– WELCOME TO A VIDEO THAT

WILL INTRODUCE FUNCTIONS OF TWO VARIABLES. THE GOALS OF THE VIDEO ARE TO

DETERMINE FUNCTION VALUES OF FUNCTIONS OF TWO VARIABLES AND ALSO TO DETERMINE THE DOMAIN

OF A FUNCTION OF TWO VARIABLES. THE FUNCTION Z=F OF XY

IS A FUNCTION OF 2 VARIABLES. SO Z IS DEFINED IN TERMS OF

BOTH X AND Y THEREFORE THE DOMAIN IS THE SET

OF ALL POSSIBLE VALUES OF X AND Y. AND AGAIN X AND Y WILL BE

THE INPUTS INTO THE FUNCTION AND Z WILL BE THE OUTPUT. SO THE RANGE IS A SET

OF ALL POSSIBLE VALUES OF Z FROM THE DOMAIN. AND SINCE THE FUNCTION CONTAINS

THREE VARIABLES X,Y,Z THE GRAPH IS THE SET

OF ALL POSSIBLE POINTS CALLED ORDER TRIPLES X,Y,Z THAT

ARE PRODUCED BY THE FUNCTION. AND GEOMETRICALLY THE GRAPH

IS A SURFACE IN SPACE. SO LET’S FIRST START BY TAKING

A LOOK AT SOME COOL GRAPHS OF FUNCTIONS

IN TERMS OF TWO VARIABLES. HERE WE SEE Z=COSIGN X TIMES

SIGN Y. NOTICE IT PRODUCES A SURFACE

WITH HILLS AND VALLEYS. Z=THE NATURAL LOG OF 1 – X

– Y, PRODUCES THIS SURFACE. HERE WE HAVE Z=THE SQUARE ROOT

OF 4 – X SQUARED – Y SQUARED. AND THEN LASTLY WE HAVE Z

=E THE POWER OF 1 – X SQUARED MINUS Y SQUARED WHICH LOOKS LIKE A PEAK. LET’S TAKE A LOOK AT THESE

IN ANOTHER WAY. AGAIN, HERE’S THE GRAPH OF Z

=COSIGN X TIMES SIGN Y. HERE’S THE GRAPH OF Z

=NATURAL LOG OF 1 – X – Y. NOW THESE GRAPHS ARE DONE

ON MAPLE AND NOTICE AS THIS GRAPH MOVES

DOWNWARD MAPLE HAS A DIFFICULT TIME

PLOTTING THE SURFACE AND THAT CAN SOMETIMES HAPPEN

ON SOME GRAPHING SOFTWARE. HERE’S Z=THE SQUARE ROOT OF 4

– X SQUARED – Y SQUARE, AGAIN HALF OF A SPHERE AND THEN LASTLY WE HAVE Z=EITHER THE POWER OF 1

– X SQUARED – Y SQUARED. LET’S GO AND DETERMINE

SOME FUNCTION VALUES. NOTICE IN ORDER TO EVALUATE

THIS FUNCTION WE NEED BOTH AN X AND A Y VALUE

WHICH ARE BOTH GIVEN. SO WE’RE GONNA REPLACE X WITH

NEGATIVE 3 AND Y WITH 3PI/2. AND NOW WE’LL JUST EVALUATE

THIS. WE’RE GONNA HAVE NEGATIVE THREE

TIMES THE COSIGN OF 3PI/2 IS=TO 0,

SO THIS FUNCTION VALUE IS 0. WHEN X IS -3 AND Y IS 3PI/2

THE Z COORDINATE WOULD BE 0. FOR PART B WE HAVE X=2

AND Y=PI/3. SO WE’LL HAVE 2 X COSIGN PI/3. COSIGN PI/3 IS THE SAME

AS COSIGN 60 DEGREES, SO THAT’S GONNA GIVE US 1/2, WELL 2 TIMES 1/2 WOULD BE=

TO 1. SO WHEN X IS 2 AND WHY IS PI/3

THE Z COORDINATE WOULD BE 1. IF WE WANTED TO WE COULD WRITE

THIS AS AN ORDERED TRIPLE THAT WOULD LOOK LIKE THIS,

2 PI/3, 1 AND FOR 2

IT’S A SIMILAR PROCESS. WE’LL BASICALLY PERFORM

SUBSTITUTION WHEN X IS 5 AND Y IS 3. SO WE HAVE 5 SQUARED + 3

DIVIDED BY 3. SO WE HAVE WHAT,

25 + 28 DIVIDED BY 3. THAT DOESN’T SIMPLIFY AND AGAIN IF WE WANTED TO THE

