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Multivariable Calculus
MATH 2110Q – David Nichols
MATH 2110Q – David Nichols Multivariable Calculus
Multivariable Calculus
This is where we start to take a look at the “multivariable” side ofmultivariable and vector calculus.
We’ll define functions where we plug in not just x, but x and y(and sometimes more). So we’ll have f(x, y) instead of f(x).
Then we’ll do calculus with these functions. The general idea isthat everything’s the same—except now we have to worry aboutmore than one variable.
Limits are still about what values f(x, y) approaches near a point.Derivatives are still about rate of change / slope. The onlydifference is that now we have both x and y to work with.
MATH 2110Q – David Nichols Multivariable Calculus
Multivariable Calculus
This is where we start to take a look at the “multivariable” side ofmultivariable and vector calculus.
We’ll define functions where we plug in not just x, but x and y(and sometimes more). So we’ll have f(x, y) instead of f(x).
Then we’ll do calculus with these functions. The general idea isthat everything’s the same—except now we have to worry aboutmore than one variable.
Limits are still about what values f(x, y) approaches near a point.Derivatives are still about rate of change / slope. The onlydifference is that now we have both x and y to work with.
MATH 2110Q – David Nichols Multivariable Calculus
Multivariable Calculus
This is where we start to take a look at the “multivariable” side ofmultivariable and vector calculus.
We’ll define functions where we plug in not just x, but x and y(and sometimes more). So we’ll have f(x, y) instead of f(x).
Then we’ll do calculus with these functions. The general idea isthat everything’s the same—except now we have to worry aboutmore than one variable.
Limits are still about what values f(x, y) approaches near a point.Derivatives are still about rate of change / slope. The onlydifference is that now we have both x and y to work with.
MATH 2110Q – David Nichols Multivariable Calculus
Multivariable Calculus
This is where we start to take a look at the “multivariable” side ofmultivariable and vector calculus.
We’ll define functions where we plug in not just x, but x and y(and sometimes more). So we’ll have f(x, y) instead of f(x).
Then we’ll do calculus with these functions. The general idea isthat everything’s the same—except now we have to worry aboutmore than one variable.
Limits are still about what values f(x, y) approaches near a point.Derivatives are still about rate of change / slope. The onlydifference is that now we have both x and y to work with.
MATH 2110Q – David Nichols Multivariable Calculus
Functions of Two Variables • Definition
A “function of two variables” is a function that takes a pair ofnumbers as input and gives assigns them one number as output.
Before
Functions take a number asinput and assign a number asoutput.
in x 7→ f(x) out
Example: f(3) = 11
Now
Functions take two numbersas input and assign a number asoutput.
in (x, y) 7→ f(x, y) out
Example: f(2,−15) = 0
MATH 2110Q – David Nichols Multivariable Calculus
Functions of Two Variables • Terminology
z = f(x, y)
“independent variables”“dependent variable”
The “domain” of f(x, y) is the set of all pairs (x, y) that we canplug in to f .
The “range” of f(x, y) is the set of all possible outputs we canget from f .
MATH 2110Q – David Nichols Multivariable Calculus
Functions of Two Variables • Example
Find and sketch the domain of f(x, y) = x ln(y2 − x).
We need to find all pairs (x, y) that we can plug in to f . What’sthe constraint?
MATH 2110Q – David Nichols Multivariable Calculus
Functions of Two Variables • Example
Find and sketch the domain of f(x, y) = x ln(y2 − x).
We need to find all pairs (x, y) that we can plug in to f . What’sthe constraint?
ln(y2 − x) is only defined if y2 − x > 0, which means y2 > x.That gives us the domain:
MATH 2110Q – David Nichols Multivariable Calculus
Functions of Two Variables • Example
Find and sketch the domain of f(x, y) = x ln(y2 − x).
The domain D = {(x, y) | x < y2}.
