MAT232 Final Exam Prep
April 2023 Exam
A
1
r → ( t ) = ⟨ t 2 + 2 t + 3 , 4 t cos t , 2 e 3 t ⟩ r → ′ ( t ) = ⟨ 2 t + 2 , 4 cos t − 4 t sin t , 6 e 3 t ⟩ r → ′ ( 0 ) = ⟨ 2 , 4 , 6 ⟩ r → ( 0 ) = ⟨ 3 , 0 , 2 ⟩ T → ( t ) = P + D → T → ( t ) = ⟨ 3 , 0 , 2 ⟩ + t ⟨ 2 , 4 , 6 ⟩ T → ( t ) = ⟨ 3 , 0 , 2 ⟩ + t ⟨ 1 , 2 , 3 ⟩
2
Arc Length
r → ( t ) = ⟨ t 2 2 , ( 2 t + 1 ) 3 2 3 ⟩ for 0 ≤ t ≤ 4
| r → ′ ( t ) | = ∫ 0 4 x ′ ( t ) 2 + y ′ ( t ) 2 \differential t x ( t ) = t 2 2 x ′ ( t ) = 2 t 2 = t y ( t ) = 1 3 ( 2 t + 1 ) 3 2 y ′ ( t ) = 1 3 3 2 ( 2 t + 1 ) 1 2 ( 2 ) y ′ ( t ) = 1 3 3 ( 2 ) 2 ( 2 t + 1 ) 1 2 y ′ ( t ) = 1 3 3 ( 2 t + 1 ) 1 2 y ′ ( t ) = ( 2 t + 1 ) 1 2 = 2 t + 1 | r → ′ ( t ) | = ∫ 0 4 t 2 + 2 t + 1 \differential t | r → ′ ( t ) | = ∫ 0 4 ( t + 1 ) 2 \differential t | r → ′ ( t ) | = ∫ 0 4 t + 1 \differential t | r → ′ ( t ) | = ( 1 2 t 2 + t ) 0 4 | r → ′ ( t ) | = ( 1 2 ( 4 ) 2 + ( 4 ) ) = 12
3
Absolute max and minimum
g ( x , y ) = x 2 − 2 x y + 2 y R = { ( x , y ) : 0 ≤ x ≤ 3 , 0 ≤ y ≤ 2 } g x ( x , y ) = 2 x − 2 y g y ( x , y ) = − 2 x + 2 g x = 0 g y = 0 g x x ( x , y ) = 2 g y y ( x , y ) = 0 g x y ( x , y ) = − 2 D ( x , y ) = g x x ( x , y ) g y y ( x , y ) − [ g x y ( x , y ) ] 2 D ( x , y ) = ( 2 ) ( 0 ) − [ ( − 2 ) ] 2 = − 4 for ( 2 , 2 ) D < 0
Then plug in points:
g ( 0 , 0 ) = ( 0 ) 2 − ( 2 ) ( 0 ) ( 0 ) + ( 2 ) ( 0 ) = 0 g ( 0 , 2 ) = ( 0 ) 2 − ( 2 ) ( 0 ) ( 2 ) + ( 2 ) ( 2 ) = 4 g ( 3 , 0 ) = ( 3 ) 2 − ( 2 ) ( 3 ) ( 0 ) + ( 2 ) ( 0 ) = 9 g ( 3 , 2 ) = ( 3 ) 2 − ( 2 ) ( 3 ) ( 2 ) + ( 2 ) ( 2 ) = 1 g ( 1 , 1 ) = ( 1 ) 2 − ( 2 ) ( 1 ) ( 1 ) + ( 2 ) ( 1 ) = 1 g ( 2 , 2 ) = ( 2 ) 2 − ( 2 ) ( 2 ) ( 2 ) + ( 2 ) ( 2 ) = 0
4
Which are true
x = t − t 2 y = t − t 3 d 2 y d x 2 = 2 − 6 t + 6 t 2 ( 1 − 2 t ) 3
\differential y \differential x = \differential y \differential t \differential x \differential t \differential y \differential t = 1 − 3 t 2 \differential x \differential t = 1 − 2 t \differential y \differential x = 1 − 3 t 2 1 − 2 t d 2 y d x 2 = \differential \differential t ( \differential y \differential x ) \differential x \differential t d 2 y d x 2 = \differential \differential t ( 1 − 3 t 2 1 − 2 t ) 1 − 2 t u = 1 − 3 t 2 u ′ = − 6 t v = 1 − 2 t v ′ = − 2 \differential \differential t ( u v ) = u ′ v − v ′ u v 2 \differential \differential t ( u v ) = ( − 6 t ) ( 1 − 2 t ) − ( − 2 ) ( 1 − 3 t 2 ) ( 1 − 2 t ) 2 \differential \differential t ( u v ) = ( − 6 t + 12 t 2 ) − ( − 2 + 6 t 2 ) ( 1 − 2 t ) 2 \differential \differential t ( u v ) = ( − 6 t + 6 t 2 + 2 ) ( 1 − 2 t ) 2 d 2 y d x 2 = ( − 6 t + 6 t 2 + 2 ) ( 1 − 2 t ) 2 1 − 2 t d 2 y d x 2 = ( − 6 t + 6 t 2 + 2 ) ( 1 − 2 t ) 3 True
r = 4 2 cos θ − sin θ m = 2 y b = − 4
2 r cos θ − r sin θ = 4 2 x − y = 4 − y = 4 − 2 x y = 2 x − 4 True
Directions of zero change at a point ( a , b ) z = f ( x , y ) ∇ → f ( a , b )
Tangent plane to the surface f ( x , y ) = ln ( 2 x + y ) ( − 1 , 3 ) 2 x + y − z − 1 = 0
f x = 2 2 x + y f y = 1 2 x + y f x ( − 1 , 3 ) = 2 ( 2 ) ( − 1 ) + 3 = 2 f y ( − 1 , 3 ) = 1 ( 2 ) ( − 1 ) + 3 = 1 f ( − 1 , 3 ) = ln ( ( 2 ) ( − 1 ) + 3 ) = 0 z − z 0 = f x ( − 1 , 3 ) ( x − x 0 ) + f y ( − 1 , 3 ) ( y − y 0 ) z = 2 ( x + 1 ) + ( y − 3 ) z = 2 x + 2 + y − 3 z = 2 x + y − 1 2 x + y − 1 − z = 0 True
5
Which are true
∫ 0 1 ∫ 0 3 x e x y \differential x \differential y = e 3 − 4
0 ≤ x ≤ 3 0 ≤ y ≤ 1 ∫ 0 3 ∫ 0 1 x e x y \differential y \differential x ∫ 0 3 x ∫ 0 1 e x y \differential y \differential x ∫ 0 3 x [ x − 1 e x y ] 0 1 \differential x ∫ 0 3 x [ ( e x x − 1 x ) ] \differential x ∫ 0 3 e x − 1 \differential x [ e x − x ] 0 3 [ e 3 − 3 ] − [ e 0 − 0 ] [ e 3 − 3 ] − [ 1 ] [ e 3 − 3 ] − 1 e 3 − 3 − 1 e 3 − 4 True
f ( x , y , z ) = 2 x y + z C r → ( t ) = ⟨ cos t , sin t , t ⟩ 0 ≤ t ≤ π ∫ C f ( x , y , z ) \differential s = 2 ∫ 0 π sin 2 t + t \differential t
r → ′ ( t ) = ⟨ − sin t , cos t , 1 ⟩ | r → ′ ( t ) | = ( − sin t ) 2 + ( cos t ) 2 + 1 | r → ′ ( t ) | = sin 2 t + cos 2 t + 1 | r → ′ ( t ) | = 2 f ( r → ( t ) ) ∫ C f ( x , y , z ) \differential s = ∫ 0 π f ( r → ( t ) ) | r → ′ ( t ) | \differential t ∫ C f ( x , y , z ) \differential s = ∫ 0 π f ( r → ( t ) ) | 2 | \differential t ∫ C f ( x , y , z ) \differential s = 2 ∫ 0 π sin 2 t + t \differential t
A line integral ∫ C F → ⋅ d r → D F → ∫ C ′ F → ⋅ d r → = 0 C ′ D
Let f ( x , y , z ) = x y z F → = ⟨ y z , x z , x y ⟩ = ∇ → f C A ( − 1 , 3 , 9 ) B ( 1 , 6 , − 4 ) ∫ C F → ⋅ d r → = 3
It's a conservative field
So we can just do f ( B ) − f ( A )
f ( B ) = ( 1 ) ( 6 ) ( − 4 ) = − 24 f ( A ) = ( − 1 ) ( 3 ) ( 9 ) = − 27 − 24 − − 27 = 3 True
B
1
2
a
f a u = a ( x + c t ) w = f ( u ) ∂ 2 w ∂ t 2 = c 2 ∂ 2 w ∂ x 2 u = a x + a c t w t t
