Chain Rule/Directional derivatives
Christopher Croke
University of Pennsylvania
Math 115
Christopher Croke
Calculus 115
Chain Rule
In the one variable case z = f (y ) and y = g (x) then
Christopher Croke
Calculus 115
dz
dx
=
dz dy
dy dx .
Chain Rule
In the one variable case z = f (y ) and y = g (x) then
dz
dx
=
dz dy
dy dx .
When there are two independent variables, say w = f (x, y ) is
differentiable and where both x and y are differentiable functions
of the same variable t then w is a function of t.
Christopher Croke
Calculus 115
Chain Rule
In the one variable case z = f (y ) and y = g (x) then
dz
dx
=
dz dy
dy dx .
When there are two independent variables, say w = f (x, y ) is
differentiable and where both x and y are differentiable functions
of the same variable t then w is a function of t. and
dw
∂w dx
∂w dy
=
+
.
dt
∂x dt
∂y dt
Christopher Croke
Calculus 115
Chain Rule
In the one variable case z = f (y ) and y = g (x) then
dz
dx
=
dz dy
dy dx .
When there are two independent variables, say w = f (x, y ) is
differentiable and where both x and y are differentiable functions
of the same variable t then w is a function of t. and
dw
∂w dx
∂w dy
=
+
.
dt
∂x dt
∂y dt
or in other words
dw
(t) = wx (x(t), y (t))x 0 (t) + wy (x(t), y (t))y 0 (t).
dt
Christopher Croke
Calculus 115
Chain Rule
In the one variable case z = f (y ) and y = g (x) then
dz
dx
=
dz dy
dy dx .
When there are two independent variables, say w = f (x, y ) is
differentiable and where both x and y are differentiable functions
of the same variable t then w is a function of t. and
dw
∂w dx
∂w dy
=
+
.
dt
∂x dt
∂y dt
or in other words
dw
(t) = wx (x(t), y (t))x 0 (t) + wy (x(t), y (t))y 0 (t).
dt
The proof is not hard and given in the text.
Christopher Croke
Calculus 115
Chain Rule
In the one variable case z = f (y ) and y = g (x) then
dz
dx
=
dz dy
dy dx .
When there are two independent variables, say w = f (x, y ) is
differentiable and where both x and y are differentiable functions
of the same variable t then w is a function of t. and
dw
∂w dx
∂w dy
=
+
.
dt
∂x dt
∂y dt
or in other words
dw
(t) = wx (x(t), y (t))x 0 (t) + wy (x(t), y (t))y 0 (t).
dt
The proof is not hard and given in the text.
Show tree diagram.
Christopher Croke
Calculus 115
Problem: Compute
and y = sin(t).
dw
dt
two ways when w = x 2 + xy , x = cos(t),
Christopher Croke
Calculus 115
Problem: Compute
and y = sin(t).
π
What is dw
dt ( 2 )?
dw
dt
two ways when w = x 2 + xy , x = cos(t),
Christopher Croke
Calculus 115
Problem: Compute
and y = sin(t).
π
What is dw
dt ( 2 )?
dw
dt
two ways when w = x 2 + xy , x = cos(t),
Do tree diagram for w=f(x,y,z) where x,y ,z are functions of t.
Christopher Croke
Calculus 115
Problem: Compute
and y = sin(t).
π
What is dw
dt ( 2 )?
dw
dt
two ways when w = x 2 + xy , x = cos(t),
Do tree diagram for w=f(x,y,z) where x,y ,z are functions of t.
What do you do if the intermediate variables are also functions of
two variables? Say w = f (x, y , z) where each of x,y ,z are
functions of r and θ. This makes w a function of r and θ.
Christopher Croke
Calculus 115
Problem: Compute
and y = sin(t).
π
What is dw
dt ( 2 )?
dw
dt
two ways when w = x 2 + xy , x = cos(t),
Do tree diagram for w=f(x,y,z) where x,y ,z are functions of t.
What do you do if the intermediate variables are also functions of
two variables? Say w = f (x, y , z) where each of x,y ,z are
functions of r and θ. This makes w a function of r and θ. For
example compute ∂w
∂r .
Christopher Croke
Calculus 115
Problem: Compute
and y = sin(t).
