- WELCOME TO A LESSON ON THE CHAIN RULE
FOR FUNCTIONS OF SEVERAL VARIABLES.
THIS IS PART ONE,
AND THIS PART WILL DEAL WITH FUNCTIONS OF TWO VARIABLES
WHERE BOTH X AND Y ARE DEFINED IN TERMS OF ONE VARIABLE T.
IN PART TWO WE'LL TAKE A LOOK AT WHEN X AND Y ARE DEFINED
IN TERMS OF TWO INDEPENDENT VARIABLES.
REMEMBER FROM CALCULUS ONE,
IF WE HAVE Y = F OF U AND U = G OF T,
TO DETERMINE DYDT WE FOUND THE PRODUCT OF DYDU AND DUDT.
AND WE CAN DO SOMETHING SIMILAR WHEN WE HAVE A FUNCTION
IN TERMS OF TWO VARIABLES WITH ONE INDEPENDENT VARIABLE.
WHAT I MEAN BY THAT IS IF Z = F OF X, Y, X = G OF T
AND Y = H OF T
WHERE Z, X, AND Y ARE DIFFERENTIABLE
THEN DZDT IS GOING TO BE EQUAL TO THE PARTIAL DERIVATIVE OF Z
WITH RESPECTS TO X x DXDT
+ THE PARTIAL DERIVATIVE OF Z WITH RESPECTS TO Y x DYDT.
IT CAN OFTEN BE HELPFUL TO USE A TREE DIAGRAM
TO REMEMBER THE FORMULA FOR THE CHAIN RULE.
FOR EXAMPLE, IF WE HAVE A FUNCTION Z
DEFINED IN TERMS OF X AND Y,
WE WOULD START OFF WITH A TREE THAT LOOKS LIKE THIS.
Z IS DEFINED IN TERMS OF X AND Y,
AND THEN X IS DEFINED IN TERMS OF T.
SO WE'D HAVE ANOTHER BRANCH HERE FOR T, AND THE SAME THING FOR Y.
AND ONCE WE HAVE THIS TREE, IF WE WANT TO DETERMINE DZDT
WE'D HAVE TO FOLLOW EACH OF THESE BRANCHES
TO HELP US DETERMINE THE FORMULA FOR THE CHAIN RULE.
WHAT I MEAN BY THAT IS WE'LL START OFF AT Z AND GO TOWARD X.
THIS SHOULD REMIND US OF THE PARTIAL OF Z WITH RESPECTS TO X.
AND WE CONTINUE DOWN THIS BRANCH AND WE ENCOUNTER X AND T,
AND THIS REMINDS US OF DERIVATIVE OF X
WITH RESPECTS TO T.
AND NOW WE STILL HAVE TO FOLLOW THE BRANCH TO THE RIGHT,
SO WE'LL HAVE PLUS HERE WE GO FROM Z TO Y
THAT'S THE PARTIAL OF Z WITH RESPECTS TO Y.
AND THEN WE'LL GO FROM Y TO T THAT WILL REMIND US OF DYDT.
SO THE SECOND HALF OF THE CHAIN RULE
IS GOING TO BE THE PARTIAL OF Z WITH RESPECTS TO Y x DYDT.
THE NICE THING ABOUT THIS IS WE CAN EXTEND THIS
SO IF WE ADD A FUNCTION IN TERMS OF THREE VARIABLES
WE'D HAVE THREE BRANCHES.
AND IF WE HAD THE VARIABLES X AND Y
IN TERMS OF TWO VARIABLES
WE'D JUST HAVE TWO BRANCHES OFF OF X AND Y,
AND WE'LL SEE THAT IN THE NEXT VIDEO.
LET'S TAKE A LOOK AT AN EXAMPLE NOW.
WE WANT TO DETERMINE DZDT BY USING THE CHAIN RULE
AND ALSO BY PERFORMING SUBSTITUTION TO DO A COMPARISON.
SO WE'RE GOING TO FIND THIS DERIVATIVE TWO WAYS,
AND THEN MAKE SURE WE GET THE SAME ANSWER.
LET'S FIRST WRITE OUT THE CHAIN RULE FOR THIS PROBLEM.
IF WE WANT DZDT--PARTIAL OF Z WITH RESPECTS TO X x DXDT
+ THE PARTIAL OF Z WITH RESPECTS TO Y x DYDT.