w t = \differential f \differential u ∂ u ∂ t w t t = ∂ ∂ t w t w t t = ∂ ∂ t [ \differential f \differential u ∂ u ∂ t ] \differential f \differential u ∂ u ∂ t = a c f ′ ( u ) ∂ ∂ t [ a c f ′ ( u ) ] a c ∂ ∂ t [ f ′ ( u ) ] a c ∂ ∂ t [ f ′ ( u ) ] = a c [ d 2 f d u 2 ∂ u ∂ t ] a c ∂ ∂ t [ f ′ ( u ) ] = a c [ a c f ″ ( u ) ] a c ∂ ∂ t [ f ′ ( u ) ] = a 2 c 2 f ″ ( u )
w x x
w x = \differential f \differential u ∂ u ∂ x w x x = ∂ ∂ x [ \differential f \differential u ∂ u ∂ x ] \differential f \differential u ∂ u ∂ x = a f ′ ( u ) ∂ ∂ x [ a f ′ ( u ) ] a ∂ ∂ x [ f ′ ( u ) ] a ∂ ∂ x [ f ′ ( u ) ] = a [ d 2 f d u 2 ∂ u ∂ x ] a ∂ ∂ x [ f ′ ( u ) ] = a [ a f ″ ( u ) ] a ∂ ∂ x [ f ′ ( u ) ] = a 2 f ″ ( u )
w t t = c 2 w x x a 2 c 2 f ″ ( u ) = c 2 a 2 f ″ ( u ) True
b
Find max and min values of the function f ( x , y ) = 3 x + 4 y x 2 + y 2 = 1
g ( x , y ) = x 2 + y 2 − 1 ∇ → f = ⟨ 3 , 4 ⟩ ∇ → g = ⟨ 2 x , 2 y ⟩ ∇ → f = λ ∇ → g 3 = λ ( 2 x ) 4 = λ ( 2 y ) 3 2 x = λ 4 2 y = λ x = 3 2 λ y = 4 2 λ 3 2 x = 4 2 y 6 y = 8 x 3 y = 4 x y = 4 3 x 3 4 y = x x 2 + ( 4 3 x ) 2 = 1 x 2 + 16 9 x 2 = 1 9 9 x 2 + 16 9 x 2 = 1 25 9 x 2 = 1 25 x 2 = 9 x 2 = 9 25 x = ± 9 25 = ± 3 5 y = 4 3 x y = 4 3 ( ± 9 25 ) y = ± 4 3 ( 9 25 ) y = ± 4 3 9 25 = ± 4 3 3 5 y = ± 12 15 f ( ± 3 5 , ± 12 15 ) f ( 3 5 , 12 15 ) = 9 5 + 48 15 = 5 f ( 3 5 , − 12 15 ) = 9 5 − 48 15 = − 7 5 f ( − 3 5 , 12 15 ) = − 9 5 + 48 15 = 7 5 f ( − 3 5 , − 12 15 ) = − 9 5 − 48 15 = − 5 Absolute minimum is − 3 5 , − 12 15
Absolute max is 3 5 , 12 15
3
4
We have the integral ∫ 0 4 ∫ y 2 y 2 + 1 2 x − y 2 \differential x \differential y
a
u = 2 x − y 2 v = y 2 Sketch the region
0 ≤ y ≤ 4 y 2 ≤ x ≤ y 2 + 1 x = y 2 x = y 2 + 1 Intersections
0 = 2 x ⟹ x = 0 4 = 2 x ⟹ x = 2 0 = 2 x − 2 ⟹ x = 1 4 = 2 x − 2 ⟹ x = 3
Transform Parallelogram
( 0 , 0 ) ( 1 , 0 ) ( 2 , 4 ) ( 3 , 4 ) ( u = 0 , v = 0 ) ( u = 1 , v = 0 ) ( u = 0 , v = 2 ) ( u = 1 , v = 2 )
left=-5; right=5;
top=5; bottom=-5;
---
y=2x
y=2x-2
y=0|dashed
y=4|dashed
(0,0)
(2,4)
(1,0)
(3,4)
left=-5; right=5;
top=5; bottom=-5;
---
(0,0)
(1,0)
(0,2)
(1,2)
b
Compute the jacobian
u = 2 x − y 2 u = 1 2 ( 2 x − y ) v = y 2 v = 1 2 ( y ) u x = 1 u y = − 1 2 v x = 0 v y = 1 2 | J | = | u x u y v x v y | | J | = | 1 − 1 2 0 1 2 | | J | = ( 1 ) ( 1 2 ) − ( − 1 2 ) ( 0 ) = 1 2
c
f u , v ( u , v ) = f x , y ( u ( x , y ) , v ( x , y ) ) | J | f ( x , y ) = 2 x − y 2 u = 2 x − y 2 v = y 2 Inverse
2 u = 2 x − 2 v 2 u + 2 v = 2 x u + v = x y = 2 v
Jacobian
f ( u , v ) = ( 2 ) ( u + v ) − 2 v 2 f ( u , v ) = 2 u + 2 v − 2 v 2 f ( u , v ) = 2 u 2 f ( u , v ) = u u | J | f u , v ( u , v ) = 2 u 0 ≤ u ≤ 1 0 ≤ v ≤ 2 ∫ 0 1 2 u ∫ 0 2 1 \differential v \differential u ∫ 0 1 2 u [ v ] 0 2 \differential u ∫ 0 1 4 u \differential u 4 ∫ 0 1 u \differential u 4 [ 1 2 u 2 ] 0 1 4 [ 1 2 ( 1 ) 2 ] 4 [ 1 2 ] = 2
5
a
b
Using the previous parameterization compute:
∫ C ( x y 2 + x 2 ) \differential x + ( 4 x − 1 ) \differential y Green's Theorem
Since ∫ C P \differential x + Q \differential y = ∬ D ( ∂ Q ∂ x − ∂ P ∂ y ) \differential A
P = x y 2 + x 2 Q = 4 x − 1 ∂ P ∂ y = 2 x y ∂ Q ∂ x = 4 ∬ D 4 − 2 x y \differential A − 3 ≤ x ≤ 0 0 ≤ y ≤ x + 3 ∫ − 3 0 ∫ 0 x + 3 4 − 2 x y \differential y \differential x ∫ − 3 0 2 ∫ 0 x + 3 2 − x y \differential y \differential x ∫ − 3 0 2 [ 2 y − x 2 y 2 ] 0 x + 3 \differential x ∫ − 3 0 2 [ 2 ( x + 3 ) − x 2 ( x + 3 ) 2 ] \differential x ∫ − 3 0 2 [ 2 x + 6 − x 2 ( x 2 + 6 x + 9 ) ] \differential x ∫ − 3 0 2 [ 2 x + 6 − ( x 3 2 + 6 x 2 2 + 9 x 2 ) ] \differential x 2 ∫ − 3 0 [ 2 x + 6 − ( x 3 2 + 6 x 2 2 + 9 x 2 ) ] \differential x 2 ∫ − 3 0 2 x + 6 − x 3 2 − 6 x 2 2 − 9 x 2 \differential x 2 [ x 2 + 6 x − 1 2 1 4 x 4 − x 3 − 9 2 1 2 x 2 ] − 3 0 2 [ x 2 + 6 x − 1 8 x 4 − x 3 − 9 4 x 2 ] − 3 0 2 [ ( ( 0 ) 2 + 6 ( 0 ) − 1 8 ( 0 ) 4 − ( 0 ) 3 − 9 4 ( 0 ) 2 ) − ( ( − 3 ) 2 + 6 ( − 3 ) − 1 8 ( − 3 ) 4 − ( − 3 ) 3 − 9 4 ( − 3 ) 2 ) ] − 2 [ ( ( − 3 ) 2 + 6 ( − 3 ) − 1 8 ( − 3 ) 4 − ( − 3 ) 3 − 9 4 ( − 3 ) 2 ) ] − 2 [ ( 9 − 18 − 81 8 − ( − 27 ) − 81 4 ) ] − 2 [ ( 9 − 18 − 81 8 − ( − 27 ) − 81 4 ) ] = 99 4
Directly
C 1
∫ C ( x y 2 + x 2 ) \differential x + ( 4 x − 1 ) \differential y x ( t ) = 0 x ′ ( t ) = 0 y ( t ) = 3 t y ′ ( t ) = 3 For 0 ≤ t ≤ 1
P = x y 2 + x 2 = ( 0 ) ( 3 ) 2 + ( 0 ) 2 = 0 Q = ( 4 ) ( 0 ) − ( 1 ) = − 1 ∫ 0 1 P \differential x \differential t + Q \differential y \differential t \differential t ∫ 0 1 0 ( 0 ) + ( − 1 ) ( 3 ) \differential t ∫ 0 1 − 3 \differential t 3 ∫ 1 0 1 \differential t 3 [ t ] 1 0 3 [ − 1 ] = − 3
C 2
∫ C ( x y 2 + x 2 ) \differential x + ( 4 x − 1 ) \differential y x ( t ) = − 3 t y ( t ) = − 3 t + 3 for 0 ≤ t ≤ 1
