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Chain Rule/Directional derivatives
Christopher Croke
University of Pennsylvania
Math 115UPenn, Fall 2011
Christopher Croke Calculus 115
Chain Rule
In the one variable case z = f (y) and y = g(x) then dzdx = dz
dydydx .
When there are two independent variables, say w = f (x , y) isdifferentiable and where both x and y are differentiable functionsof the same variable t then w is a function of t. and
dw
dt=∂w
∂x
dx
dt+∂w
∂y
dy
dt.
or in other words
dw
dt(t) = wx(x(t), y(t))x ′(t) + wy (x(t), y(t))y ′(t).
The proof is not hard and given in the text.Show tree diagram.
Christopher Croke Calculus 115
Chain Rule
In the one variable case z = f (y) and y = g(x) then dzdx = dz
dydydx .
When there are two independent variables, say w = f (x , y) isdifferentiable and where both x and y are differentiable functionsof the same variable t then w is a function of t.
and
dw
dt=∂w
∂x
dx
dt+∂w
∂y
dy
dt.
or in other words
dw
dt(t) = wx(x(t), y(t))x ′(t) + wy (x(t), y(t))y ′(t).
The proof is not hard and given in the text.Show tree diagram.
Christopher Croke Calculus 115
Chain Rule
In the one variable case z = f (y) and y = g(x) then dzdx = dz
dydydx .
When there are two independent variables, say w = f (x , y) isdifferentiable and where both x and y are differentiable functionsof the same variable t then w is a function of t. and
dw
dt=∂w
∂x
dx
dt+∂w
∂y
dy
dt.
or in other words
dw
dt(t) = wx(x(t), y(t))x ′(t) + wy (x(t), y(t))y ′(t).
The proof is not hard and given in the text.Show tree diagram.
Christopher Croke Calculus 115
Chain Rule
In the one variable case z = f (y) and y = g(x) then dzdx = dz
dydydx .
When there are two independent variables, say w = f (x , y) isdifferentiable and where both x and y are differentiable functionsof the same variable t then w is a function of t. and
dw
dt=∂w
∂x
dx
dt+∂w
∂y
dy
dt.
or in other words
dw
dt(t) = wx(x(t), y(t))x ′(t) + wy (x(t), y(t))y ′(t).
The proof is not hard and given in the text.Show tree diagram.
Christopher Croke Calculus 115
Chain Rule
In the one variable case z = f (y) and y = g(x) then dzdx = dz
dydydx .
When there are two independent variables, say w = f (x , y) isdifferentiable and where both x and y are differentiable functionsof the same variable t then w is a function of t. and
dw
dt=∂w
∂x
dx
dt+∂w
∂y
dy
dt.
or in other words
dw
dt(t) = wx(x(t), y(t))x ′(t) + wy (x(t), y(t))y ′(t).
The proof is not hard and given in the text.
Show tree diagram.
Christopher Croke Calculus 115
Chain Rule
In the one variable case z = f (y) and y = g(x) then dzdx = dz
dydydx .
When there are two independent variables, say w = f (x , y) isdifferentiable and where both x and y are differentiable functionsof the same variable t then w is a function of t. and
dw
dt=∂w
∂x
dx
dt+∂w
∂y
dy
dt.
or in other words
dw
dt(t) = wx(x(t), y(t))x ′(t) + wy (x(t), y(t))y ′(t).
The proof is not hard and given in the text.Show tree diagram.
Christopher Croke Calculus 115
Problem: Compute dwdt two ways when w = x2 + xy , x = cos(t),
and y = sin(t).
What is dwdt (π2 )?
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 oftwo variables? Say w = f (x , y , z) where each of x ,y ,z arefunctions of r and θ. This makes w a function of r and θ. Forexample compute ∂w
∂r .
Problem: Compute ∂w∂r and ∂w
∂s in terms of r and s whenw = x + y − z2, x = rs, y = r + s, and z = ers .
Christopher Croke Calculus 115
Problem: Compute dwdt two ways when w = x2 + xy , x = cos(t),
and y = sin(t).What is dw
dt (π2 )?
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 oftwo variables? Say w = f (x , y , z) where each of x ,y ,z arefunctions of r and θ. This makes w a function of r and θ. Forexample compute ∂w
∂r .
Problem: Compute ∂w∂r and ∂w
∂s in terms of r and s whenw = x + y − z2, x = rs, y = r + s, and z = ers .
