁡
Is conservative,
Then
𝜕𝑃⁡
𝜕𝑄
=⁡
𝜕𝑦
𝜕𝑥
13.4 GREEN's THEOREM
𝜕𝑄 𝜕𝑃
∫ 𝑃⁡𝑑𝑥 + ⁡⁡𝑄⁡𝑑𝑦 = ⁡ ∬ ( − ⁡ )⁡
𝜕𝑥
𝜕𝑦
𝑐
for positively oriented, smooth, closed curves.
13.5
For
𝐹 = 𝑃⁡𝑖 + 𝑄⁡𝑗 + 𝑅⁡𝑘⁡
Curl of F =
×f =
𝑖
𝑗
𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑃
𝑄
𝑅
Curl (
f) = 0 -> Curl of a conservative F is 0.
Div (curl F) = 0
SUMMARY OF LINE INTEGRALS
13.6
A surface can be defined as in terms of two parameters (u,v)
as 𝑟(𝑢, 𝑣) = 𝑥(𝑢, 𝑣)𝑖 + 𝑦(𝑢, 𝑣)⁡𝑗 + 𝑧(𝑢, 𝑣)𝑘
For 𝑟 = ⁡⁡𝑥⁡𝑖 + 𝑦⁡𝑗 + 𝑧⁡𝑘
Tangent plane at (u0,v0) is given by 𝑟𝑢 × 𝑟𝑣
Where
𝑟𝑢 = ⁡
𝜕𝑥
𝜕𝑦
𝜕𝑧
⁡𝑖 + ⁡ ⁡𝑗 + ⁡ ⁡𝑘
𝜕𝑢
𝜕𝑢
𝜕𝑢
𝑟𝑢 = ⁡
𝜕𝑥
𝜕𝑦
𝜕𝑧
⁡𝑖 + ⁡ ⁡𝑗 + ⁡ ⁡𝑘
𝜕𝑣
𝜕𝑣
𝜕𝑣
At 𝑢 = ⁡ 𝑢0, ⁡𝑣 = 𝑣0
At 𝑢 = ⁡ 𝑢0, ⁡𝑣 = 𝑣0
●Area of surface = ∬𝐷 |𝑟𝑎 ⁡ × ⁡ 𝑟𝑏 | ⁡𝑑𝐴
● If z = f(x,y)
𝜕𝑧 2
𝜕𝑧 2
𝐴(𝑠) = ⁡ ∬ √1 + ⁡ ( ) + ⁡ ( ) ⁡𝑑𝐴
𝜕𝑥
𝜕𝑦
13.7 Surface integral
∬ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑠 = ⁡ ∬ 𝑓(𝑟)|𝑟𝑎 × 𝑟𝑏 |⁡𝑑𝐴
EXAMPLE PROBLEMS:
1.
2.
3.
4.
⟨2,4⟩ ∙ ⟨3, −1⟩
⟨1,3,4⟩ × ⟨2,7, −5⟩
Find the equation of a plane that passes through the points P(1,3,2), Q(3,-1,6), and R(5,2,0)
𝐹(𝑥, 𝑦) = 4 − 𝑥 2 − 2𝑦 2 ⁡𝑓𝑖𝑛𝑑⁡𝐹𝑥 (1,1)𝑎𝑛𝑑⁡𝐹𝑦 (1,1)
𝑥
5. 𝐹(𝑥, 𝑦) = 𝑠𝑖𝑛 (1+𝑦) find the partial derivatives
𝑑𝑧
6. If 𝑧 = 𝑥 2 𝑦 + 3𝑥𝑦 4 𝑤ℎ𝑒𝑟𝑒⁡𝑥 = 𝑠𝑖𝑛2𝑡⁡𝑛𝑑⁡𝑦 = 𝑐𝑜𝑠𝑡⁡𝑓𝑖𝑛𝑑 𝑑𝑡 when t=0
7. Find the direction the function 𝐹(𝑥, 𝑦) = 𝑥𝑒 𝑦 increases the fastest at the point P(2,0). What is
the maximum rate of change?
8. Find the absolute maximum and minimum values of the function 𝐹(𝑥, 𝑦) = 𝑥 2 − 2𝑥𝑦 + 2𝑦 on
the rectangle 𝐷 = {(𝑥, 𝑦)|0 ≤ 𝑥 ≤ 3, 0 ≤ 𝑦 ≤ 2}
2
9. Use polar co-ordinates to find the volume of the solid above the cone 𝑧 = ⁡ √𝑥 2 + ⁡ 𝑦 2 and
below the sphere 𝑥 2 + ⁡ 𝑦 2 + ⁡ 𝑧 2 = 1
10. Evaluate ∭𝐸 𝑧⁡𝑑𝑉 is the solid tetrahedron bounded by the four planes x=0, y=0,z=0 and
x+y+y=1
11. Use Green’s Theorem to evaluate line integral along the positively oriented curve ∫𝐶 ⁡𝑦 3 𝑑𝑥 −
𝑥 3 𝑑𝑦 where C is the circle 𝑥 2 + ⁡ 𝑦 2 = 4
12. Determine whether or not the vector field is conservative. If it is conservative, find a function f
such that F =
F(x,y,z) = yz i + xz j + xy k