From Math 2220 Class 6 Tangent Lines Planes in R 3 Lines in R 3 Dr. - - PowerPoint PPT Presentation

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

From Math 2220 Class 6

• Dr. Allen Back
• Sep. 10, 2014

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

The chain rule in multivariable calculus is in some ways very

• simple. But it can lead to extremely intricate sorts of

relationships (try thermodynamics in physical chemistry . . . ) as well as counter-intuitive looking formulas like ∂y ∂x = –

∂z ∂x ∂z ∂y

. (The above in a context where f (x, y, z) = C.)

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

First let’s try the conceptually simple point of view, using the fact that derivatives of functions are linear transformations. (Matrices.)

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Think about differentiable functions g : U ⊂ Rn → Rm and f : V ⊂ Rm → Rp where the image of f, namely (f (U)) is a subset of the domain V of g. The chain rule is about the derivative of the composition f ◦ g.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Here’s a picture:

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Fix a point p0 with g(p0) = q0 and f (q0) = r0. Let the derivatives 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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Fix a point p0 with g(p0) = q0 and f (q0) = r0. Let the derivatives 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 second gives the approximation ∆r ∼ S(T∆p) = (ST)∆p.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

∆r ∼ (ST)∆p. What is this saying?

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 derivative at p0 of g ◦ f .

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

∆q = T∆p ∆r = S∆q ∆r = ST∆p

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 also differentiable at p0 with derivative the matrix product (Dg(q0)) (Df (p0)) .

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Problem: Suppose we have the polar coordinate map g(r, θ) = (r cos θ, r sin θ) and (r, θ) = f (u, v) is given by f (u, v) = (uv, v). Find the derivative of g ◦ f .

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

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

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

e.g. z = z(x, y) x = x(t) y = y(t)

• r explicitly

z =

• x2 + y2

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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

c(t) = (x(t), y(t)) x = cos t y = 2 sin t (A vector valued function with 1 dimensional domain is sometimes interpreted as a path c. It’s image is a curve; the above c(t) could parametrize the ellipse 4x2 + y2 = 4. )

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Df =

• ∂f

∂x ∂f ∂y

• Dc =

dx

dt dy dt

• D(f ◦ c) =
• ∂f

∂x ∂f ∂y

dx

dt dy dt

• =

∂f ∂x dx dt + ∂f ∂y dy dt

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Df =

• ∂f

∂x ∂f ∂y

• Dc =

dx

dt dy dt

• D(f ◦ c) =
• ∂f

∂x ∂f ∂y

dx

dt dy dt

• =

∂f ∂x dx dt + ∂f ∂y dy dt

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Df =

• ∂f

∂x ∂f ∂y

• Dc =

dx

dt dy dt

• D(c ◦ f ) =

dx

dt dy dt ∂f ∂x ∂f ∂y

• =

dx

dt ∂f ∂x dx dt ∂f ∂y dy dt ∂f ∂x dy dt ∂f ∂y

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

If we use t to denote both scalars in the domain of c and the range of f (instead of z for the latter), the above might more intuitively be written as D(c ◦ f ) = dx

dt ∂t ∂x dx dt ∂t ∂y dy dt ∂t ∂x dy dt ∂t ∂y

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

If we use t to denote both scalars in the domain of c and the range of f (instead of z for the latter), the above might more intuitively be written as D(c ◦ f ) = dx

dt ∂t ∂x dx dt ∂t ∂y dy dt ∂t ∂x dy dt ∂t ∂y

• where more confusingly, using t = t(x, y) instead of

z = f (x, y) we have c(f (x, y)) = (x(t(x, y), y(t(x, y)).

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

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 why we have so many notations for derivatives.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Tree diagrams can be helpful in showing the dependencies for chain rule applications: z = z(x, y) x = x(t) y = y(t) z = z(x(t), y(t)).

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

z = z(x(t), y(t)) ∂z ∂x dx dt + ∂z ∂y dy dt

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

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 multiplier

dx 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 dt times ∆t.

3 Similarly for the y part.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

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

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

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

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

Problem: z = ex2y w = cos (x + y) x = u2 − v2 y = 2uv ∂z ∂u ?

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

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

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

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)).

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

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)). And Dh = Df ◦ Dk = . . . .