Christopher Croke Calculus 115
Problem: Compute dwdt two ways when w = x2 + xy , x = cos(t),
and y = sin(t).What is dw
dt (π2 )?
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 oftwo variables? Say w = f (x , y , z) where each of x ,y ,z arefunctions of r and θ. This makes w a function of r and θ. Forexample compute ∂w
∂r .
Problem: Compute ∂w∂r and ∂w
∂s in terms of r and s whenw = x + y − z2, x = rs, y = r + s, and z = ers .
Christopher Croke Calculus 115
Problem: Compute dwdt two ways when w = x2 + xy , x = cos(t),
and y = sin(t).What is dw
dt (π2 )?
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 oftwo variables? Say w = f (x , y , z) where each of x ,y ,z arefunctions of r and θ. This makes w a function of r and θ.
Forexample compute ∂w
∂r .
Problem: Compute ∂w∂r and ∂w
∂s in terms of r and s whenw = x + y − z2, x = rs, y = r + s, and z = ers .
Christopher Croke Calculus 115
Problem: Compute dwdt two ways when w = x2 + xy , x = cos(t),
and y = sin(t).What is dw
dt (π2 )?
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 oftwo variables? Say w = f (x , y , z) where each of x ,y ,z arefunctions of r and θ. This makes w a function of r and θ. Forexample compute ∂w
∂r .
Problem: Compute ∂w∂r and ∂w
∂s in terms of r and s whenw = x + y − z2, x = rs, y = r + s, and z = ers .
Christopher Croke Calculus 115
Problem: Compute dwdt two ways when w = x2 + xy , x = cos(t),
and y = sin(t).What is dw
dt (π2 )?
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 oftwo variables? Say w = f (x , y , z) where each of x ,y ,z arefunctions of r and θ. This makes w a function of r and θ. Forexample compute ∂w
∂r .
Problem: Compute ∂w∂r and ∂w
∂s in terms of r and s whenw = x + y − z2, x = rs, y = r + s, and z = ers .
Christopher Croke Calculus 115
Implicit Differentiation
If F (x , y) = 0 defines y implicitly as a function of x , that isy = h(x), and if F is differentiable then:
0 =dF
dx= Fx · 1 + Fy ·
dy
dx.
Thusdy
dx= −Fx
Fy
whenever Fy 6= 0
Problem: Find dydx if x2 + y2x + sin(y) = 0.
Christopher Croke Calculus 115
Implicit Differentiation
If F (x , y) = 0 defines y implicitly as a function of x , that isy = h(x), and if F is differentiable then:
0 =dF
dx= Fx · 1 + Fy ·
dy
dx.
Thusdy
dx= −Fx
Fy
whenever Fy 6= 0
Problem: Find dydx if x2 + y2x + sin(y) = 0.
Christopher Croke Calculus 115
Implicit Differentiation
If F (x , y) = 0 defines y implicitly as a function of x , that isy = h(x), and if F is differentiable then:
0 =dF
dx= Fx · 1 + Fy ·
dy
dx.
Thusdy
dx= −Fx
Fy
whenever Fy 6= 0
Problem: Find dydx if x2 + y2x + sin(y) = 0.
Christopher Croke Calculus 115
Directional Derivatives
Given a function f (x , y) then the rate of change (w.r.t.t) along acurve (x(t), y(t)) is (by the chain rule)
fxdx
dt+ fy
dy
dt.
If ~u is a unit vector we can write ~u = u1~i + u2~j (whereu21 + u22 = 1). So the line through (x0, y0) in the direction ~u witharc length parameter is (x(s), y(s)) where:
x(s) = u1s + x0 y(s) = u2s + y0.
The directional derivative in the direction ~u is thus(dfds
)~u,(x0,y0)
= fx(x0, y0)u1 + fy (x0, y0)u2 ≡(D~uf
)(x0,y0)
.
Christopher Croke Calculus 115
Directional Derivatives
Given a function f (x , y) then the rate of change (w.r.t.t) along acurve (x(t), y(t)) is (by the chain rule)
fxdx
dt+ fy
dy
dt.
If ~u is a unit vector we can write ~u = u1~i + u2~j (whereu21 + u22 = 1).
So the line through (x0, y0) in the direction ~u witharc length parameter is (x(s), y(s)) where:
x(s) = u1s + x0 y(s) = u2s + y0.