MATH 2110Q – David Nichols Multivariable Calculus
Functions of Two Variables • Graphing
The “graph” of z = f(x, y) is the set of all points (x, y, z) in R3
where z = f(x, y) and (x, y) is in the domain of f .
MATH 2110Q – David Nichols Multivariable Calculus
Functions of Two Variables • Graphing
The “graph” of z = f(x, y) is the set of all points (x, y, z) in R3
where z = f(x, y) and (x, y) is in the domain of f .
Before
x
y
D
MATH 2110Q – David Nichols Multivariable Calculus
Functions of Two Variables • Graphing
The “graph” of z = f(x, y) is the set of all points (x, y, z) in R3
where z = f(x, y) and (x, y) is in the domain of f .
Now
MATH 2110Q – David Nichols Multivariable Calculus
Example • Domain and Range
Find the domain and range of g(x, y) =√
9− x2 − y2.
MATH 2110Q – David Nichols Multivariable Calculus
Example • Domain and Range
Find the domain and range of g(x, y) =√
9− x2 − y2.
For the domain, we need 9− x2 − y2 ≥ 0, i.e. x2 + y2 ≤ 9. So
D = {(x, y) | x2 + y2 ≤ 9}
MATH 2110Q – David Nichols Multivariable Calculus
Example • Domain and Range
Find the domain and range of g(x, y) =√
9− x2 − y2.
For the domain, we need 9− x2 − y2 ≥ 0, i.e. x2 + y2 ≤ 9. So
D = {(x, y) | x2 + y2 ≤ 9}
For the range:√9− x2 − y2 is smallest when we subtract as much as
possible:√9− 9 = 0.√
9− x2 − y2 is biggest when we subtract as little aspossible:
√9− 0 = 3.
R = {z | 0 ≤ z ≤ 3} = [0, 3]
MATH 2110Q – David Nichols Multivariable Calculus
Example • Graphing
Sketch the graph of g(x, y) =√9− x2 − y2.
MATH 2110Q – David Nichols Multivariable Calculus
Example • Graphing
Sketch the graph of g(x, y) =√9− x2 − y2.
z =√9− x2 − y2
z2 = 9− x2 − y2
x2 + y2 + z2 = 9
MATH 2110Q – David Nichols Multivariable Calculus
Example • Graphing
Sketch the graph of g(x, y) =√9− x2 − y2.
z =√9− x2 − y2
z2 = 9− x2 − y2
x2 + y2 + z2 = 9
Equation of a sphere of radius 3. But z =√
9− x2 − y2 ≥ 0, sowe have only the top half of the sphere.
MATH 2110Q – David Nichols Multivariable Calculus
Example • Graphing
Sketch the graph of g(x, y) =√9− x2 − y2.
MATH 2110Q – David Nichols Multivariable Calculus
Another Example
Sketch the graph of f(x, y) = 6− 3x− 2y.
MATH 2110Q – David Nichols Multivariable Calculus
Another Example
Sketch the graph of f(x, y) = 6− 3x− 2y.
This is a linear function. We know from before that its graphmust be a plane.
How do we graph it?
MATH 2110Q – David Nichols Multivariable Calculus
Another Example
Sketch the graph of f(x, y) = 6− 3x− 2y.
This is a linear function. We know from before that its graphmust be a plane.
How do we graph it?
Same as in algebra: to graph a linear function, find the intercepts.
MATH 2110Q – David Nichols Multivariable Calculus
Another Example
Sketch the graph of f(x, y) = 6− 3x− 2y.
This is a linear function. We know from before that its graphmust be a plane.
How do we graph it?
Same as in algebra: to graph a linear function, find the intercepts.
z = 6− 3x− 2y
Plug in x = 0, y = 0. z-intercept is 6.
Plug in x = 0, z = 0. y-intercept is 3.
Plug in y = 0, z = 0. x-intercept is 2.