x ′ ( t ) = − 3 y ′ ( t ) = − 3 P | x ( t ) , y ( t ) = ( − 3 t ) ( − 3 t + 3 ) 2 + ( − 3 t )
( − 3 t ) ( − 3 t + 3 ) ( − 3 t + 3 ) + ( − 3 t ) ( − 3 t ) ( 9 t 2 − 18 t + 9 ) + ( − 3 t ) ( − 27 t 3 + 54 t 2 − 27 t ) + ( − 3 t ) − 27 t 3 + 54 t 2 − 27 t − 3 t − 27 t 3 + 54 t 2 − 30 t
Q | x ( t ) , y ( t ) = 4 ( − 3 t ) − 1 = − 12 t − 1 ∫ 0 1 P \differential x \differential t + Q \differential y \differential t \differential t ∫ 0 1 P ( − 3 ) + Q ( − 3 ) \differential t − 3 ∫ 0 1 P + Q \differential t − 3 ∫ 0 1 − 27 t 3 + 54 t 2 − 30 t − 12 t − 1 \differential t − 3 ∫ 0 1 − 27 t 3 + 54 t 2 − 42 t − 1 \differential t − 3 [ − 27 1 4 t 4 + 54 1 3 t 3 − 42 1 2 t 2 − t ] 0 1 − 3 [ − 27 1 4 + 54 1 3 − 42 1 2 − 1 ] = 129 4
C 3
∫ C ( x y 2 + x 2 ) \differential x + ( 4 x − 1 ) \differential y x ( t ) = 3 t − 3 y ( t ) = 0 for 0 ≤ t ≤ 1
x ′ ( t ) = 3 y ′ ( t ) = 0 P | x ( t ) , y ( t ) = ( 3 t − 3 ) 2 = 9 t 2 − 18 t 2 + 9 Q | x ( t ) , y ( t ) = 4 ( 3 t − 3 ) − 1 = 12 t − 13 ∫ 0 1 P \differential x \differential t + Q \differential y \differential t \differential t 3 ∫ 0 1 9 t 2 − 18 t 2 + 9 \differential t 27 ∫ 0 1 t 2 − 2 t 2 + 1 \differential t 27 [ 1 3 t 3 − 2 3 t 3 + t ] 0 1 27 [ 1 3 − 2 3 + 1 ] = 18
C 1 + C 2 + C 3
Exam 2022
A
1
C x = t 2 y = t 3 − 3 t Find where C
x ′ ( t ) = 2 t y ′ ( t ) = 3 t 2 − 3 Since \differential x \differential y = \differential x \differential t \differential y \differential t
Then when x ′ ( t ) = 0
When y ′ ( t ) = 0
So there are vertical tangent lines at t = 0 t = ± 1
2
Find arc length
r → ( t ) = ⟨ t 2 , 2 t , ln t ⟩ 1 ≤ t ≤ 2 | r → ( t ) | = ∫ 1 2 \differential x \differential t 2 + \differential y \differential t 2 + \differential z \differential t 2 \differential t \differential x \differential t = 2 t \differential y \differential t = 2 \differential z \differential t = 1 t | r → ( t ) | = ∫ 1 2 4 t 2 + 4 + 1 t 2 \differential t | r → ( t ) | = ∫ 1 2 t − 2 ( 4 t 4 + 4 t 2 + 1 ) \differential t | r → ( t ) | = ∫ 1 2 t − 2 ( 2 t 2 + 1 ) 2 \differential t | r → ( t ) | = ∫ 1 2 t − 1 ( 2 t 2 + 1 ) \differential t | r → ( t ) | = ∫ 1 2 2 t + t − 1 \differential t | r → ( t ) | = [ t 2 + ln t ] 1 2 | r → ( t ) | = [ 4 + ln 2 ] − [ 1 ] | r → ( t ) | = 3 + ln 2
3
A , B , C B > 0 C ≠ 0 Find all values of α T ( x , t ) = A + e − B t sin ( C x ) ∂ T ∂ t = α ∂ 2 T ∂ x 2
∂ T ∂ t = − B e − B t sin ( C x ) ∂ T ∂ x = C cos ( C x ) e − B t ∂ 2 T ∂ x 2 = − e − B t C 2 sin ( C x ) − B e − B t sin ( C x ) = ( α ) ( − e − B t C 2 sin ( C x ) ) − B e − B t = ( α ) ( − e − B t C 2 ) B e − B t = ( α ) ( e − B t C 2 ) B = ( α ) ( C 2 ) B C 2 = α
4
Find absolute max and minimum
f ( x , y ) = x y − 3 x + y D ( 0 , 0 ) , ( 2 , 0 ) , ( 0 , 2 ) f x = y − 3 f y = x − 1 f x x = 0 f y y = 0 f x y = 1 f x = 0 f y = 0 ( 1 , 3 ) Boundary
D ( x , y ) = f x x f y y − f x y 2 D ( x , y ) = − 1
D < 0 Saddle point on interior
5
Which are true?
1: f ( x , y ) = 2 x y u → = ⟨ 4 5 , 3 5 ⟩ D u → f ( 1 , 3 ) = 4
| u → | = ( 4 5 ) 2 + ( 3 5 ) 2 | u → | = 16 25 + 9 25 = 1 This is a unit vector
D u → f ( x 0 , y 0 ) = ∇ → f ( x 0 , y 0 ) ⋅ u → ∇ → f = ⟨ 2 y , 2 x ⟩ ∇ → f ( 1 , 3 ) = ⟨ 2 , 6 ⟩ ⟨ 2 , 6 ⟩ ⋅ ⟨ 4 5 , 3 5 ⟩ 8 5 + 18 5 26 5 ≠ 4
2: the function F → ( x , y ) = ⟨ x 2 y , cos y , e ⟩
P = x 2 y Q = cos y R = e P y = Q x x 2 = 0 Not a gradient field
3: the curve r → ( t ) = ⟨ 0 , sin 2 t , 4 sin t − 3 ⟩
x = 0 y = sin 2 t z = 4 sin t − 3 z − 3 = 4 sin t z − 3 4 = sin t y = ( sin t ) 2 y = ( z − 3 4 ) 2 y = 1 16 ( z − 3 ) 2 This is a parabola
4: ∫ C ∇ → f ⋅ d r → = 4 f ( x , y , z ) = cos π x + sin π y − x y z C ( 1 , 1 2 , 2 ) → ( 2 , 1 , − 1 )
If we're path independent
f ( B ) − f ( A ) f ( 2 , 1 , − 1 ) − f ( 1 , 1 2 , 2 ) ( cos 2 π + sin π + 2 ) − ( cos π + sin π 2 − 1 ) ( 3 ) − ( − 1 ) = 4 To prove it's path independent we need to know if it's a vector field
I can assume it's probably one, as we're given ∇ → f ⋅ d r →
B
1
a
z x = x ( t ) y = y ( t ) z x ( 1 , 2 ) = 4 z y ( 1 , 2 ) = 2 If x y x ( 0 ) = 1 y ( 0 ) = 1 x ′ ( 0 ) = 0.5 z ′ ( 0 ) = 2 y ′ ( 0 )
z ( x ( t ) , y ( t ) ) ∂ z ∂ x \differential x \differential t + ∂ z ∂ y \differential y \differential t z ′ ( 0 ) = z x ( x ( 0 ) , y ( 0 ) ) x ′ ( 0 ) + z y ( x ( 0 ) , y ( 0 ) ) y ′ ( 0 ) z ′ ( 0 ) = ( 4 ) ( 0.5 ) + ( 2 ) y ′ ( 0 ) 2 = ( 4 ) ( 0.5 ) + ( 2 ) y ′ ( 0 ) 2 = 2 + 2 y ′ ( 0 ) 1 = 1 + y ′ ( 0 ) 0 = y ′ ( 0 )
b
ψ x 2 + x y 2 − y z 2 + 8 z = − 4 A ( 2 , 0 , − 1 ) Find the equation of the tangent plane ψ ( 0 , 0 , 0 )
f ( x , y , z ) = x 2 + x y 2 − y z 2 + 8 z f ( x , y , z ) = − 4 f x = 2 x + y 2 f y = 2 x y − z 2 f z = 2 y z + 8 ∇ → f ( x , y , z ) = ⟨ 2 x + y 2 , 2 x y − z 2 , 2 y z + 8 ⟩ This is our normal vector