The directional derivative in the direction ~u is thus(dfds
)~u,(x0,y0)
= fx(x0, y0)u1 + fy (x0, y0)u2 ≡(D~uf
)(x0,y0)
.
Christopher Croke Calculus 115
Directional Derivatives
Given a function f (x , y) then the rate of change (w.r.t.t) along acurve (x(t), y(t)) is (by the chain rule)
fxdx
dt+ fy
dy
dt.
If ~u is a unit vector we can write ~u = u1~i + u2~j (whereu21 + u22 = 1). So the line through (x0, y0) in the direction ~u witharc length parameter is (x(s), y(s)) where:
x(s) = u1s + x0 y(s) = u2s + y0.
The directional derivative in the direction ~u is thus(dfds
)~u,(x0,y0)
= fx(x0, y0)u1 + fy (x0, y0)u2 ≡(D~uf
)(x0,y0)
.
Christopher Croke Calculus 115
Directional Derivatives
Given a function f (x , y) then the rate of change (w.r.t.t) along acurve (x(t), y(t)) is (by the chain rule)
fxdx
dt+ fy
dy
dt.
If ~u is a unit vector we can write ~u = u1~i + u2~j (whereu21 + u22 = 1). So the line through (x0, y0) in the direction ~u witharc length parameter is (x(s), y(s)) where:
x(s) = u1s + x0 y(s) = u2s + y0.
The directional derivative in the direction ~u is thus(dfds
)~u,(x0,y0)
= fx(x0, y0)u1 + fy (x0, y0)u2 ≡(D~uf
)(x0,y0)
.
Christopher Croke Calculus 115
Definition:(D~uf
)(x0,y0)
= lims→0
f (x0 + u1s, y0 + u2s)− f (x0, y0)
s.
What is geometric meaning?
If you look at the formula you note that(D~uf
)(x0,y0)
=(fx(x0, y0)~i + fy (x0, y0)~j
)· ~u
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 .
andD~uf = ∇f · ~u.
Christopher Croke Calculus 115
Definition:(D~uf
)(x0,y0)
= lims→0
f (x0 + u1s, y0 + u2s)− f (x0, y0)
s.
What is geometric meaning?
If you look at the formula you note that(D~uf
)(x0,y0)
=(fx(x0, y0)~i + fy (x0, y0)~j
)· ~u
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 .
andD~uf = ∇f · ~u.
Christopher Croke Calculus 115
Definition:(D~uf
)(x0,y0)
= lims→0
f (x0 + u1s, y0 + u2s)− f (x0, y0)
s.
What is geometric meaning?
If you look at the formula you note that(D~uf
)(x0,y0)
=(fx(x0, y0)~i + fy (x0, y0)~j
)· ~u
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 .
andD~uf = ∇f · ~u.
Christopher Croke Calculus 115
Definition:(D~uf
)(x0,y0)
= lims→0
f (x0 + u1s, y0 + u2s)− f (x0, y0)
s.
What is geometric meaning?
If you look at the formula you note that(D~uf
)(x0,y0)
=(fx(x0, y0)~i + fy (x0, y0)~j
)· ~u
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 .
andD~uf = ∇f · ~u.
Christopher Croke Calculus 115
Definition:(D~uf
)(x0,y0)
= lims→0
f (x0 + u1s, y0 + u2s)− f (x0, y0)
s.
What is geometric meaning?
If you look at the formula you note that(D~uf
)(x0,y0)
=(fx(x0, y0)~i + fy (x0, y0)~j
)· ~u
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 .
andD~uf = ∇f · ~u.
Christopher Croke Calculus 115
Definition:(D~uf
)(x0,y0)
= lims→0
f (x0 + u1s, y0 + u2s)− f (x0, y0)
s.
What is geometric meaning?
If you look at the formula you note that(D~uf
)(x0,y0)
=(fx(x0, y0)~i + fy (x0, y0)~j
)· ~u
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 .
andD~uf = ∇f · ~u.
Christopher Croke Calculus 115
Problem: Compute D~uf where f (x , y) = exy + x2 at the point(1, 0) in the direction ~u = 2√
13~i − 3√
13~j .
If θ is the angle between ∇f and ~u then
D~uf = ∇f · ~u = |∇f ||~u| cos(θ) = |∇f | cos(θ).
