Download Partial Derivatives

Multivariable Calculus
Summary 1 - Partial Derivatives
Limits: when dealing with a function of two variables, we see that (x,y) can approach
(a,b) along many different paths. In order for a limit to exist, we must get the same value
for the limit no matter what path is used in a approaching (a,b). A convenient method to
show that a given limit does not exist is to show that we get different results when we
approach (a, b) along different paths.
Note: limits of continuous functions can be found by substitution provided the point in
question is a limit point of the domain of the function.
Example: Show that lim
x0
y0
x2 y
Proof: lim
x 0
y 0
x4  y2
x2y
x 4  y2
 lim
x 0
y 0
 0 along any line y=mx for each m.
x 2 mx
x 4  (mx) 2
mx3
 lim
x 4  (m 2 x 2 )
x 0
y 0
 lim
x 0
y 0
mx
x 2  m2
0
2
What is the limit along y  x ? Does the above limit exist.?
lim
x 0
y 0
x2 y
x4  y2
 lim
x 0
y 0
x2 x2
x 4  ( x) 2
 lim
x 0
y 0
x4
x4  x4
 lim
x4
x 0 2 x 4
y 0

1
0
2
Therefore, the limit does not exist.
u ( x, t )  e  n kt sin nx where n and k are constants,
u
 2u
satisfies the one-dimensional heat equation
k 2
t
x
2
Problem: Show that the function
Proof:


u
 e  n kt sin nx   n 2 k
t
u
 2u
 n kt
e
cos nx  n  and
 e  n kt sin nx   n 2
2
x
x
 2u 1  n kt
1 u
 2  e
sin nx   n 2 k 
k
k t
x
u
 2u

k 2
t
x
2
2
2
2


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

 m 2  n 2 kt
Problem: Show that the function u ( x, y, t )  e
the two-dimensional heat equation for an insulated plane
sin mx cos ny satisfies
  2u  2u 
u
 k  2  2 
t
y 
 x
Problem: show that the equation y=sin(x + at) satisfies the one-dimensional wave
equation
2
2 y
2  y
a
t 2
x 2
Problem: Show that the ideal gas law pV=nRT (n is the number of moles of the gas, R is
a constant, p is pressure, V is volume, T is temperature) satisfies the equation
p V T


 1
V T p
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Problem: Show that the equation
equation
u( x, y)  e x sin y
satisfies the Laplace’s
 2u  2u
 2 0
2
x
y
Problem: the electric potential field of a point charge q (coulombs) is defined by
 ( x, y, z ) 
q
x y z
2
2
2
. Show that  satisfy the three-dimensional Laplace
equation
 2  2  2


0
x 2 y 2 z 2
i
Problem: Does the equation

u ( x, y)  ln x 2  y 2
 2u  2u
 2 0
2
x
y
-3-

satisfy the Laplace’s equation
Tangent plane to a surface and normal line:
Method I : if z=f(x,y) and the partial derivatives of z are continuous at ( x 0 , y 0 ), then
equation of the plane tangent and the line normal are given respectively by:
x  x 0 y  y 0 z  z0


a
b
1
f
f
where a  ( x0 , y0 ) and b  ( x0 , y0 )
x
y
(z- z 0 ) = a(x- x 0 ) +b(y- y 0 ) and
Proof:
f
f x
If z  f ( x, y ), then at ( x0 , y0 ), a 

gives the instantaneous rate of
x 1
change of z with respect to x in a fixed y-direction.
f
for each unit that x changes.
x
f
Therefore if y 2  y1  0 then x2  x1  1 and z 2  z1  a 
x

 u  1, 0, a is tangent to the surface at the point ( x0 , y0 ) .
Therefore at the given point z is changing a 
Similarly,
f
f y
gives the instantaneous rate of
If z  f ( x, y ), then at ( x0 , y0 ), b 

y 1
change of z with respect to y in a fixed x-direction.
f
for each unit that x changes.
y
f
Therefore if x2  x1  0 then y2  y1  1 and z 2  z1  b 
y

 v  0, 1, b is tangent to the surface at the point ( x0 , y0 ) .
 
This will mean that u  v is normal to the plane tangent to the surface z=f(x, y)
Therefore at the given point z is changing b 
i j k
 
u  v  1 0 a  ai  bj  k  a, b,  1 is normal to the plane
0 1 b
Therefore, the equation of the plane is given by
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a( x  x0 )  b( y  y0   1( z  z0 )  0
or ( z  z0 )  a( x  x0 )  b( y  y0  where a 
f
f
and b 
at ( x0 , y0 )
x
y
Method II : if F(x,y, z)=0) and the partial derivatives of F are continuous at
( x 0 , y 0 , z0 ), then equation of the plane tangent and the line normal are given
respectively by:
a(x- x 0 ) +b(y- y 0 ) +c (z- z 0 )=0 and
where a 
x  x 0 y  y 0 z  z0


a
b
c
F
F
F
( x0 , y0 , z0 ) and b 
( x0 , y0 , z0 ) and c 
( x0 , y0 , z0 )
x
y
z
Tangent line and normal plane to curve
If x=f(t), y=g(t) and z=h(t) are parametric equations of a curve in space, then the
equation of the normal plane and the tangent line at the point ( x 0 , y 0 , z0 ) are given
by
a(x- x 0 ) +b(y- y 0 ) +c (z- z 0 )=0 and
where a 
x  x 0 y  y 0 z  z0


a
b
c
x
 y
z
(t 0 ) and b 
(t 0 ) and c 
(t 0 )
t
t
t
Problems
1. Find the equation of the plane tangent and the line normal to the surface
z  x 2 y  3xy 2  x 3at (1,2,13)
Solution:
 z
2
2
 x  2 xy  3 y  3 x  13 at (1,  2,  13)  a  13
 z
 x 2  6 xy  13 at (1,  2,  13)  b  13

 y
The equation of the plane is given by -13(x - 1)+13(y + 2) -1(z+13)=0
or 13x -13y + x -26 =0
answer  Plane : 13x  13 y  z  26  0
Line :
x  1 y  2 z  13


13
 13
1
2. Find the equation of the plane tangent and the line normal to the surface
F ( x, y, z )  x 2 z  4 xy 2  z 3  14 at (1,2,3)
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Solution:
 F
2
 x  2 xz  4 y  10 at (1,  2, 3)  a  10
 F

 y  8 xy  16 at (1,  2, 3)  b  16

 F
2
2
 z  x  3 z  28 at (1,  2, 3)  c  28
A vector normal to the plane is given by  10, 16, 28 or 5,  8,  14
The equation of the plane is given by 5(x - 1)-8(y + 2) -14(z-3)=0
or 5x -8y -14z +21 =0
answer  Plane : 5 x  8 y  14 z  21  0
Line :
x 1 y  2 z  3


5
8
 14
3. Find the equation of the line tangent and the plane normal to the curve
x  t 2  3, y  5t  3, andz  t 2  2t  5 at t=0
Solution:
 dx
 t  2t  0 at t  0  a  0 and x  3
 dy

 5 at t  0  b  5 and y  3

dt

 dz  2t  2  2 at t  0  c  2 and z  5

t
A vector tan gent to the curve is given by 0, 5,  2
The equation of the plane normal is given by 0(x - 3)+5(y -3) -2(z-5)=0
or 5y -2z- 5 =0
answer  Plane : 5 y  2 z  5  0
Line :
y 3 z 5

, x  3
5
2
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