0 = f x ( x 0 , y 0 , z 0 ) ( x − x 0 ) + f y ( x 0 , y 0 , z 0 ) ( y − y 0 ) + f z ( x 0 , y 0 , z 0 ) ( z − z 0 ) 0 = f x ( 2 , 0 , − 1 ) ( x − 2 ) + f y ( 2 , 0 , − 1 ) ( y ) + f z ( 2 , 0 , − 1 ) ( z + 1 ) 0 = ( ( 2 ) ( 2 ) ) ( x − 2 ) + ( − ( − 1 ) 2 ) ( y ) + ( 8 ) ( z + 1 ) = 4 x − y + 8 z 0 = 4 ( 0 ) − ( 0 ) + 8 ( 0 ) = 0 Origin lies on ψ
2
a
f ( x , y ) = 2 x y In what directions at the point ( 1 , 2 ) f = 4
∇ → f ( x , y ) = ⟨ 2 y , 2 x ⟩ D u → f ( x , y ) = ∇ → f ( x , y ) ⋅ u → D u → f ( x , y ) = ⟨ 2 y , 2 x ⟩ ⋅ u → 4 = ⟨ 2 y , 2 x ⟩ ⋅ u → 4 = 2 y u 1 + 2 x u 2 4 = 2 ( 2 ) u 1 + 2 ( 1 ) u 2 2 = 2 u 1 + u 2 1 = u 1 + 1 2 u 2 1 − 1 2 u 2 = u 1 u → = ⟨ 1 − 1 2 u 2 , u 2 ⟩ ( 1 − 1 2 u 2 ) 2 + u 2 2 = 1 ( 1 − 1 2 u 2 ) 2 + u 2 2 = 1 1 − 1 2 u 2 − 1 2 u 2 + 1 4 u 2 2 + u 2 2 = 1 1 − u 2 + 1 4 u 2 2 + 4 4 u 2 2 = 1 1 − u 2 + 5 4 u 2 2 = 1 5 4 u 2 2 − u 2 = 0 u 2 ( 5 4 u 2 − 1 ) = 0 5 4 u 2 = 1
u 2 = 4 5 u 1 = 1 − 1 2 u 2 u 1 = 1 − 1 2 4 5 u 1 = 1 − 4 10 u 1 = 10 10 − 4 10 u 1 = 6 10 u 1 = 3 5 u → = ⟨ 3 5 , 4 5 ⟩ Verify
u → = ⟨ − 3 5 , − 4 5 ⟩
u 2 = 0
⟹ u 1 = 1 Clearly unit vector
u → = ⟨ 1 , 0 ⟩
b
Determine the largest value of the product x y z
x , y , z ∈ R x + y + z = 100 f ( x , y , z ) = 100 f ( x , y , z ) = x + y + z g ( x , y , z ) = x y z ∇ → f ( x , y , z ) = ⟨ 1 , 1 , 1 ⟩ ∇ → g ( x , y , z ) = ⟨ y z , x z , x y ⟩ 1 = λ y z 1 = λ x z 1 = λ x y λ = 1 y z λ = 1 x z λ = 1 x y 1 y z = 1 x z = 1 x y x x y z = y x y z = z x y z x = y = z x + y + z = 100 3 x = 100 x = 100 3 = y = z x y z ( 100 3 ) ( 100 3 ) ( 100 3 ) = 1000000 27 However if x , y , z ∈ R ∞
3
4
∬ R x y 3 \differential A R y = x y = 3 x x y = 1 x y = 4 Use the change of variables u = y x v = x y
a
Sketch the region R
x y = 1 x y = 1
3 x 2 = 1 x 2 = 1 3 x = ± 1 3 x = 1 3 ⟹ 1 = y 1 3 ⟹ 1 1 3 = y
x y = 4 x y = 4
3 x 2 = 4 x 2 = 4 3 x = ± 4 3 ⟹ 4 3 ⟹ 4 3 y = 4 ⟹ y = 4 4 3
left=-5; right=5;
top=5; bottom=-5;
---
y=x
y=3x
xy=1
xy=4
(1,1)
(\sqrt{1/3},1/\sqrt{1/3})
(2,2)
(\sqrt{4/3},4/\sqrt{4/3})
( 1 , 1 ) ( 1 3 , 1 3 − 1 )
u = 1 3 − 1 1 3 u = 1 1 3 1 3 u = 1 1 3 = 3 v = x y v = 1 3 1 3 = 1
( 2 , 2 ) ( 4 3 , 4 4 3 )
u = 4 4 3 − 1 4 3 u = 4 4 3 4 3 u = 4 4 3 u = 4 ÷ 4 3 u = 4 ⋅ 3 4 u = 3 v = x y v = 4
( 1 , 1 ) ( 3 , 1 ) ( 1 , 4 ) ( 3 , 4 ) left=-5; right=5;
top=5; bottom=-5;
---
(1,1)
(3,1)
(1,4)
(3,4)
b
Compute the jacobian
u = y x − 1 v = x y | J | = | u x u y v x v y | | J | = | − y x − 2 x − 1 y x | | J | = ( x ) ( − y x − 2 ) − ( x − 1 ) ( y ) | J | = ( x ) ( − y x 2 ) − ( y x ) | J | = ( − y x ) − ( y x ) | J | = − 2 y x | J | = 2 y x
c
u = y x u x = y v = x y v = u x 2 v u = x 2 x = v u y = u x y = u v u | J | − 1 = − x 2 y | J | − 1 = − v u 2 u v u | J | − 1 = − 1 2 u | J | − 1 = 1 2 u f x , y ( x , y ) = x y 3 f u , v ( u , v ) = f x , y ( x ( u , v ) , y ( u , v ) ) | J | − 1 f u , v ( u , v ) = ( 1 2 u ) f x , y ( v u , u v u ) f u , v ( u , v ) = ( 1 2 u ) ( v u ( u v u ) 3 ) f u , v ( u , v ) = ( 1 2 u ) ( v u ( u v ) ( u v u ) ) f u , v ( u , v ) = ( 1 2 u ) ( v 2 ) f u , v ( u , v ) = ( v 2 2 u ) ∫ 1 4 ∫ 1 3 v 2 2 u \differential u \differential v ∫ 1 4 v 2 ∫ 1 3 1 2 u \differential u \differential v ∫ 1 4 v 2 [ 1 2 ln 2 u ] 1 3 \differential v ∫ 1 4 v 2 2 [ ln 2 ( 3 ) − ln 2 ] \differential v ∫ 1 4 v 2 2 [ ln 3 ] \differential v ln 3 2 ∫ 1 4 v 2 \differential v ln 3 2 [ 1 3 v 3 ] 1 4 ln 3 2 1 3 [ 4 3 − 4 ] ln 3 2 1 3 [ ( 4 ) ( 4 2 − 1 ) ] ln 3 2 1 3 [ ( 4 ) ( 16 − 1 ) ] ln 3 2 1 3 [ ( 4 ) ( 15 ) ] ln 3 2 1 3 [ ( 4 ) ( 3 ) ( 5 ) ] ln 3 2 [ ( 4 ) ( 5 ) ] ln 3 2 [ ( 2 ) ( 2 ) ( 5 ) ] ln 3 [ ( 2 ) ( 5 ) ] 10 ln 3
5
Vector field F → ( x , y ) = ⟨ y , − x ⟩
C ( 2 , 0 ) → ( 2 , 2 ) → ( 0 , 2 ) Connecting ( 2 , 0 ) → ( 0 , 2 ) x 2 + y 2 = 2 2
a
Parameterize C
C 1 C 2
x ( t ) = 2 − 2 t y ( t ) = 2 for 0 ≤ t ≤ 1
C 3
x 2 = 2 2 − y 2 x 2 = 4 − y 2 x = 4 − y 2 y = 4 − x 2 x ( t ) = 2 sin t y ( t ) = 2 cos t for 0 ≤ t ≤ π 2
b
Compute the line integral ∫ C F → ⋅ d r →
C 1
∫ C 1 F → ⋅ d r → = ∫ 0 1 P \differential x \differential t + Q \differential y \differential t \differential t P = y Q = − x x ( t ) = 2 x ′ ( t ) = 0 y ( t ) = 2 t y ′ ( t ) = 2 ∫ 0 1 − 2 ( 2 ) \differential t − 4 ∫ 0 1 1 \differential t − 4 [ t ] 0 1 − 4 [ 1 ] = − 4
C 2
∫ C 2 F → ⋅ d r → = ∫ 0 1 P \differential x \differential t + Q \differential y \differential t \differential t x ( t ) = 2 − 2 t x ′ ( t ) = − 2 y ( t ) = 2 y ′ ( t ) = 0 ∫ 0 1 − 2 y \differential t ∫ 0 1 − ( 2 ) ( 2 ) \differential t = − 4
C 3
∫ C 3 F → ⋅ d r → = ∫ 0 π 2 P \differential x \differential t + Q \differential y \differential t \differential t x ( t ) = 2 sin t x ′ ( t ) = 2 cos t y ( t ) = 2 cos t y ′ ( t ) = − 2 sin t P = y Q = − x for 0 ≤ t ≤ π 2
