MA119-A Applied Calculus for Business Fall Homework 5 Solutions Due 10/4/ :30AM

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MA9-A Applie Calculus for Business 2006 Fall Homework 5 Solutions Due 0/4/2006 0:0AM. #0 Fin the erivative of the function by using the rules of i erentiation. r f (r) r. #4 Fin the erivative of the function f (r) 4 r 4 r 4 r r 4 r2 4r 2. f (u) 2 p u by using the rules of i erentiation. u f (u) 2 pu 2u 2 2 u 2 2 u u u 2 u 2. #8 Fin the erivative of the function by using the rules of i erentiation. f () 2 + p u. f (). #24 Fin the erivative of the function 2 + 2 + () 2 2 + 0 2 (2) 2 6. f () + 2 2 + by using the rules of i erentiation.

2 First, we can rewrite f asf () 2 + 2 +. f () 2 + 2 + 2 + (2) + () 2 + 2 + 0 2 2 + 2 + 2. [Alternative Solution] By the quotient rule, the erivative is f () + 2 2 + ( + 2 2 + ) () ( + 2 2 + ) () () 2 + 22 +. #28 Fin the erivative of the function () ( + 2 2 + ) () 2 (2 + 4 + 0) () ( + 2 2 + ) 2 2 + 2 2 + 2. by using the rules of i erentiation. f () 6 f () 6 6 4 4 + 2 5. 6 6 5. #4 Fin the erivative of the function by using the rules of i erentiation. f () + 4 p +

f () + 4 p + + + 4 ( ) 4 + 4 9 4 2 2. 4 2 2 2 + 4 2 + + () + 0 2. #8 Fin the limit 5 lim! by evaluating the erivative of s suitable function at an appropriate point. Consier f () 5. Then f (). By letting h. So, + h. Then the limit becomes lim! 5 lim (h + ) 5 5 h!0 h lim h!0 f ( + h) f () h f 0 (). We know that f 0 () 5 4. Thus, f 0 () 5 4 5. Hence, lim! 5 f 0 () 5.. #42 Fin the slope an an equation of the tangent line to the graph of the function f () 5 2 + 2 + 2 5 at the point ;. Note that f ( ) 5. So, ; 5 is on the graph of f when. To n the slope of the tangent line, we calculate the erivative rst. Thus, f 0 () 0 + 2. This tells us that f 0 ( ) 0 ( ) + 2 6, the slope. So, the tangent line equation is 5 y 6 ( ( )), or, y + 5 6 ( + ).. #56 E ect of Stopping on Average Spee Accoring to ata from a stuy, the average spee if your trip A (in mph) is relate to the number of stops/mile you make on the trip by the equation A 26:5 0:45. Compute A for 0:25 an 2. How is the rate of change of the average spee of your trip a ecte by the number of stops/mile?

4 We have A 26:5 0:45 :925 :45. 26:5 0:45 26:5 0:45 26:5 ( 0:45) :45 Thus, when 0:25, A :925 (0:25) :45 89:0 an when 2, A :925 (2) :45 4:648. When your number of stops/mile increases, the rate of change of the average spee of your trip is ecreasing..2 #4 Fin the erivative of f () (2 + ) ( 4). By the prouct rule, the erivative is f () [(2 + ) ( 4)] (2 + ) ( 4) + (2 + ) 2 ( 4) + (2 + ) 6 8 + 6 + 9 2 +. ( 4) [Alternative Solution] We notice that f () (2 + ) ( 4) 6 2 + 2. Thus, the erivative is f () 62 + 2 62 + 2 2 +..2 #8 Fin the erivative of By the quotient rule, the erivative is f (t) t 2t t + t f (t) 2t + t. ( t 2t) ( + t) + ( 2t) ( + t) t ( + t) 2 2 ( + t) + ( 2t) ( + t) 2 2t ( + t) 2..2 #24 Fin the erivative of f () 2 + p.

5 By the quotient rule, the erivative is f () 2 + p.2 #28 Fin the erivative of (2) ( p ) + ( 2 + ) 2 5 2 2 + 2 5 2. (2 + ) ( p ) + ( 2 + ) (p ) 2 p ( p ) 2 f () 2 2. 2 2 + 2 + 2 2 2 2 f () By the prouct rule, the erivative is 2 2 2 (6) 2 + 2 2 2 2 2 2 + 2 2 2..2 #2 Suppose f an g are functions that are i erentiable at an that g () 2, an g 0 (). Fin the value of h 0 () where h 0 () h () 2 + g (). By the prouct rule, the erivative h 0 () is h () 2 + g () 2 + (g ()) + 2 + (g ()) (2) (g ()) + 2 + (g ()). So, when, we have h 0 () (2 ()) (g ())+ () 2 + (g 0 ()) 2( 2)+2 2..2 #6 Fin the erivative of the function an evaluate f 0 () at 2. f () 2 + 2

6 By the quotient rule, the erivative is f () 2 + (2 + ) (2 2 2 (2 ) + (2 + ) 2 (2 ) 2 8 (2 ) 2. 4 2 + 4 + 2 (2 ) 2 ) + (2 + ) (2 ) (2 ) 2.2 #40 Fin the slope an an equation of the tangent line to the graph of the function f () 2 + at the point 2; 4. Note that f (2) 4. So, 2; 4 is on the graph of f when 2. To n the slope of the tangent line, we calculate the erivative rst. By the quotient rule, the erivative is f () 2 + (2 ) ( + ) + ( 2 ) ( + ) ( + ) 2 2 ( + ) + (2 ) ( + ) 2 22 + 2 + 2 ( + ) 2 2 + 2 ( + ) 2. This tells us that f 0 (2) (2)2 +2(2) 6, the slope. So, the tangent line equation is ((2)+) 2 9 4 y 6 ( 2), or, 6 9y 20 0. 9.2 #48 Fin the point(s) on the graph of the function f () + where the slope of the tangent line is equal to. 2 To n the slope of the tangent line, we calculate the erivative rst. By the quotient rule, the erivative is f () + ( + ) ( ( ) + ( + ) ( ) 2 + + ( ) 2 2 ( ) 2. ) + ( + ) ( ) ( ) 2 We are looking for the point(s) whose slope is equal to 2. Thus, we set 2 ( ) 2 2. This implies that ( ) 2 4, that is, 2 + 2 + 0. Therefore, we have. So, the point ( ; 0) has the slope of the tangent line 2.

.2 #54 Deman Functions The eman function for the Sicar wristwatch is given by 50 () (0 20) 0:0 2 + where (measure in units of a thousan) is the quantity emane per week an () is the unit price in ollars. (a) Fin 0 (). (b) Fin 0 (5), 0 (0), an 0 (5) an interpret your results. (a) By the quotient rule, the erivative is 0 () 50 (50) (0:0 2 + ) + (50) (0:02 + ) 0:0 2 + (0:0 2 + ) 2 0 (0:02 + ) + (50) (0:02) (0:0 2 + ) 2 (b) Thus, we have 0 (5) (5) (5) (0:0(5) 2 +) 2 :2, 0 (0) (0:0 2 + ) 2. 7 (0) (0:0(0) 2 +) 2 2:5, an 0 (5) (0:0(5) 2 +) 2 :420. So, when the quantity emane per week is increasing, the rate of change of the unit price is ecreasing. That means the more eman, the less change of the unit price.