Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
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Theorem 1.15. If z : R n → R is differentiable, then all directional derivatives exist.<br />
Furthermore, the directional derivative of z at x0 in the direction of v is given by:<br />
(1.15) ∇z(x0) · v<br />
where · denotes the dot product of two vectors.<br />
Proof. Let l(t) = x0 + vt. Then l(t) = (l1(t), . . . , ln(t)); that is, l(t) is a vector function<br />
+ vit.<br />
Apply the chain rule:<br />
whose i th component is given by li(t) = x0i<br />
(1.16)<br />
Thus:<br />
(1.17)<br />
dz(l(t))<br />
dt<br />
= ∂z dl1 ∂z dln<br />
+ · · · +<br />
∂l1 dt ∂ln dt<br />
d<br />
dl<br />
z(l(t)) = ∇z ·<br />
dt dt<br />
Clearly dl/dt = v. We have l(0) = x0. Thus:<br />
d<br />
(1.18)<br />
dt z(x0<br />
<br />
<br />
+ tv) = ∇z(x0) · v<br />
t=0<br />
We now come to the two most important results about gradients, (i) the fact that they<br />
always point in the direction of steepest ascent with respect to the level curves of a function<br />
and (ii) that they are perpendicular (normal) to the level curves of a function. We can<br />
exploit this fact as we seek to maximize (or minimize) functions.<br />
Theorem 1.16. Let z : R n → R be differentiable, x0 ∈ R n . If ∇z(x0) = 0, then ∇z(x0)<br />
points in the direction in which z is increasing fastest.<br />
Proof. Recall ∇z(x0) · v is the directional derivative of z in direction v at x0. Assume<br />
that v is a unit vector. We know that:<br />
(1.19) ∇z(x0) · v = ||∇z(x0)|| cos θ<br />
(because we assumed v was a unit vector) where θ is the angle between the vectors ∇z(x0)<br />
and v. The function cos θ is largest when θ = 0, that is when v and ∇z(x0) are parallel<br />
vectors. (If ∇z(x0) = 0, then the directional derivative is zero in all directions.) <br />
Theorem 1.17. Let z : R n → R be differentiable and let x0 lie in the level set S defined<br />
by z(x) = k for fixed k ∈ R. Then ∇z(x0) is normal to the set S in the sense that if v<br />
is a tangent vector at t = 0 of a path c(t) contained entirely in S with c(0) = x0, then<br />
∇z(x0) · v = 0.<br />
Remark 1.18. Before giving the proof, we illustrate this theorem in Figure 1.5. The<br />
function is z(x, y) = x 4 +y 2 +2xy and x0 = (1, 1). At this point ∇z(x0) = (6, 4). We include<br />
the tangent line to the level set at the point (1,1) to illustrate the normality of the gradient<br />
to the level curve at the point.<br />
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