- WELCOME TO THE FIRST OF SEVERAL VIDEOS
ON APPLICATIONS OF EXTREMA OF FUNCTIONS OF TWO VARIABLES.
THE GOAL OF THIS VIDEO IS TO DETERMINE
THE MINIMUM DISTANCE FROM A POINT TO A PLANE.
IN ORDER TO SOLVE THESE TYPES OF PROBLEMS ONCE WE DETERMINE
OUR FUNCTION IN TERMS OF TWO VARIABLES
WE'LL FOLLOW THE SAME STEPS WE FOLLOWED TO DETERMINE
THE RELATIVE EXTREMA OF FUNCTIONS OF TWO VARIABLES
IN A PREVIOUS VIDEO.
SO YOU MAY WANT TO PAUSE THE VIDEO HERE TO REVIEW
THESE STEPS.
FOR OUR EXAMPLE HERE, WE WANT TO DETERMINE THE MINIMUM
DISTANCE FROM THE PLANE 3X - 2Y + Z = 6
TO THE POINT (0, 0, 0) WHICH IS THE ORIGIN.
SO WE'RE TRYING TO DETERMINE THE MINIMUM DISTANCE
FROM THE ORIGIN TO THIS PLANE WHICH WE COULD REPRESENT
AS THIS RED SEGMENT HERE.
REMEMBER THE DISTANCE BETWEEN TWO POINTS
IN SPACE IS GIVEN BY THIS DISTANCE FORMULA
SO WE WANT TO MINIMIZE THIS DISTANCE BUT INSTEAD
OF USING THIS DISTANCE FORMULA IF WE WERE TO SQUARE BOTH SIDES
OF THIS AND THEN CALL LITTLE D BIG D WE COULD MINIMIZE
THE SQUARE OF THIS DISTANCE FORMULA AND THEREFORE ELIMINATE
THE SQUARE ROOT.
SO WE'RE GOING TO USE THE FORMULA BIG D
IS EQUAL TO X2 - X1 SQUARED + Y2 - Y1 SQUARED
+ Z2 - Z1 SQUARED INSTEAD.
SO LET'S SEE IF WE CAN COME UP WITH A FUNCTION IN TERMS
OF THREE VARIABLES AND THEN SOMEHOW REWRITE IT IN TERMS
OF TWO VARIABLES GIVEN THAT ONE OF THE POINTS
HAS TO LIE IN THIS PLANE.
WHAT I MEAN BY THAT IS WE COULD EASILY
SOLVE THIS EQUATION FOR Z IN TERMS OF X AND Y
AND THEN WE COULD HAVE ALL THREE COORDINATES IN TERMS OF X AND Y.
SO IF WE SOLVE THIS EQUATION FOR Z
WE WOULD HAVE Z = 6 - 3X + 2Y.
NOW THE REASON THAT'S HELPFUL IS GOING BACK TO THE SQUARE
OF THE DISTANCE FORMULA WE CAN NOW REWRITE THIS
AS BIG D = X - 0 SQUARED + Y - 0 SQUARED
+ Z - 0 SQUARED BUT Z = 6 - 3X + 2Y - 0 SQUARED.
SO NOW WE HAVE A FUNCTION BIG D IN TERMS
OF TWO VARIABLES X AND Y.
LET'S GO AHEAD AND CLEAN THIS UP A LITTLE BIT.
SO WE'LL HAVE BIG D = X SQUARED
+ Y SQUARED + 6 - 3X + 2Y SQUARED.
SO NOW WE'LL TAKE THIS FUNCTION AND DETERMINE
THE RELATIVE EXTREMA.
REMEMBER WHAT WE'RE GOING TO DO IS DETERMINE WHERE
THE FIRST ORDER PARTIAL DERIVATIVES ARE EQUAL TO ZERO
OR UNDEFINED AND THEN DETERMINE THE VALUE OF D AND BASED UPON
THOSE VALUES WE CAN DETERMINE WHETHER WE HAVE A RELATIVE MIN,
RELATIVE MAX, A SADDLE POINT OR THE TEST IS INCONCLUSIVE.
