Calculus Cheat Sheet Derivatives - Pauls Online Math Notes

Calculus Cheat SheetRelated RatesSketch picture and identify known/unknown quantities. Write down equation relating quantitiesand differentiate with respect to t using implicit differentiation (i.e. add on a derivative every timeyou differentiate a function of t). Plug in known quantities and solve for the unknown quantity.Ex. A 15 foot ladder is resting against a wall.The bottom is initially 10 ft away and is beingpushed towards the wall at 1 4ft/sec. How fastis the top moving after 12 sec?Ex. Two people are 50 ft apart when onestarts walking north. The angleθ changes at0.01 rad/min. At what rate is the distancebetween them changing when θ = 0.5 rad?x¢ is negative because x is decreasing. UsingPythagorean Theorem and differentiating,2 2 2x + y = 15 Þ 2xx¢ + 2yy¢= 01After 12 sec we have ( )x = 10- 12 = 7and4We have θ¢ = 0.01 rad/min. and want to findx¢ . We can use various trig fcns but easiest is,xx¢secθ = Þ secθ tanθθ¢=50 50We knowθ = 0.05 so plug in θ¢ and solve.x¢sec( 0.5) tan( 0.5)( 0.01)=50x¢ = 0.3112 ft/secRemember to have calculator in radians!2 2so y = 15 - 7 = 176 . Plug in and solvefor y¢.717( -4 ) + 176 y¢ = 0 Þ y¢= ft/sec4 176OptimizationSketch picture if needed, write down equation to be optimized and constraint. Solve constraint forone of the two variables and plug into first equation. Find critical points of equation in range ofvariables and verify that they are min/max as needed.Ex. We’re enclosing a rectangular field with500 ft of fence material and one side of thefield is a building. Determine dimensions thatwill maximize the enclosed area.Ex. Determine point(s) on yclosest to (0,2).2= x + 1 that areMaximize A= xy subject to constraint ofx+ 2y= 500 . Solve constraint for x and pluginto area.A= y( 500-2y)x= 500-2yÞ=2500y- 2yDifferentiate and find critical point(s).A¢ = 500-4y Þ y = 125By 2 nd deriv. test this is a rel. max. and so isthe answer we’re after. Finally, find x.x = 500- 2125 = 250( )The dimensions are then 250 x 125.2 22Minimize f d ( x 0) ( y 2)constraint is= = - + - and they2= x + 1. Solve constraint for2x and plug into the function.2 2x y f x y( )= -1 Þ = + -22( )2 2= y- 1+ y- 2 = y - 3y+3Differentiate and find critical point(s).3f¢ = 2y-3Þ y =2By the 2 nd derivative test this is a rel. min. andso all we need to do is find x value(s).2 3 1 1x = - 1 = Þ x=±2 2 21 3The 2 points are then ( , 2 2 ) and ( - 1, 32 2 )Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.© 2005 Paul Dawkins