MATH 2110Q – David Nichols Multivariable Calculus
Another Example
Sketch the graph of f(x, y) = 6− 3x− 2y.
z-intercept is 6. y-intercept is 3. x-intercept is 2.
MATH 2110Q – David Nichols Multivariable Calculus
Level Curves • Definition
The “level curves” or “contour lines” of a function f(x, y) of twovariables are the curves with equations
f(x, y) = constant k
(Of course, k has to be in the range of f .)
MATH 2110Q – David Nichols Multivariable Calculus
Level Curves • Idea
Level curves are like topographical maps for graphs of functions.
MATH 2110Q – David Nichols Multivariable Calculus
Level Curves • Illustration
MATH 2110Q – David Nichols Multivariable Calculus
Level Curves • Example
Sketch the level curves of g(x, y) =√9− x2 − y2 for
k = 0, 1, 2, 3.
MATH 2110Q – David Nichols Multivariable Calculus
Level Curves • Example
Sketch the level curves of g(x, y) =√9− x2 − y2 for
k = 0, 1, 2, 3.
Level curves look like √9− x2 − y2 = k.
We can square both sides and rearrange to get
x2 + y2 = 9− k2,
which is the equation of a circle.
MATH 2110Q – David Nichols Multivariable Calculus
Level Curves • Example
Sketch the level curves of g(x, y) =√9− x2 − y2 for
k = 0, 1, 2, 3.
x2 + y2 = 9− k2
MATH 2110Q – David Nichols Multivariable Calculus
Functions of 3+ Variables
All of this generalizes to functions of more variables, like f(x, y, z).
When plugging in many variables, we sometimes use vectornotation:
x = 〈x1, x2, . . . , xn〉
f(x) = f(x1, x2, . . . , xn)
MATH 2110Q – David Nichols Multivariable Calculus
Functions of 3+ Variables
Equation Dimensions
y = f(x) 2
z = f(x, y) 3
w = f(x, y, z) 4
Setting f(x, y) = const. gave us a level curve.
Setting f(x, y, z) = const. gives us a level surface.
MATH 2110Q – David Nichols Multivariable Calculus
Functions of 3+ Variables
f(x, y, z) = x2 + y2 + z2
MATH 2110Q – David Nichols Multivariable Calculus
Limits • Before
limx→a f(x) = L means that for any desired accuracy ε, there is amargin of error δ so that
whenever x is within δ of a,f(x) is within ε of L.
MATH 2110Q – David Nichols Limits
Limits • Before
limx→a f(x) = L means that for any desired accuracy ε, there is amargin of error δ so that
whenever x is within δ of a,f(x) is within ε of L.
x
y
y = f(x)
−δ a + δ
L
+ε
−ε
MATH 2110Q – David Nichols Limits
Limits • Now
lim(x,y)→(a,b) f(x, y) = L means that for any desired accuracy ε,there is a margin of error δ so that
whenever (x, y) is within δ of (a, b),f(x, y) is within ε of L.
MATH 2110Q – David Nichols Limits
Limits • Now
lim(x,y)→(a,b) f(x, y) = L means that for any desired accuracy ε,there is a margin of error δ so that
whenever (x, y) is within δ of (a, b),f(x, y) is within ε of L.
MATH 2110Q – David Nichols Limits
Limits • Now
lim(x,y)→(a,b) f(x, y) = L means that for any desired accuracy ε,there is a margin of error δ so that
whenever (x, y) is within δ of (a, b),f(x, y) is within ε of L.
MATH 2110Q – David Nichols Limits
How Limits Fail
In Calculus I, we had left- and right-hand limits:
limx→a−
f(x) and limx→a+
f(x).
MATH 2110Q – David Nichols Limits
How Limits Fail
In Calculus I, we had left- and right-hand limits:
limx→a−
f(x) and limx→a+
f(x).
If these aren’t the same, the limit doesn’t exist.
x
y
a
MATH 2110Q – David Nichols Limits
How Limits Fail
In multivariable calculus, this problem is way bigger.
