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- WHEN AIR EXPANDS ADIABATICALLY,
MEANING WITHOUT GAINING OR LOSING HEAT,
ITS PRESSURE AND VOLUME ARE RELATED BY THE EQUATION P x V
RAISED TO THE POWER OF 1.4 = C.
WHERE C IS A CONSTANT P IS THE PRESSURE AND V IS THE VOLUME.
SUPPOSE AT A CERTAIN INSTANT THE VOLUME IS 350 CUBIC CENTIMETERS
AND THE PRESSURE IS 70 KILOPASCALS
AND IS DECREASING AT A RATE OF 10 KILOPASCALS PER MINUTE.
AT WHAT RATE IN CUBIC CENTIMETERS PER MINUTE
IS THE VOLUME INCREASING AT THAT INSTANT?
SO TO SOLVE THIS PROBLEM WE'LL BE USING THE GIVEN EQUATION
P x V RAISED TO THE POWER OF 1.4 = C.
WE ALSO WANT TO CONSIDER
WHEN THE VOLUME IS EXACTLY 350 CUBIC CENTIMETERS,
SO WE KNOW V = 350, WE ALSO KNOW THE PRESSURE IS 70 KILOPASCALS
BUT THE PRESSURE IS ALSO DECREASING
AT A RATE OF 10 KILOPASCALS PER MINUTE
WHICH MEANS DP/DT WITH THE CHANGE IN THE PRESSURE
WITH RESPECT TO TIME IS EQUAL TO -10 KILOPASCALS PER MINUTE.
AND OUR GOAL IS TO FIND THE RATE OF CHANGE OF THE VOLUME
AT THIS INSTANT SO WE WANT TO FIND DV/DT.
SO TO SOLVE THIS PROBLEM
WE'RE GOING TO DIFFERENTIATE THIS EQUATION HERE
WITH RESPECT TO T.
NOTICE BECAUSE WERE DIFFERENTIATING
WITH RESPECT TO T
WE'LL HAVE TO PERFORM IMPLICIT DIFFERENTIATION
AND ALSO ON THIS LEFT SIDE HERE
SINCE WE HAVE A PRODUCT OF P AND V OF THE 1/4
WE'LL HAVE TO APPLY THEIR PRODUCT RULE
IN ORDER TO FIND THIS DERIVATIVE WITH RESPECT TO T.
AND THEN ONCE WE FIND THE DERIVATIVE WITH RESPECT TO T
WE CAN PERFORM SUBSTITUTION WITH THE KNOWN VALUES TO FIND DV/DT.
LET'S GO AND FIND THIS DERIVATIVE ON THE NEXT SLIDE.
SO AGAIN, TO FIND THE DERIVATIVE OF HERE
WE'LL HAVE TO APPLY THEIR PRODUCT RULE.
SO WE'LL HAVE THE FIRST FUNCTION P
x THE DERIVATIVE OF THE SECOND FUNCTION WITH RESPECT TO T
PLUS THE SECOND FUNCTION
TIMES THE DERIVATIVE OF THE FIRST FUNCTION
WITH RESPECT TO T.
THIS IS STILL EQUAL TO THE DERIVATIVE OF A CONSTANT
WITH RESPECT TO T.
NOW WE'LL GO BACK AND FIND THE DERIVATIVES WITH RESPECT TO T.
SO WE'LL HAVE P TIMES THE DERIVATIVE OF V
RAISED TO THE POWER OF 1.4 WITH RESPECT TO T
WE'LL APPLY THE POWER RULE WITH THE CHAIN RULE
SO WE'LL HAVE 1.4 FEET TO THE POWER OF 1.4 - 1
OR 0.4 x 4 x DV/DT.
AGAIN, WE HAVE THIS EXTRA FACTOR HERE
BECAUSE THIS IS GIVEN IN TERMS OF V
AND WE'RE DIFFERENTIATING WITH RESPECT TO T.
AND WE'LL HAVE PLUS V TO THE POWER OF 1.4
x THE DERIVATIVE OF P TO RESPECT TO T
WHICH WOULD BE 1 x DP/DT OR JUST DP/DT
= THE DERIVATIVE OF A CONSTANT IN RESPECT TO T WOULD JUST BE 0.
REMEMBER, OUR GOAL HERE IS TO SOLVE FOR DV/DT.
AND NOW WE'LL SOLVE THE EQUATION FOR DV/DT.
LET'S START BY SUBTRACTING THIS PRODUCT HERE
ON BOTH SIDES OF THE EQUATION.
WE CAN WRITE THIS AS 1.4 P V TO THE POWER OF 0.4 x DV/DT
= WE'D HAVE -V TO THE 1.4 POWER TIMES DP/DT.
AND NOW TO SOLVE FOR DP/DT
WE'LL DIVIDE BOTH SIDES BY 1.4 P V TO THE POWER OF 0.4.
WITH THE CHANGE IN THE VOLUME RESPECT TO TIME
IS = TO -V TO THE POWER OF 1.4 x DP/DT
DIVIDED BY 1.4 x P V TO THE POWER OF 0.4.
AND NOW TO FIND DV/DT WE CAN SUBSTITUTE 350 FOR V
70 FOR P -10 FOR DP/DT AND THEN SIMPLIFY DEFINE DV/DT.
LET'S GO AHEAD AND DO THIS ON THE NEXT SLIDE.
SO WE WOULD HAVE DV/DT = -350
RAISED TO THE POWER OF 1.4 X DP/DT
WHICH IS -10 DIVIDED BY 1.4 x P
WHICH IS 70 x V RAISED TO THE POWER OF 0.4.
TO SAVE SOME TIME I'VE ALREADY DETERMINED THIS VALUE DV/DT
IS APPROXIMATELY EQUAL TO 35.714.
AGAIN, THIS IS THE CHANGE IN THE VOLUME RESPECT TO TIME
THIS IS GOING TO BE CUBIC CENTIMETERS PER MINUTE.
SO THE VOLUME IS INCREASING BECAUSE THIS IS POSITIVE
AT A RATE OF APPROXIMATELY 35.714 CUBIC CENTIMETERS
PER MINUTE
WHEN THE VOLUME IS 350 CUBIC CENTIMETERS,
THE PRESSURE IS 70 KILOPASCALS
AND THE PRESSURE IS DROPPING AT A RATE
OF 10 KILOPASCALS PER MINUTE.
THIS IS IMPORTANT TO RECOGNIZE
THAT THE VOLUME IS ONLY CHANGING AT THIS RATE
GIVEN THESE CONDITIONS.
IF THESE CONDITIONS WERE DIFFERENT
IT WOULD CHANGE THE RATE AT WHICH THE VOLUME IS CHANGING.
I HOPE YOU FOUND THIS EXPLANATION HELPFUL.