Chapter 15

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Transcript Chapter 15

Chapter 15
Section 15.4
Partial Derivatives
Partial Derivatives
The partial derivative of the variable 𝑧 = 𝑓 π‘₯, 𝑦
with respect to either the variables x or y is given
by each of the limits to the right.
πœ•π‘§
𝑓 π‘₯ + β„Ž, 𝑦 βˆ’ 𝑓 π‘₯, 𝑦
= 𝑓π‘₯ π‘₯, 𝑦 = lim
β„Žβ†’0
πœ•π‘₯
β„Ž
πœ•π‘§
𝑓 π‘₯, 𝑦 + β„Ž βˆ’ 𝑓 π‘₯, 𝑦
= 𝑓𝑦 π‘₯, 𝑦 = lim
β„Žβ†’0
πœ•π‘¦
β„Ž
Computing Partial Derivatives
To make use of the differentiation rules learned
in earlier calculus courses we realize that these
derivatives only deal with 1 β€œvariable” at a time.
To calculate the derivative of the dependent
variable (z for 𝑧 = 𝑓 π‘₯, 𝑦 ) with respect to an
independent variable (x or y for 𝑧 = 𝑓 π‘₯, 𝑦 )
view the independent variable as the β€œonly
variable” and all other variables as constants.
For 𝑧 = 𝑓 π‘₯, 𝑦
Example
πœ•π‘§
πœ•π‘§
πœ•π‘§
To find πœ•π‘₯ take the derivative of z viewing x
as the variable and y and any expression
dependent on y as a constant.
πœ•π‘§
To find πœ•π‘¦ take the derivative of z viewing y
as the variable and x and any expression
dependent on x as a constant.
Find πœ•π‘₯ and 𝑑𝑦 for the function 𝑧 = 𝑓 π‘₯, 𝑦 = π‘₯ 5 cos 𝑦 + ln 𝑦 βˆ’ π‘₯𝑒 π‘₯𝑦
πœ•π‘§
πœ• 5
πœ•
πœ•
2
=
π‘₯ cos 𝑦 +
ln 𝑦 βˆ’
π‘₯𝑒 π‘₯𝑦
πœ•π‘₯ πœ•π‘₯
πœ•π‘₯
πœ•π‘₯
2
= 5π‘₯ 4 cos 𝑦 + 0 βˆ’ 𝑒 π‘₯𝑦 + π‘₯𝑦 2 𝑒 π‘₯𝑦
2
= 5π‘₯ 4 cos 𝑦 βˆ’ 𝑒 π‘₯𝑦 βˆ’ π‘₯𝑦 2 𝑒 π‘₯𝑦
2
2
2
πœ•π‘§
πœ• 5
πœ•
πœ•
2
=
π‘₯ cos 𝑦 +
ln 𝑦 βˆ’
π‘₯𝑒 π‘₯𝑦
πœ•π‘¦ πœ•π‘¦
πœ•π‘¦
πœ•π‘¦
1
2
= βˆ’π‘₯ 5 sin 𝑦 + βˆ’ π‘₯𝑒 π‘₯𝑦 2π‘₯𝑦
𝑦
1
2
= βˆ’π‘₯ 5 sin 𝑦 + βˆ’ 2π‘₯ 2 𝑦𝑒 π‘₯𝑦
𝑦
Partial Derivatives with more Variables
If there are more independent variables for
the function than x and y view all the
independent variables that you are not taking
the derivative with respect to as constants.
Example
Find
πœ•π‘€
πœ•π‘§
For 𝑀 = 𝑓 π‘₯, 𝑦, 𝑧
πœ•π‘€
Find πœ•π‘§ by taking the derivative of w viewing z
as the variable and x and y and all expressions
depending only on x or y as constants
for the function 𝑀 = 𝑓 π‘₯, 𝑦, 𝑧 = 𝑒 βˆ’2π‘₯ sec 𝑦 βˆ’ 𝑦 sinh 𝑧 + π‘₯ 2 arctan 𝑧
πœ•π‘€
πœ• βˆ’2π‘₯
πœ•
πœ• 2
π‘₯2
=
𝑒
sec 𝑦 βˆ’
𝑦 sinh 𝑧 +
π‘₯ arctan 𝑧 = βˆ’π‘¦ cosh 𝑧 +
πœ•π‘§ πœ•π‘§
πœ•π‘§
πœ•π‘§
1 + 𝑧2
Implicit Partial Derivatives
Some expression are very complicated or
even impossible to solve for the dependent
variable, but it still might be useful to know
the partial derivative. To do this we can take
the partial derivative implicitly.
Example
For the sphere π‘₯ 2 + 𝑦 2 + 𝑧 2 = 16 find
Solve for z : 𝑧 = ± 16 βˆ’ π‘₯ 2 βˆ’ 𝑦 2
πœ•π‘§
βˆ’2π‘₯
βˆ’π‘₯
=
=
πœ•π‘₯ ±2 16 βˆ’ π‘₯ 2 βˆ’ 𝑦 2
𝑧
πœ•π‘§
.
πœ•π‘₯
Steps for implicit derivatives
1. Take derivative of both sides of equation
keeping in mind independent variable,
multiplying by derivative where appropriate.
2. Solve for derivative
Implicitly, keeping in mind 𝑧 = 𝑓 π‘₯, 𝑦 .
πœ• 2 πœ• 2
πœ• 2
πœ•
π‘₯ +
𝑦 +
𝑧 =
16
πœ•π‘₯
πœ•π‘₯
πœ•π‘₯
πœ•π‘₯
πœ•π‘§
2π‘₯ + 0 + 2𝑧
=0
πœ•π‘₯
πœ•π‘§ βˆ’2π‘₯ βˆ’π‘₯
=
=
πœ•π‘₯
2𝑧
𝑧
Geometric View of Derivative
Remember for a function of 1 variable the derivative
at a point π‘₯0 represents the slope of the tangent line
or if you go 1 unit in the x direction you go up or
down on the tangent line by the amount of the
derivative 𝑑𝑦
.
𝑑π‘₯
y
1
π‘₯0
π‘₯=π‘₯0
𝑑𝑦
𝑑π‘₯ π‘₯=π‘₯
0
x
Geometric View of Partial Derivative
For a surface 𝑧 = 𝑓 π‘₯, 𝑦 if we fix a value for x or y say π‘₯0 and 𝑦0 this makes a curve where
the surface intersects π‘₯ = π‘₯0 and 𝑦 = 𝑦0 . These curves have tangent lines at the point
π‘₯0 , 𝑦0 . The partial derivative tells you if you go 1 unit forward in the x direction or 1 unit
right in the y direction how much you go up or down in the z direction.
z
z
1
πœ•π‘§
πœ•π‘¦
1
πœ•π‘§
πœ•π‘₯
π‘₯0
π‘₯=π‘₯0
𝑦=𝑦0
𝑦0
𝑦0
y
π‘₯0 , 𝑦0
x
Direction Vector of Tangent line:
πœ•π‘§
1,0,
πœ•π‘₯
π‘₯0
π‘₯=π‘₯0
𝑦=𝑦0
y
π‘₯0 , 𝑦0
x
Direction Vector of Tangent line:
πœ•π‘§
0,1,
πœ•π‘¦