AMS261 Recitation Course 7
Recitation Instructor: Yiyang Yang
Department of Applied Mathematics and Statistics
State University of New York at Stony Brook
Spring 2013
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
1 / 14
Outline
1
Review
Differentials
Chain Rules for Functions of Several Variables
Directional Derivatives and Gradients
Tangent Planes and Normal Lines
2
Recitation Problems
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
2 / 14
Differentials
Definition of Total Differential
If z = f (x, y ) and 4x and 4y are increments of x and y , then the differentials
of the independent variables x and y are
dx = 4x, dy = 4y , 4z = f (x + 4x, y + 4y ) − f (x, y )
and the total differential of the dependent variable z is
dz =
∂z
∂z
dx +
dy = fx (x, y ) dx + fy (x, y ) dy .
∂x
∂y
Definition of Differentiability
A function f given by z = f (x, y ) is differentiable at (x0 , y0 ) if 4z can be written
in the form
4z = fx (x0 , y0 ) 4x + fy (x0 , y0 ) 4y + 1 4x + 2 4y
where both 1 and 2 → 0 as (4x, 4y ) → (0, 0). The function f is differentiable
in a region R if it is differentiable at each point in R.
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
3 / 14
Chain Rules for Functions of Several Variables
Chain Rule: One Independent Variable
Let w = f (x, y ), where f is a differentiable function of x and y . If x and y are
differentiable functions of t, then w is a differentiable function of t and
dw
∂w dx
∂w dy
=
+
.
dt
∂x dt
∂y dt
Chain Rule: Two Independent Variables
Let w = f (x, y ), where f is a differentiable function of x and y . If x and y are
∂x ∂y ∂y
differentiable functions of s, t and ∂x
∂s , ∂t , ∂s , ∂t all exist, then
∂w
∂w ∂x
∂w ∂y ∂w
∂w ∂x
∂w ∂y
=
+
,
=
+
.
∂s
∂x ∂s
∂y ∂s ∂t
∂x ∂t
∂y ∂t
Chain Rule: Implicit Differentiation
If F (x, y ) = 0 defines y implicitly as a function of x, then
dy
dx
)
= − FFyx (x,y
(x,y ) .
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
4 / 14
Directional Derivatives and Gradients
Definition of Directional Derivative
→
−
→
−
−
Let f be a function of x and y and let →
u = cos θ i + sin θ j be a unit vector.
→
−
−
Then the directional derivative of f in the direction of u , denoted by D→
u f is
−
D→
u f (x, y ) = lim
t→0
f (x + t cos θ, y + t sin θ) − f (x, y )
.
t
Directional Derivative
If f is a differentiable function of x and y , then the directional derivative of f in
−
→
−
→
→
the direction of the unit vector −
u = cos θ i + sin θ j is
−
D→
u f (x, y ) = fx (x, y ) cos θ + fy (x, y ) sin θ.
Definition of Gradient of a Function of Two variables
Let z = f (x, y ) be a function of x and y such that fx and fy exist. Then the
−
→
−
→
gradient of f denoted by ∇f (x, y ) is the vector ∇f (x, y ) = fx (x, y ) i + fy (x, y ) j .
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
5 / 14
Tangent Planes and Normal Lines
Definitions of Tangent Plane and Normal Line
Let F be differentiable at the point P (x0 , y0 , z0 ) on the surface S given by
→
−
F (x, y , z) = 0 such that ∇F (x0 , y0 , z0 ) 6= 0 .
1
The Plane through P that is normal to ∇F (x0 , y0 , z0 ) is called the tangent
plane to S at P.
2
The line through P having the direction of ∇F (x0 , y0 , z0 ) is called the
normal line to S at P.
Let S be the surface F (x, y , z) = 0 and P (x0 , y0 , z0 ) be a point on S. Let C be a curve on
−
→
−
→
−
→
→
S through P that is defined by vector-valued function −
r (t) = x (t) i + y (t) j + z (t) k .
Then for all t, F (x (t) , y (t) , z (t)) = 0
0 = F 0 (x (t) , y (t) , z (t)) = Fx x 0 (t) + Fy y 0 (t) + Fz z 0 (t)
Gradient is Normal to Level Surface
→
−
If F is differential at (x0 , y0 , z0 ) and ∇F (x0 , y0 , z0 ) 6= 0 , then ∇F (x0 , y0 , z0 ) is
normal to the level surface through (x0 , y0 , z0 ).
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
6 / 14
Textbook Problems
13.5.53 on p.932
The radius of a right circular cylinder is increasing at a rate of 6 inches per
minute, and the height is decreasing at a rate of 4 inches per minute. What are
the rates of change of the volume and surface area when the radius is 12 inches
and the height is 36 inches.
Solution:
The volume of a right circular cylinder is defined by
V = πr 2 h
The rates of change at (r , h) = (12, 36) is
dV
dr
2 dh
|
= 2πrh + πr
|
dt (12,36)
dt
dt (12,36)
= 2π · 12 · 36 · 6 + π · 122 · (−4) = 4608π
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
7 / 14
Textbook Problems (Cont.)
