Lecture 14 September 26, Today. WH 3 now posted Due Tues. Oct. 2, 2018 Quiz 4 tomorrow. Differentiation summary Related rates

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1 Lecture 4 September 6, 08 WH 3 now poste Due Tues. Oct., 08 Quiz 4 tomorrow Toay Differentiation summary Relate rates

2 Differentiation Summary Basic erivatives (memorize) x x x c = 0 sin x = cos x cos x = sin x x x r = rx r if x > 0 x tan x = sec x x cot x = csc x x x x ex = e x sec x = sec x tan x csc x = csc x cot x x arcsin x = x x arccos x = x x arctan x = +x x sec x = x x x cot x = +x x csc x = x x x ln x = x Derivative rules If f an g are ifferentiable at x then: Sum rule Prouct rule Quotient rule Chain rule Inverse function rule x x [f (x) + g(x)] = f (x) + g (x) x [f (x)g(x)] = f (x)g(x) + f (x)g (x) f (x) g(x) = f (x)g(x) f (x)g (x) [g(x)] assuming g(x) 0 x f (g(x)) = f (g(x))f (x) assuming f is iff. at g(x) x f (x) = f (f (x)) assuming f is iff. at f (x) an f (f (x)) 0

3 Miscellaneous techniques Implicit ifferentiation. take x of both sies of equation. put terms with y on one sie 3. solve for y Definition of a b a b = e ln(ab) = e b ln a if a > 0 Definition of log a b log a b = ln b ln a if a > 0 an b > 0 Logarithmic ifferentiation. start with equation y = f (x). take ln of both sies of equation 3. take x 4. solve for y of both sies of equation Relate Rates Problems Relate Rates If y is a quantity of interest relate to a quantity x, how oes the rate of change of y epen on x an the rate of change of x? Give equation relating x an y. Take erivative with respect to proper variable (generally t for time) Rate of change of x is x t Rate of change of y is y t Tough part of these problems is getting correct equation relating y an x

4 A trough has a triangular cross section which is m high, m wie. The trough is 4m long. Water is poure in at a rate of m3 /min. How fast is the water level rising when the epth is m? Solution: Draw picture with changing quantities labele with variables. Give equation relating relevant quantities. V = bhl l = 4 is not changing. using similar triangles h b = V = hh4 V = h V t ( ) V t = 4h h t so h t = 4h Now plug in h = m an V h t = ( ) V t t 4h = ( ) 4 = 8 m/min so h = b so b = h = m3 /min Helpful geometry Helpful geometry Circumference of circle πr Area of circle πr Length of sector r θ Area of sector r θ Area of rectangle l w Area of triangle bh Area of trapezoi (a + b)h Area of annulus π(r r ) Volume of cyliner πr h Volume of prism Ah Volume of cone 3 πr h Volume of cone 3 Ah Surface area of sphere 4πr Volume of sphere 4 3 πr 3

5 Helpful Trigonometry Helpful Trigonometry Similar triangles Law of sines sin α A = sin β B = sin γ C Law of cosines A = B + C BC cos α An inverte conical tank with height 3m an raius m is fille at a rate of 3 m3 /min. How fast is the water level rising when m3 has been poure into the tank? Solution: Draw picture with changing quantities labele with variables. V = 3 πr h using similar triangles r h = 3 so r = h 3 V = 3 π( h 3 ) h = π 7 h3 h = 3 7 π V = 3 3 π V 3 h t = 3 3 π 3 V /3 V t = 3 π V /3 V t Now plug in V = m3 an V = 3 m3 /min h t = ( 3 /3 ( π ) ) 3 = 3 t 4 7π m/min

6 A rotating wheel of raius m is connecte to a 3m crankshaft an then to a piston. The wheel is rotating clockwise at a rate of 5 rpm. What is the spee of the piston when the angle mae by the the axis of the piston, the center of the flywheel an the connection of the flywheel to the crankshaft is π 4? Draw picture with changing quantities labele with variables. x = cos θ + 3 sin θ = cos θ + (9 sin θ) / x t = sin θ θ t + (9 sin θ) / (0 sin θ cos θ θ t ) What is θ t? 5 rpm (clockwise) = π 5 ra/min = 0π ra/min x t = sin π 4 ( 0π) + (9 sin π 4 ) / ( sin π 4 cos π 4 ) ( 0π) x t = 0π + (9 ( ) ) / (( ) ) (0π) x t = 5 π + ( ) ( 7 ) / 0π = 5 π + 7 5π spee = x t = 5π + 5π 7 m/min 7.6 m/min An actor m tall walks at 3m/s away from a wall an towar a floor light 4m from the wall. How fast oes his shaow on the wall behin him grow when he is 3m from the wall? Solution: Draw picture with changing quantities labele with variables. using similar triangles 4 x = h 4 h t = 8(4 x) ( ) x t Now plug in x = 3m an x t = 3 m/s h t = 8(4 x) x t = 8(4 3) 3 = so h = 8(4 x) = 8(4 x) x t 8 3 m/s

7 The hour han on a clock has length m an minute han has length m. How fast is istance between the tips of the hans changing at 3pm? θ = angle between hour han an minute han x = istance between tip of hour han an tip of minute han By Law of Cosines x = + cos θ = 5 4 cos θ x x θ t = 4 sin θ t θ = θ m θ h θ t = θ m t θ h t = π ( ) π = π 6 ra/h At 3pm θ = π. At 3pm x = + = 5 So x t = 4 sin θ θ t x = 4(sin π π )( 6 5 ) = π 3 5 m/h