From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
From Math 2220 Class 7
Dr. Allen Back
Sep. 12, 2014
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Lines
The tangent vector to the path c(t) = (x(t), y(t), z(t)) att = t0 is defined to be the vector
Dc(t0) = c ′(t0) =
dxdt
∣∣t=t0
dydt
∣∣∣t=t0
dzdt
∣∣t=t0
which we will sometimes write more informally as
~c ′(t0) = (x ′(t0), y ′(t0), z ′(t0)).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Lines
The line given by~r = c(t0) + t~c ′(t0)
is called the tangent line to the path c at t = t0.
(The approximation ∆c ∼ ~c ′∆t is replaced by an exactequality on the tangent line.)
Here
~r =
xyz
is the “position vector” of a general point on the line and t isany real number.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Lines
Tangent line to the helix c(t) = (4 cos t, 4 sin t, 3t) at t = π.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Chain Rule
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Chain Rule
For g : U ⊂ Rn → Rm and f : V ⊂ Rm → Rp, let’s use
p to denote a point of Rn
q to denote a point of Rm
r to denote a point of Rp.
So more colloquially, we might write
q = g(p)
r = f (q)
and so of course f ◦ g gives the relationship r = f (g(p)).(The latter is (f ◦ g)(p).)
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Chain Rule
Fix a point p0 with g(p0) = q0 and f (q0) = r0. Let thederivatives of g and f at the relevant points be
T = Dg(p0) S = Df (q0).
How are the changes in p, q, and r related?
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Chain Rule
Fix a point p0 with g(p0) = q0 and f (q0) = r0. Let thederivatives of g and f at the relevant points be
T = Dg(p0) S = Df (q0).
How are the changes in p, q, and r related?By the linear approximation properties of the derivative,
∆q ∼ T ∆p ∆r ∼ S∆q
And so plugging the first approximate equality into the secondgives the approximation
∆r ∼ S(T ∆p) = (ST )∆p.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Chain Rule
∆r ∼ (ST )∆p.
What is this saying?
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Chain Rule
∆r ∼ (ST )∆p.
What is this saying?For g : U ⊂ Rn → Rm and f : V ⊂ Rm → Rp,
T = Df (p0) is an m × n matrix
S = Dg(q0) is an p ×m matrix
So the product ST is a p × n matrix representing the derivativeat p0 of g ◦ f .
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Chain Rule
∆q ∼ T ∆p ∆r ∼ S∆q ∆r = ST ∆p
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Chain Rule
So the chain rule theorem says that if f is differentiable at p0
with f (p0) = q0 and g is differentiable at q0, then g ◦ f is alsodifferentiable at p0 with derivative the matrix product
(Dg(q0)) (Df (p0)) .
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Planes
The tangent plane to the graph of z = f (x , y) at(x , y) = (x0, y0) is defined to be the plane given by
z − z0 = fx(x0, y0)(x − x0) + fy (x0, y0)(y − y0).
(The approximation ∆z ∼ fx∆x + fy ∆y is replaced by an exactequality on the tangent plane.)
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Planes
Tangent plane to z = x2 − y 2 at (−1, 0, 1).
Note the tangent plane needn’t meet the surface in just onepoint.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Planes
The tangent plane to the graph of z = f (x , y) at(x , y) = (x0, y0) is defined to be the plane given by
z − z0 = fx(x0, y0)(x − x0) + fy (x0, y0)(y − y0).
Formula for tangent plane to z = x2 − y 2 at (−1, 0, 1)?
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Planes
The tangent plane to the graph of z = f (x , y) at(x , y) = (x0, y0) is defined to be the plane given by
z − z0 = fx(x0, y0)(x − x0) + fy (x0, y0)(y − y0).
Formula for tangent plane to z = x2 − y 2 at (−1, 0, 1)?
f (−1, 0) = 1 fx(−1, 0) = −2 fy (−1, 0) = 0.
So the tangent plane is
(z − 1) = −2(x + 1) + 0(y − 0).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Planes
Actually there is an “easier” way to compute tangent planes forboth graphs and level surfaces in the same way.
It starts out by observing that the graph of f (x , y) (thinkz = f (x , y)) is the same as the level set of
g(x , y , z) = f (x , y)− z = 0.