π
What is dw
dt ( 2 )?
dw
dt
two ways when w = x 2 + xy , x = cos(t),
Do tree diagram for w=f(x,y,z) where x,y ,z are functions of t.
What do you do if the intermediate variables are also functions of
two variables? Say w = f (x, y , z) where each of x,y ,z are
functions of r and θ. This makes w a function of r and θ. For
example compute ∂w
∂r .
∂w
Problem: Compute ∂w
∂r and ∂s in terms of r and s when
2
w = x + y − z , x = rs, y = r + s, and z = e rs .
Christopher Croke
Calculus 115
Implicit Differentiation
Let F (x, y ) = 0 define y implicitly as a function of x, that is
y = h(x).
Christopher Croke
Calculus 115
Implicit Differentiation
Let F (x, y ) = 0 define y implicitly as a function of x, that is
y = h(x). Then w (x) = F (x, h(x)) = 0:
Christopher Croke
Calculus 115
Implicit Differentiation
Let F (x, y ) = 0 define y implicitly as a function of x, that is
y = h(x). Then w (x) = F (x, h(x)) = 0:
If F is differentiable:
0=
dw
dy
= Fx · 1 + Fy ·
.
dx
dx
Christopher Croke
Calculus 115
Implicit Differentiation
Let F (x, y ) = 0 define y implicitly as a function of x, that is
y = h(x). Then w (x) = F (x, h(x)) = 0:
If F is differentiable:
0=
Thus
dw
dy
= Fx · 1 + Fy ·
.
dx
dx
dy
Fx
=−
dx
Fy
whenever Fy 6= 0
Christopher Croke
Calculus 115
Implicit Differentiation
Let F (x, y ) = 0 define y implicitly as a function of x, that is
y = h(x). Then w (x) = F (x, h(x)) = 0:
If F is differentiable:
0=
Thus
dw
dy
= Fx · 1 + Fy ·
.
dx
dx
dy
Fx
=−
dx
Fy
whenever Fy 6= 0
Problem: Find
dy
dx
if x 2 + y 2 x + sin(y ) = 0.
Christopher Croke
Calculus 115
VECTORS in the PLANE- Chapter 12
~i is the unit vector in the positive x direction and ~j is the unit
vector in the positive y direction.
Christopher Croke
Calculus 115
VECTORS in the PLANE- Chapter 12
~i is the unit vector in the positive x direction and ~j is the unit
vector in the positive y direction.
Any vector can be written as ~v = a~i + b~j.
Christopher Croke
Calculus 115
VECTORS in the PLANE- Chapter 12
~i is the unit vector in the positive x direction and ~j is the unit
vector in the positive y direction.
Any vector can be written as ~v = a~i + b~j. It has length
p
|~v | = a2 + b 2
and points in the direction
u~ =
~v
a
~i + √ b
~j.
=√
2
2
|~v |
a +b
a2 + b 2
Christopher Croke
Calculus 115
VECTORS in the PLANE- Chapter 12
~i is the unit vector in the positive x direction and ~j is the unit
vector in the positive y direction.
Any vector can be written as ~v = a~i + b~j. It has length
p
|~v | = a2 + b 2
and points in the direction
u~ =
~v
a
~i + √ b
~j.
=√
2
2
|~v |
a +b
a2 + b 2
If ~v1 = a1~i + b1~j and ~v2 = a2~i + b2~j then
~v1 · ~v2 = a1 a2 + b1 b2
.
Christopher Croke
Calculus 115
VECTORS in the PLANE- Chapter 12
~i is the unit vector in the positive x direction and ~j is the unit
vector in the positive y direction.
Any vector can be written as ~v = a~i + b~j. It has length
p
|~v | = a2 + b 2
and points in the direction
u~ =
~v
a
~i + √ b
~j.
=√
2
2
|~v |
a +b
a2 + b 2
If ~v1 = a1~i + b1~j and ~v2 = a2~i + b2~j then
~v1 · ~v2 = a1 a2 + b1 b2
.
√
Note that |~v | = ~v · ~v .
Christopher Croke
Calculus 115
VECTORS in the PLANE- Chapter 12
~i is the unit vector in the positive x direction and ~j is the unit
vector in the positive y direction.
Any vector can be written as ~v = a~i + b~j. It has length
p
|~v | = a2 + b 2
and points in the direction
u~ =
~v
a
~i + √ b
~j.