SO WE'LL HAVE DZDT EQUALS--
THE PARTIAL DERIVATIVE OF Z WITH RESPECTS TO X.
SO WE'RE TREATING Y AS A CONSTANT, THAT'D BE 2X x DXDT.
WELL, X = 2 COSINE T, SO WE'LL HAVE -2 SINE T
+ THE PARTIAL OF Z WITH RESPECTS TO Y
IS GOING TO BE 2Y x DERIVATIVE OF Y WITH RESPECTS TO T.
DERIVATIVE OF SINE T WOULD BE COSINE T.
NOW, WE WANT A DERIVATIVE IN TERMS OF T,
SO LET'S GO AHEAD AND PERFORM A SUBSTITUTION
WHERE WE'RE GOING TO REPLACE X WITH 2 COSINE T
AND REPLACE Y WITH SINE T.
NOW, LET'S TAKE A CLOSE AT THIS. THESE ARE ACTUALLY LIKE TERMS.
HERE WE'RE GOING TO HAVE -8 COSINE T SINE T
+ 2 COSINE T SINE T.
THAT'S GOING TO GIVE US -6 COSINE T SINE T.
NOW, JUST FOR AN ILLUSTRATION,
I WENT AHEAD AND EVALUATED THIS DERIVATIVE AT T = PI/4.
AND THAT CAME OUT TO -3.
NOW, THE REASON I WANTED TO DO THIS
IS BECAUSE YOU COULD THINK OF DZDT
AS MEASURING THE CHANGE IN THE HEIGHT
WITH RESPECTS TO THE CHANGE IN TIME AT T = PI/4
ON THE SURFACE GENERATED BY THIS FUNCTION.
AND YOU CAN THINK OF X OF T AND Y AND T
AS THE PATH IN THE X, Y PLANE, THE PARTICLE IS TAKING.
LET'S TAKE A LOOK AT THIS.
IN YELLOW WE HAVE THE SURFACE GENERATED BY THE FUNCTION.
AND IF I ANGLE THIS CORRECTLY
YOU CAN SEE THE ELLIPTICAL PATH OF THE PARTICLE ON THE SURFACE
GIVEN BY X = 2 COSINE T AND Y = SINE T.
IT'S AN ELLIPSE.
SO AS T CHANGES YOU CAN VIEW THIS BLUE VECTOR TRACING OUT
WHAT COULD BE THE PATH OF A PARTICLE ON THE SURFACE.
AND T = PI/4 IS APPROXIMATELY 0.785.
IF I GO BACK TO 0.785 OR 0.78,
REMEMBER DZDT WAS EQUAL TO -3 AT THIS POINT,
MEANING YOU COULD INTERPRET THIS AS THE HEIGHT DROPPING
3 FEET OR 3 INCHES PER SECOND AS THE PARTICLE MOVES.
NOTICE AS IT INCREASE T THE PARTICLE IS MOVING DOWNWARD.
SO THIS WOULD BE ONE INTERPRETATION
OF WHAT DZDT ACTUALLY REPRESENTS.
NOW LET'S GO BACK AND TAKE A LOOK AT FINDING DZDT
USING BASIC SUBSTITUTION
TO MAKE SURE THAT WE GET THE SAME DERIVATIVE.
WHAT I MEAN BY THAT IS WE'RE GOING TO REPLACE X
WITH TWO COSINE T AND Y WITH SINE T,
AND THEN FIND DZDT.
AND THE RESULT SHOULD BE THE SAME.
SO Z'S GOING TO BE EQUAL TO 4 COSINE SQUARE T
+ SINE SQUARED T + 1.
LET'S GO AHEAD AND DETERMINE THIS DERIVATIVE HERE.
WE'LL HAVE TO APPLY THE CHAIN RULE HERE.
MULTIPLY BY 2 THAT'S GOING TO BE 8 COSINE T
x THE DERIVATIVE OF COSINE, WHICH IS -SINE T
+ THE DERIVATIVE OF SINE SQUARED T IS GOING TO BE 2 SINE T
x THE DERIVATIVE OF SINE T IS GOING TO BE COSINE T.
SO DZDT IS GOING TO EQUAL TO-- NOW, THESE ARE LIKE TERMS AGAIN.
THIS IS -8 COSINE T SINE T, THIS IS +2 COSINE T SINE T.
SO OUR DERIVATIVE OF COURSE IS THE SAME.
WE HAVE -6 COSINE T SINE T.