For example
z = x2 − y 2 ⇔ x2 − y 2 − z = 0.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Planes
It uses the concept of the gradient ∇g of a (real valued)function; g , namely, just its derivative, viewed as a vector.For example
g(x , y , z) = x2 − y 2 − z
∇g =
2x−2y−1
.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Planes
The key observation is that the gradient of g evaluated at anypoint p0 is perpendicular to the tangent plane to the levelsurface g(x , y , z) = c at the point p0.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Planes
In these terms, the tangent plane to x2 − y 2 − z = 0 at(−1, 0, 1) becomes
∇g =
2x−2y−1
∇g |(−1,0,1) =
2x−2y−1
∣∣∣∣∣∣(−1,0,1)
=
−20−1
.So the tangent plane thru (−1, 0, 1) is−2
0−1
·x − (−1)
y − 0z − 1
= 0
or − 2(x + 1) + 0y − (z − 1) = 0.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Tangent Planes
Why perpendicular?
Think about paths c(t) on a level surface g(x , y , z) = c orthink aout ∆g for points on the level surface.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Geometry of the Chain Rule
Consider f : R2 → R2 defined by(u, v) = f (x , y) = (x2 − y 2, 2xy).Given any path c(t) = (x(t), y(t)) in the xy plane, we can view
d = f ◦ c
as a curve in the uv plane. (i.e. (u(t), v(t)) = f (x(t), y(t)).)
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Geometry of the Chain Rule
Consider f : R2 → R2 defined by(u, v) = f (x , y) = (x2 − y 2, 2xy).Given any path c(t) = (x(t), y(t)) in the xy plane, we can view
d = f ◦ c
as a curve in the uv plane. (i.e. (u(t), v(t)) = f (x(t), y(t)).)
How are the tangents to the paths c and d related?
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Geometry of the Chain Rule
By the chain rule, if d = f ◦ c , then the derivatives are relatedby
Dd(t0) = Df (c(t0)) ◦ Dc(t0)
or using c ′ and d ′ to denote the tangents, we see that thematrix T = Df (t0) transforms (e.g. by matrix multiplication)the tangent c ′(t0) to the tangent d ′(t0).
d ′(t0) = Df (c(t0))c ′(t0).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Geometry of the Chain Rule
Explicitly
x(t) =et + e−t
2
y(t) =et − e−t
2
“parameterizes” the right hand half of the hyperbolax2 − y 2 = 1 and the curve d(t) = (u(t), v(t)) =
f (x(t), y(t)) = ((x(t))2 − (y(t))2, 2x(t)y(t))
explicitly isu(t) = 1
v(t) =e2t − e−2t
2.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Geometry of the Chain Rule
(u, v) = f (x , y) = (x2 − y 2, 2xy).
x(t) =et + e−t
2u(t) = 1
y(t) =et − e−t
2v(t) =
e2t − e−2t
2For example
c(0) = (1, 0), c ′(0) = (0, 1), d(0) = (1, 0), and d ′(0) = (0, 2).
Df =
[2x −2y2y 2x
]and at (1, 0),
Df (1, 0) =
[2 00 2
].
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Geometry of the Chain Rule
For example
c(0) = (1, 0), c ′(0) = (0, 1), d(0) = (1, 0), and d ′(0) = (0, 2).
Df =
[2x −2y2y 2x
]and at (1, 0),
Df (1, 0) =
[2 00 2
].
And d ′(0) = Df (c(0))c ′(0) checks:[02
]=
[2 00 2
] [01
].
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Geometry of the Chain Rule
And d ′(0) = Df (c(0))c ′(0) checks:[02
]=
[2 00 2
] [01
].
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Geometry of the Chain Rule
By the way, those formulas
x(t) =et + e−t
2
y(t) =et − e−t
2
could be more nicely expressed using the hyperbolic functions
cosh t =et + e−t
2
sinh t =et − e−t
2
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Geometry of the Chain Rule
cosh t =et + e−t
2
sinh t =et − e−t
2
The algebra and differentiation we did come down to theformulas
cosh2 t − sinh2 t = 1
d
dt(cosh t) = sinh t
d
dt(sinh t) = cosh t
(The hyperbolic functions are closely related to cos and sin .)