=√
2
2
|~v |
a +b
a2 + b 2
If ~v1 = a1~i + b1~j and ~v2 = a2~i + b2~j then
~v1 · ~v2 = a1 a2 + b1 b2
.
√
Note that |~v | = ~v · ~v .
Christopher Croke
Calculus 115
VECTORS in the PLANE and SPACE-Chapter 12
Also we have ~v1 · ~v2 = |v1 ||v2 |cos(θ) where θ is the angle between
~v1 and ~v2 .
Christopher Croke
Calculus 115
VECTORS in the PLANE and SPACE-Chapter 12
Also we have ~v1 · ~v2 = |v1 ||v2 |cos(θ) where θ is the angle between
~v1 and ~v2 .
Problem: Find the angle between the vectors ~v1 = ~i and
~v2 = ~i + ~j.
Christopher Croke
Calculus 115
VECTORS in the PLANE and SPACE-Chapter 12
Also we have ~v1 · ~v2 = |v1 ||v2 |cos(θ) where θ is the angle between
~v1 and ~v2 .
Problem: Find the angle between the vectors ~v1 = ~i and
~v2 = ~i + ~j.
A parametric equation for the line in the plane parallel to the
vector a~i + b~j through the point (x0 , y0 ) is given by
x(t) = x0 + at
Christopher Croke
y (t) = y0 + bt
Calculus 115
VECTORS in the PLANE and SPACE-Chapter 12
Also we have ~v1 · ~v2 = |v1 ||v2 |cos(θ) where θ is the angle between
~v1 and ~v2 .
Problem: Find the angle between the vectors ~v1 = ~i and
~v2 = ~i + ~j.
A parametric equation for the line in the plane parallel to the
vector a~i + b~j through the point (x0 , y0 ) is given by
x(t) = x0 + at
y (t) = y0 + bt
Things in 3D space are similar with vectors ~v = a~i + b~j + c ~k where
~k is the unit vector in the positive z direction.
Christopher Croke
Calculus 115
VECTORS in the PLANE and SPACE-Chapter 12
Also we have ~v1 · ~v2 = |v1 ||v2 |cos(θ) where θ is the angle between
~v1 and ~v2 .
Problem: Find the angle between the vectors ~v1 = ~i and
~v2 = ~i + ~j.
A parametric equation for the line in the plane parallel to the
vector a~i + b~j through the point (x0 , y0 ) is given by
x(t) = x0 + at
y (t) = y0 + bt
Things in 3D space are similar with vectors ~v = a~i + b~j + c ~k where
~k is the unit vector in the positive z direction.
(a1~i + b1~j + c1 ~k) · (a2~i + b2~j + c2 ~k) = a1 a2 + b1 b2 + c1 c2 .
Christopher Croke
Calculus 115
VECTORS in the PLANE and SPACE-Chapter 12
Also we have ~v1 · ~v2 = |v1 ||v2 |cos(θ) where θ is the angle between
~v1 and ~v2 .
Problem: Find the angle between the vectors ~v1 = ~i and
~v2 = ~i + ~j.
A parametric equation for the line in the plane parallel to the
vector a~i + b~j through the point (x0 , y0 ) is given by
x(t) = x0 + at
y (t) = y0 + bt
Things in 3D space are similar with vectors ~v = a~i + b~j + c ~k where
~k is the unit vector in the positive z direction.
(a1~i + b1~j + c1 ~k) · (a2~i + b2~j + c2 ~k) = a1 a2 + b1 b2 + c1 c2 .
√
√
|~v | = ~v · ~v = a2 + b 2 + c 2
Christopher Croke
Calculus 115
VECTORS in the PLANE and SPACE-Chapter 12
Also we have ~v1 · ~v2 = |v1 ||v2 |cos(θ) where θ is the angle between
~v1 and ~v2 .
Problem: Find the angle between the vectors ~v1 = ~i and
~v2 = ~i + ~j.
A parametric equation for the line in the plane parallel to the
vector a~i + b~j through the point (x0 , y0 ) is given by
x(t) = x0 + at
y (t) = y0 + bt
Things in 3D space are similar with vectors ~v = a~i + b~j + c ~k where
~k is the unit vector in the positive z direction.
(a1~i + b1~j + c1 ~k) · (a2~i + b2~j + c2 ~k) = a1 a2 + b1 b2 + c1 c2 .
