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Applications of Derivatives Flash Cards
Absolute Extrema(Max and Min)
Absolute Extrema:MUST BE WITHIN A CONTINUOUS INTERVALABS max: the highest point the graph reaches ABS min:  the lowest point the graph reaches
Local Extema(Max and Min)
Local ExtremaTHERE CAN BE AN INFINITE NUMBER OF LOCAL MAXS/MINSLocal max: the highest point the graph reaches where the slope of the line equals zeroLocal min: the lowest point the graph reaches where the slope of the line equals zero
Position, Velocity, Acceleration(relationships)
s”(t) = v’(t) = a(t)
Related Rates(steps)
1. Determine the rates and the variables2. Find the missing variable using the Pythagorean theorem3. Once all of the variables are found, determine which rate needs to be found.4. Use the derivative of the Pythagorean theorem to find the missing rate5. Plug all variables and rates into the derivative and solve for the missing rate. That's your answer! Rates: ,[object Object],dt ,[object Object],dt ,[object Object],dt Variables: ,[object Object]
X  =
C  =dc dt dy dt C Y dx dt X Use the Pythagorean theorem to find the missing variable:  X2 + Y2 = C2 Use the derivative of the Pythagorean theorem to find the missing variable:  2x dx + 2y dy = 2c dc dtdtdt

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Final review flash cards

  • 3. Absolute Extrema:MUST BE WITHIN A CONTINUOUS INTERVALABS max: the highest point the graph reaches ABS min: the lowest point the graph reaches
  • 5. Local ExtremaTHERE CAN BE AN INFINITE NUMBER OF LOCAL MAXS/MINSLocal max: the highest point the graph reaches where the slope of the line equals zeroLocal min: the lowest point the graph reaches where the slope of the line equals zero
  • 9.
  • 10. X =
  • 11. C =dc dt dy dt C Y dx dt X Use the Pythagorean theorem to find the missing variable: X2 + Y2 = C2 Use the derivative of the Pythagorean theorem to find the missing variable: 2x dx + 2y dy = 2c dc dtdtdt
  • 13. 1. Find 2 equations that correspond with the problem [one equation will be a substitution equation (the variable found will be substituted into the other equation), and the other will be minimized or maximized].2. Solve for one variable (in the substitution equation) and plug that into the maximized/minimized equation(answer will have a variable in it).3. Take the derivative of the minimized/maximized equation (simplify before if possible).4. Set the derivative equal to zero and solve for the variable (will be an actual value). That’s your answer!