From Math V3 Tangent Lines From Math 2220 Class...

From Math2220 Class 7

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

Tangent Lines

Chain Rule

TangentPlanes

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) =

∣∣t=t0

∣∣∣t=t0

∣∣t=t0

which we will sometimes write more informally as

~c ′(t0) = (x ′(t0), y ′(t0), z ′(t0)).

Tangent Lines

Chain Rule

TangentPlanes

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.)

is the “position vector” of a general point on the line and t isany real number.

Tangent Lines

Chain Rule

TangentPlanes

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

Tangent Lines

Chain Rule

TangentPlanes

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).)

Tangent Lines

Chain Rule

TangentPlanes

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?

Tangent Lines

Chain Rule

TangentPlanes

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

∆r ∼ (ST )∆p.

What is this saying?

Tangent Lines

Chain Rule

TangentPlanes

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 .

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

∆q ∼ T ∆p ∆r ∼ S∆q ∆r = ST ∆p

Tangent Lines

Chain Rule

TangentPlanes

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)) .

Tangent Lines

Chain Rule

TangentPlanes

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.)

Tangent Lines

Chain Rule

TangentPlanes

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Tangent Planes

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)?

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Tangent Planes

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).

Tangent Lines

Chain Rule

TangentPlanes

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.

Tangent Lines

Chain Rule

TangentPlanes

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

Tangent Lines

Chain Rule

TangentPlanes

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.

Tangent Lines

Chain Rule

TangentPlanes

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

·x − (−1)

y − 0z − 1

or − 2(x + 1) + 0y − (z − 1) = 0.

Tangent Lines

Chain Rule

TangentPlanes

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.

Tangent Lines

Chain Rule

TangentPlanes

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)).)

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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?

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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).

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Explicitly

x(t) =et + e−t

y(t) =et − e−t

“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

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

(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).

[2x −2y2y 2x

]and at (1, 0),

Df (1, 0) =

[2 00 2

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

For example

c(0) = (1, 0), c ′(0) = (0, 1), d(0) = (1, 0), and d ′(0) = (0, 2).

[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

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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

[2 00 2

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

By the way, those formulas

x(t) =et + e−t

y(t) =et − e−t

could be more nicely expressed using the hyperbolic functions

cosh t =et + e−t

sinh t =et − e−t

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

cosh t =et + e−t

sinh t =et − e−t

The algebra and differentiation we did come down to theformulas

cosh2 t − sinh2 t = 1

dt(cosh t) = sinh t

dt(sinh t) = cosh t

(The hyperbolic functions are closely related to cos and sin .)

Tangent Lines

Chain Rule

TangentPlanes

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 ?

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

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

Tangent Lines

Chain Rule

TangentPlanes

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. )

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

∂f∂x

∂f∂y

[dxdtdydt

]D(f ◦ c) =

[∂f∂x

∂f∂y

] [dxdtdydt

dt+∂f

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

∂f∂x

∂f∂y

[dxdtdydt

]D(f ◦ c) =

[∂f∂x

∂f∂y

] [dxdtdydt

dt+∂f

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

∂f∂x

∂f∂y

[dxdtdydt

]D(c ◦ f ) =

[dxdtdydt

] [∂f∂x

∂f∂y

∂f∂x

∂f∂y

∂f∂x

∂f∂y

Tangent Lines

Chain Rule

TangentPlanes

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 ) =

∂t∂x

∂t∂y

∂t∂x

∂t∂y

Tangent Lines

Chain Rule

TangentPlanes

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 ) =

∂t∂x

∂t∂y

∂t∂x

∂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)).

Tangent Lines

Chain Rule

TangentPlanes

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.

Tangent Lines

Chain Rule

TangentPlanes

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)).

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

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

dt+∂z

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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

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

dttimes ∆t.

3 Similarly for the y part.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

z = z(x , y)

x = x(u, v)

y = y(u, v)

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

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

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

∂u=∂z

∂u+∂z

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

Problem:z = ex2y

w = cos (x + y)

x = u2 − v 2

y = 2uv

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

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

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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))

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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))

Write h as a composition h = f ◦ k for k : R2 → R3.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

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)).

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

So k should be defined as

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

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

AndDh = Df ◦ Dk = . . . .

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

More Chain Rule

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

∂x=∂f

∂x+∂f

Notice expressions like

∂xor

have some ambiguity here that D1f does not.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

First although the customary labeling of the entries of a matixis different, we can label a general 2× 2 matrix A as[

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).

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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.”)

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

We compute

f (A0 + h)− f (A0) = (A0 + h)2 − A20

= (A0 + h)(A0 + h)− A20

= A0(A0 + h) + h(A0 + h) −A20

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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

= A0h + hA0 + h2.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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).

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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.

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.

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‖=

‖h‖= h

‖h‖

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

But this is because

(A0h + hA0 + h2)− (A0h + hA0)

‖h‖=

‖h‖= h

‖h‖

And as h→ 0, the product

‖h‖

)→ 0

sinceh

‖h‖is a “unit vector.” (So each entry is bounded by 1 in

absolute value.)

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Planes in R3

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.

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).

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

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

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

∣∣∣∣ 1 12 2

∣∣∣∣− j

∣∣∣∣ −2 10 2

∣∣∣∣+ k

∣∣∣∣ −2 10 2

∣∣∣∣ =< 0, 4,−4 > .

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).

∣∣∣∣ 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

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

0(x − 1) + 4(y − 0)− 4(z − 1) = 0 or 4y − 4z + 4 = 0.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Lines in R3

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).

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Lines in R3

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.

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.

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

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

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

The cross product of vectors in R3 is another vector.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

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)

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

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.”

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

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

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

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

∣∣∣∣− j

∣∣∣∣ v1 v3

∣∣∣∣+ k

∣∣∣∣ v1 v2

∣∣∣∣ .where ∣∣∣∣ a b

∣∣∣∣ = ad − bc

and i =< 1, 0, 0 >, j =< 0, 1, 0 >, and k =< 0, 0, 1 >,

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

~v × ~w = −~w × ~v

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

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.

Tangent Lines

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

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 .

From Math V3 Tangent Lines From Math 2220 Class...

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