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X
- WE WANT TO FIND THE EQUATION OF THE TANGENT LINE TO THE CURVE
4 x THE QUANTITY X SQUARED + Y SQUARED
TO THE 2nd = 50 x THE QUANTITY X SQUARED - Y SQUARED
AT THE POINT 3,1.
WELL, TO FIND THE EQUATION OF THE TANGENT LINE
WE'LL HAVE TO FIND THE SLOPE OF THE TANGENT LINE,
WHICH WE CAN FIND BY DETERMINING DYDX.
AND SINCE WE HAVE AN IMPLICIT EQUATION HERE,
WE'LL HAVE TO PERFORM IMPLICIT DIFFERENTIATION
IN ORDER TO FIND DYDX.
SO THE FIRST STEP IS TO DIFFERENTIATE
BOTH SIDES OF THE EQUATION WITH RESPECT TO X.
SO WE'LL HAVE THE DERIVATIVE
OF 4 x THE QUANTITY X SQUARED + Y SQUARED TO THE 2nd
WITH RESPECT TO X MUST = THE DERIVATIVE OF 50
x THE QUANTITY X SQUARED - Y SQUARED,
AGAIN WITH RESPECT TO X.
NOW, WE'LL DIFFERENTIATE AS WE NORMALLY WOULD,
BUT IF THE TERM CONTAINS Y,
WE'LL HAVE AN EXTRA FACTOR OF DYDX,
BECAUSE WE'LL HAVE TO APPLY THE CHAIN RULE.
SO LOOKING AT THE LEFT SIDE,
NOTICE HOW THIS IS A COMPOSITE FUNCTION.
SO WE'LL HAVE TO APPLY THE CHAIN RULE,
WHERE THE INNER FUNCTION WOULD BE X SQUARED + Y SQUARED.
SO IF IT'S HELPFUL, WE CAN LET THE INNER FUNCTION
BE EQUAL TO U,
AND THEREFORE WE CAN APPLY THE GENERAL POWER RULE
GIVEN HERE WITH THE CHAIN RULE,
THINKING OF THIS AS 4U TO THE 2nd.
SO WE WOULD MULTIPLY BY 2,
THAT WOULD GIVE US 8 x U TO THE 1st
WHERE U IS X SQUARED + Y SQUARED x U PRIME,
WHICH IS THE DERIVATIVE OF THE INNER FUNCTION.
SO WE'D HAVE DERIVATIVE OF X SQUARED + Y SQUARED
WITH RESPECT TO X.
WE'LL FIND THIS DERIVATIVE IN THE NEXT STEP.
AND NOW ON THE RIGHT WE WOULD HAVE
50 x THE DERIVATIVE OF X SQUARED WITH RESPECT TO X.
THAT WOULD JUST BE 2X - THE DERIVATIVE OF Y SQUARED,
WHICH WOULD BE 2Y x DYDX.
AGAIN WE HAVE A FACTOR OF DYDX,
BECAUSE THIS TERM CONTAINS A Y, NOT X.
NOW WE'LL FIND THE DERIVATIVE HERE.
SO WE HAVE 8 x THE QUANTITY X SQUARED + Y SQUARED.
WE CAN DROP THE POWER OF 1 HERE x THE DERIVATIVE OF X SQUARED
WITH RESPECT TO X.
IT'D BE 2X + THE DERIVATIVE OF Y SQUARED
WITH RESPECT TO X WOULD BE 2Y x DYDX.
AGAIN WE HAVE A FACTOR OF DYDX, BECAUSE THIS IS A Y TERM,
NOT AN X TERM.
LET'S GO AHEAD AND DISTRIBUTE HERE ON THE RIGHT.
SO WE HAVE 100X - 100Y DYDX.
REMEMBER, OUR ULTIMATE GOAL IS TO HAVE
ALL THE DYDX TERMS ON ONE SIDE,
SO THAT WE CAN FACTOR OUT DYDX AND SOLVE FOR DYDX.
SO LOOKING AT THE LEFT SIDE,
LET'S GO AHEAD AND CLEAR THESE PARENTHESES.
WE'LL HAVE FOUR PRODUCTS HERE, ONE, TWO, THREE, AND FOUR.
SO WE'LL HAVE 8 x X SQUARED x 2X = 2X TO THE 3rd
+ THE NEXT PRODUCT WOULD BE 2X SQUARED Y DYDX,
AND THEN WE HAVE + 2XY SQUARED,
AND THEN WE HAVE + 2Y TO THE 3rd DYDX.
ON THE RIGHT SIDE WE STILL HAVE 100X - 100Y DYDX.
NOW WE'LL GO AHEAD AND DISTRIBUTE 8.
NOTICE HOW WE HAVE THREE TERMS THAT CONTAIN A FACTOR OF DYDX
HERE, HERE, AND HERE.
WE NEED TO HAVE ALL OF THESE TERMS
ON ONE SIDE OF THE EQUATION,
AND ALL THE OTHER TERMS ON THE OTHER SIDE.