ORDER TRIPLE WOULD BE 5 3 28/3. AND LET’S GO AHEAD AND TAKE

A LOOK AT 1 MORE, F OF -44, SO WE GOTTA BE CAREFUL

WITH OUR SIGNS HERE. WE’RE GONNA SQUARE -4

AND THEN ADD 4 DIVIDED BY +4. SO WE’RE GONNA HAVE POSITIVE 16

+ 4=20 DIVIDED BY 4 IT’S GONNA GIVE US 5. SO THE ORDER TRIPLE WOULD BE -4 4 5 WHICH REPRESENTS 1 POINT

ON THIS SURFACE. NOW, LET’S TALK ABOUT

DETERMINING A DOMAIN OF A FUNCTION

IN TERMS OF 2 VARIABLES. WELL, IF THIS FUNCTION

HAS 2 INPUTS, BOTH X AND Y, ALL THE POSSIBLE VALUES OF X

AND Y WILL MAKE UP THE DOMAIN. AND THE MORE THAT

WE CAN REMEMBER ABOUT OUR BASIC FUNCTIONS LIKE

Y=SIGN X AS WE SEE HERE AND Y=COSIGN X AS WE SEE HERE WILL HELP US DETERMINE

THE DOMAIN OF THIS FUNCTION AND 2 VARIABLES. WHAT I MEAN BY THAT IS FOR BOTH

OF THESE FUNCTIONS DOWN BELOW X CAN BE ANY REAL NUMBER. SO LOOKING AT THIS FUNCTION

HERE, X CAN BE ANY REAL NUMBER AND SO CAN Y. SO THAT TELLS US THE DOMAIN

IS GOING TO BE ALL REAL NUMBERS FOR BOTH X AND Y. LET’S GO AND TAKE A LOOK

AT ANOTHER ONE. HERE WE HAVE Z=

THE NATURAL LOG OF 1 – X – Y. NOW, LOOKING AT THIS FUNCTION

HERE AND TWO VARIABLES SHOULD REMIND US OF THE FUNCTION

AND 1 VARIABLE OF Y=NATURAL LOG X. IF YOU REMEMBER

HERE’S THE GRAPH. THE DOMAIN OF THIS FUNCTION

WOULD BE X GREATER THAN 0 AND THE REASON THAT’S HELPFUL IS IF YOU TAKE A LOOK

AT THE GIVEN FUNCTION AND TWO VARIABLES, THE ARGUMENT 1 – X – Y

MUST BE GREATER THAN 0 FOR THE SAME REASON WE HAVE A

DOMAIN OF X IS GREATER THAN 0 HERE. SO AGAIN THE DOMAIN WOULD BE 1

1 – X – Y IS GREATER THAN 0. NOW, THIS IS THE DOMAIN BUT IT’S

OFTEN BETTER TO MANIPULATE THIS TO MAKE IT LOOK LIKE

A MORE FAMILIAR FORM. WHAT I MEAN BY THAT IS

WE CAN MOVE THE X AND THE Y TERM TO THE RIGHT SIDE. SO WE WOULD HAVE 1 GREATER THAN

X + Y. FROM HERE’S IT’S PROBABLY BETTER

TO GO AHEAD AND SWITCH THIS AROUND TO STATE THAT X + Y

WOULD BE LESS THAN ONE. AND OFTEN WHEN THE DOMAIN

IS EXPRESSED IN TERMS OF TWO VARIABLES IT’S HELPFUL IF WE GRAPH THIS

ON THE X,Y PLANE TO SHOW THE REGION

OF THE DOMAIN. LET’S GO AHEAD AND DO THAT. SO X + Y LESS THAN 1

IS GONNA BE A DASHED LINE WITH AN X AND Y INTERCEPT OF +1. THEN IF WE TEST THE ORIGIN 0,0

IN THIS INEQUALITY THE 0 LESS THAN 1 IS TRUE THEREFORE WE SHADE THIS HALF

OF THE PLANE. SO ANY POINT IN THIS REGION WOULD BE PART OF THE DOMAIN

OF THE GIVEN FUNCTION IN TERMS OF 2 VARIABLES. LET’S GO AND TAKE A LOOK

AT ONE MORE. HERE WE HAVE Z=THE SQUARE ROOT

OF 4 – X SQUARED – Y SQUARED. LOOKING AT THIS WE SHOULD BE

