Lecture 12 (Oct. 8)
Implicit Di↵erentiation
Example: If x4 + y 4 = 1, find dy/dx.
Revisit this example: write the equation as F (x, y) = 0, where F (x, y) := x4 + y 4 1.
We assume that this equation defines y = y(x) implicitly. Now use the chain rule:
(Justification of our “assumption”: “Implicit Function Theorem”:
if F (x0 , y0 ) = 0, Fx (x, y) and Fy (x, y) are continuous near (x0 , y0 ), and Fy (x0 , y0 ) 6=
0, then for x near x0 , there is a di↵erentiable function y = y(x) with y(x) = y0 , and
F (x, y(x)) = 0.)
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Example: If ex + ey + ez = 1, find @z/@x.
Revisit: F (x, y, z) := ex + ey + ez
1 = 0.
15.6: Directional Derivatives and the Gradient
Definition: Let f (x, y) be a function of 2 variables, and let u =< a, b > be a unit
vector. The directional derivative of f at (x0 , y0 ) in the direction u is
f (x0 + ha, y0 + hb)
h!0
h
Du f (x0 , y0 ) = lim
f (x0 , y0 )
,
if this limit exists.
Remark: Du f (x, y) is the rate of change of f in the direction u. Note that Dî f =
@f /@x and Dĵ f = @f /@y.
Definition: The gradient of f (x, y) is the vector-function (of 2 variables)
rf (x, y) =< fx (x, y), fy (x, y) >=
@f
@f
î +
ĵ.
@x
@y
Remark: the gradient is just a convenient way of recording the partial derivatives.
Theorem: if f (x, y) is di↵erentiable, then f has a directional derivative in the direction
u for any unit vector u, and
Du f (x, y) = rf (x, y) · u.
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Proof:
Example: find the rate of change of f (x, y) = xy in the direction u = (cos(✓), sin(✓))
for any 0  ✓ < 2⇡. In what direction is f changing the fastest at (1, 1)?
Remark: the gradient and directional derivatives carry over to functions of
ables in the obvious way.
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3 vari-
Question: in what direction is a function f increasing/decreasing most rapidly?
Answer:
Theorem: suppose f is di↵erentiable. Then Du is maximized when u points in the
direction of rf .
Re-phrase: rf points in the direction in which f is most rapidly increasing.
Example: find the maximum rate of change of f (x, y, z) =
direction in which it is attained.
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x
y
+
y
z
at (4, 2, 1), and the
Tangent Planes to Level Surfaces
Let F (x, y, z) be a function of 3 variables. Let S be the level surface given by
F (x, y, z) = k (a two-dimensional surface in R3 ).
Let
r(t) = hx(t), y(t), z(t)i
describe a path on S, with r(0) = hx(0), y(0), z(0)i = hx0 , y0 , z0 i. Di↵erentiate
F (x(t), y(t), z(t)) = 0 with respect to t at t = 0:
d
F (x(t), y(t), z(t))|t=0
dt
= Fx (x0 , y0 , z0 )x0 (0) + Fy (x0 , y0 , z0 )y 0 (0) + Fz (x0 , y0 , z0 )z 0 (0)
= rF (x0 , y0 , z0 ) · r0 (0)
0=
(here we used the chain rule). Recall that the vector r0 (0) is tangent to S.
Conclusion: the gradient rF (x0 , y0 , z0 ) is normal to the tangent plane to S at
(x0 , y0 , z0 ). In other words, the equation of the tangent plane to S at (x0 , y0 , z0 ) is
rF (x0 , y0 , z0 ) · hx
x0 , y
y0 , z
z0 i = 0.
Also, the normal line to S at (x0 , y0 , z0 ) (the line through (x0 , y0 , z0 ) in the direction
perpendicular to the tangent plane to S) is given by
x x0
y y0
z z0
=
=
.
Fx (x0 , y0 , z0 )
Fy (x0 , y0 , z0 )
Fz (x0 , y0 , z0 )
Example: Let S be the graph z = f (x, y). Reproduce our familiar equation for the
tangent plane.
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Example: Find the tangent plane and normal line at a point on the sphere x2 +y 2 +z 2 =
r2 .
The situation in 2 dimensions is similar: rF (x, y) is perpendicular to the level
curve f (x, y) = k.
Summarize properties of the gradient:
• rf points in the direction of fastest increase of f
• rf is perpendicular to the level curves/surfaces of f .
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