∫ 0 π 2 2 y cos t + 2 x sin t \differential t ∫ 0 π 2 4 cos 2 t + 4 sin 2 t \differential t 4 ∫ 0 π 2 cos 2 t + sin 2 t \differential t 4 ∫ 0 π 2 1 \differential t 4 [ t ] 0 π 2 4 π 2 = 2 π
C 1 + C 2 + C 3
c
Use Green's Theorem to evaluate ∫ C F → ⋅ d r →
∫ C F → ⋅ d r → = ∫ C P \differential x + Q \differential y P = y Q = − x ∫ C y \differential x − x \differential y ∬ D ( ∂ Q ∂ x − ∂ P ∂ y ) \differential A Q x = − 1 P y = 1 ∬ D ( − 1 − 1 ) \differential A ∬ D ( − 2 ) \differential A ∬ D − 2 \differential A D \differential x \differential y x = 4 − y 2 4 − y 2 ≤ x ≤ 2 0 ≤ y ≤ 2 ∫ 0 2 ∫ 4 − y 2 2 − 2 \differential x \differential y ∫ 0 2 − 2 ∫ 4 − y 2 2 1 \differential x \differential y ∫ 0 2 − 2 [ x ] 4 − y 2 2 \differential y ∫ 0 2 − 2 [ 2 − 4 − y 2 ] \differential y ∫ 0 2 − 4 + 2 4 − y 2 \differential y − 2 ∫ 0 2 2 − 4 − y 2 \differential y − 2 ∫ 0 2 2 − ( 4 − y 2 ) 1 2 \differential y ∫ 0 2 − 4 \differential y − ∫ 0 2 ( 4 − y 2 ) 1 2 \differential y [ − 4 y ] 0 2 − ∫ 0 2 ( 4 − y 2 ) 1 2 \differential y [ − ( 4 ) ( 2 ) ] − ∫ 0 2 ( 4 − y 2 ) 1 2 \differential y − 8 − ∫ 0 2 ( 4 − y 2 ) 1 2 \differential y x = 4 − y 2 x 2 = 4 − y 2 x 2 + y 2 = 4 Now we evaluate this from 0 ≤ y ≤ 2
π r 2 We need a quarter
1 4 π r 2 1 4 π 2 2 = π − 8 − 2 π
Exam Summer 2023
1
Write the equation of the plane that passes through the line of intersection of the planes P 1 P 2 P 3
P 1 : − 2 x + y − z = 0 P 2 : x = 2 P 3 : x + y + z = 0 We want ( P 1 × P 2 ) × P 3
P 1 : ⟨ − 2 , 1 , − 1 ⟩ P 2 : ⟨ 1 , 0 , 0 ⟩ P 3 : ⟨ 1 , 1 , 1 ⟩ P 1 × P 2 = | i → j → k → − 2 1 − 1 1 0 0 | P 1 × P 2 = ⟨ ( 1 ) ( 0 ) − ( − 1 ) ( 0 ) , ( − 2 ) ( 0 ) − ( − 1 ) ( 1 ) , ( − 2 ) ( 0 ) − ( 1 ) ( 1 ) ⟩ P 1 × P 2 = ⟨ 0 , 1 , − 1 ⟩ ( P 1 × P 2 ) × P 3 = | i → j → k → 0 1 − 1 1 1 1 | ( P 1 × P 2 ) × P 3 = ⟨ ( 1 ) ( 1 ) − ( − 1 ) ( 1 ) , ( 0 ) ( 1 ) − ( − 1 ) ( 1 ) , ( 0 ) ( 1 ) − ( 1 ) ( 1 ) ⟩ ( P 1 × P 2 ) × P 3 = ⟨ 2 , 1 , − 1 ⟩
2
Find minimum distance from ( 1 , 1 , 1 ) S = { ( x , y , z ) ∈ R 3 : x 2 + y 2 4 − z 2 2 = 1 }
x 2 + 1 4 y 2 − 1 2 z 2 = 1 f ( x , y , z ) = ( x − 1 ) 2 + ( y − 1 ) 2 + ( z − 1 ) 2 g ( x , y , z ) = x 2 + y 2 4 − z 2 2 − 1 ∇ → f = ⟨ 2 ( x − 1 ) , 2 ( y − 1 ) , 2 ( z − 1 ) ⟩ ∇ → g = ⟨ 2 x , 1 2 y , − z ⟩ ∇ → f = λ ∇ → g 2 x − 2 = λ 2 x 2 y − 2 = λ 2 y 2 z − 2 = − λ z 2 x − 2 2 x = λ 4 y − 4 y = λ − 2 z + 2 z = λ 2 x − 2 2 x = 4 y − 4 y = − 2 z + 2 z 2 ( x − 1 ) 2 x = 4 y − 4 y = − 2 z + 2 z x − 1 x = 4 y − 4 y = − 2 z + 2 z y z ( x − 1 ) x y z = 4 x z ( y − 1 ) x y z = − 2 x y ( z − 1 ) x y z y z x − y z = 4 x y z − 4 x z = − 2 x y z + 2 x y y z ( x − 1 ) = 4 x z ( y − 1 ) = − 2 x y ( z − 1 ) y z = 0 4 x z = 0 − 2 x y = 0 2 x − 2 = λ 2 x
2 ( x − 1 ) = λ 2 x x − 1 = λ x − 1 = λ x − x − 1 = x ( λ − 1 ) λ = 1
2 y − 2 = λ 2 y
2 ( y − 1 ) = 1 2 λ y 4 ( y − 1 ) = λ y 4 y − 4 = λ y − 4 = λ y − 4 y − 4 = y ( λ − 4 ) λ = 4
2 z − 2 = − λ z
− 2 = − λ z − 2 z − 2 = − z ( λ + 2 ) λ = − 2
3
F R 2 A flow line of F → r → ( t ) : R → R 2 r → ′ ( t ) = F → ( r → ( t ) )
The velocity vector of r → ( t ) F → ( r → ( t ) )
Consider the vector field F ( x , y ) → [ − y x ] T
a
For each of the following curves, determine whether it's a flow line of the vector field F
r → 1 ( t ) = ( e − 3 t , e 3 t )
F → ( r → 1 ( t ) ) F → ( e − 3 t , e 3 t ) ( − e 3 t e − 3 t ) T
4
f ( x , y ) = { C exp ( − ( x 2 + y 2 ) k ) x ≥ 0 , y ≥ 0 0 otherwise C e − ( x 2 + y 2 ) k To have it be a valid PDF
∫ 0 ∞ f ( x ) \differential x = 1 ∫ 0 ∞ ∫ 0 ∞ C e − x 2 + y 2 k \differential x \differential y = 1 ∫ 0 ∞ C ∫ 0 ∞ e − x 2 + y 2 k \differential x \differential y = 1 ∫ 0 ∞ C [ − k 2 x e − x 2 + y 2 k ] 0 ∞ \differential y = 1 ∫ 0 ∞ C [ lim x → ∞ − k 2 x e − x 2 + y 2 k + lim x → 0 k 2 x e − x 2 + y 2 k ] \differential y = 1 ∫ 0 ∞ C [ lim x → ∞ − k 2 x e − x 2 + y 2 k + lim x → 0 1 2 k x − 1 e − x 2 + y 2 k ] \differential y = 1 l'hôpital's rule.
∫ 0 ∞ C [ lim x → ∞ − k 2 x e − x 2 + y 2 k + lim x → 0 − 2 x k 2 x 2 e − x 2 + y 2 k ] \differential y = 1 ∫ 0 ∞ C [ lim x → ∞ − k 2 x e − x 2 + y 2 k + lim x → 0 − 2 x ( k ) 2 x ( x ) e − x 2 + y 2 k ] \differential y = 1 ∫ 0 ∞ C [ lim x → ∞ − k 2 x e − x 2 + y 2 k + lim x → 0 − k x e − x 2 + y 2 k ] \differential y = 1 l'hôpital's rule.
∫ 0 ∞ C [ lim x → ∞ − k 2 x e − x 2 + y 2 k + lim x → 0 − k x e − x 2 + y 2 k ] \differential y = 1 \differential \differential x ( − k x − 1 ) = k x − 2 = k x 2 k x 2 ( \differential \differential x [ − x 2 + y 2 k ] ) k x 2 ( 2 x ) x ( 2 k ) x ( x ) 2 k x l'hôpital's rule.
\differential \differential x 2 k x − 1 = − 2 k x − 2 − 2 k x − 2 2 x − 4 k x x 2 − 4 k x ????