REMINDED OF THE FUNCTION Y=THE SQUARE ROOT OF X WHEN WE HAVE A FUNCTION

IN TERMS OF 1 VARIABLE. AND AGAIN WE CAN SEE

FROM THIS GRAPH THE DOMAIN OF THIS FUNCTION

WOULD BE WHEN X IS GREATER THAN

OR=TO 0. REMEMBER, THE SQUARE ROOT OF 0

IS DEFINED. SO WE’RE RELATED THIS DOMAIN

TO THE GIVEN FUNCTION. THIS SHOULD HELP US DETERMINE

THAT 4 – X SQUARED – Y SQUARED MUST BE GREATER THAN OR EQUAL TO

ZERO AND THAT WOULD BE OUR DOMAIN. BUT NORMALLY IT’S NOT LEFT

IN THIS FORM. LET’S GO AHEAD AND MOVE THE X

SQUARED AND Y SQUARED TERMS TO THE OTHER SIDE AND THAT WOULD GIVE US 4

GREATER THAN OR=TO X SQUARED + Y SQUARED. AND NOW WE’LL SWITCH THIS AROUND MEANING WE HAVE X SQUARED

+ Y SQUARED MUST BE LESS THAN OR=TO 4. AND AGAIN LET’S GO AHEAD

AND GRAPH THIS DOMAIN ON THE X,Y PLANE. NOTICE IF WE GRABBED X SQUARE

+ Y SQUARED=4 THAT WOULD BE A CIRCLE

CENTERED AT THE ORIGIN WITH A RADIUS OF TWO. AND SINCE IT CAN BE LESS THAN

OR=TO FOUR WE’LL MAKE A SOLID CIRCLE. AGAIN IF WE TEST THE POINT 0 0, 0 IS LESS THAN OR=TO 4

IS TRUE. SO FOR THIS PROBLEM THE DOMAIN

WOULD BE ALL THE VALUES ON THE CIRCLE

OR INSIDE THE CIRCLE. THIS GRAPH REPRESENTS THE DOMAIN

OF THE GIVEN FUNCTION AND TWO VARIABLES. OKAY. THAT’S GONNA DO IT

FOR THIS INTRODUCTORY VIDEO. THANK YOU FOR WATCHING.

My god! I've been looking for you!

Very well explained.

thanks you for posting dude, very useful…my Professors teach in Dutch and my textbooks are in english..i've just learned that language last year, so it was very frustating. Thanks God i found such a video!

amazingggg, love you :)))

i really donâ€™t know why i go to my math class when i have you

Is there a second part to this video where it shows more about the range, level curves, and graphs of two variable functions? our videos are helpful, but I can never tell what order they are in.

Yes, I know. The best way to view them in order is to check out my website mathispower4u

Thanks for the comment.

Thanks. simple way of explaining

nice circle btw

not helping

is there a way to see the maple documents ( i have maple)

Thank you for this helpful video

Seriously, if you're taking multivariable calculus & you don't this stuff, then you must have slept through every math class that came before it. I wish these videos would examine more complex equations (then they might not be a waste of time).

helpful

thanks

thanks very wel explained.Would u please tell me which software did u use to draw the graph in 3-d?

Thank u so much bro. very useful.

excellent explanation

z=cosx*siny Looks like an egg box

how did you find the radius of 2?

Nice