The surface area of a right circular cylinder is defined by
S = 2πr 2 + 2πrh
The rates of change at (r , h) = (12, 36) is
dS
dr
dh
|
= (4πr + 2πh)
+ 2πr
|
dt (12,36)
dt
dt (12,36)
= (4π · 12 + 2π · 36) · 6 + 2π · 12 · (−4) = 624π
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
8 / 14
Textbook Problems (Cont.)
13.7.57 on p.952
Show that the surfaces x 2 + 2y 2 + 3z 2 = 3, x 2 + y 2 + z 2 + 6x − 10y + 14 = 0 are
tangent to each other at the given point by showing that the surfaces have the
same tangent plane at point (−1, 1, 0).
Solution:
Assume F (x, y , z) = x 2 + 2y 2 + 3z 2 − 3, then
Fx (x, y , z) = 2x, Fy (x, y , z) = 4y , Fz (x, y , z) = 6z
at (−1, 1, 0)
Fx (−1, 1, 0) = −2, Fy (−1, 1, 0) = 4, Fz (−1, 1, 0) = 0
thus the tangent plane is
−2 (x + 1) + 4 (y − 1) + 0 (z − 0) = 0
⇒ x − 2y + 3 = 0
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
9 / 14
Textbook Problems (Cont.)
Assume G (x, y , z) = x 2 + y 2 + z 2 + 6x − 10y + 14, then
Gx (x, y , z) = 2x + 6, Gy (x, y , z) = 2y − 10, Gz (x, y , z) = 2z
at (−1, 1, 0)
Fx (−1, 1, 0) = 4, Fy (−1, 1, 0) = −8, Fz (−1, 1, 0) = 0
thus the tangent plane is
4 (x + 1) − 8 (y − 1) + 0 (z − 0) = 0
⇒ x − 2y + 3 = 0
In sum, two tangent planes are the same.
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
10 / 14
Textbook Problems (Cont.)
13.8.43 on p.961
2
2
Find the critical points of the function f (x, y , z) = x 2 + (y − 3) + (z + 1) and
from the form of the function, determine whether a relative maximum or a
relative minimum occurs at each point.
Solution:
2
2
Given f (x, y , z) = x 2 + (y − 3) + (z + 1) , we have
fx (x, y , z) = 2x, fy (x, y , z) = 2 (y − 3) , fz (x, y , z) = 2 (z + 1)
thus the system is
2x = 0, 2 (y − 3) = 0, 2 (z + 1) = 0
⇒ x = 0, y = 3, z = −1
and the critical point (0, −3, 1) determine the relative minimum.
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
11 / 14
Textbook Problems (Cont.)
13.9.17 on p.967
Common blood types are determined genetically by three alleles A, B, and O. (An
allele is any of a group of possible mutational forms of a gene.) A person whose
blood type is AA, BB, or OO is homozygous. The Hardy-Weinberg Law states
that the proportion P of heterozygous individuals in any given population is
P (p, q, r ) = 2pq + 2pr + 2qr
where p represents the percent of the allele A in the population, q represents the
percent of allele B in the population, and r represents the percent of allele O in
the population. Use the fact that p + q + r = 1 to show that the maximum
proportion of heterozygous individuals in any population is 23 .
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
12 / 14
Textbook Problems (Cont.)
Solution:
Given P (p, q, r ) = 2pq + 2pr + 2qr and constraint p + q + r = 1
P (p, q) = 2pq + 2p (1 − p − q) + 2q (1 − p − q)
= −2pq − 2p 2 − 2q 2 + 2p + 2q
to obtain critical points
∂P
∂P
= −2q − 4p + 2 = 0,
= −2p − 4q + 2 = 0
∂p
∂q
thus
1
3
2
2
2
1 1
1
1
1
1
1
2
P
,
= −2
−2
−2
+2
+2
=
3 3
3
3
3
3
3
3
p=q=
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
13 / 14
Textbook Problems (Cont.)
13.10.47 on p.977
Use Lagrange multipliers to find the dimensions of a right circular cylinder with
volume V0 cubic units and minimum surface area.
Solution:
To minimize surface area 2πrh + 2πr 2 , with constraint πr 2 h = V0 , then
L (r , h, λ) = 2πrh + 2πr 2 + λ πr 2 h − V0
the critical points can be obtained from
∂L
∂L
∂L
= 2πh + 4πr + 2λπhr = 0,
= 2πr + λπr 2 = 0,
= πr 2 h − V0 = 0
∂r
∂h
∂λ
r
r
3 V0
3 V0
⇒r =
, h=2
2π
2π
q
q
V0
V0
the dimensions are r = 3 2π
and h = 2 3 2π
.
Yiyang Yang (Department of Applied Mathematics and
AMS261
Statistics
Recitation
State University
Course 7 of New York at Stony Brook)
Spring 2013
14 / 14