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
f : U ⊂ R2 → R and g : R → R2. Derivatives/Partialderivatives of f ◦ g and g ◦ f ?
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
e.g.
z = z(x , y) x = x(t) y = y(t)
or explicitly
z =√
x2 + y 2 x = cos t y = 2 sin t
f : R2 → R
c : R → R2
c(t) =(x(t), y(t))
f ◦ c :R → R
c ◦ f :R2 → R2
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
c(t) = (x(t), y(t)) x = cos t y = 2 sin t
(A vector valued function with 1 dimensional domain issometimes interpreted as a path c . It’s image is a curve; theabove c(t) could parametrize the ellipse 4x2 + y 2 = 4. )
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
Df =[
∂f∂x
∂f∂y
]Dc =
[dxdtdydt
]D(f ◦ c) =
[∂f∂x
∂f∂y
] [dxdtdydt
]=
∂f
∂x
dx
dt+∂f
∂y
dy
dt
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
Df =[
∂f∂x
∂f∂y
]Dc =
[dxdtdydt
]D(f ◦ c) =
[∂f∂x
∂f∂y
] [dxdtdydt
]=
∂f
∂x
dx
dt+∂f
∂y
dy
dt
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
Df =[
∂f∂x
∂f∂y
]Dc =
[dxdtdydt
]D(c ◦ f ) =
[dxdtdydt
] [∂f∂x
∂f∂y
]=
[dxdt
∂f∂x
dxdt
∂f∂y
dydt
∂f∂x
dydt
∂f∂y
]
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
If we use t to denote both scalars in the domain of c and therange of f (instead of z for the latter), the above might moreintuitively be written as
D(c ◦ f ) =
[dxdt
∂t∂x
dxdt
∂t∂y
dydt
∂t∂x
dydt
∂t∂y
]
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
If we use t to denote both scalars in the domain of c and therange of f (instead of z for the latter), the above might moreintuitively be written as
D(c ◦ f ) =
[dxdt
∂t∂x
dxdt
∂t∂y
dydt
∂t∂x
dydt
∂t∂y
]
where more confusingly, using t = t(x , y) instead ofz = f (x , y) we have
c(f (x , y)) = (x(t(x , y), y(t(x , y)).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
Alternatively
z = f (x1, x2) t = (c1(z), c2(z)) t = (c1(f (x1, x2)), c2(f (x1, x2)))
looks quite sensible.Tradeoffs among naturality, intuitiveness, and precision are whywe have so many notations for derivatives.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
More Chain Rule
Tree diagrams can be helpful in showing the dependencies forchain rule applications:
z = z(x , y)
x = x(t)
y = y(t)
z = z(x(t), y(t)).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
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More Chain Rule
z = z(x(t), y(t))
∂z
∂x
dx
dt+∂z
∂y
dy
dt
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
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More Chain Rule
z = z(x(t), y(t))
Intuitively, one might think:
1 A change ∆t in t causes a change ∆x in x with multiplierdx
dt.
2 The change ∆x in x contributes to a further change ∆z in
z with multiplier∂z
∂x. So the overall contribution to the
change in z from the x part has multiplier
∂z
∂x
dx
dttimes ∆t.
3 Similarly for the y part.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
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More Chain Rule
z = z(x , y)
x = x(u, v)
y = y(u, v)
z = z(x(u, v), y(u, v)).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
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More Chain Rule
z = z(x(u, v), y(u, v)).
∂z
∂u=∂z
∂x
∂x
∂u+∂z
∂y
∂y
∂u.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
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DifferentiatingThe MatrixSquaredFunction
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Problem:z = ex2y
w = cos (x + y)
x = u2 − v 2
y = 2uv
∂z
∂u?
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
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More Chain Rule
Cases like z = f (x , u(x , y), v(y)).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
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DifferentiatingThe MatrixSquaredFunction
Planes in R3
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More Chain Rule
More formal approach:
f : R3 → R
u : R2 → R
v : R → R
h : R2 → R
h(x , y) =f (x , u(x , y), v(y))
Dh =?
