- WE WANT TO FIND THE DERIVATIVE OF F OF X = 4 SINE X,
AND THEN FIND THE EQUATION OF THE TANGENT LINE
AT THE POINT (PI/6,2).
SO IF F OF X = 4 SINE X
THEN THE DERIVATIVE FUNCTION F PRIME OF X
IS GOING TO BE EQUAL TO 4 x THE DERIVATIVE OF SINE X,
WHICH IS EQUAL TO COSINE X.
SO OUR DERIVATIVE FUNCTION IS 4 COSINE X.
AND NOW TO FIND THE EQUATION OF THE TANGENT LINE
AT THE GIVEN POINT,
WE NEED TO FIND THE SLOPE OF THE TANGENT LINE,
WHICH WE CAN FIND BY EVALUATING THE DERIVATIVE FUNCTION
AT X = PI/6.
SO THE SLOPE OF THE TANGENT LINE
WOULD BE EQUAL TO F PRIME OF PI/6,
WHICH IS EQUAL TO 4 x COSINE PI/6.
WE SHOULD RECOGNIZE THIS COSINE FUNCTION VALUE AS
SQUARE ROOT 3 DIVIDED BY 2,
BUT JUST IN CASE WE DON'T LET'S TAKE A LOOK AT THE UNIT CIRCLE.
HERE'S THE TERMINAL SIDE OF PI/6 RADIANS.
ON THE UNIT CIRCLE
THE X COORDINATE IS EQUAL TO COSINE THETA.
SO COSINE PI/6 RADIANS IS EQUAL TO SQUARE ROOT 3 DIVIDED BY 2.
SO WE HAVE 4 x SQUARE ROOT 3 DIVIDED BY 2.
THIS IS THE SAME AS 4/1.
2 AND THE 4 SIMPLIFY,
SO F PRIME OF PI/6 IS EQUAL TO 2 SQUARE ROOT 3,
WHICH SHOULD BE THE SLOPE OF THE TANGENT LINE.
SO NOW OUR GOAL IS TO FIND THE EQUATION OF THE TANGENT LINE
THAT HAS A SLOPE OF 2 SQUARE ROOT 3
AND PASSES THROUGH THE POINT (PI/6,2).
SO TO FIND THE EQUATION OF THIS LINE,
LET'S GO AHEAD AND USE THE POINT SLOPE FORM OF THE LINE
OR THE FORM Y - Y SUB 1 = M x X - X SUB 1
WHERE M IS THE SLOPE OF THE LINE
AND X SUB 1, Y SUB 1 ARE THE COORDINATES OF THE POINT.
SO THE EQUATION OF THE TANGENT LINE WOULD BE
Y - 2 = 2 SQUARE ROOT 3 x THE QUANTITY X - PI/6.
SO THIS IS THE EQUATION OF THE TANGENT LINE
IN POINT SLOPE FORM.
LET'S GO AHEAD AND SOLVE THIS FOR Y
AND PUT IT IN SLOPE INTERCEPT FORM.
TO DO THIS, WE'LL DISTRIBUTE AND THEN ADD 2 TO BOTH SIDES.
LOOKS LIKE IT MIGHT BE A LITTLE MESSY
BUT LET'S GO AHEAD AND DO IT.
WE HAVE Y - 2 = THIS WOULD BE 2 SQUARE ROOT 3 x X.
AND THEN - 2 SQUARE ROOT 3 x PI/6.
SEE, 2 SQUARE ROOT 3/1 x PI/6.
NOTICE HOW THE 2 AND THE 6 WOULD SIMPLIFY.
THIS SIMPLIFIES TO 1, THIS SIMPLIFIES TO 3,
SO WE'LL HAVE - PI SQUARE ROOT 3 DIVIDED BY 3.
NOW LET'S GO AHEAD AND ADD 2 TO BOTH SIDES.
SO WE WOULD HAVE Y = 2 SQUARE ROOT 3 x X.
I COULD WRITE THIS AS 2X SQUARE ROOT 3,
BUT IN THIS FORM WE CAN SEE OUR SLOPE IS 2 SQUARE ROOT 3.
SO WE HAVE - PI SQUARE ROOT 3/3 + 2,
WHICH I'M GOING TO WRITE AS + 2/1
SO WE CAN ADD THESE TWO TERMS.
NOTICE OUR COMMON NOMINATOR HERE WOULD BE 3--
NOTICE THE COMMON NOMINATOR OF THESE TWO TERMS WOULD BE 3,
SO WE'RE GOING TO MULTIPLY THIS FRACTION HERE BY 3/3.
SO FINALLY WE HAVE Y = 2 SQUARE ROOT 3 x X.
AND THEN HERE WE WOULD HAVE 6 - PI SQUARE ROOT 3/3.
SO THIS WOULD BE THE EQUATION OF OUR TANGENT LINE
IN SLOPE INTERCEPT FORM.
NOW WE'LL GO AHEAD AND CHECK THIS
BY GRAPHING THE FUNCTION AND OUR TANGENT LINE
TO MAKE SURE THIS LINE IS TANGENT AT THIS POINT.
BUT TO MAKE IT EASIER TO CHECK GRAPHICALLY,
2 SQUARE ROOT 3 IS APPROXIMATELY 3.46,
AND THIS QUADRANT HERE IS APPROXIMATELY 0.19.
SO IF YOU LOOK AT THE GRAPH,
THE RED GRAPH IS A GRAPH OF F OF X = 4 SINE X.
THE POINT (PI/6,2) WOULD BE THIS POINT HERE.
NOTICE HOW THE BLUE LINE IS THE TANGENT LINE AT THAT POINT.
AND IT DOES LOOK LIKE THE SLOPE IS APPROXIMATELY 3.46
AND THE Y INTERCEPT IS APPROXIMATELY 0.19.
SO THIS IS GOOD ENOUGH TO VERIFY THAT OUR WORK LOOKS GOOD.
I HOPE YOU FOUND THIS HELPFUL.