√
√
|~v | = ~v · ~v = a2 + b 2 + c 2
~v1 · ~v2 = |~v1 ||~v2 |cos(θ)
Christopher Croke
Calculus 115
VECTORS in the PLANE and SPACE-Chapter 12
Also we have ~v1 · ~v2 = |v1 ||v2 |cos(θ) where θ is the angle between
~v1 and ~v2 .
Problem: Find the angle between the vectors ~v1 = ~i and
~v2 = ~i + ~j.
A parametric equation for the line in the plane parallel to the
vector a~i + b~j through the point (x0 , y0 ) is given by
x(t) = x0 + at
y (t) = y0 + bt
Things in 3D space are similar with vectors ~v = a~i + b~j + c ~k where
~k is the unit vector in the positive z direction.
(a1~i + b1~j + c1 ~k) · (a2~i + b2~j + c2 ~k) = a1 a2 + b1 b2 + c1 c2 .
√
√
|~v | = ~v · ~v = a2 + b 2 + c 2
~v1 · ~v2 = |~v1 ||~v2 |cos(θ)
x(t) = x0 + at,
y (t) = y0 + bt,
Christopher Croke
z(t) = z0 + ct.
Calculus 115
VECTORS in SPACE-Chapter 12
Some things are different though. The Plane through the point
(x0 , y0 , z0 ) which is perpendicular to the vector ~v = a~i + b~j + c ~k is
given by the equation;
0 = a(x − x0 ) + b(y − y0 ) + c(z − z0 )
Christopher Croke
Calculus 115
VECTORS in SPACE-Chapter 12
Some things are different though. The Plane through the point
(x0 , y0 , z0 ) which is perpendicular to the vector ~v = a~i + b~j + c ~k is
given by the equation;
0 = a(x − x0 ) + b(y − y0 ) + c(z − z0 )
This is since we know that ~v is perpendicular to the vector
pointing from (x0 , y0 , z0 ) to (x, y , z).
Christopher Croke
Calculus 115
VECTORS in SPACE-Chapter 12
Some things are different though. The Plane through the point
(x0 , y0 , z0 ) which is perpendicular to the vector ~v = a~i + b~j + c ~k is
given by the equation;
0 = a(x − x0 ) + b(y − y0 ) + c(z − z0 )
This is since we know that ~v is perpendicular to the vector
pointing from (x0 , y0 , z0 ) to (x, y , z).
There is one other operation in three dimensions, the cross
product: ~v1 × ~v2 . It is a vector that is perpendicular to both ~v1
and ~v2 and has length |~v1 ||~v2 |sin(θ).
Christopher Croke
Calculus 115
VECTORS in SPACE-Chapter 12
Some things are different though. The Plane through the point
(x0 , y0 , z0 ) which is perpendicular to the vector ~v = a~i + b~j + c ~k is
given by the equation;
0 = a(x − x0 ) + b(y − y0 ) + c(z − z0 )
This is since we know that ~v is perpendicular to the vector
pointing from (x0 , y0 , z0 ) to (x, y , z).
There is one other operation in three dimensions, the cross
product: ~v1 × ~v2 . It is a vector that is perpendicular to both ~v1
and ~v2 and has length |~v1 ||~v2 |sin(θ).
The formula is (a1~i + b1~j + c1 ~k) × (a2~i + b2~j + c2 ~k)=
(b1 c2 − c1 b2 )~i + (c1 a2 − a1 c2 )~j + (a1 b2 − b1 a2 )~k.
Christopher Croke
Calculus 115
Directional Derivatives
Given a function f (x, y ) then the rate of change (w.r.t.t) along a
curve (x(t), y (t)) is (by the chain rule)
dy
dx
+ fy .
fx
dt
dt
Christopher Croke
Calculus 115
Directional Derivatives
Given a function f (x, y ) then the rate of change (w.r.t.t) along a
curve (x(t), y (t)) is (by the chain rule)
dy
dx
+ fy .
fx
dt
dt
If u~ is a unit vector we can write u~ = u1~i + u2~j (where
u12 + u22 = 1).