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Chain Rule

More informally, e.g. thinking about a tree diagram for z = f (x, u(x, y), v(y)) and thinking of the underlying f as f (x, u, v), we’d have ∂z ∂x = ∂f ∂x + ∂f ∂u ∂u ∂x . Notice expressions like ∂f ∂x

• r

∂z ∂x have some ambiguity here that D1f does not.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 exact equality on the tangent plane.)

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Tangent Planes

Tangent plane to z = x2 − y2 at (−1, 0, 1). Note the tangent plane needn’t meet the surface in just one point.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Tangent Lines

The tangent vector to the path c(t) = (x(t), y(t), z(t)) at t = t0 is defined to be the vector Dc(t0) = c′(t0) =    

dx dt

• t=t0

dy dt

• t=t0

dz dt

• t=t0

    which we will sometimes write more informally as

• c′(t0) = (x′(t0), y′(t0), z′(t0)).

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 exact equality on the tangent line.) Here

• r =

  x y z   is the “position vector” of a general point on the line and t is any real number.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Planes in R3

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 point and P0 = (x0, y0, z0), then n · ( r − P0) = 0 becomes a(x − x0) + b(y − y0) + c(z − z0) = 0.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Planes in R3

Find the equation of the plane through the three points P0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Planes in R3

Find the equation of the plane through the three points P0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3). Solution: First find the normal

• n = −

− − → P0P1 × − − − → P0P2

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Planes in R3

Find the equation of the plane through the three points P0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3). Solution: First find the normal

• n = −

− − → P0P1 × − − − → P0P2 The cross product:

• ˆ

i ˆ j ˆ k −2 1 1 2 2

• which is

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Planes in R3

Find the equation of the plane through the three points P0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3). Solution: First find the normal

• n = −

− − → P0P1 × − − − → P0P2 The cross product:

• ˆ

i ˆ j ˆ k −2 1 1 2 2

• which is

ˆ i

• 1

1 2 2

• − ˆ

j

• −2

1 2

• + ˆ

k

• −2

1 2

• =< 0, 4, −4 > .

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Planes in R3

Find the equation of the plane through the three points P0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3). ˆ i

• 1

1 2 2

• − ˆ

j

• −2

1 2

• + ˆ

k

• −2

1 2

• =< 0, 4, −4 > .

So our plane is < 0, 4, −4 > ·( r− < 1, 0, 1 >) = 0

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Planes in R3

Find the equation of the plane through the three points P0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3). So our plane is < 0, 4, −4 > ·( r− < 1, 0, 1 >) = 0

• r

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

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Lines in R3

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 can express this as the parametric form: x = 1 + t y = 1 + t z = 2t

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Lines in R3

Thinking our general position vector r =< x, y, z >, we can express 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 = y and z = 2(y − 1) but there are many other pairs of planes containing this line.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Cross product

The cross product of vectors in R3 is another vector.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 spanned by v and w. Choice from the remaining two possibilities is now made based on the “right hand rule.”

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 spanned by v and w. Choice from the remaining two possibilities is now made based on the “right hand rule.”

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 spanned by v and w. Choice from the remaining two possibilities is now made based on the “right hand rule.”

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 spanned by v and w. Choice from the remaining two possibilities is now made based on the “right hand rule.”

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Cross product

The algebraic definition of the cross product is based on the “determinant”

• v ×

w =

• ˆ

i ˆ j ˆ k v1 v2 v3 w1 w2 w3

• which means
• v ×

w = ˆ i

• v2

v3 w2 w3

• − ˆ

j

• v1

v3 w1 w3

• + ˆ

k

• v1

v2 w1 w2

• .

where

• a

b c d

• = ad − bc

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

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Cross product

• v ×

w = − w × v

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Cross product

Since ˆ i × ˆ j = ˆ k and cyclic (so ˆ j × ˆ k = ˆ i and ˆ k ׈ i = ˆ j) it is sometimes easiest to use that algebra or comparison with the picture below to determine cross products or use the right hand rule.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines Planes in R3 Lines in R3 Cross product

Cross product

For example < 1, 1, 0 > × < 0, 0, 1 >= (ˆ i + ˆ j) × ˆ k = −ˆ j +ˆ i =< 1, −1, 0 > is easier than writing out the 3 × 3 determinant.

From Math 2220 Class 6 V1c Chain Rule Tangent Planes Tangent Lines 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 two vectors in R3. find the volume of a parallelopiped using the scalar triple product

• u · (

v × w) = ( u × v) · w.