- THE CIRCUMFERENCE OF A SPHERE
WAS MEASURED TO BE 75 CENTIMETERS
WITH A POSSIBLE ERROR OF 0.5 CENTIMETERS.
WE WANT TO USE A LINEAR APPROXIMATION
TO ESTIMATE THE MAXIMUM ERROR IN CALCULATING THE SURFACE AREA.
BECAUSE WE'RE USING A LINEAR APPROXIMATION
WE'LL BE USING DIFFERENTIALS,
WHICH USES THE CHANGE OF Y ALONG THE TANGENT LINE
RATHER THAN THE CHANGE OF Y OF THE ACTUAL FUNCTION.
AND THEN WE'LL ALSO FIND THE RELATIVE ERROR
IN CALCULATING THE SURFACE AREA.
LET'S BEGIN BY REVIEWING THE FORMULAS THAT WE NEED.
THE SURFACE AREA OF A SPHERE IS EQUAL TO 4PI R SQUARED,
AND THE CIRCUMFERENCE OF A CIRCLE,
WHICH WOULD BE THE PERIMETER OF THE CIRCLE,
IS EQUAL TO PI x D OR 2PI R.
SO AGAIN, THE SURFACE AREA OF A SPHERE
IS EQUAL TO 4PI R SQUARED,
AND THE CIRCUMFERENCE OF A CIRCLE IS EQUAL TO 2PI R.
NOW, BECAUSE WE'RE LOOKING FOR THE ERROR IN THE SURFACE AREA,
AND WE'RE USING A LINEAR APPROXIMATION,
WE'RE GOING TO USE DIFFERENTIAL S
TO CALCULATE THIS MAXIMUM ERROR.
WHERE DIFFERENTIAL S = DERIVATIVE OF 4PI R SQUARED
WITH RESPECT TO R x DR.
SO AGAIN, DIFFERENTIAL S,
WHICH WOULD BE THE ERROR IN THE SURFACE AREA,
= 8PI R x DIFFERENTIAL R.
BUT NOTICE HOW THE ERROR DEALS WITH THE MEASUREMENT
OF THE CIRCUMFERENCE,
WHERE THE ERROR IS +/- 0.5 CENTIMETERS.
SO THIS TELLS US THAT DELTA C, OR DIFFERENTIAL C,
= +/- 0.5 CENTIMETERS.
BUT NOTICE TO FIND DIFFERENTIAL S WE NEED R AND DIFFERENTIAL R,
NOT DIFFERENTIAL C.
SO IF WE TAKE OUR CIRCUMFERENCE FORMULA C = 2PI R,
AND WE DIVIDED BOTH SIDES BY 2PI,
NOTICE THAT R = THE CIRCUMFERENCE DIVIDED BY 2PI.
AND THEREFORE IF DIFFERENTIAL C = +/- 0.5 CENTIMETERS,
THEN IT FOLLOWS
THAT DIFFERENTIAL R = +/- 0.5 CENTIMETERS DIVIDED BY 2PI.
SO WE'LL BE USING THIS TO FIND DIFFERENTIAL S
AND THIS TO FIND DIFFERENTIAL S, WHERE DIFFERENTIAL S = 8,
WHICH I'M GOING TO WRITE AS 8/1 x PI x R.
BUT NOTICE THAT R = CIRCUMFERENCE DIVIDED BY 2PI,
WHERE THE CIRCUMFERENCE IS 76 CENTIMETERS.
SO R = 76 DIVIDED BY 2PI,
AND THEN FOR DIFFERENTIAL R WE'LL HAVE +/- 0.5
DIVIDED BY 2PI.
THIS PRODUCT WILL GIVE US THE ERROR
WHEN CALCULATING THE SURFACE AREA,
IF THE ERROR IN MEASURING THE CIRCUMFERENCE
IS +/- 0.5 CENTIMETERS.
NOTICE ONE FACTOR OF PI SIMPLIFIES OUT,
AND THEN 2 x 2 = 4.
THAT SIMPLIFIES WITH THE 8, SO THESE TWO SIMPLIFY TO 1,
AND THE 8 SIMPLIFIES TO 2.
SO THAT LEAVES US WITH DIFFERENTIAL S
= 2 x 76 x +/- 0.5 DIVIDED BY PI.
AND THIS DOES DEAL WITH SURFACE AREA,
SO THIS WOULD BE CENTIMETERS SQUARED.
LET'S GET A DECIMAL APPROXIMATION FOR THIS
USING THE CALCULATOR.
SO THE NUMERATOR IS GOING TO BE 2 x 76 x .5
AND DIVIDE THIS BY PI.
LET'S GO AHEAD AND ROUND THIS TO FOUR DECIMAL PLACES.
THIS WOULD BE APPROXIMATELY 24.1916 SQUARE CENTIMETERS,
WHICH MEANS THE POSSIBLE ERROR
WHEN DETERMINING THE SURFACE AREA
IS +/- 24.1916 SQUARE CENTIMETERS.
NOW TO DETERMINE THE RELATIVE AREA
WHEN CALCULATING THE SURFACE AREA.
THE RELATIVE ERROR WOULD BE EQUAL TO DIFFERENTIAL S
DIVIDED BY S,
OR THE ERROR DETERMINING THE SURFACE AREA
DIVIDED BY THE SURFACE AREA.
SO DIFFERENTIAL S WAS, AGAIN, 8PI R x DR,
AND THE SURFACE AREA = 4PI R SQUARED.
LET'S GO AHEAD AND SIMPLIFY THIS,
AND THEN WE'LL MAKE OUR SUBSTITUTIONS.
SO NOTICE 8 AND 4 SHARE A COMMON FACTOR OF 4.
THERE'S 1 FOUR IN 4 AND 2 FOURS IN 8.
THE PI SIMPLIFIES TO 1, AND ALSO ONE FACTOR OF R SIMPLIFIES OUT.
THIS SIMPLIFIES TO 1.
THIS SIMPLIFIES TO R TO THE 1st.
SO THE RELATIVE ERROR = 2 x DIFFERENTIAL R DIVIDED BY R.
SO WE'D HAVE 2 x AGAIN,
DIFFERENTIAL R WAS +/- 0.5 DIVIDED BY 2PI,
AND R = 76 DIVIDED BY 2PI.
THESE TWO PI WOULD SIMPLIFY OUT, AND 2 x +/- 0.5 = +/- 1.
SO THIS SIMPLIFIES TO +/- 1 DIVIDED BY 76.
LET'S GO AHEAD AND GET A DECIMAL APPROXIMATION.
1 DIVIDED BY 76 IS APPROXIMATELY 0.0132.
AGAIN, THIS WOULD BE PLUS OR MINUS.
IF WE WERE ASKED TO CONVERT THIS TO A PERCENTAGE,
WE'D HAVE TO MULTIPLY BY 100 AND ADD A PERCENT SIGN.
SO THIS WOULD BE EQUAL TO 1.32% OR +/- 1.32%.
I HOPE YOU FOUND THIS HELPFUL.