Multivariable Calculusmariano/math2110s16/slides14.pdf · Multivariable Calculus This is where we start to take a look at the \multivariable" side of multivariable and vector calculus

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Multivariable Calculus

MATH 2110Q – David Nichols

MATH 2110Q – David Nichols 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.




















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


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 .


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?




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:




The domain D = {(x, y) | x < y2}.


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





Now


Example • Domain and Range

Find the domain and range of g(x, y) =√

9− x2 − y2.




9− x2 − y2.

For the domain, we need 9− x2 − y2 ≥ 0, i.e. x2 + y2 ≤ 9. So

D = {(x, y) | x2 + y2 ≤ 9}




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]


Example • Graphing

Sketch the graph of g(x, y) =√9− x2 − y2.




z =√9− x2 − y2

z2 = 9− x2 − y2

x2 + y2 + z2 = 9




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.





Another Example

Sketch the graph of f(x, y) = 6− 3x− 2y.


Another Example


This is a linear function. We know from before that its graphmust be a plane.

How do we graph it?


Another Example



How do we graph it?

Same as in algebra: to graph a linear function, find the intercepts.


Another Example



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.


Another Example


z-intercept is 6. y-intercept is 3. x-intercept is 2.


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 .)


Level Curves • Idea

Level curves are like topographical maps for graphs of functions.


Level Curves • Illustration


Level Curves • Example

Sketch the level curves of g(x, y) =√9− x2 − y2 for

k = 0, 1, 2, 3.




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.




k = 0, 1, 2, 3.

x2 + y2 = 9− k2


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)



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.



f(x, y, z) = x2 + y2 + z2


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


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.


Limits • Now




Limits • Now




How Limits Fail

In Calculus I, we had left- and right-hand limits:

limx→a−

f(x) and limx→a+

f(x).


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


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.


How Limits Fail




How Limits Fail




Examples

Show that lim(x,y)→(0,0)x2 − y2

x2 + y2does not exist.


Examples



If f(x, y) = xy/(x2 + y2), does lim(x,y)→(0,0) f(x, y) exist?


Examples





Examples




If f(x, y) =xy2

x2 + y4, does lim(x,y)→(0,0) f(x, y) exist?


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.


What’s Continuous?

power functions (where defined)

polynomials

rational functions (where defined)

exponentials

logarithms (where defined)

trig functions (where defined)

etc.


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.


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.



The slope of f(x, y) = 4− x2 − 2y2 at the point (1, 1, 1) could bethis...



The slope of f(x, y) = 4− x2 − 2y2 at the point (1, 1, 1) could bethis......or this:



The slope of f(x, y) = 4− x2 − 2y2 at the point (1, 1, 1) could beeither of these:



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.”


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.


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.


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.


Another Example

Find∂f

∂xand

∂f

∂ywhen f(x, y) = y2 + sin(2x).


Another Example

Find∂f

∂xand

∂f


∂f

∂x=

∂

∂x(y2) +

∂

∂x(sin(2x)) = 0 + 2 cos(2x) = 2 cos(2x)


Another Example

Find∂f

∂xand

∂f


∂f

∂x=

∂

∂x(y2) +

∂

∂x(sin(2x)) = 0 + 2 cos(2x) = 2 cos(2x)

∂f

∂y=

∂

∂y(y2) +

∂

∂y(sin(2x)) = 2y + 0 = 2y


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)




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)




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)




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)




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)




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)


Higher Derivatives

We can take more than one derivative.


Higher Derivatives



Higher Derivatives



Higher Derivatives



Higher 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.




fxx =∂

∂x

(∂

∂x

(x3 + x2y3 − 2y2

))=

∂

∂x

(3x2 + 2xy3 − 0

)= 6x+ 2y3.




fxx =∂

∂x

(∂

∂x

(x3 + x2y3 − 2y2

))=

∂

∂x

(3x2 + 2xy3 − 0

)= 6x+ 2y3.




fxx =∂

∂x

(∂

∂x

(x3 + x2y3 − 2y2

))=

∂

∂x

(3x2 + 2xy3 − 0

)= 6x+ 2y3.




fxx =∂

∂x

(∂

∂x

(x3 + x2y3 − 2y2

))=

∂

∂x

(3x2 + 2xy3 − 0

)= 6x+ 2y3.


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.


Derivatives with 3+ Variables

Everything we’ve done generalizes to 3+ variables. Example:

f(x, y, z) = xy2z3

fx = y2z3, fy = 2xyz3, fz = 3xy2z2


Documents

Multivariable Calculusmariano/math2110s16/slides14.pdf · Multivariable Calculus This is where we start to take a look at the \multivariable" side of multivariable and vector calculus