From Math V3 Tangent Lines From Math 2220 Class 7pi.math.cornell.edu/~back/m222_f14/slides/sep12_v3.pdf · 2015. 1. 15. · From Math 2220 Class 7 V3 Tangent Lines Chain Rule Tangent

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


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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)).


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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.


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Tangent line to the helix c(t) = (4 cos t, 4 sin t, 3t) at t = π.


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Chain Rule


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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).)


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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?


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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.


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∆r ∼ (ST )∆p.

What is this saying?


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∆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 .


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Chain Rule

∆q ∼ T ∆p ∆r ∼ S∆q ∆r = ST ∆p


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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)) .


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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.)


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Tangent plane to z = x2 − y 2 at (−1, 0, 1).

Note the tangent plane needn’t meet the surface in just onepoint.


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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)?


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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).


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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.


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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

.


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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.


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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.


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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.


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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)).)


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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?


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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).


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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.


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(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

].


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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

].


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And d ′(0) = Df (c(0))c ′(0) checks:[02

]=

[2 00 2

] [01

].


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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


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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 .)


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More Chain Rule

f : U ⊂ R2 → R and g : R → R2. Derivatives/Partialderivatives of f ◦ g and g ◦ f ?


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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


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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. )


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Df =[

∂f∂x

∂f∂y

]Dc =

[dxdtdydt

]D(f ◦ c) =

[∂f∂x

∂f∂y

] [dxdtdydt

]=

∂f

∂x

dx

dt+∂f

∂y

dy

dt


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Df =[

∂f∂x

∂f∂y

]Dc =

[dxdtdydt

]D(f ◦ c) =

[∂f∂x

∂f∂y

] [dxdtdydt

]=

∂f

∂x

dx

dt+∂f

∂y

dy

dt


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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

]


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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

]


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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)).


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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.


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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)).


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z = z(x(t), y(t))

∂z

∂x

dx

dt+∂z

∂y

dy

dt


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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.


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z = z(x , y)

x = x(u, v)

y = y(u, v)

z = z(x(u, v), y(u, v)).


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z = z(x(u, v), y(u, v)).

∂z

∂u=∂z

∂x

∂x

∂u+∂z

∂y

∂y

∂u.


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Problem:z = ex2y

w = cos (x + y)

x = u2 − v 2

y = 2uv

∂z

∂u?


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Cases like z = f (x , u(x , y), v(y)).


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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 =?


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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.


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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)).


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So k should be defined as

k(x , y) = (x , u(x , y), v(y))


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AndDh = Df ◦ Dk = . . . .


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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.


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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.


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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).


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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.


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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.”)


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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.


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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


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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


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We compute

f (A0 + h)− f (A0) = (A0 + h)2 − A20

= (A0 + h)(A0 + h)− A20

= A0(A0 + h) + h(A0 + h) −A20


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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


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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.


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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.


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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).


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Chain Rule

TangentPlanes



Planes in R3

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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.


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product


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.


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product


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‖

).


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product


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.)


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product


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.


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Planes in R3


V3

Tangent Lines

Chain Rule

TangentPlanes



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.


V3

Tangent Lines

Chain Rule

TangentPlanes



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).


V3

Tangent Lines

Chain Rule

TangentPlanes



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


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Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Planes in R3


~n =−−−→P0P1 ×

−−−→P0P2

The cross product: ∣∣∣∣∣∣i j k−2 1 10 2 2

∣∣∣∣∣∣which is


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Planes in R3


~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 > .


V3

Tangent Lines

Chain Rule

TangentPlanes



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


V3

Tangent Lines

Chain Rule

TangentPlanes



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.


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Lines in R3


V3

Tangent Lines

Chain Rule

TangentPlanes



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).


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Lines in R3



V3

Tangent Lines

Chain Rule

TangentPlanes



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.


V3

Tangent Lines

Chain Rule

TangentPlanes



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.


V3

Tangent Lines

Chain Rule

TangentPlanes



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


V3

Tangent Lines

Chain Rule

TangentPlanes



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.


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Cross product

The cross product of vectors in R3 is another vector.


V3

Tangent Lines

Chain Rule

TangentPlanes



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)


V3

Tangent Lines

Chain Rule

TangentPlanes



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.”


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Cross product

~u = ~v × ~w is geometrically determined by the properties:





V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Cross product






V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Cross product






V3

Tangent Lines

Chain Rule

TangentPlanes



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 >,


V3

Tangent Lines

Chain Rule

TangentPlanes



Planes in R3

Lines in R3

Cross product

Cross product

~v × ~w = −~w × ~v


V3

Tangent Lines

Chain Rule

TangentPlanes



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.


V3

Tangent Lines

Chain Rule

TangentPlanes



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.


V3

Tangent Lines

Chain Rule

TangentPlanes



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 .

Documents

From Math V3 Tangent Lines From Math 2220 Class 7pi.math.cornell.edu/~back/m222_f14/slides/sep12_v3.pdf · 2015. 1. 15. · From Math 2220 Class 7 V3 Tangent Lines Chain Rule Tangent