Exam Fall 2024
MAT232H5 F24 - Final Exam.pdf
A
1
Which of these parametric equations is an equation of the tangent line to the curve r → ( t ) = ⟨ ln t , t − 1 t − 2 , t ln t ⟩ t = 1
r → ′ ( t ) = ⟨ 1 t , ⟩ u = t − 1 v = t − 2 u ′ = 1 v ′ = 1 u ′ v − v ′ u v 2 t − 2 − t + 1 ( t − 2 ) 2 − 1 ( t − 2 ) 2 − 1 ( t − 2 ) 2 r → ′ ( t ) = ⟨ 1 t , − 1 ( t − 2 ) 2 , ⟩ \differential \differential t t ln t = ln t + 1 r → ′ ( t ) = ⟨ 1 t , − 1 ( t − 2 ) 2 , ln t + 1 ⟩ r → ′ ( 1 ) = ⟨ 1 , − 1 ( 1 − 2 ) 2 , 1 ⟩ r → ′ ( 1 ) = ⟨ 1 , − 1 ( − 1 ) 2 , 1 ⟩ r → ′ ( 1 ) = ⟨ 1 , − 1 , 1 ⟩ This is our direction vector
Point at t = 1
r → ( 1 ) = ⟨ 0 , 0 , 0 ⟩ L = Point + t Vector L = 0 → + t ⟨ 1 , − 1 , 1 ⟩ x = t y = − t z = t
2
Find the length of the curve r → ( t ) = ⟨ t cos t , t sin t , 2 2 3 t 3 2 ⟩
From Origin to ( − π , 2 , 2 2 π 3 2 3 )
2 2 3 t 3 2 = 2 2 π 3 2 3 2 2 t 3 2 = 2 2 π 3 2 t 3 2 = π 3 2 t = π 2 2 3 t 3 2 = 0 t 3 2 = 0 t = 0 0 ≤ t ≤ π ∫ 0 π x t 2 + y t 2 + z t 2 \differential t r → ′ ( t ) = ⟨ cos t − t sin t , sin t + t cos t , 2 t ⟩ ( cos t − t sin t ) 2 + ( sin t + t cos t ) 2 + ( 2 t ) 2 cos 2 t − 2 t sin t cos t + t 2 sin 2 t + sin 2 t + 2 t cos t sin t + t 2 cos 2 t + 2 t cos 2 t + sin 2 t = 1 cos 2 t + sin 2 t − 2 t sin t cos t + t 2 sin 2 t + 2 t cos t sin t + t 2 cos 2 t + 2 t 1 − t 2 sin t cos t + t 2 sin 2 t + t 2 cos t sin t + t 2 cos 2 t + 2 t 1 − t sin 2 t + t sin 2 t + t 2 sin 2 t + t 2 cos 2 t + 2 t 1 + t 2 sin 2 t + t 2 cos 2 t + 2 t 1 + t 2 + 2 t t 2 + 2 t + 1 p = 1 s = 2 f = 1 , 1 ( t + 1 ) 2 ∫ 0 π t + 1 \differential t [ 1 2 t 2 + t ] 0 π [ 1 2 π 2 + π ]
3
4
Which are true:
1: The cartesian equation ( x − 1 2 ) 2 + y 2 = 1 4 r = cos θ
( x − 1 2 ) 2 x 2 − x + 1 4 + y 2 = 1 4 x 2 − x + y 2 = 0 x 2 + y 2 = x r 2 = r cos θ r = cos θ True
2: The polar equation r = 2 sin θ − 3 cos θ y = 3 x + 2
r sin θ − 3 r cos θ = 2 y − 3 x = 2 y = 2 + 3 x True
3: For the polar curve r = 1 + cos θ θ = π
\differential y \differential x = \differential r \differential θ sin θ + r cos θ \differential r \differential θ cos θ − r sin θ x = r cos θ r = f ( θ ) x = f ( θ ) cos θ y = f ( θ ) sin θ \differential y \differential x = \differential r \differential θ sin θ + f ( θ ) cos θ \differential r \differential θ cos θ − f ( θ ) sin θ \differential y \differential θ = − f ′ ( θ ) sin θ \differential x \differential θ = f ′ ( θ ) cos θ \differential r \differential θ = − sin θ \differential y \differential x = − sin 2 θ +
5
Which are true
1:
x = t − sin t y = 1 − cos t t ∈ R d 2 y \differential x 2 = − 1 ( 1 − cos t ) 2 \differential y \differential x = \differential r \differential θ sin θ + cos θ \differential r \differential θ sin θ − cos θ \differential x \differential t = 1 − cos t \differential y \differential t = sin t \differential y \differential x = sin t 1 − cos t d 2 y \differential x 2 = d d t ( \differential y \differential x ) \differential x \differential t \differential \differential t ( \differential y \differential x ) = u ′ v − v ′ u v 2
u = 1 − cos t u ′ = sin t v = sin t v ′ = cos t
\differential \differential t ( \differential y \differential x ) = sin 2 t − cos t ( 1 − cos t ) sin 2 t \differential \differential t ( \differential y \differential x ) = sin 2 t − cos t + cos 2 t sin 2 t \differential \differential t ( \differential y \differential x ) = − cos t + sin 2 t + cos 2 t sin 2 t \differential \differential t ( \differential y \differential x ) = − cos t + 1 sin 2 t \differential \differential t ( \differential y \differential x ) = − cos t − 1 sin 2 t \differential \differential t ( \differential y \differential x ) = 1 − cos t sin 2 t
2: Given f ( x , y ) = e x cos y f ( x , y ) ( 1 , π 4 ) π 6 D u → f ( 1 , π 4 ) = e 2 ( 3 − 1 ) 4
θ = π 6 u → = ⟨ cos θ , sin θ ⟩ u → = ⟨ cos π 6 , sin π 6 ⟩ cos π 6 = 3 2 sin π 6 = 1 2 u → = ⟨ 3 2 , 1 2 ⟩ f x = cos y e x f y = − e x sin y D u → f = ∇ → f ⋅ u → D u → f = ⟨ cos y e x , − e x sin y ⟩ ⋅ ⟨ 3 2 , 1 2 ⟩ D u → f = 3 cos y e x 2 − e x sin y 2 D u → f = e x ( 3 cos y − sin y ) 2 D u → ( 1 , π 4 ) f = e ( 3 cos π 4 − sin π 4 ) 2 = e ( − 2 2 + 6 2 ) 2 D u → ( 1 , π 4 ) f = e ( − 6 − 2 2 ) 2 D u → ( 1 , π 4 ) f = e 2 ( 3 − 1 ) 4
3: The tangent plane to the surface x 2 + y 2 + z 2 = 4 ( 1 , 1 , 2 ) z = 2 − x − 1 2 − y − 1 2
f ( x , y , z ) = x 2 + y 2 + z 2 − 4 ∇ → f ( x , y , z ) = ⟨ 2 x , 2 y , 2 z ⟩ Point: ( 1 , 1 , 2 )
f x ( x − x 0 ) + f y ( y − y 0 ) + f z ( z − z 0 ) = 0 f x ( x − 1 ) + f y ( y − 1 ) + f z ( z − 2 ) = 0 2 x ( x − 1 ) + 2 y ( y − 1 ) + 2 z ( z − 2 ) = 0 2 ( x − 1 ) + 2 ( y − 1 ) + 2 2 ( z − 2 ) = 0 2 x − 2 + 2 y − 2 + 2 2 z − 4 = 0 2 x − 2 + 2 y − 2 − 4 = − 2 2 z 2 x + 2 y − 8 = − 2 2 z − 2 2 z = 2 x + 2 y − 8 z = − 2 x 2 2 − 2 y 2 2 + 8 2 2 z = − x 2 − y 2 + 4 2 True
6
Which are true
1: C r → ( t ) a → = r → ( a ) b → = r → ( b ) ∫ C ∇ → f ⋅ d r → = f ( b → ) − f ( a → ) = 0
If the field is conservative
Since we're given ∇ → f
a → ( a 1 , a 2 , a 3 ) Since it's conservative it's path independent, so we can take any two end points.
However we don't know that C a b
f ( b → ) − f ( a → ) = 0 I'll say it's not true, we don't know for sure.