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
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Planes in R3
Lines in R3
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More Chain Rule
More formal approach:
f : R3 → R
u : R2 → R
v : R → R
h : R2 → R
h(x , y) =f (x , u(x , y), v(y))
Dh =?
Write h as a composition h = f ◦ k for k : R2 → R3.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
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Write h as a composition h = f ◦ k for k : R2 → R3.What should k be ?
(Recall h(x , y) = f (x , u(x , y), v(y)).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
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More Chain Rule
Write h as a composition h = f ◦ k for k : R2 → R3.
(Recall h(x , y) = f (x , u(x , y), v(y)).
So k should be defined as
k(x , y) = (x , u(x , y), v(y))
From Math2220 Class 7
V3
Tangent Lines
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Write h as a composition h = f ◦ k for k : R2 → R3.
(Recall h(x , y) = f (x , u(x , y), v(y)).
AndDh = Df ◦ Dk = . . . .
From Math2220 Class 7
V3
Tangent Lines
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More informally, e.g. thinking about a tree diagram forz = f (x , u(x , y), v(y)) and thinking of the underlying f asf (x , u, v), we’d have
∂z
∂x=∂f
∂x+∂f
∂u
∂u
∂x.
Notice expressions like
∂f
∂xor
∂z
∂x
have some ambiguity here that D1f does not.
From Math2220 Class 7
V3
Tangent Lines
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Differentiating f (A) = A2, A a 2× 2 matrix
This problem is “just like” the differentiating the cross productproblem on Supplementary Problems B, and we present thesolution as a model in the same style as the three partsrequested on that problem.Although we think of the matrix here as 2× 2, only notationalchanges would be needed to handle the n × n case.
From Math2220 Class 7
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Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
First although the customary labeling of the entries of a matixis different, we can label a general 2× 2 matrix A as[
x1 x2
x3 x4
].
So the matrix squaring function f (A) = A2 can be viewed as afunction f : R4 → R4 with f (x1, x2, x3, x4) giving the fourentries of the matrix A2 in terms of (x1, x2, x3, x4).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
The definition we gave in class of the derivative T of a functionf : R4 → R4 is a linear transformation (think T (v) = Av for amatrix A) T so that
limp→p0
f (p)− f (p0)− T (p − p0)
‖p − p0‖= 0.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
The definition we gave in class of the derivative T of a functionf : R4 → R4 is a linear transformation (think T (v) = Av for amatrix A) T so that
limp→p0
f (p)− f (p0)− T (p − p0)
‖p − p0‖= 0.
We will try to work out the derivative of f (A) = A2 at aparticular matrix A0.So in the definition above we think of p0 = A0 and p = A.(A is a more typical notation for a matrix than p which tendspsychologically to stand for “point.”)
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
Let’s use h = A− A0 so the denominator ‖p − p0‖ becomes‖h‖ and limp→p0 becomes limh→0 . The definition of thederivative T becomes the requirement that T is a lineartransformation satisfying
limh→0
f (A0 + h)− f (A0)− T (h)
‖h‖= 0.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
Let’s use h = A− A0 so the denominator ‖p − p0‖ becomes‖h‖ and limp→p0 becomes limh→0 . The definition of thederivative T becomes the requirement that T is a lineartransformation satisfying
limh→0
f (A0 + h)− f (A0)− T (h)
‖h‖= 0.
Looking at the first two terms of the numerator (as in part (a)of your homework problem) we compute
f (A0 + h)− f (A0) = (A0 + h)2 − A20
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
Looking at the first two terms of the numerator (as in part (a)of your homework problem) we compute
f (A0 + h)− f (A0) = (A0 + h)2 − A20
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
We compute
f (A0 + h)− f (A0) = (A0 + h)2 − A20
= (A0 + h)(A0 + h)− A20
= A0(A0 + h) + h(A0 + h) −A20
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
We compute
f (A0 + h)− f (A0) = (A0 + h)2 − A20
= (A0 + h)(A0 + h)− A20
= A0(A0 + h) + h(A0 + h) −A20
= A20 + A0h + hA0 + h2 −A2
0
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
We compute
f (A0 + h)− f (A0) = (A0 + h)2 − A20
= (A0 + h)(A0 + h)− A20
= A0(A0 + h) + h(A0 + h) −A20
= A20 + A0h + hA0 + h2 −A2
0
= A0h + hA0 + h2.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
f (A0 + h)− f (A0) = A0h + hA0 + h2
suggests what the derivative should be.