SO THE PARTIAL DERIVATIVE OF D WITH RESPECTS TO X WOULD BE 2X.
THIS WOULD BE ZERO.
HERE WE'LL HAVE TO APPLY THE CHAIN RULE.
WE'RE GOING TO HAVE 2 x 6 - 3X + 2Y x U PRIME
WHICH IS GOING TO BE -3.
LET'S GO AHEAD AND SIMPLIFY THIS.
WE'LL HAVE 2X.
HERE WE HAVE -3 x 2 THAT'S -6 SO WE'LL DISTRIBUTE -6.
IT'LL BE - 36 + 18X - 12Y.
SO THE PARTIAL WITH RESPECTS TO X IS GOING TO BE EQUAL
TO 20X - 12Y - 36 AND WE WANT TO KNOW
WHEN THIS WILL BE EQUAL TO ZERO.
NOW LET'S DETERMINE THE PARTIAL DERIVATIVE OF D
WITH RESPECTS TO Y.
SO WE'RE GOING TO HAVE 0 + 2Y
+ AGAIN WE HAVE 2 x 6 - 3X + 2Y TO THE 1st AND THEN x 2.
LET'S SIMPLIFY THIS.
WE GOING TO HAVE 2Y.
HERE WE HAVE 2 x 2 SO WE'LL DISTRIBUTE 4 SO + 24 - 12X + 8Y.
SO WE HAVE THE PARTIAL DERIVATIVE WITH RESPECTS
TO Y WOULD BE EQUAL TO -12X + 10Y + 24
AND WE ALSO WANT TO KNOW WHEN THIS IS EQUAL TO ZERO.
SO WHAT WE'LL DO NOW IS TREAT THESE AS A SYSTEM
OF EQUATIONS AND SOLVE FOR X AND Y.
LET'S GO AHEAD AND DO THAT ON THE NEXT SCREEN.
OKAY. SO WHAT I'M GOING TO DO HERE IS MOVE THE CONSTANTS
TO THE RIGHT SIDE AND I'M ALSO GOING TO DIVIDE
THIS FIRST EQUATION BY 4 AND DIVIDE
THE SECOND EQUATION BY 2.
SO IF WE DIVIDE BY 4 WE'RE GOING TO HAVE 5X - 3Y
DIVIDE THIS BY 4 IT'LL GIVE US 9 SO WE'LL HAVE A +9
ON THE RIGHT SIDE AND NOW WE'LL DIVIDE THIS EQUATION BY 2
AND MOVE THE CONSTANT OVER.
SO WE'LL HAVE -6X + 5Y = -12.
NOW WE'LL USE ELIMINATION TO SOLVE THIS SYSTEM.
WE'LL ELIMINATE THE Y TERMS BY MULTIPLYING
THE FIRST EQUATION BY 5 AND THE SECOND EQUATION BY 3.
THAT'LL MAKE THE Y TERMS -15 AND +15.
SO WE'LL HAVE 25X - 15Y = 45.
THE SECOND EQUATION WILL BE -18X + 15Y = -36.
THAT'S ZERO.
WE'LL HAVE 7X = THIS WILL GIVE US 9.
SO X = 9/7.
NOW LET'S GO AHEAD AND DETERMINE Y BY SUBSTITUTING X = 9/7
INTO THIS EQUATION HERE.
SO IF WE MOVE THE X TERM TO THE OTHER SIDE
WE'LL HAVE 5Y = 6 x 9/7 - 12.
THIS WILL GIVE US 5Y = -30/7.
MULTIPLY BOTH SIDES BY 1/5.
WE HAVE Y = -6/7.
OKAY. SO WE HAVE X = 9/7.
Y = -6/7.
WE SHOULD PROBABLY GO AHEAD AND DETERMINE Z.
REMEMBER WE HAD AN EQUATION FOR Z ON THE OTHER SLIDE.