2: given F → = ⟨ x 2 y 3 , x y 4 ⟩ f ( x , y ) F → = ∇ → f
Let's check if it's conservative
P = x 2 y 3 Q = x y 4 ∂ P ∂ y = ∂ Q ∂ x ∂ P ∂ y = 3 y 2 x 2 ∂ Q ∂ x = y 4 Not equal, not conservative
3: Given a triangular region with ( 0 , 0 ) , ( 1 , 0 ) , ( 0 , 1 ) ∫ ∂ D x 4 \differential x + x y \differential y = − ∬ D y \differential A ∂ D D
y = − x + 1 P = x 4 Q = x y ∂ P ∂ y = 0 ∂ Q ∂ x = y ∬ D y \differential A − ∬ D y \differential A Just means that we can re arrange the boundaries of x , y
So true
B
1
2
a
z = e x 2 y x ( u , v ) = u v x ( u , v ) = ( u v ) 1 2 y ( u , v ) = 1 v y ( u , v ) = v − 1 Find ∂ z ∂ u ∂ z ∂ v
∂ z ∂ u = ∂ z ∂ x ∂ x ∂ u + ∂ z ∂ y ∂ y ∂ u
z x = 2 x y e x 2 y x u = v 2 u v z y = x 2 e x 2 y y u = 0 ∂ z ∂ u = z x x u + z y y u ∂ z ∂ u = ( 2 x y e x 2 y ) ( v 2 u v ) + ( x 2 e x 2 y ) ( 0 ) ∂ z ∂ u = ( 2 x y e x 2 y ) ( v 2 u v )
∂ z ∂ v = ∂ z ∂ x ∂ x ∂ v + ∂ z ∂ y ∂ y ∂ v
z x = 2 x y e x 2 y x v = u 2 u v z y = x 2 e x 2 y y v = − v − 2 ∂ z ∂ v = z x x v + z y y v ∂ z ∂ v = ( 2 x y e x 2 y ) ( u 2 u v ) + ( x 2 e x 2 y ) ( − 1 v 2 )
b
Find points on x 2 − z 2 − 1 = 0
f ( x , y , z ) = x 2 + y 2 + z 2 g ( x , y , z ) = x 2 − z 2 − 1 ∇ → f = ⟨ 2 x , 2 y , 2 z ⟩ ∇ → g = ⟨ 2 x , 0 , − 2 z ⟩ 2 x = λ 2 x 2 y = λ 0 2 z = λ ( − 2 z )
0 = − 2 z λ − 2 z 0 = − 2 z ( λ + 1 ) z = 0 λ = − 1
λ = ± 1 x 2 − z 2 − 1 = 0 x = 0
− z 2 − 1 = 0 − z 2 = 1 z 2 = − 1 z = − i ∨ z = i
z = 0 y = 0
x 2 − z 2 = 1 ( x − z ) ( x + z ) = 1 x − z = 1 x + z = 1 x = 1 + z z = 1 + x ( 1 + z ) 2 − z 2 = 1 x 2 − ( 1 + x ) 2 = 1
x 2 − ( x 2 + 2 x + 1 ) = 1 − 2 x − 1 = 1 − 2 x = 2 x = − 1
x = ± 1 z = 0 y = 0
3
a
b
Convert polar integral into cartesian
∫ 0 π 4 ∫ 1 2 sec θ r 5 sin 2 θ \differential r \differential θ
Region
Function
r 4 sin 2 θ ( x 4 + 2 x 2 y 2 + y 4 ) sin 2 θ x 4 sin 2 θ + 2 x 2 y 2 sin 2 θ + y 4 sin 2 θ sin θ = y r sin 2 θ = y x 2 + y 2 x 4 y x 2 + y 2 + 2 x 2 y 3 x 2 + y 2 + y 5 x 2 + y 2 f ( x , y ) = x 4 y + 2 x 2 y 3 + y 5 x 2 + y 2
x 2 + y 2 = 1 x = 1 − y 2 y = x y = 1 − y 2 y 2 = 1 − y 2 0 = 1 − 2 y 2 1 = 2 y 2 1 2 = y 2 1 2 = y x 2 + 1 2 = 1 x 2 = 1 2 x = 1 2 ∫ 1 2 ∫ 0 x f ( x , y ) \differential y \differential x + ∫ 0 1 2 ∫ 1 − y 2 x f ( x , y ) \differential x \differential x
∫ 0 2 ∫ 2 4 − x 2 x + 2 y \differential y \differential x
Transform region
Transform Function
∫ π 4 π 2 ∫ 2 sin θ 2 r 2 cos θ + 2 r 2 sin θ \differential r \differential θ
4
∫ 1 2 ∫ 1 y y y x e x y \differential x \differential y Change of variables as u = x y v = y x
a
Sketch region
1 y ≤ x ≤ y x = 1 y y = x 1 ≤ y ≤ 2 left=0; right=3;
top=3; bottom=0;
---
xy=1
y=x
y=1
y=2
(1,1)
(2,2)
(0.5,2)
( 1 , 1 ) ( 2 , 2 ) y = 1 x y = 2 2 = 1 x 2 x = 1 x = 1 2 y = 1 1 2 = 2 ( 1 2 , 2 ) ( 1 , 1 ) ( 2 , 2 ) ( 1 2 , 2 ) left=-2; right=4;
top=4; bottom=-2;
---
(1,1)
(2,1)
(1,2)
y=1
x=1
y=-x+3
v = 1 u = 1 m = v 2 − v 1 u 2 − u 1 m = 1 − 2 2 − 1 = − 1 v = − 1 u + b ( 2 ) = − 1 + b 3 = + b v = − u + 3
b
Compute jacobian
u = ( x y ) 1 2 u x = y 2 x y u y = x 2 x y v = y x 1 2 v = ( y x − 1 ) 1 2 v x = 1 2 y x − 1 ( − y x − 2 ) v x = y x − 2 2 y x − 1 v x = y x − 2 2 y x − 1 2 v x = y x − 2 x 1 2 2 y v x = y x 1 2 − 2 2 y v x = y x 1 2 − 4 2 2 y v x = y x − 3 2 2 y v x = − y 2 y x 3 2 v y = 1 2 ( y x ) − 1 2 1 x | J | = | u x u y v x v y | | J | = | y 2 x y x 2 x y − y 2 x 3 y 1 2 x y x |
5
a
Conservative field
F → = ⟨ y z , x z , x y ⟩ = ∇ → f f ( x , y , z ) = x y z Find ∫ C F → ⋅ d r → C ( − 1 , 3 , 9 ) → ( 1 , 6 , − 4 )
As it's conservative, it's path independent
f ( 1 , 6 , − 4 ) − f ( − 1 , 3 , 9 ) ( 6 ) ( − 4 ) + ( 3 ) ( 9 ) ( 2 ) ( 3 ) ( − 4 ) + ( 3 ) ( 9 ) ( 2 ) ( − 12 ) + 27 ( − 24 ) + 27 3
b
R = { ( x , y ) : − 1 ≤ x ≤ 1 , − 1 ≤ y ≤ 1 } Parameterize the boundary of the region R ∂ R
Evaluate the line integral as ∫ ∂ R x y \differential y − y 2 \differential x
( − 1 , − 1 ) ( − 1 , 1 ) ( 1 , − 1 ) ( 1 , 1 ) C 1 : ( − 1 , − 1 ) → ( − 1 , 1 ) C 2 : ( − 1 , − 1 ) → ( 1 , − 1 ) C 3 : ( − 1 , 1 ) → ( 1 , 1 ) C 4 : ( 1 , − 1 ) → ( 1 , 1 ) C 1
P = − y 2 Q = x y \differential x = 0 \differential y = 2 ∫ 0 1 P \differential x + Q \differential y \differential t ∫ 0 1 2 x y \differential t ∫ 0 1 2 ( − 1 ) ( − 1 + 2 t ) \differential t ∫ 0 1 − 2 ( − 1 + 2 t ) \differential t − 2 ∫ 0 1 − 1 + 2 t \differential t − 2 [ − 1 t + t 2 ] 0 1 − 2 [ − 1 + 1 ] = 0
C 2
x ( t ) = − 1 + 2 t y ( t ) = − 1 P = − y 2 Q = x y \differential x = 2 \differential y = 0 ∫ 0 1 − 2 y 2 \differential x ∫ 0 1 − 2 ( − 1 ) 2 \differential x ∫ 0 1 − 2 \differential x [ − 2 x ] 0 1 = − 2
C 3
x ( t ) = − 1 + 2 t y ( t ) = 1 P = − y 2 Q = x y ∫ C 3 ? \differential t = − 2
C 4 C 1 + C 2 + C 3 + C 4 0 + 0 − 2 − 2 = − 4
Use Green's Theorem
∫ ∂ R x y \differential y − y 2 \differential x P = − y 2 Q = x y ∂ Q ∂ x = y ∂ P ∂ y = − 2 y y − ( − 2 y ) ∫ − 1 1 ∫ − 1 1 y + 2 y \differential x \differential y ∫ − 1 1 y + 2 y ∫ − 1 1 1 \differential x \differential y ∫ − 1 1 y + 2 y [ x ] 1 − 1 \differential y ∫ − 1 1 y + 2 y [ 1 − ( − 1 ) ] \differential y ∫ − 1 1 y + 2 y [ 2 ] \differential y 2 ∫ − 1 1 y + 2 y \differential y 2 [ 1 2 y 2 + y 2 ] − 1 1 2 [ 3 2 y 2 ] − 1 1 2 [ 3 2 ( 1 ) 2 − 3 2 ( − 1 ) 2 ] 2 [ 3 2 − 3 2 ] = 0
Exam April 2023
MAT232H S23 - Final Exam (1).pdf
A
1
Which of these is an equation of the tangent line to the curve:
r → ( t ) = ⟨ t 2 + 2 t + 3 , 4 t cos t , 2 e 3 t ⟩ t = 0 r → ( 0 ) = ⟨ 3 , 0 , 2 ⟩ r → ′ ( t ) = ⟨ 2 t + 2 , 4 cos t − 4 t sin t , 6 e t ⟩ r → ′ ( 0 ) = ⟨ 2 , 4 , 6 ⟩ ⟨ x , y , z ⟩ = ⟨ 3 , 0 , 2 ⟩ + t ⟨ 1 , 2 , 3 ⟩
2
Find arc length
