Math 241: Multivariable Calculus, Lecture 12

Math 241: Multivariable calculus, Lecture 12

Geometry of the Gradient Section 14.6

go.illinois.edu/math241fa17

Monday, September 25th, 2017[1cm] Exam I Monday Evening, No class on Wednesday

Math 241: Problem of the day Implicit Diﬀerentiation. Find the equation of the line tangent to the curve y2 − x3 + 5x2 = 37 at the point (2, 5).

Two solutions...

Directional Derivative in Arbitrary dimension n For f : R → R and a unit vector u = hu1, . . . , uni we deﬁne the Directional Derivative as

d Duf(a1, . . . , an) = f(a1 + tu1, . . . , an + tun). dt t=0

This is calculated as a dot product with the gradient ∇f = hfx1 , . . . , fxn i:

Duf(a1, . . . , an) = ∇f(a1, . . . , an) · u

1 Gradient and the chain rule revisited n f : R → R and x1 = x1(t), . . . , xn = xn(t). Write

r(t) = hx1(t), . . . , xn(t)i

0 0 0 a vector valued function. Set r (t) = hx1(t), . . . , xn(t)i

The chain rule becomes: d f(r(t)) = ∇f(r(t)) · r0(t). dt This is the rate of change of f along the curve r(t).

Vector Fields. n n A Vector in R is ~v = hv1, v2, . . . , vni.A Vector Field on R is for each point x ∈ Rn a vector

V~ (x) = hV1(x),V2(x),...,Vn(x)i.

Think of the wind vector ﬁeld W~ (x) in Illinois: for each point x on the map you get a vector describing the direction and magnitude of the wind that is blowing at x. If f : R2 → R is a function of two variables, f(x, y), then we get Gradient Vector Field ∂f ∂f ∇f(a, b) = h (a, b), (a, b)i ∂x ∂y

Similarly we can of course look at f(x1, x2, . . . , xn). This gives a gradient vector ﬁeld on Rn. We want to understand what this vector ﬁeld ∇f tells us about the function f.

Meaning of the gradient Questions Let f : R2 → R be a diﬀerentiable function, and ∇f the corresponding gradient vector ﬁeld

• At (a, b) for which ~u is the directional derivative D~uf(a, b) maximal? • What is the relation between level curves f(x, y) = c and the gradient vector ﬁeld ∇f? • What does it mean when ∇f(a, b) = 0?

Question: For f : R2 → R and (a, b) ∈ R2, what is the direction of greatest increase of f at (a, b)? I.e. for what unit vector u is Duf(a, b) largest? What is this largest value?

2 We know that

Duf(a, b) = ∇f(a, b) · u = k∇f(a, b)kkuk cos(θ) = k∇f(a, b)k cos(θ)

This is biggest when θ = 0, i.e., when u points in the same direction as as the gradient vector. Answer:

∇f(a,b) • The directional derivative is maximal in direction of u = k∇f(a,b)k , and • Max increase is k∇f(a, b)k.

Here we are assuming ∇f(a, b) 6= 0 (otherwise Duf(a, b) = 0 for all u – why? )

Implicit Function Theorem Theorem. Suppose f : R2 → R is diﬀerentiable and f(a, b) = 0.

• If fy(a, b) 6= 0, then there is a diﬀerentiable function g(x) deﬁned on an interval (a − , a + ) so that g(a) = b and for every x ∈ (a − , a + ), we have f(x, g(x)) = 0. Furthermore,

f (x, g(x)) dy ∂f g0(x) = − x or = − ∂x f (x, g(x)) dx ∂f y ∂y

• If fx(a, b) 6= 0, then there is a diﬀerentiable function h(y) deﬁned on an interval (b − , b + ) so that h(b) = a and for every y ∈ (b − , b + ), we have f(h(y), y) = 0. Furthermore,

∂f f (h(y), y) dx h0(y) = − y or = − ∂y f (h(y), y) dy ∂f x ∂x

The tangent line to a level curve. Goal: Understand relation Gradient vector field and level curves. First we find the tangent line to a level curve. If (a, b) is a point, and f(x, y) a (differentiable) function (with ∇f(a, b) 6= ~0), then we get a level curve through (a, b) with equation f(x, y) = f(a, b)

Assume that ∂yf(a, b) 6= 0. By Implicit Function Thm we can near (a, b) think ∂f dy ∂x of y as a function y = g(x), and dx = − ∂f . Then the tangent line at (a, b) is ∂y dy y − b = (x − b). dx Rewrite this as ∂f ∂f (a, b)(x − a) + (a, b)(y − b) = 0. ∂x ∂y

3 Gradient and level curves Conclusion: Tangent line of the level curve at (a, b) is

hx − a, y − bi · ∇f(a, b) = 0.

In other words: ∇f(a, b) is orthogonal to the tangent line of the level curve f(x, y) = f(a, b) at (a, b).

Higher dimension Implicit Function Theorem. If f is diﬀerentiable and f(a1, . . . , an) = 0 with fxk (a1, . . . , an) 6= 0 some k = 1, . . . , n, then near (a1, . . . , an), f(x1, . . . , xn) = 0 is the graph of a diﬀerentiable function of the other n − 1 variables.

E.g. If ∇f(a, b, c) 6= 0, then near (a, b, c), the level surface f(x, y, z) = 0 is the graph of a function of two variables, and the tangent plane is

hx − a, y − b, z − ci · ∇f(a, b, c) = 0.

In general ∇f(a1, . . . , an) is orthogonal to “tangent hyperplane” to level set f(x1, . . . , xn) = 0.

What if ∇f(a, b) = ~0? Then we can not solve (using f(x, y) = f(a, b)) y = g(x) or x = h(y) (Implicit Function Thm does not work.) We will see that f will have a “critical point” at (a, b), that is a maximum, or a minimum, or . . . . GOOD LUCK TONIGHT! No class Wednesday.