WE HAVE Z = 6 - 3X + 2Y SO LET'S GO AHEAD
AND DETERMINE Z WHILE WE'RE HERE.
FOR THE SAKE OF TIME, I'VE ALREADY DETERMINED THIS.
IT COMES OUT TO 3/7.
SO WE HAVE SO FAR ARE THE THREE COORDINATES THAT WE THINK
WILL MINIMIZE OR MAXIMIZE THE DISTANCE
BETWEEN THE PLANE AND THE ORIGIN.
BUT TO BE SURE WE SHOULD TAKE THE TIME
TO DETERMINE THE VALUE OF D AS WE SEE HERE
AND MAKE SURE THAT IT SATISFIES THE CONDITION SUCH THAT D
IS POSITIVE AND THE SECOND ORDER PARTIAL WITH RESPECTS TO X
IS POSITIVE AS WELL.
LET'S GO AHEAD AND DO THAT AND ONCE WE VERIFY THIS
WE CAN THEN DETERMINE WHAT THAT MINIMUM DISTANCE WOULD BE.
SO AGAIN, WE'RE GOING TO GO BACK TO OUR FIRST ORDER OF PARTIALS
AS WE SEE HERE AND WRITE THOSE OUT AGAIN.
SO HERE OUR FIRST ORDER OF PARTIALS THAT WE DETERMINED
ON THE PREVIOUS SLIDE.
LET'S GO BACK AND JUST MAKE SURE.
HERE THEY ARE HIGHLIGHTED IN BLUE.
SO NOW WE'RE GOING TO DETERMINE THE SECOND ORDER OF PARTIALS
TO DETERMINE THE VALUE OF D.
SO THIS WOULD BE THE PARTIAL DERIVATIVE OF THIS
WITH RESPECTS TO X.
THAT'LL GIVE US 20, THE SECOND ORDER OF PARTIAL
WITH RESPECTS TO Y WOULD BE +10 AND THEN THE MIXED SECOND ORDER
PARTIAL WOULD BE THE PARTIAL DERIVATIVE OF THIS
WITH RESPECTS TO Y.
THAT WILL BE -12.
OKAY. SO D = 20 x 10 - -12 SQUARED.
IT'S GOING TO BE 200 - 144 WHICH IS 56 WHICH IS POSITIVE
AND THEN AGAIN THE SECOND ORDER PARTIAL WITH RESPECTS TO X
IS ALSO POSITIVE.
SO THIS DOES VERIFY THAT WE DID FIND A RELATIVE MINIMUM
FOR OUR FUNCTION.
SO THAT TELLS US THAT WHEN X IS 9/7, Y IS -6/7 AND Z IS 3/7
WE ARE MINIMIZING THE DISTANCE FROM THE PLANE TO THE ORIGIN.
SO THE LAST STEP IS TO DETERMINE WHAT THAT MINIMUM DISTANCE
WOULD BE.
SO TO DETERMINE THAT MINIMUM DISTANCE WE DO HAVE TO GO BACK
AND USE THE DISTANCE FORMULA BETWEEN TWO POINTS IN SPACE
SO LET'S GO AHEAD AND PERFORM THE SUBSTITUTION.
REMEMBER THE SECOND POINT WAS THE ORIGIN
SO WE'LL HAVE 9/7 - 0 SQUARED OR JUST 9/7 SQUARED
+ -6/7 - 0 SQUARED
AND THEN 3/7 - 0 SQUARED.
WE CAN SEE OUR DENOMINATORS ARE ALL GOING TO BE 49,
OUR NUMERATORS ARE GOING TO BE 81 + 36 + 9.
IT'S GOING TO BE EQUAL TO THE SQUARE ROOT OF 126/7.
126 SIMPLIFIES TO 3 SQUARE ROOT 14.
THIS IS APPROXIMATELY 1.6.
SO THIS WOULD BE THE MINIMUM DISTANCE BETWEEN THE ORIGIN
AND THE PLANE AS WE SEE HERE.