Now, instead of left- and right-hand limits, we have limits fromevery direction, and all of these have to be equal for there to be alimit.
MATH 2110Q – David Nichols Limits
How Limits Fail
In multivariable calculus, this problem is way bigger.
Now, instead of left- and right-hand limits, we have limits fromevery direction, and all of these have to be equal for there to be alimit.
MATH 2110Q – David Nichols Limits
How Limits Fail
In multivariable calculus, this problem is way bigger.
Now, instead of left- and right-hand limits, we have limits fromevery direction, and all of these have to be equal for there to be alimit.
MATH 2110Q – David Nichols Limits
Examples
Show that lim(x,y)→(0,0)x2 − y2
x2 + y2does not exist.
MATH 2110Q – David Nichols Limits
Examples
Show that lim(x,y)→(0,0)x2 − y2
x2 + y2does not exist.
If f(x, y) = xy/(x2 + y2), does lim(x,y)→(0,0) f(x, y) exist?
MATH 2110Q – David Nichols Limits
Examples
Show that lim(x,y)→(0,0)x2 − y2
x2 + y2does not exist.
If f(x, y) = xy/(x2 + y2), does lim(x,y)→(0,0) f(x, y) exist?
MATH 2110Q – David Nichols Limits
Examples
Show that lim(x,y)→(0,0)x2 − y2
x2 + y2does not exist.
If f(x, y) = xy/(x2 + y2), does lim(x,y)→(0,0) f(x, y) exist?
If f(x, y) =xy2
x2 + y4, does lim(x,y)→(0,0) f(x, y) exist?
MATH 2110Q – David Nichols Limits
Continuity
Just like in Calculus I:
A function f(x, y) is “continuous at (a, b)” if
lim(x,y)→(a,b)
f(x, y) = f(a, b).
A function f(x, y) is “continuous” if it is continuous at eachpoint in its domain.
MATH 2110Q – David Nichols Limits
What’s Continuous?
power functions (where defined)
polynomials
rational functions (where defined)
exponentials
logarithms (where defined)
trig functions (where defined)
etc.
MATH 2110Q – David Nichols Limits
Continuity • Example
What is lim(x,y)→(1,1)
y4 lnx
x+ y?
The function is continuous (except when y = −x), so
lim(x,y)→(1,1)
y4 lnx
x+ y=
14 ln 1
1 + 1=
0
2= 0.
MATH 2110Q – David Nichols Limits
Partial Derivatives
MATH 2110Q – David Nichols
MATH 2110Q – David Nichols Partial Derivatives
Functions of Several Variables • Derivatives
Where there are functions, there are derivatives. How do we takethe derivative of f(x, y)?
Graphically, derivative = slope. But in 3D, which slope?
Take a look at f(x, y) = 4− x2 − 2y2.
MATH 2110Q – David Nichols Partial Derivatives
Functions of Several Variables • Derivatives
The slope of f(x, y) = 4− x2 − 2y2 at the point (1, 1, 1) could bethis...
MATH 2110Q – David Nichols Partial Derivatives
Functions of Several Variables • Derivatives
The slope of f(x, y) = 4− x2 − 2y2 at the point (1, 1, 1) could bethis......or this:
MATH 2110Q – David Nichols Partial Derivatives
Functions of Several Variables • Derivatives
The slope of f(x, y) = 4− x2 − 2y2 at the point (1, 1, 1) could beeither of these:
MATH 2110Q – David Nichols Partial Derivatives
Functions of Several Variables • Derivatives
In theory, we could measure the slope of f(x, y) in any direction.
In practice, the slopes in the x-direction and y-direction areenough.
These derivatives are called the “partial derivatives” of f(x, y).
The derivative in the x direction is called the “partial derivativewith respect to x.”
The derivative in the y direction is called the “partial derivativewith respect to y.”