SO LET'S CONTINUE ON THE NEXT SLIDE.
AND NOW TO HAVE ALL OF THE DYDX TERMS ON THE LEFT SIDE,
WE'LL GO AHEAD AND ADD 100Y DYDX ON BOTH SIDES.
AND THEN TO MOVE THESE TWO TERMS
ON THE RIGHT SIDE OF THE EQUATION,
WE'LL SUBTRACT 16X TO THE 3rd ON BOTH SIDES,
AND SUBTRACT 16XY SQUARED ON BOTH SIDES.
SO WE'D STILL HAVE 16X SQUARED Y DYDX + 16Y TO THE 3rd DYDX,
BUT THEN WE'D HAVE + 100Y DYDX = 100X - 16X TO THE 3rd
AND THEN - 16XY SQUARED.
NOW THAT ALL THE DYDX TERMS ARE ON THE LEFT,
WE CAN GO AHEAD AND FACTOR OUT DYDX FROM EACH TERM.
SO WE'D HAVE DYDX x THE QUANTITY 16X SQUARED
Y + 16Y TO THE 3rd + 100Y.
AND THE RIGHT SIDE STAYS THE SAME.
AND NOW TO SOLVE FOR DYDX WE'LL DIVIDE BOTH SIDES
BY THIS QUANTITY HERE.
SO NOW WE HAVE THIS SOLVED FOR DYDX,
BUT THE RIGHT SIDE IS GOING TO SIMPLIFY.
LET'S GO AHEAD AND FACTOR BOTH THE NUMERATOR AND DENOMINATOR.
THE GREATEST COMMON FACTOR IN THE NUMERATOR WOULD BE 4X.
WE'LL FACTOR OUT 4X.
WE'LL ALSO PUT THE TERMS IN DESCENDING ORDER.
SO THIS WOULD BE -4X TO THE 2nd, THIS WOULD BE - 4Y TO THE 2nd,
AND THEN WE'D HAVE + 25.
THE GREATEST COMMON FACTOR OF THE DENOMINATOR IS 4Y,
SO WE'LL FACTOR OUT 4Y.
THAT WOULD LEAVE US WITH 4X TO THE 2nd + 4Y SQUARED + 25.
NOTICE IN THIS FORM 4/4 SIMPLIFIES TO 1,
SO DYDX = X x THE QUANTITY -4X TO THE 2nd - 4Y TO THE 2nd + 25.
I GUESS WE COULD'VE FACTORED OUT -X, BUT THAT'S OKAY.
OUR DENOMINATOR IS Y x THE QUANTITY 4X TO THE 2nd + 4Y
TO THE 2nd + 25.
REMEMBER, OUR ULTIMATE GOAL HERE
WAS TO FIND THE SLOPE OF THE TANGENT LINE AT THE GIVEN POINT,
SO NOW WE'LL EVALUATE THIS DERIVATIVE AT THE POINT 3,1.
SO WE'LL SUBSTITUTE 3 FOR X AND 1 FOR Y.
THIS COMES OUT TO -45 DIVIDED BY 65, WHICH SIMPLIFIES TO -9/13.
SO NOW THAT WE KNOW THE SLOPE OF THE TANGENT LINE,
AND WE ALSO KNOW THE LINE PASSES THROUGH THE POINT 3,1,
WE CAN FIND THE EQUATION OF THE TANGENT LINE.
WE'LL GO AHEAD AND USE THE POINT SLOPE FORM OF A LINE,
WHICH IS THE FORM Y - Y1 = M x THE QUANTITY X - X SUB 1,
WHERE M IS THE SLOPE AND X SUB 1, Y SUB 1
IS THE POINT ON THE LINE.
SO THIS IS X SUB 1, AND THIS IS Y SUB 1.
SO WE WOULD HAVE Y - 1 = -9/13 x THE QUANTITY X - 3.
LET'S GO AHEAD AND SOLVE THIS FOR Y.
SO NOW WE'LL CLEAR THE PARENTHESES
BY DISTRIBUTING HERE.
SO WE HAVE Y - 1 = -9/13X, AND THEN WE HAVE + 27/13.
AND NOW WE'LL ADD 1 TO BOTH SIDES TO SOLVE FOR Y,
SO WE'LL HAVE Y = -9/13X + 27/13.
AND WE'RE ADDING 1 TO BOTH SIDES,
BUT 1 IS THE SAME AS 13/13.
SO FINALLY WE HAVE THE EQUATION OF OUR TANGENT LINE.
WE HAVE Y = -9/13X + 40/13.
LET'S TAKE A LOOK AT THE GRAPH OF THIS.
HERE IS THE POINT OF TANGENCY, THE POINT 3,1,
AND HERE IS THE EQUATION OF THE TANGENT LINE THAT WE JUST FOUND.
I HOPE YOU FOUND THIS HELPFUL.