1.2 Directional Derivatives

Functions and their properties 9
y ∂z
∂y
x ∂z ∂z
+ y
∂x ∂y
=
x 2
x 2 + y 2
1
= y ·
2 (
1
x2 ) · (2y)
+ y2 =
= 1
y 2
x 2 + y 2
1.2 Directional Derivatives
Let z = f(x, y) define a surface S in R 3 and P0(x0,y0) apointonS.
The gradient of z (or f) atP0 is a vector defined by:
∇f(P0) = ∂f
∂x |P0 i + ∂f
|P0
j
∂y
(this vector, if it is not equal to zero, is always perpendicular to the surface S
(ornormalto it)). Why??
If F (x, y, z) = 0 defines a surface S ∗ then, for P0 on S ∗ , the gradient of F is:
∇F |P0 = ∂f
∂x |P0 i + ∂f
∂y |P0 j + ∂f
∂z |P0 k
Let u be any vector on the xy−plane, and z = f(x, y) define the surface
S. The directional derivative of f in the direction u at the point P0 on S:(u is
a unit vector here)
∂f
∂s |P0 = ∇F |P0 · u
[It represents the rate of change of f along the ”curve,” f(x, y) onS where (x, y)
runs along u]
Note: The directional derivative is a maximum when the angle Θ between ∇f|p0
and u is 0, because:
∇f|p0 · u = |∇f||p0 ·|u| cos Θ
Figure 1
ie. when ∇f|P0 and u are parallel.
So, directional derivative is maximum in the direction of ∇f|P0 of f at P0
The directional derivative of z = f(x, y) atP0(x0,y0) onS in the direction Θ
(Θ in radians) means in the direction of the unit vector u =(cosθ)i +(sinΘ)j

10 The ABC’s of Calculus
Example 9
Find ∂f
∂s |P0
Here f(x, y) =x 2 + xy + y 2
z = x 2 + xy + y 2 , P0 =(3, 1), Θ= π
3
∇f = ∂f
∂x i + ∂f
∂y j
= (2x + y)i +(x +2y)j
∇f|P0 = ∇f| (3,1)
= (2· 3+1)i +(3+2· 1)j
= 7i +5j
Also u = cosΘi +sinΘj
= cos π
3 i +sin π
3 j
= 1
2 i
√
3
+
2 j
Therefore ∂f
∂s |P0 = ∇f|P0 · u
= (7i +5j) · ( 1
2 i +
= 7
2 + 5√3 2
Example 10 z =2x 2 +3xy − y 2 at (1, 1) toward (2, 1).
√ 3
2 j)
Here f(x, y) =2x 2 +3xy − y 2 , ∇f =(4x +3y, 3x − 2y) or ∇f| (1,−1) =(4· 1+
3(−1), 3(1) − 2(−1)) = (1, 5)
To find u we need to find a vector from (1, 1) to (2, 1). This is given by v =
√(2, 1) − (1, −1) √ = (1, 2). Tomakevinto a unit vector we divide v by its length:
12 +22 = 5.
Therefore u = v (1, 2)
= √ =(
|v| 5 1
√ ,
5 2
√ )
5
∂f
∂s | (1,−1) = ∇f| (1,−1) · u
= (1, 5) · ( 1
√ ,
5 2 √ )
5
=
1
√ +2
5 √ 5
√
5
=
5 +2√5 = 11√
5
5