r → ′ ( t ) = ⟨ t 2 2 , ( 2 t + 1 ) 3 2 3 ⟩ 0 ≤ t ≤ 4 x ( t ) = t 2 2 x ′ ( t ) = 1 2 2 t = t y ( t ) = 1 3 ( 2 t + 1 ) 3 2 y ′ ( t ) = 1 3 3 2 ( 2 ) ( 2 t + 1 ) 1 2 y ′ ( t ) = ( 2 t + 1 ) 1 2 ∫ 0 4 x ′ ( t ) 2 + y ′ ( t ) 2 \differential t ∫ 0 4 t 2 + 2 t + 1 \differential t x y = 1 x + y = 2 ∫ 0 4 t + 1 \differential t [ 1 2 t 2 + t ] 0 4 [ 1 2 4 2 + 4 ] [ ( 2 2 ) 2 2 + 4 ] [ ( 4 ) 2 2 + 4 ] [ 16 2 + 4 ] [ 8 + 4 ] = 12
3
Find max and min
g ( x , y ) = x 2 − 2 x y + 2 y R = { ( x , y ) : 0 ≤ x ≤ 3 , 0 ≤ y ≤ 2 } Interior
g x = 2 x − 2 y g y = − 2 x + 2 g x = 0 g y = 0 ( 1 , 1 )
Boundary
( 0 , 0 ) ( 0 , 2 ) ( 3 , 0 ) ( 3 , 2 ) ( 0 , 0 ) → ( 0 , 2 )
g ( 0 , y ) = 2 y g x ( 0 , y ) = 2 Nothing on here
( 0 , 2 ) → ( 3 , 2 )
g ( x , 2 ) = x 2 − 4 x + 4 g x ( x , 2 ) = 2 x − 4 2 x = 4 x = 2 y = 2 ( 2 , 2 )
( 3 , 2 ) → ( 3 , 0 )
g ( 3 , y ) = 9 − 4 y g x ( 3 , y ) = − 4 Nothing on here
( 3 , 0 ) → ( 0 , 0 )
g ( x , 0 ) = x 2 g x ( x , 0 ) = 2 x x = 0 y = 0 ( 0 , 0 )
Check
Point Value Result ( 0 , 0 ) 0 Min ( 0 , 2 ) 4 ( 3 , 0 ) 9 Max ( 3 , 2 ) 1 ( 2 , 2 ) 0 Min ( 1 , 1 ) 1
4
Which are true
1:
x = t − t 2 y = t − t 3 d 2 y d x 2 = 2 − 6 t − 6 t 2 ( 1 − 2 t ) 3 x ′ = 1 − 2 t y ′ = 1 − 3 t 2 \differential y \differential x = \differential y \differential t \differential x \differential t \differential y \differential x = 1 − 3 t 2 1 − 2 t d 2 y d x 2 = d d t ( \differential y \differential x ) \differential x \differential t \differential \differential t ( 1 − 3 t 2 1 − 2 t ) u = 1 − 3 t 2 v = 1 − 2 t u ′ = − 6 t v ′ = − 2 u ′ v − v ′ u v 2 − 6 t ( 1 − 2 t ) − ( − 2 ) ( 1 − 3 t 2 ) ( 1 − 2 t ) 2 ( − 6 t + 12 t 2 ) + 2 ( 1 − 3 t 2 ) ( 1 − 2 t ) 2 − 6 t + 12 t 2 + 2 − 6 t 2 ( 1 − 2 t ) 2 6 t 2 − 6 t + 2 ( 1 − 2 t ) 2 d 2 y d x 2 = 6 t 2 − 6 t + 2 ( 1 − 2 t ) 2 1 − 2 t = − 6 t 2 + 6 t − 2 ( 2 t − 1 ) 3 True
2:
r = 4 2 cos θ − sin θ m = 2 b = − 4 2 r cos θ − r sin θ = 4 2 x − y = 4 − y = 4 − 2 x y = 2 x − 4 True
3:
The direction of zero change at ( a , b ) z = f ( x , y ) ∇ → f ( a , b )
True because ∇ → f f
When we have 0 f
Still orthogonal
4:
Tangent plane to f ( x , y ) = ln ( 2 x + y ) ( − 1 , 3 ) 2 x + y − z − 1 = 0
f ( − 1 , 3 ) = ln ( − 2 + 3 ) = ln 1 = 0 f x = 2 2 x + y f x ( − 1 , 3 ) = 2 − 2 + 3 = 2 1 f y = 1 2 x + y f y ( − 1 , 3 ) = 1 2 ( − 1 ) + 3 = 1 1 = 1 z = 2 ( x + 1 ) + ( y − 3 ) z = 2 x + 2 + y − 3 z = 2 x + y − 1 True
5
Which are true
1:∫ 0 1 ∫ 0 3 x e x y \differential x \differential y = e 3 − 4
∫ 0 3 x ∫ 0 1 e x y \differential y \differential x ∫ 0 3 x [ x − 1 e x y ] 0 1 \differential x ∫ 0 3 x [ x − 1 e x − x − 1 ] \differential x ∫ 0 3 e x − 1 \differential x [ e x − x ] 0 3 [ e 3 − 3 ] − e 0 e 3 − 4 True
2:
f ( x , y , z ) = 2 x y + z C r → ( t ) = ⟨ cos t , sin t , t ⟩ 0 ≤ t ≤ π ∫ C f ( x , y , z ) \differential s = 2 ∫ 0 π sin 2 t + t \differential t
3: A line integral ∫ C F → ⋅ d r → D F → ∫ C ′ F → ⋅ d r → = 0 C ′ D
I would say false, we don't know if it's a conservative field by that.
4:
B
Exam 2022 April
MAT232H S22 - Final Exam (2).pdf
1
a
If x = f ( t ) y = g ( t )
Then d 2 y d x 2 = d 2 y d t 2 d 2 x d t 2
No, the chain rule says it's d 2 y d x 2 = \differential \differential t ( \differential y \differential x ) \differential x \differential t
b
Curve with r → ( t ) = ⟨ 3 − t 4 , 5 t 4 − 2 , t 4 ⟩
Check if the derivative is constant
r → ′ ( t ) = ⟨ − 4 t 3 , 20 t 3 , 4 t 3 ⟩ Not constant, not a line
c
f ( x , y ) = 2 x y D u → f ( 1 , 2 ) = 4 u → = ⟨ 3 5 , 4 5 ⟩ Check if that's a unit vector
9 25 + 16 25 = 1 ∇ → f ( x , y ) = ⟨ 2 y , 2 x ⟩ ∇ → f ( 1 , 2 ) = ⟨ 2 , 4 ⟩ = ⟨ 1 , 2 ⟩ D u → f = ⟨ 2 , 4 ⟩ ⋅ ⟨ 3 5 , 4 5 ⟩ 6 5 + 16 5 22 5 ≠ 4
d
The function F → ( x , y ) = ⟨ x 2 y , cos y , e ⟩
No because z z ( x , y , z )
e
Evaluating ∫ C ∇ → f ⋅ d r → = 4 f ( x , y , z ) = cos π x + sin π y − x y z C ( 1 , 1 2 , 2 ) ( 2 , 1 , − 1 )
This is path independent because we have a gradient field
f ( 2 , 1 , − 1 ) − f ( 1 , 1 2 , 2 ) ( cos 2 π + sin 2 π − ( 2 ) ( 1 ) ( − 1 ) ) − ( cos π + sin π 2 − ( 1 ) ( 1 2 ) ( 2 ) ) ( 3 ) − ( − 1 ) 4 True
2
a
Determine if f ( x , y ) = e x sin y + e y cos x
∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 = 0 f x = e x sin y − e y sin x f y = e x cos y + e y cos x f x x = e x sin y − e y cos x f y y = − e x sin y + e y cos x e x sin y − e y cos x − e x sin y + e y cos x = 0 True
b
w = x y + z 2 x = t 2 e s y = t cos s z = sin t find ∂ w ∂ t | s = 0 , t = π
∂ w ∂ t = ∂ w ∂ x ∂ x ∂ t + ∂ w ∂ y ∂ y ∂ t + ∂ w ∂ z ∂ z ∂ t w x = y w y = x w z = 2 z x t = 2 t e s y t = cos s z t = cos t ∂ w ∂ t = 2 t y e s + x cos s + 2 z cos t ∂ w ∂ t | s = 0 , t = π = 2 t y + x − 2 z ∂ w ∂ t | s = 0 , t = π = 2 π 2 cos ( 0 ) + π 2 − 2 sin π ∂ w ∂ t | s = 0 , t = π = 2 π 2 + π 2 ∂ w ∂ t | s = 0 , t = π = 3 π 2
c
Find critical points and classify
f ( x , y ) = x 5 y + x y 5 + x y f x = 5 x 4 y + y 5 + y f y = x 5 + 5 x y 4 + x f x x = 20 x 3 y f y y = 20 x y 3 f x y = 5 x 4 + 5 y 4 + 1 f x = 0 f y = 0
5 x 4 y + y 5 + y = 0 y ( 5 x 4 + y 4 + 1 ) = 0 x 5 + 5 x y 4 + x = 0 x ( x 4 + 5 y 4 + 1 ) = 0 y = 0 x = 0 ( 0 , 0 ) 5 x 4 + y 4 + 1 = 0 x 4 + 5 y 4 + 1 = 0 4 x 4 = 4 y 4 x 4 = y 4 ± x 2 = ± y 2 Since negative won't work
x 2 = y 2 ± x = ± y
left=-3; right=5;
top=5; bottom=-5;
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x=y^2
x=y
y=0
y=2
y=1
left=-3; right=5;
top=5; bottom=-5;
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y=x
y=2x
y=2x-2
y=x+1
x = u − v y = 2 u − v
left=-3; right=5;
top=5; bottom=-5;
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(1,0)
(0,0)
(4/3,-2/3)
(7/3,-2/3)