2.5 Related Rates 1A) Equation: 4 3x y+ = Given that 1dxdt=, find dydtwhen 2x =Given) 1dxdt=; Find) dydt; When) 2x = Equation) 4 3x y+ =Work: We will need to take the derivative with respect to time and solve for the “find”.4 3x y+ =( )4 3x yd ddt dt+ =4 0dx dydt dt+ � =4dy dxdt dt=-14dy dxdt dt=- *Note that x did not appear, so dydtdid not depend on When.( )1 114 4dydt=- = -2B) Given that 3,dydt=Find dxdtwhen ( )2 2 2, ,2 2x y� �+=� �� �� �.Given) 3dydt=; Find) dxdt; When) ( )2 2 2, ,2 2x y� �+=� �� �� � Equation) 2 22x y x+ =Work: 2 22x y x+ =( )( )2 22d ddtx y xdt+ =2 2 2dx dy dxx ydt dt dt� + � = �2dy dx xdxydt dt dt-� =( )1dy dxy xdt dt= -1y dy dxx dt dt� =-2232 2 22 2dxdt�=+� �-� �� �22322dxdt�=( )1 3 3dx dxdt dt�= =3) Oil from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2 ft/sec. How fast is the area of the spill increasing when the radius of the spill is 60 ft? rGiven) 2secdr ftdt=; Find) dAdt; When) 60r ft= Equation) 2A rp= �Work:2dA drrdt dtp= � �( )2 60 2secftftp= � �2240secdA ftdtp=Example: 4) A baseball diamond is a square whose sides are 90ft long. Suppose that a player running from second base to third base has a speed of 30 ft/sec at the instant when he is 20 ft from third base. At what rate is the players distance from home plate changing at that instant?W XY 90ftWork: Given: 30secdx ftdt=-Find: dydtWhen: 20x ft=Equation: 2 2 290x y+ =( )2 2 290d dx ydt dt+ =2 0 2dx dyx ydt dt+ =x dx dyy dt dt=2030sec10 85ft ft dydtft� ��- =� �� �*Note you have to solve the original equation for Y. 2 2 22 2 2229020 90400 81008500850010 8510 85x yyyyyyy+ =+ =+ ==� =� ==Thus:60sec85ft dydt=- =6.51secdy ftdt�-***Note: Do not substitute a number in for a variable if the variable is changing. Wait until you have isolated your “find” after youhave taken the derivative.5) A camera 3000 ft away from the launching pad of a rocket is photographing the takeoff. If the rocket is rising vertically at 880 ft/sec when it is 4000 ft above the launch pad, how fast must the camera angle change at that instant to keep the camera aimed atthe rocket? h θ 3000 ftWork: Given: 880secdh ftdt=Find: ddtqWhen: 4000h ft=Equation: tan3000hq =[ ]tan3000d d hdt dtq� �=� �� �21sec 13000d dhdt dtqq � = ��21 13000 secd dhdt dtqq= � �( )21 1880300053ddtq= � �� �� �� �1 98803000 25ddtq= � �66625 secd raddtq=180 6.05.1056sec secd raddt radqp� � �o o5) Alternate method for solving #5Work: Given: 880secdh ftdt=Find: ddtqWhen: 4000h ft=Equation: tan3000hq =tan3000hq =arctan3000hq =21 1300013000d dhdt dthq= �� �+� �� �21 18803000400013000ddtq= � �� �+� �� �1 1881630019= � �+1 12225759= � �3 1 662225 25 625 secrad= � � =Example:6) As shown in the figure above, water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that is changing at the rate of (h-12) feet per minute. The volume V of a cone with radius r and height h is 213V r hp= Work:Given: 12mindh fthdt= -Find:dvdtWhen:3h ft=Equation: 213V r hp=A) Write an expression for the volume of water in the conical tank as a function of h.4 * using similar triangles we find what r is equal to.213V r hp= 1221 13 3V h hp� �=� �� �4 112 3rr hh= =3127V hp=B) At what rate is the volume of water in the conical tank changing when h=3? Indicate units of measure.21327d dhV hdt dtp= � �( )213 3 927 mindv ftftdtp� �= � �-� �� �39mindv ftdtp=-C) Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h=3 ? Indicate units of measure.Given: 39mindv ftdtp=Find:dydtWhen:3h ft=Equation: V Area of Base Height= �2400V ft yp= �2400dv dyftdt dtp= �21400dv dyft d t dtp= � =321 9400 minft dyft dtpp= � =9400 minft dydt= =Angular Rate of Change: In a free-fall experiment, an object is dropped from a height of 256 feet. A camera on the ground 500 feet from the point of impact records the fall of the object. S 256ft θ500ftA) Find the position function giving the height of the object at time t, assuming the object is released at time t = 0. At what time will the object reach ground level?( )( )( )20 0221616 0 25616 256s t t V t Ss t t ts t t=- + +=- + �+=- +(* Position function)When will the object reach the ground?2220 16 25616 256164tttt=- +===�4sect =B) Find the rates of change of the angle of elevation of camera when t = 1 and t = 2.Find ddtqWhen 1sec; 2sect t= =Equation:tan500sq =arctan500sq =216 256arctan500tq+= -( )221 13250016 2561500dtdttq= � �-� �- ++� �� �When t = 1:( )( )( )221 132 1 0.052050016 1 2561500ddtq= � �- =-� �- ++� �� �� �When t = 2:( )( )( )221 132 2 0.111551118350016 2 2561500ddtq= � �- =-� �- ++� �� ��