Functions and their properties 11
What is its maximum value? When:
u = ∇f
|∇f| |P0 : ∂f
∂s | (1,−1) = 1 2 +5 2 = √ 26
1.3 Maxima and Minima Subject to Constraints
To look for maximum and minimun of functions of two variables, z = f(x, y)
we set:
∂f
∂f
=0 and
∂x ∂y =0
and look for those points P (x, y) atwhichbothof these hold. If, at such P ,
( ∂2f ∂x∂y )2 − ( ∂2f ∂x2 )(∂2 f
) < 0
∂y2 and
1) ∂2f ∂x2 + ∂2f ∂y2 > 0
OR:
at p ⇒ relative minimum at p
2) ∂2f ∂x2 + ∂2f ∂y2 < 0 at p ⇒ relative maximum at p
3) ∂2 f
∂x 2 + ∂2 f
∂y 2 =0 atp ⇒ Undetermined: need more info.
NOTE : The boundary of the domain of f must always be checked separately for
relative maximum or minimum values.
Example 11 z =2x +4y − x 2 − y 2 − 3 (= f(x, y))
(Domain of f is the whole plane R 2 ), (so boundary of domain is at ∞)
So we need:
(1, 2) is only possible candidate
∂z 0= ∂x =2− 2x ⇒ x =1
=4− 2y ⇒ y =2
0= ∂z
∂y
so x =1, y =2. This means that (1, 2) is the only critical point inside dom(f).
At P0:
[( ∂2 f
∂x∂y )2 +( ∂2 f
∂x 2 ) · (∂2 f
∂y 2 )](1, 2) = (0)2 − (−2)(−2) = −4 < 0 O.K.
So go to the next step, At P0:
[( ∂2f ∂x2 )+(∂2 f
)](1, 2) = (−2) + (−2) = −4 < 0
∂y2 By this second derivative test, therefore relative maximum at (1, 2).

12 The ABC’s of Calculus
Example 12 Find positive numbers x, y, z such that x + y + z =12and xy 2 z 3
is a maximum.
So, here we want to maximize the function xy 2 z 3 subject to the constraint x +
y + z =12.
But then z =12−x−y and so we wnat to maximize the function of two variables
f(x, y) =xy 2 (12 − x − y) 3 over those x>0,y >0.
Now,
0= ∂f
∂x = y2 (12 − x − y) 3 + xy 2 · 3(12 − x − y) 2 (−1)
= (12−x−y) 2 y 2 [(12 − x − y)+x(−3)]
= (12−x−y) 2 y 2 next
(12 − 4x − y)
0= ∂f
∂y = 2xy(12 − x − y)3 + xy 2 · 3(12 − x − y) 2 (−1)
= xy(12 − x − y) 2 [2(12 − x − y) − 3y]
= xy(12 − x − y) 2 (24 − 2x − 5y)
so we need to find those x, y for which,
and
(12 − x − y) 2 y 2 (12 − 4x − y) =0
(12 − x − y) 2 xy(24 − 2x − 5y) =0
there are many possiblilities – Each one must be considered in turn.
1) One possibility is that (12 − x − y) 2 =0⇒ x + y =12
In this case, since x+y+z =12and x+y =12⇒ z =0and so xy 2 z 3 =0
is a ”minimum” (as x, y, z are all assumed positive).
2) Another possibility is tah y =0but again this gives a minimum.
3) Yet another possibility is that 12 − 4x − y =0Now from the second equation
there are two sub-cases (Why?):
a) x =0
b) 24 − 2x − 5y =0
a) x =0As before this gives xy2z 3 =0, a minimum.
b) 24 − 2x − 5y =0and 12 − 4x − y =0have the common solution
(x, y) =(2, 4).
So the only possible critical point which can give a maximum is: (x, y) =(2, 4)
For (x, y) =(2, 4), z=12− x − y =12− 2 − 4=6.

Functions and their properties 13
Now at (2, 4) =
[( ∂2 f
∂x∂y )2 +( ∂2 f
∂x 2 ) · ( ∂2 f
∂y 2 )] < 0
And calculations show that the above is a maximum.
Maximum value
xy 2 z 3 = (2)(4) 2 (6) 3 =32· 6 3
Remarks: In many of these problems we are faced with solving systems of
equations of the form:-
ABC =0
Where A, B, C are functions of several variables.
Then the cases are:
A = 0 B = 0 C =0
A = 0 B =0 C = 0
A = 0 B =0 C =0
A =0 B = 0 C =0
A =0 B =0 C = 0
A =0 B =0 C =0
A =0 B = 0 C = 0
and each one must be treated separately.
1.4 Constrained Maxima and Minima: Method
of Lagrange
This techmique is used for finding the maximum and minimum of functions
z = f(x, y) subject to some additional constraint, like:
The idea is then:
1) Form the function
h(x, y) = f(x, y)
Origional
φ(x, y) =0
+ λ
Lagrange multiplierf(x,y)
2) Solve ∂h ∂h
∂x =0, ∂x = 0 for all possible x, y and λ.
constraint
φ(x, y)

