MATH 251 – LECTURE 10
JENS FORSGÅRD
http://www.math.tamu.edu/~jensf/
This week: 12.4–6
webAssign: 12.4–6, due 2/15 11:55 p.m.
Friday:
Kevin.
Next week: 12.7
webAssign: 12.7, opens 2/15 12 a.m.
Friday 2/19:
Midterm 1: Covering chapters 11 and 12.
Help Sessions:
M W 5.30–8 p.m. in BLOC 161
Office Hours:
BLOC 641C
M 12:30–2:30 p.m.
W –
or by appointment.
The chain rule
Let y = f (x) and x = g(t). Then y(t) = f (x(t)), and y 0(t) = f 0(g(t))g 0(t). In other words
dy dy dx
=
.
dt dx dt
Definition 1. Let z = f (x, y) be a differentiable function, where x = x(t) and y = y(t) are differentiable
function of the variable t. Then
dz ∂f dx ∂f dy
=
+
.
dt
∂x dt ∂y dt
The chain rule
Exercise 2. Let z = f (x, y) = x2y + x4y 4, where x(t) = ln(t) and y(t) = sin(t). Find dz/dt.
The chain rule
Definition 3. Let z = f (x, y) be a differentiable function, where x = x(s, t) and y = y(s, t) are differentiable
function of the variables s and t. Then
∂z ∂f ∂x ∂f ∂y
=
+
,
∂s ∂x ∂s ∂y ∂s
∂z ∂f ∂x ∂f ∂y
=
+
.
∂t
∂x ∂t ∂y ∂t
The chain rule
Exercise 4. Let z = f (x, y) = x7 sin(y), where x(s, t) = t/s and y(s, t) = sin(st). Find ∂z/∂s and ∂z/∂t.
The chain rule
Exercise 5. The radius of a right circular cone is increasing at a rate of 1cm/s while its height is decreasing at
a rate of 2 cm/s. At what rate is the volume of the cone changing when the radius is 100cm and the height is
132cm?
The chain rule
Exercise 6. Tom is walking down a hill conveniently described by the surface z = 5 − x2 − y 2. If he’s walking
along the plane curve r(t) = ht2, t5i, how steep is his decline at the time t?
The Gradient
Definition 7. Let f (x, y) be a function. We define the gradient of f to be
∇f (x, y) = hfx0 (x, y), fy0 (x, y)i.
Directional derivatives
Let z = f (x, y) be a surface. We want to find the rate of change of the function f (x, y) at some point (x0, y0)
and in the direction of a unit vector u = hu1, u2i.
Directional derivatives
Definition 8. The directional derivative of f in the direction u = hu1, u2i, where |u| = 1, is given by
Duf (x, y) = ∇f (x, y) · u = fx0 (x, y)u1 + fy0 (x, y)u2.
Exercise 9. Find the directional derivative of the function f (x, y) = ln(x + ln(y)) at the point (π/2, 1) in the
direction of the vector v = h1, −1i.
Directional derivatives
Notice that since |u| = 1 we have that u = hcos(θ), sin(θ)i for some θ, so that
Duf (x, y) = fx0 (x, y) sin(θ) + fy0 (x, y) cos(θ).
Directional derivatives
Exercise 10. Tom is in a landscape defined by f (x, y) = sin(x) + cos(2yx), at the point P = P (1, 1). In which
direction should Tom walk if he wants to ascend as fast as possible?