Christopher Croke
Calculus 115
Directional Derivatives
Given a function f (x, y ) then the rate of change (w.r.t.t) along a
curve (x(t), y (t)) is (by the chain rule)
dy
dx
+ fy .
fx
dt
dt
If u~ is a unit vector we can write u~ = u1~i + u2~j (where
u12 + u22 = 1). So the line through (x0 , y0 ) in the direction u~ with
arc length parameter is (x(s), y (s)) where:
x(s) = u1 s + x0
Christopher Croke
y (s) = u2 s + y0 .
Calculus 115
Directional Derivatives
Given a function f (x, y ) then the rate of change (w.r.t.t) along a
curve (x(t), y (t)) is (by the chain rule)
dy
dx
+ fy .
fx
dt
dt
If u~ is a unit vector we can write u~ = u1~i + u2~j (where
u12 + u22 = 1). So the line through (x0 , y0 ) in the direction u~ with
arc length parameter is (x(s), y (s)) where:
x(s) = u1 s + x0
Christopher Croke
y (s) = u2 s + y0 .
Calculus 115
The directional derivative in the direction u~ (i.e. the rate of change
of f as one moves with unit speed along the line in the u~ direction)
is thus
df = fx (x0 , y0 )u1 + fy (x0 , y0 )u2 ≡ Du~ f (x ,y ) .
~
u
,(x
,y
)
0 0
0 0
ds
Definition:
Du~ f
(x0 ,y0 )
f (x0 + u1 s, y0 + u2 s) − f (x0 , y0 )
.
s→0
s
= lim
Christopher Croke
Calculus 115
The directional derivative in the direction u~ (i.e. the rate of change
of f as one moves with unit speed along the line in the u~ direction)
is thus
df = fx (x0 , y0 )u1 + fy (x0 , y0 )u2 ≡ Du~ f (x ,y ) .
~
u
,(x
,y
)
0 0
0 0
ds
Definition:
Du~ f
(x0 ,y0 )
f (x0 + u1 s, y0 + u2 s) − f (x0 , y0 )
.
s→0
s
= lim
What is geometric meaning?
Christopher Croke
Calculus 115
The directional derivative in the direction u~ (i.e. the rate of change
of f as one moves with unit speed along the line in the u~ direction)
is thus
df = fx (x0 , y0 )u1 + fy (x0 , y0 )u2 ≡ Du~ f (x ,y ) .
~
u
,(x
,y
)
0 0
0 0
ds
Definition:
Du~ f
(x0 ,y0 )
f (x0 + u1 s, y0 + u2 s) − f (x0 , y0 )
.
s→0
s
= lim
What is geometric meaning?
Christopher Croke
Calculus 115
Christopher Croke
Calculus 115
If you look at the formula you note that
Du~ f (x ,y ) = fx (x0 , y0 )~i + fy (x0 , y0 )~j · u~
0
0
Christopher Croke
Calculus 115
If you look at the formula you note that
Du~ f (x ,y ) = fx (x0 , y0 )~i + fy (x0 , y0 )~j · u~
0
0
We call the vector fx (x0 , y0 )~i + fy (x0 , y0 )~j the gradient of f at
(x0 , y0 ) and we denote it ∇f .
Christopher Croke
Calculus 115
If you look at the formula you note that
Du~ f (x ,y ) = fx (x0 , y0 )~i + fy (x0 , y0 )~j · u~
0
0
We call the vector fx (x0 , y0 )~i + fy (x0 , y0 )~j the gradient of f at
(x0 , y0 ) and we denote it ∇f .
So
∇f = fx ~i + fy ~j.
Christopher Croke
Calculus 115
If you look at the formula you note that
Du~ f (x ,y ) = fx (x0 , y0 )~i + fy (x0 , y0 )~j · u~
0
0
We call the vector fx (x0 , y0 )~i + fy (x0 , y0 )~j the gradient of f at
(x0 , y0 ) and we denote it ∇f .
So
∇f = fx ~i + fy ~j.
and
Du~ f = ∇f · u~.
Christopher Croke
Calculus 115
Christopher Croke
Calculus 115
Problem: Compute Du~ f where f (x, y ) = e xy + x 2 at the point
(1, 0) in the direction u~ = √213 ~i − √313 ~j.
Christopher Croke
Calculus 115
Problem: Compute Du~ f where f (x, y ) = e xy + x 2 at the point
(1, 0) in the direction u~ = √213 ~i − √313 ~j.