This means that f increases most rapidly in the direction of ∇f .(And least rapidly in the direction −∇f .)In this direction D~uf = |∇f | (or −|∇f |).
Another conclusion is that if ~u⊥∇f then D~uf = 0
Christopher Croke Calculus 115
Problem: Compute D~uf where f (x , y) = exy + x2 at the point(1, 0) in the direction ~u = 2√
13~i − 3√
13~j .
If θ is the angle between ∇f and ~u then
D~uf = ∇f · ~u = |∇f ||~u| cos(θ) = |∇f | cos(θ).
This means that f increases most rapidly in the direction of ∇f .(And least rapidly in the direction −∇f .)In this direction D~uf = |∇f | (or −|∇f |).
Another conclusion is that if ~u⊥∇f then D~uf = 0
Christopher Croke Calculus 115
Problem: Compute D~uf where f (x , y) = exy + x2 at the point(1, 0) in the direction ~u = 2√
13~i − 3√
13~j .
If θ is the angle between ∇f and ~u then
D~uf = ∇f · ~u = |∇f ||~u| cos(θ) = |∇f | cos(θ).
This means that f increases most rapidly in the direction of ∇f .
(And least rapidly in the direction −∇f .)In this direction D~uf = |∇f | (or −|∇f |).
Another conclusion is that if ~u⊥∇f then D~uf = 0
Christopher Croke Calculus 115
Problem: Compute D~uf where f (x , y) = exy + x2 at the point(1, 0) in the direction ~u = 2√
13~i − 3√
13~j .
If θ is the angle between ∇f and ~u then
D~uf = ∇f · ~u = |∇f ||~u| cos(θ) = |∇f | cos(θ).
This means that f increases most rapidly in the direction of ∇f .(And least rapidly in the direction −∇f .)
In this direction D~uf = |∇f | (or −|∇f |).
Another conclusion is that if ~u⊥∇f then D~uf = 0
Christopher Croke Calculus 115
Problem: Compute D~uf where f (x , y) = exy + x2 at the point(1, 0) in the direction ~u = 2√
13~i − 3√
13~j .
If θ is the angle between ∇f and ~u then
D~uf = ∇f · ~u = |∇f ||~u| cos(θ) = |∇f | cos(θ).
This means that f increases most rapidly in the direction of ∇f .(And least rapidly in the direction −∇f .)In this direction D~uf = |∇f | (or −|∇f |).
Another conclusion is that if ~u⊥∇f then D~uf = 0
Christopher Croke Calculus 115
Problem: Compute D~uf where f (x , y) = exy + x2 at the point(1, 0) in the direction ~u = 2√
13~i − 3√
13~j .
If θ is the angle between ∇f and ~u then
D~uf = ∇f · ~u = |∇f ||~u| cos(θ) = |∇f | cos(θ).
This means that f increases most rapidly in the direction of ∇f .(And least rapidly in the direction −∇f .)In this direction D~uf = |∇f | (or −|∇f |).
Another conclusion is that if ~u⊥∇f then D~uf = 0
Christopher Croke Calculus 115
Problem: Find the directions of most rapid increase and decreaseat (x , y) = (1,−1) of f (x , y) = x2 − 2xy .What are the directions of 0 change?
Now f (x , y) does not change along the level curves c = f (x , y) forany constant c . If we write the curve as (x(t), y(t)) and use thechain rule we see:
d
dtf (x(t), y(t)) = fx ·
dx
dt+ fy ·
dy
dt= 0.
In other words:
∇f · (dxdt~i +
dy
dt~j) = 0.
Which says that ∇f is perpendicular to the level curve. So thetangent line to the level curve is the line perpendicular to ∇f .
Problem: Find tangent line to the curve
x + sin(y) + exy = 2
at the point (1, 0).
Christopher Croke Calculus 115
Problem: Find the directions of most rapid increase and decreaseat (x , y) = (1,−1) of f (x , y) = x2 − 2xy .What are the directions of 0 change?
Now f (x , y) does not change along the level curves c = f (x , y) forany constant c . If we write the curve as (x(t), y(t)) and use thechain rule we see:
d
dtf (x(t), y(t)) = fx ·
dx
dt+ fy ·
dy
dt= 0.
In other words:
∇f · (dxdt~i +
dy
dt~j) = 0.