There is a linear (in h) part A0h + hA0 and a quadratic part h2.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
This is easy to make precise if you’ve studied linear algebra.There one learns that although in the end all lineartransformations are given by matrices times vectors, thedefinition of a linear transformation L : R4 → R4 is merely afunction satisfying
1 L(v1 + v2) = L(v1) + L(v2).
2 L(cv) = cL(v).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
So, as in part b) of your homework problem, based on
f (A0 + h)− f (A0) = A0h + hA0 + h2
define the linear part by
T (h) = A0h + hA0
for h a 2× 2 matrix h.It is clear that this function T satisfies the two conditionsabove of being a linear transformation.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
As in part c) of the homework, to show T is the derivative of hwe just have to explain why
limh→0
f (A0 + h)− f (A0)− T (h)
‖h‖= 0.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
As in part c) of the homework, to show T is the derivative of hwe just have to explain why
limh→0
f (A0 + h)− f (A0)− T (h)
‖h‖= 0.
But this is because
(A0h + hA0 + h2)− (A0h + hA0)
‖h‖=
h2
‖h‖= h
(h
‖h‖
).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
But this is because
(A0h + hA0 + h2)− (A0h + hA0)
‖h‖=
h2
‖h‖= h
(h
‖h‖
).
And as h→ 0, the product
h
(h
‖h‖
)→ 0
sinceh
‖h‖is a “unit vector.” (So each entry is bounded by 1 in
absolute value.)
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Differentiating f (A) = A2, A a 2× 2 matrix
Thus our conclusion is that the linear transformationT : R4 → R4 defined by
T (h) = A0h + hA0
is the derivative of the function f (A) = A2 at the point(matrix) A = A0.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Planes in R3
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Planes in R3
If the normal ~n =< a, b, c >, ~r =< x , y , z > is a general pointand P0 = (x0, y0, z0), then ~n · (~r − P0) = 0 becomes
a(x − x0) + b(y − y0) + c(z − z0) = 0.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Planes in R3
Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Planes in R3
Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).Solution: First find the normal
~n =−−−→P0P1 ×
−−−→P0P2
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Planes in R3
Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).Solution: First find the normal
~n =−−−→P0P1 ×
−−−→P0P2
The cross product: ∣∣∣∣∣∣i j k−2 1 10 2 2
∣∣∣∣∣∣which is
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Planes in R3
Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).Solution: First find the normal
~n =−−−→P0P1 ×
−−−→P0P2
The cross product: ∣∣∣∣∣∣i j k−2 1 10 2 2
∣∣∣∣∣∣which is
i
∣∣∣∣ 1 12 2
∣∣∣∣− j
∣∣∣∣ −2 10 2
∣∣∣∣+ k
∣∣∣∣ −2 10 2
∣∣∣∣ =< 0, 4,−4 > .
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Planes in R3
Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).
i
∣∣∣∣ 1 12 2
∣∣∣∣− j
∣∣∣∣ −2 10 2
∣∣∣∣+ k
∣∣∣∣ −2 10 2
∣∣∣∣ =< 0, 4,−4 > .
So our plane is
< 0, 4,−4 > ·(~r− < 1, 0, 1 >) = 0
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Planes in R3
Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).So our plane is
< 0, 4,−4 > ·(~r− < 1, 0, 1 >) = 0
or
0(x − 1) + 4(y − 0)− 4(z − 1) = 0 or 4y − 4z + 4 = 0.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Lines in R3
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Lines in R3
Find the equation of the line through the points P0 = (1, 1, 0)and P1 = (2, 2, 2).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Lines in R3
Find the equation of the line through the points P0 = (1, 1, 0)and P1 = (2, 2, 2).
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Lines in R3
Find the equation of the line through the points P0 = (1, 1, 0)and P1 = (2, 2, 2).
Solution:−−−→P0P1 = (2, 2, 2)− (1, 1, 0) =< 1, 1, 2 > .