If θ is the angle between ∇f and u~ then
u | cos(θ) = |∇f | cos(θ).
Du~ f = ∇f · u~ = |∇f ||~
Christopher Croke
Calculus 115
Problem: Compute Du~ f where f (x, y ) = e xy + x 2 at the point
(1, 0) in the direction u~ = √213 ~i − √313 ~j.
If θ is the angle between ∇f and u~ then
u | cos(θ) = |∇f | cos(θ).
Du~ f = ∇f · u~ = |∇f ||~
This means that f increases most rapidly in the direction of ∇f .
Christopher Croke
Calculus 115
Problem: Compute Du~ f where f (x, y ) = e xy + x 2 at the point
(1, 0) in the direction u~ = √213 ~i − √313 ~j.
If θ is the angle between ∇f and u~ then
u | cos(θ) = |∇f | cos(θ).
Du~ f = ∇f · u~ = |∇f ||~
This means that f increases most rapidly in the direction of ∇f .
(And least rapidly in the direction −∇f .)
Christopher Croke
Calculus 115
Problem: Compute Du~ f where f (x, y ) = e xy + x 2 at the point
(1, 0) in the direction u~ = √213 ~i − √313 ~j.
If θ is the angle between ∇f and u~ then
u | cos(θ) = |∇f | cos(θ).
Du~ f = ∇f · u~ = |∇f ||~
This means that f increases most rapidly in the direction of ∇f .
(And least rapidly in the direction −∇f .)
In this direction Du~ f = |∇f | (or −|∇f |).
Christopher Croke
Calculus 115
Problem: Compute Du~ f where f (x, y ) = e xy + x 2 at the point
(1, 0) in the direction u~ = √213 ~i − √313 ~j.
If θ is the angle between ∇f and u~ then
u | cos(θ) = |∇f | cos(θ).
Du~ f = ∇f · u~ = |∇f ||~
This means that f increases most rapidly in the direction of ∇f .
(And least rapidly in the direction −∇f .)
In this direction Du~ f = |∇f | (or −|∇f |).
Another conclusion is that if u~⊥∇f then Du~ f = 0
Christopher Croke
Calculus 115
Problem: Find the directions of most rapid increase and decrease
at (x, y ) = (1, −1) of f (x, y ) = x 2 − 2xy .
What are the directions of 0 change?
Christopher Croke
Calculus 115
Problem: Find the directions of most rapid increase and decrease
at (x, y ) = (1, −1) of f (x, y ) = x 2 − 2xy .
What are the directions of 0 change?
Now f (x, y ) does not change along the level curves c = f (x, y ) for
any constant c. If we write the curve as (x(t), y (t)) and use the
chain rule we see:
dx
dy
d
f (x(t), y (t)) = fx ·
+ fy ·
= 0.
dt
dt
dt
Christopher Croke
Calculus 115
Problem: Find the directions of most rapid increase and decrease
at (x, y ) = (1, −1) of f (x, y ) = x 2 − 2xy .
What are the directions of 0 change?
Now f (x, y ) does not change along the level curves c = f (x, y ) for
any constant c. If we write the curve as (x(t), y (t)) and use the
chain rule we see:
dx
dy
d
f (x(t), y (t)) = fx ·
+ fy ·
= 0.
dt
dt
dt
In other words:
dy
dx
∇f · ( ~i + ~j) = 0.
dt
dt
Christopher Croke
Calculus 115
Problem: Find the directions of most rapid increase and decrease
at (x, y ) = (1, −1) of f (x, y ) = x 2 − 2xy .
What are the directions of 0 change?
Now f (x, y ) does not change along the level curves c = f (x, y ) for
any constant c. If we write the curve as (x(t), y (t)) and use the
chain rule we see:
dx
dy
d
f (x(t), y (t)) = fx ·
+ fy ·
= 0.
dt
dt
dt
In other words:
dy
dx
∇f · ( ~i + ~j) = 0.
dt
dt
Which says that ∇f is perpendicular to the level curve.
Christopher Croke
Calculus 115
Problem: Find the directions of most rapid increase and decrease
at (x, y ) = (1, −1) of f (x, y ) = x 2 − 2xy .
What are the directions of 0 change?