Which says that ∇f is perpendicular to the level curve. So thetangent line to the level curve is the line perpendicular to ∇f .
Problem: Find tangent line to the curve
x + sin(y) + exy = 2
at the point (1, 0).
Christopher Croke Calculus 115
Problem: Find the directions of most rapid increase and decreaseat (x , y) = (1,−1) of f (x , y) = x2 − 2xy .What are the directions of 0 change?
Now f (x , y) does not change along the level curves c = f (x , y) forany constant c . If we write the curve as (x(t), y(t)) and use thechain rule we see:
d
dtf (x(t), y(t)) = fx ·
dx
dt+ fy ·
dy
dt= 0.
In other words:
∇f · (dxdt~i +
dy
dt~j) = 0.
Which says that ∇f is perpendicular to the level curve. So thetangent line to the level curve is the line perpendicular to ∇f .
Problem: Find tangent line to the curve
x + sin(y) + exy = 2
at the point (1, 0).
Christopher Croke Calculus 115
Problem: Find the directions of most rapid increase and decreaseat (x , y) = (1,−1) of f (x , y) = x2 − 2xy .What are the directions of 0 change?
Now f (x , y) does not change along the level curves c = f (x , y) forany constant c . If we write the curve as (x(t), y(t)) and use thechain rule we see:
d
dtf (x(t), y(t)) = fx ·
dx
dt+ fy ·
dy
dt= 0.
In other words:
∇f · (dxdt~i +
dy
dt~j) = 0.
Which says that ∇f is perpendicular to the level curve.
So thetangent line to the level curve is the line perpendicular to ∇f .
Problem: Find tangent line to the curve
x + sin(y) + exy = 2
at the point (1, 0).
Christopher Croke Calculus 115
Problem: Find the directions of most rapid increase and decreaseat (x , y) = (1,−1) of f (x , y) = x2 − 2xy .What are the directions of 0 change?
Now f (x , y) does not change along the level curves c = f (x , y) forany constant c . If we write the curve as (x(t), y(t)) and use thechain rule we see:
d
dtf (x(t), y(t)) = fx ·
dx
dt+ fy ·
dy
dt= 0.
In other words:
∇f · (dxdt~i +
dy
dt~j) = 0.
Which says that ∇f is perpendicular to the level curve. So thetangent line to the level curve is the line perpendicular to ∇f .
Problem: Find tangent line to the curve
x + sin(y) + exy = 2
at the point (1, 0).
Christopher Croke Calculus 115
Problem: Find the directions of most rapid increase and decreaseat (x , y) = (1,−1) of f (x , y) = x2 − 2xy .What are the directions of 0 change?
Now f (x , y) does not change along the level curves c = f (x , y) forany constant c . If we write the curve as (x(t), y(t)) and use thechain rule we see:
d
dtf (x(t), y(t)) = fx ·
dx
dt+ fy ·
dy
dt= 0.
In other words:
∇f · (dxdt~i +
dy
dt~j) = 0.
Which says that ∇f is perpendicular to the level curve. So thetangent line to the level curve is the line perpendicular to ∇f .
Problem: Find tangent line to the curve
x + sin(y) + exy = 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 .
andD~uf = ∇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 surfacef (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
Three variables
Lets look at three independent variables, i.e. f (x , y , z). Then
∇f = fx~i + fy~j + fz~k .
andD~uf = ∇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 surfacef (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
Three variables
Lets look at three independent variables, i.e. f (x , y , z). Then
∇f = fx~i + fy~j + fz~k .
andD~uf = ∇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 surfacef (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
Three variables
Lets look at three independent variables, i.e. f (x , y , z). Then
∇f = fx~i + fy~j + fz~k .
andD~uf = ∇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 surfacef (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
Three variables
Lets look at three independent variables, i.e. f (x , y , z). Then
∇f = fx~i + fy~j + fz~k .
andD~uf = ∇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 surfacef (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
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 ofz = f (x , y) = x2 − 3xy at the point (1, 2,−5).
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, z = z0 + fz t.
Problem: Find the tangent plane and normal line to the graph ofz = f (x , y) = x2 − 3xy at the point (1, 2,−5).
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, z = z0 + fz t.
Problem: Find the tangent plane and normal line to the graph ofz = f (x , y) = x2 − 3xy at the point (1, 2,−5).
Christopher Croke Calculus 115
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