AND THAT SHOULD MAKE SENSE WHETHER WE USE THE CHAIN RULE
OR WE USE SUBSTITUTION,
OUR DERIVATIVE SHOULD BE THE SAME.
LET'S GO AND TAKE A LOOK AT ONE MORE.
WHAT I WANT YOU TO NOTICE ON THIS ONE IS WE HAVE W
DEFINED IN TERMS OF THREE VARIABLES,
AND WE HAVE A SINGLE PERIMETER FOR X, Y, AND Z.
SO WHEN THERE ARE ADDITIONAL VARIABLES
WE CAN JUST EXTEND THE CHAIN RULE.
WE WOULD HAVE DWDT = THE PARTIAL OF W WITH RESPECTS TO X x DXDT
+ THE PARTIAL DERIVATIVE OF W WITH RESPECTS TO Y x DYDT
+ THE PARTIAL DERIVATIVE OF W WITH RESPECTS TO Z x DZDT.
LET'S GO AHEAD AND APPLY THE CHAIN RULE HERE
AND SEE WHAT WE GET.
SO THE PARTIAL DERIVATIVE OF W WITH RESPECTS TO X,
TREATING Y AND Z AS CONSTANTS,
WE'LL HAVE Z + Y + 0 OR Z + Y x DXDT.
BE CAREFUL HERE, U IS GOING TO BE -T,
SO WE'LL HAVE E TO THE -T x -1
OR -E TO THE -T + THE PARTIAL OF W WITH RESPECTS TO Y.
THIS WOULD BE 0, THIS WOULD BE X, AND THIS WOULD BE Z.
SO WE HAVE X + Z x DYDT THAT'S GOING TO BE E TO THE T
+ THE PARTIAL OF W WITH RESPECTS TO Z.
SO WE'LL HAVE X + 0 + Y OR X + Y.
FOR DZDT WE'LL HAVE THE DERIVATIVE OF E TO THE T
MINUS E TO THE -T.
THAT'S GOING TO BE E TO THE T AND THEN - E TO THE -T x -1.
THAT'LL BECOME + E TO THE -T.
NOW THAT WE HAVE DWDT
WE'RE GOING TO GO AHEAD AND WRITE IT IN TERMS OF T NOW.
SO WE'LL PERFORM SUBSTITUTION FOR X, Y, AND Z.
Z + Y IS GOING TO GIVE US 2E TO THE T - E TO THE -T.
MULTIPLY THIS BY -E TO THE -T + X + Z x E TO THE T.
WELL, X + Z WE HAVE TWO TERMS THAT ARE OPPOSITE,
SO WE'LL JUT HAVE E TO THE T x E TO THE T.
AND THEN X + Y IS GOING TO BE E TO THE -T + E TO THE T
+ E TO THE T + E TO THE -T.
NOW LET'S GO AHEAD AND MULTIPLY THIS OUT AND THEN SIMPLIFY IT.
NOW, LOOKING AT THE EXPONENTIAL PARTS,
WE'RE GOING TO ADD THE EXPONENTS.
THAT'S GOING TO BE E TO THE 0, THAT'S GOING TO BE 1.
SO WE HAVE -1 x 2 THAT'S -2.
+ E TO THE -T + -T IS -2T.
+ E TO THE 2T + AND HERE WE'RE GOING TO HAVE TO FOIL THIS.
SO LETS SEE WHAT WE GET.
WE HAVE E TO THE -T x E TO THE T, THAT'S 1.
AND THEN WE HAVE E TO THE -T + E TO THE -T THAT'S + E TO THE -2T.
AND THEN WE HAVE + E TO THE +2T, AND THEN WE HAVE ANOTHER 1.
SO LET'S GO AHEAD AND WRITE IT--
LET'S GO AHEAD AND FINISH THIS UP.
A LOT OF THINGS SIMPLIFY OUT HERE, WE HAVE + 1 + 1 THAT'S 2.
THERE'S -2, AND THEN WE HAVE SOME LIKE TERMS.
WE HAVE TWO OF THESE AND TWO OF THESE.
SO DWDT = 2E TO THE 2T + 2E TO THE -2T.
AND THAT'S ALL WE HAVE TIME FOR IN THIS VIDEO.
IN PART TWO WE'LL TALK ABOUT WHEN X AND Y AND EVEN Z
ARE DEFINED IN TERMS OF TWO VARIABLES.
THANK YOU FOR WATCHING.