Now f (x, y ) does not change along the level curves c = f (x, y ) for
any constant c. If we write the curve as (x(t), y (t)) and use the
chain rule we see:
dx
dy
d
f (x(t), y (t)) = fx ·
+ fy ·
= 0.
dt
dt
dt
In other words:
dy
dx
∇f · ( ~i + ~j) = 0.
dt
dt
Which says that ∇f is perpendicular to the level curve. So the
tangent line to the level curve is the line perpendicular to ∇f .
Christopher Croke
Calculus 115
Problem: Find the directions of most rapid increase and decrease
at (x, y ) = (1, −1) of f (x, y ) = x 2 − 2xy .
What are the directions of 0 change?
Now f (x, y ) does not change along the level curves c = f (x, y ) for
any constant c. If we write the curve as (x(t), y (t)) and use the
chain rule we see:
dx
dy
d
f (x(t), y (t)) = fx ·
+ fy ·
= 0.
dt
dt
dt
In other words:
dy
dx
∇f · ( ~i + ~j) = 0.
dt
dt
Which says that ∇f is perpendicular to the level curve. So the
tangent line to the level curve is the line perpendicular to ∇f .
Christopher Croke
Calculus 115
Christopher Croke
Calculus 115
Problem: Find tangent line to the curve
x + sin(y ) + e xy = 2
at the point (1, 0).
Christopher Croke
Calculus 115
Three variables
Lets look at three independent variables, i.e. f (x, y , z). Then
∇f = fx ~i + fy ~j + fz ~k.
Christopher Croke
Calculus 115
Three variables
Lets look at three independent variables, i.e. f (x, y , z). Then
∇f = fx ~i + fy ~j + fz ~k.
and
Du~ f = ∇f · u~ = |∇f | cos(θ).
where θ is the angle between ∇f and u~.
Christopher Croke
Calculus 115
Three variables
Lets look at three independent variables, i.e. f (x, y , z). Then
∇f = fx ~i + fy ~j + fz ~k.
and
Du~ f = ∇f · u~ = |∇f | cos(θ).
where θ is the angle between ∇f and u~.
This means that ∇f is perpendicular to the level surfaces of f .
(i..e it is normal to all curves in the level surface.)
Christopher Croke
Calculus 115
Three variables
Lets look at three independent variables, i.e. f (x, y , z). Then
∇f = fx ~i + fy ~j + fz ~k.
and
Du~ f = ∇f · u~ = |∇f | cos(θ).
where θ is the angle between ∇f and u~.
This means that ∇f is perpendicular to the level surfaces of f .
(i..e it is normal to all curves in the level surface.)
The Tangent Plane at (x0 , y0 , z0 ) to the level surface
f (x, y , z) = c is the plane through the point (x0 , y0 , z0 ) normal to
∇f (x0 , y0 , z0 ).
Christopher Croke
Calculus 115
Three variables
Lets look at three independent variables, i.e. f (x, y , z). Then
∇f = fx ~i + fy ~j + fz ~k.
and
Du~ f = ∇f · u~ = |∇f | cos(θ).
where θ is the angle between ∇f and u~.
This means that ∇f is perpendicular to the level surfaces of f .
(i..e it is normal to all curves in the level surface.)
The Tangent Plane at (x0 , y0 , z0 ) to the level surface
f (x, y , z) = c is the plane through the point (x0 , y0 , z0 ) normal to
∇f (x0 , y0 , z0 ).
Thus the equation is:
fx (x − x0 ) + fy (y − y0 ) + fz (z − z0 ) = 0.
Christopher Croke
Calculus 115
Christopher Croke
Calculus 115
The Normal Line of the surface at (x0 , y0 , z0 ) is the line through
(x0 , y0 , z0 ) which is parallel to ∇f .
Christopher Croke
Calculus 115
The Normal Line of the surface at (x0 , y0 , z0 ) is the line through
(x0 , y0 , z0 ) which is parallel to ∇f .
x = x0 + fx t,
y = y0 + fy t,
Christopher Croke
Calculus 115
z = z0 + fz t.
The Normal Line of the surface at (x0 , y0 , z0 ) is the line through
(x0 , y0 , z0 ) which is parallel to ∇f .
x = x0 + fx t,
y = y0 + fy t,
z = z0 + fz t.
Problem: Find the tangent plane and normal line to the graph of
z = f (x, y ) = x 2 − 3xy at the point (1, 2, −5).
Christopher Croke
Calculus 115