So our line is
~r =< 1, 1, 0 > +t < 1, 1, 2 >=< 1 + t, 1 + t, 2t > .
where t is any real number.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Lines in R3
Solution:−−−→P0P1 = (2, 2, 2)− (1, 1, 0) =< 1, 1, 2 > .
So our line is
~r =< 1, 1, 0 > +t < 1, 1, 2 >=< 1 + t, 1 + t, 2t > .
where t is any real number.This is called the vector form of the equation of a line.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Lines in R3
Solution:−−−→P0P1 = (2, 2, 2)− (1, 1, 0) =< 1, 1, 2 > .
So our line is
~r =< 1, 1, 0 > +t < 1, 1, 2 >=< 1 + t, 1 + t, 2t > .
where t is any real number.This is called the vector form of the equation of a line.Thinking our general position vector ~r =< x , y , z >, we canexpress this as the parametric form:
x = 1 + t
y = 1 + t
z = 2t
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Lines in R3
Thinking our general position vector ~r =< x , y , z >, we canexpress this as the parametric form:
x = 1 + t
y = 1 + t
z = 2t
Solving for t shows
t = x − 1 = y − 1 =z
2
which realizes this line as the intersection of the planes x = yand z = 2(y − 1) but there are many other pairs of planescontaining this line.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
The cross product of vectors in R3 is another vector.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
The cross product of vectors in R3 is another vector.It is good because:
it is geometrically meaningful
it is straightforward to calculate
it is useful (e.g. torque, angular momentum)
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
The cross product of vectors in R3 is another vector.~u = ~v × ~w is geometrically determined by the properties:
~u is perpendicular to both ~v and ~w .
~u| is the |~v ||~w | sin θ, the area of the parallelogram spannedby ~v and ~w .
Choice from the remaining two possibilities is now madebased on the “right hand rule.”
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
~u = ~v × ~w is geometrically determined by the properties:
~u is perpendicular to both ~v and ~w .
~u| is the |~v ||~w | sin θ, the area of the parallelogram spannedby ~v and ~w .
Choice from the remaining two possibilities is now madebased on the “right hand rule.”
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
~u = ~v × ~w is geometrically determined by the properties:
~u is perpendicular to both ~v and ~w .
~u| is the |~v ||~w | sin θ, the area of the parallelogram spannedby ~v and ~w .
Choice from the remaining two possibilities is now madebased on the “right hand rule.”
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
~u = ~v × ~w is geometrically determined by the properties:
~u is perpendicular to both ~v and ~w .
~u| is the |~v ||~w | sin θ, the area of the parallelogram spannedby ~v and ~w .
Choice from the remaining two possibilities is now madebased on the “right hand rule.”
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
The algebraic definition of the cross product is based on the“determinant”
~v × ~w =
∣∣∣∣∣∣i j k
v1 v2 v3
w1 w2 w3
∣∣∣∣∣∣which means
~v × ~w = i
∣∣∣∣ v2 v3
w2 w3
∣∣∣∣− j
∣∣∣∣ v1 v3
w1 w3
∣∣∣∣+ k
∣∣∣∣ v1 v2
w1 w2
∣∣∣∣ .where ∣∣∣∣ a b
c d
∣∣∣∣ = ad − bc
and i =< 1, 0, 0 >, j =< 0, 1, 0 >, and k =< 0, 0, 1 >,
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
~v × ~w = −~w × ~v
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
Sincei × j = k
and cyclic (so j × k = i and k × i = j) it is sometimes easiestto use that algebra or comparison with the picture below todetermine cross products or use the right hand rule.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
For example
< 1, 1, 0 > × < 0, 0, 1 >= (i + j)× k = −j + i =< 1,−1, 0 >
is easier than writing out the 3× 3 determinant.
From Math2220 Class 7
V3
Tangent Lines
Chain Rule
TangentPlanes
Geometry ofthe Chain Rule
DifferentiatingThe MatrixSquaredFunction
Planes in R3
Lines in R3
Cross product
Cross product
Cross products can be used to
find the area of a parallelogram or triangle spanned by twovectors in R3.
find the volume of a parallelopiped using the scalar tripleproduct
~u · (~v × ~w) = (~u × ~v) · ~w .