Introduction to Functions of Two Variables

– 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.

19 Replies to “Introduction to Functions of Two Variables”

My god! I've been looking for you!

2. matherrrmatician says:

Very well explained.

3. Karina Anggelia says:

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!

4. Tatevik Barakazyan says:

amazingggg, love you :)))

5. femiairboy94 says:

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

6. PeglegSJ says:

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.

7. Mathispower4u says:

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

8. BigGus87 says:

Thanks. simple way of explaining

nice circle btw

not helping

10. spring1 says:

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

11. Daniel O'Connell says:

Thank you for this helpful video

12. Frank J says:

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).

13. Mohammed Alfusail says:

thanks

14. shorab hossain says:

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

15. Jamil Amri says:

Thank u so much bro. very useful.

16. Wassim Boudja says:

excellent explanation

17. Arkane Betha says:

z=cosx*siny Looks like an egg box

18. Joel Jr Casasiempre says:

how did you find the radius of 2?