MATH 2110Q – David Nichols Partial Derivatives
How to Write Partial Derivatives
There are many notations for partial derivatives. If z = f(x, y),then
∂z
∂x=∂f
∂x=
∂
∂xf = fx = Dxf = f1 = D1f
∂z
∂y=∂f
∂y=
∂
∂yf = fy = Dyf = f2 = D2f
See page 927.
MATH 2110Q – David Nichols Partial Derivatives
How to Compute Partial Derivatives
fx(x, y) measures how f(x, y) changes as x varies.
So treat x as the only variable. Treat y as just a constant andcalculate the derivative with respect to x:
fx(x, y) = limh→0
f(x+ h, y)− f(x, y)h
Example: f(x, y) = x3y.
fx(x, y) =∂
∂x(x3y) = 3x2y.
MATH 2110Q – David Nichols Partial Derivatives
How to Compute Partial Derivatives
fy(x, y) measures how f(x, y) changes as y varies.
So treat y as the only variable. Treat x as just a constant andcalculate the derivative with respect to y:
fy(x, y) = limh→0
f(x, y + h)− f(x, y)h
Example: f(x, y) = x3y.
fy(x, y) =∂
∂y(x3y) = x3.
MATH 2110Q – David Nichols Partial Derivatives
Another Example
Find∂f
∂xand
∂f
∂ywhen f(x, y) = y2 + sin(2x).
MATH 2110Q – David Nichols Partial Derivatives
Another Example
Find∂f
∂xand
∂f
∂ywhen f(x, y) = y2 + sin(2x).
∂f
∂x=
∂
∂x(y2) +
∂
∂x(sin(2x)) = 0 + 2 cos(2x) = 2 cos(2x)
MATH 2110Q – David Nichols Partial Derivatives
Another Example
Find∂f
∂xand
∂f
∂ywhen f(x, y) = y2 + sin(2x).
∂f
∂x=
∂
∂x(y2) +
∂
∂x(sin(2x)) = 0 + 2 cos(2x) = 2 cos(2x)
∂f
∂y=
∂
∂y(y2) +
∂
∂y(sin(2x)) = 2y + 0 = 2y
MATH 2110Q – David Nichols Partial Derivatives
Implicit Differentiation
Implicit differentiation works the same as it did in Calculus I. Takea look:
eyz + x2 + y2 + z2 = 0
∂
∂x
(eyz + x2 + y2 + z2
)=
∂
∂x(0)
eyzy∂z
∂x+ 2x+ 0 + 2z
∂z
∂x= 0
(eyz + 2z)∂z
∂x+ 2x = 0
(eyz + 2z)∂z
∂x= −2x
∂z
∂x=
−2x(eyz + 2z)
MATH 2110Q – David Nichols Partial Derivatives
Implicit Differentiation
Implicit differentiation works the same as it did in Calculus I. Takea look:
eyz + x2 + y2 + z2 = 0
∂
∂x
(eyz + x2 + y2 + z2
)=
∂
∂x(0)
eyzy∂z
∂x+ 2x+ 0 + 2z
∂z
∂x= 0
(eyz + 2z)∂z
∂x+ 2x = 0
(eyz + 2z)∂z
∂x= −2x
∂z
∂x=
−2x(eyz + 2z)
MATH 2110Q – David Nichols Partial Derivatives
Implicit Differentiation
Implicit differentiation works the same as it did in Calculus I. Takea look:
eyz + x2 + y2 + z2 = 0
∂
∂x
(eyz + x2 + y2 + z2
)=
∂
∂x(0)
eyzy∂z
∂x+ 2x+ 0 + 2z
∂z
∂x= 0
(eyz + 2z)∂z
∂x+ 2x = 0
(eyz + 2z)∂z
∂x= −2x
∂z
∂x=
−2x(eyz + 2z)
MATH 2110Q – David Nichols Partial Derivatives
Implicit Differentiation
Implicit differentiation works the same as it did in Calculus I. Takea look:
eyz + x2 + y2 + z2 = 0
∂
∂x
(eyz + x2 + y2 + z2
)=
∂
∂x(0)
eyzy∂z
∂x+ 2x+ 0 + 2z
∂z
∂x= 0
(eyz + 2z)∂z
∂x+ 2x = 0
(eyz + 2z)∂z
∂x= −2x
∂z
∂x=
−2x(eyz + 2z)
MATH 2110Q – David Nichols Partial Derivatives
Implicit Differentiation