14 The ABC’s of Calculus
3) Continue as in previous section.
Example 13 Find area of largest rectangle which can be inscribed in a semicircle
of radius a with base on its diameter.
Denote rectangle’s sides by: 2x, y
Figure 2
We want to maximize the area of the rectangle, i.e. A(x, y) =sxy subject to
constraint that ”rectangle is inscribed in semicircle”. This means that x 2 + y 2 =
a 2 (see diagram) or, φ(x, y) =x 2 + y 2 − a 2 =0.
UseLagrangemultipliers:So let h(x, y) =2xy + λ(x 2 + y 2 − a 2 )
Then:
Also
∂h
=0⇒ 2y + λ(2x) =0 ie. λx + y =0
∂x
∂h
=0⇒ 2x + λ(2y) =0 ie. λy + x =0
∂y
From the first of these two we get λ = −y
x (if x = 0).
(an assumption which is good, since if x =0⇒ no area)
Substitute this value of λ = −y
x
−y 2
into the second equation, to find:
x + x =0 or x2 = y 2 ⇒ x = ±y
Since we can assume x>0,y >0, thismeansx = y.
Now:
x 2 + y 2
or 2x 2 =7a 2
or x =
= a 2
⇒ x 2 + x 2 = a 2
a
√ 2
Thus x = a √ 2 ,y = a √ 2 is an extreme point and:
Area = 2xy
= 2· a √ ·
2 a √
2
= a 2

Functions and their properties 15
Example 14 Find the shortest and largest distance from the origin to the curve
x 2 + xy + y 2 =16and give a geometric iterpretation.
HINT: find the maximum of x 2 + y 2 .
CONSTRAINT: x 2 + xy + y 2 =16
[h(x, y) =x 2 + y 2 + λ(x 2 + xy + y 2 − 16)]
∂h
∂x
= 2x + λ(2x + y) =0 1
∂h
∂y
= 2x + λ(x +2y) =0 2
∂h
∂λ = x2 + xy + y 2 − 16 = 0 3
Now: if 2x + y =0,thenx =0too! but then y = −2x =0too! so x =0, y =0
is only possibility here for 1 .
NOTE: x =0, y =0satisfied 1 , 2 ,butnot 3 so 2x + y = 0can solve for λ
λ = −( 2x
2x+y ) from 1
so we feed this value of λ into 2
2y +(− 2x
)(x +2y)
2x + y
= 0
2y(2x + y) − 2x(x +2y)
2x + y
= 0
so y(2x + y) − x(x +2y) = 0
y 2 − x 2 = 0
⇒ y = ±x
finally set y = ±x into constraint equation x 2 + xy + y 2 =16.
Which leaves two cases: y = x and y = −x
case 1: y = x ⇒ x 2 + x 2 + x 2 =16or 3x 4 =16
x = ± 4 √ 3 But y = x ⇒ y = ± 4 √ 3 too!

oy-
16 The ABC’s of Calculus
case2: y = −x ⇒ x 2 − x 2 + x 2 =16or x 2 =16.
So we get four critical points:
x 2 + y 2
32
3
32
3
32
32
x 2 + y 2
32
3
32
3
√
√
32
32
x = ±4 But y = −x ⇒ y = ±4 too!
x = 4 √ 3
check critical points for max/min.
, y = 4 √ 3
x = − 4 √
3
, y = − 4 √
3
x =4 , y = −4
x = −4 , y =4
Note that maximum is at (4, −4) and (−4, 4) and the value at this point is √ 32
Note that minimum is at (± 4 √ 3 , ± 4 √ 3 ) and the value here is
1.5 Double and Iterated Integrals
32
3
Let domain f = R and write z = f(x, y) continuous over a finite region R of
the x, y plane. We divide R into n subregions area ∆A1, ∆A2, ......, ∆An in any
fashion what so ever.
Figure 4
Next we pick a point (x1,y1), (x2,y2), ......, (xn,yn) in each one of these subregions
and form the sum:
n
f(xi,yi)∆Ai = f(x1,y1)∆A1 + ...... + f(xn,yn)∆An
i=1
Think about f(xi,yi)∆Ai : This is the signed volume of a parallelepipeds of
base area ∆Ai and height f(xi,yi).