Implicit differentiation works the same as it did in Calculus I. Takea look:
eyz + x2 + y2 + z2 = 0
∂
∂x
(eyz + x2 + y2 + z2
)=
∂
∂x(0)
eyzy∂z
∂x+ 2x+ 0 + 2z
∂z
∂x= 0
(eyz + 2z)∂z
∂x+ 2x = 0
(eyz + 2z)∂z
∂x= −2x
∂z
∂x=
−2x(eyz + 2z)
MATH 2110Q – David Nichols Partial Derivatives
Implicit Differentiation
Implicit differentiation works the same as it did in Calculus I. Takea look:
eyz + x2 + y2 + z2 = 0
∂
∂x
(eyz + x2 + y2 + z2
)=
∂
∂x(0)
eyzy∂z
∂x+ 2x+ 0 + 2z
∂z
∂x= 0
(eyz + 2z)∂z
∂x+ 2x = 0
(eyz + 2z)∂z
∂x= −2x
∂z
∂x=
−2x(eyz + 2z)
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives
We can take more than one derivative.
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives
We can take more than one derivative.
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives
We can take more than one derivative.
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives
We can take more than one derivative.
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives
We can take more than one derivative.
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives • Example
Find the second partial derivatives of f(x, y) = x3 + x2y3 − 2y2.
fxx =∂
∂x
(∂
∂x
(x3 + x2y3 − 2y2
))=
∂
∂x
(3x2 + 2xy3 − 0
)= 6x+ 2y3.
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives • Example
Find the second partial derivatives of f(x, y) = x3 + x2y3 − 2y2.
fxx =∂
∂x
(∂
∂x
(x3 + x2y3 − 2y2
))=
∂
∂x
(3x2 + 2xy3 − 0
)= 6x+ 2y3.
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives • Example
Find the second partial derivatives of f(x, y) = x3 + x2y3 − 2y2.
fxx =∂
∂x
(∂
∂x
(x3 + x2y3 − 2y2
))=
∂
∂x
(3x2 + 2xy3 − 0
)= 6x+ 2y3.
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives • Example
Find the second partial derivatives of f(x, y) = x3 + x2y3 − 2y2.
fxx =∂
∂x
(∂
∂x
(x3 + x2y3 − 2y2
))=
∂
∂x
(3x2 + 2xy3 − 0
)= 6x+ 2y3.
MATH 2110Q – David Nichols Partial Derivatives
Higher Derivatives • Example
Find the second partial derivatives of f(x, y) = x3 + x2y3 − 2y2.
fxx =∂
∂x
(∂
∂x
(x3 + x2y3 − 2y2
))=
∂
∂x
(3x2 + 2xy3 − 0
)= 6x+ 2y3.
MATH 2110Q – David Nichols Partial Derivatives
Clairaut’s Theorem
When is fxy = fyx ?
The answer is given by Clairaut’s Theorem:
Suppose f is defined in a disk D around (a, b), and that fxy andfyx are continuous in the disk D. Then fxy(a, b) = fyx(a, b).
The moral of the story is that
mixed partials are equal where they are continuous.
MATH 2110Q – David Nichols Partial Derivatives
Derivatives with 3+ Variables
Everything we’ve done generalizes to 3+ variables. Example:
f(x, y, z) = xy2z3
fx = y2z3, fy = 2xyz3, fz = 3xy2z2
MATH 2110Q – David Nichols Partial Derivatives
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