1. Derivative as a Rate of Change

2. Derivative as a Rate of Change If y = f(x) and if x changes from the value x 1 to x 2 , then y changes from f(x 1 ) to f(x 2 ). So, the change in y, which we denote by  y, is f(x 2 ) - f(x 1 ) when the change in x is  x = x 2 – x 1 . The average rate of change of y with respect to x, over the interval [x 1 , x 2 ] , is then This can also be interpreted as the slope of the secant line.

3. Definition of Instantaneous Rate of Change Instantaneous Rate of Change= f’(x 1 ) is the instantaneous rate of change at x 1 . Note that a positive rate means a quantity increases with respect to the other quantity, that is y increases with x. If it is negative, then the quantity decreases with respect to other quantity. Note that heat, velocity, density, current, temperature, pressure, molar concentration, fluid flow, bacterial growth, reaction rate, blood flow and cost are just some of the few quantities that maybe analyzed through derivatives. We consider the average rate of change over a smaller and smaller intervals by letting x 2 approaches x 1 and therefore letting  x approach 0. The limit of this average rate of change is called the (instantaneous) rate of change of y with respect to x at x = x 1 , which is interpreted as the slope of the tangent line to the curve y = f(x) at x = x 1 .

4. Given a set of data, we may approximate instantaneous rate of change in values using average values or the graph representing the set of data. For example: Page 147 number 25 The table below shows the estimated population (in percent) of Europe that use cell phones (midyear estimates are given). Find the average rate of cell phone users growth i. From 2000 to 2002 ii. From 1999 to 2000 iii. From 2000 to 2001 Year 1998 1999 2000 2001 2002 2003 P 28 39 55 68 77 83

5. Solution: Find the average rate of cell phone users growth Here, we will use the formula i. From 2000 to 2002 ii. From 1999 to 2000 iii. From 2000 to 2001

6. 1. p.147 #28 If a cylindrical tank holds 100,000 liters of water, which can be drained from the bottom of the tank in an hour, then Torricelli’s law gives the volume V of water remaining in the tank after t minutes as Find the rate of change at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t) as a function of t.

7. 1. Using Ohm’s Law where V volts is the electromotive force, R ohms is the resistance and I amperes is the current in an electric circuit, find the rate of change in I with respect to R and find the instantaneous rate of change of I with respect to R in an electric circuit having 120 volts when the resistance is 20 ohms. (Take that V is constant) Example Solution: Ohm’s Law states that V = IR. Thus I = V/R = VR -1 . So, I= 120R -1 . We have And so when R = 20 ohms, then = -0.30 ampere/ohm The negative signs implies that current is decreasing at this conditions.

8. 2. A solid consists of a right circular cylinder and a hemisphere on each end, and the length of the cylinder is twice its radius. Let r units be the radius of the cylinder and the two hemispheres, and V(r) cubic units be the volume of the solid. Find the instantaneous rate of change in V(r) with respect to r. Solution: If the height of the cylinder is twice its radius r, then h = 2r. Thus, the volume of the cylinder is V 1 =  r 2 h =  r 2 (2r) = 2  r 3 . Since two hemispheres are equal to a sphere, the volume is V 2 = 4/3  r 3 . So the volume of the solid is V(r) = 2  r 3 + 4/3  r 3 = 10/3  r 3 . The instantaneous rate of change is V ’(r) = 10  r 2 . Note that the rate of change of the volume with respect to r can be obtained when r is given.

9. 3.Sand is being dropped onto a conical pile in such a way that the height of the pile is always twice the base radius. Find the rate of change of the volume of the pile with respect to the radius when the height of the pile is (a) 4 m and (b) 8 m. Solution: If the height of the cone is twice its radius, then h = 2r. So, the volume of the cone is V = 1/3  r 2 h V = 1/3  r 2 (2r) So, V(r) = 2/3  r 3 . The rate of change of the volume with respect to r is V ’(r) = 2  r 2 . When a) h = 4m, r = 2 then v’(2) = 2  (2) 2 = 8  m 2 . b) h = 8m, r = 4, then v’(4) = 2  (4) 2 = 32  m 2 .

11. In manufacturing companies, the costs of producing their products is a major concern. The following terms are useful in dealing with problems involving costs. Total Cost Function, C(x) – expression giving the total amount needed to produce x units of a certain product. Marginal Cost Function, C’(x) – Rate of change in cost when x units of product is produced. In a similar sense, we may consider the revenue of the company. Thus, we have Total Revenue Function, R(x) – expression giving the total amount earned in the sales of x units of a certain product. Marginal Revenue Function, R’( x) – Rate of change in revenue when x units of product is sold.

12. Solution: a. Since C(x) = 1500 + 3x + x 2 then C’(x) = 3 + 2x b. W h en x = 40, C’(x) = 3 + 2(40) = 83 dollars/watch. 5. The number of dollars in the total cost of manufacturing x watches in a certain plant is given by C(x) = 1500 + 3x + x 2 . Find (a) the marginal cost function and (b) the marginal cost when x = 40.

17. Vertical Motion Here, the velocity is still the derivative of position function y = f(t), given by v = If a particle is projected straight upward from an initial height y 0 (ft) above the ground at time t = 0 (sec) and with initial velocity v 0 (ft/sec) and if air resistance is negligible, then its height y = f(t) (in feet above the ground) at time t is given by a formula known from Physics, y = f(t) = ½ gt 2 + v 0 t + y 0 . where g denotes the acceleration due to the force of gravity. At the surface of the earth, g  -32 ft/s 2 (or -9.8 m/s 2 ). Thus, y= f(t) = -16t 2 + v 0 t + y 0 . (or y= f(t) = -4.9t 2 + v 0 t + y 0 ) y is increasing y is decreasing v > 0, decreasing v < 0, increasing y=f(t) Ground level y = 0

20. The acceleration a of the particle is defined to be the instantaneous time rate of change of its velocity: a = We define the instantaneous velocity v of the particle at the time t to be the limit of the average velocity as  t  0. That is, v = The velocity of the moving particle may be positive or negative depending on whether the particle is moving in the positive or negative direction along the line of motion. We define the speed of the particle to be the absolute value of the velocity.

21. Rectilinear Motion

24. Next, we describe the position and motion of the particle in a table that includes the intervals of time when the particle is moving to the left, when it is moving to the right, and when the velocity is increasing or decreasing. t = 0.75 t = 2 t = - 0.5 Position function: Velocity function: v = Acceleration function: a =

25. First, we find t when v = 0 and a = 0. From the preceding solution, we found that t = -0.5 and t = 2 when v = 0, and t = 0.75 when a = 0. From these, we obtain the intervals: t < -0.5, -0.5 < t < 0.75, 0.57 < t < 2, and t > 2. We describe the position and motion of the particle during these intervals by constructing a table as follows:

26. t=-0.5 t< -0.5 -0.5<t<0.75 t=0.75 t = 2 0.75 < t < 2 t=-0.5 t = 2 t > 2

29. 2. Vertical Motion Thus, the equation of motion of an object moving in a vertical line is y(t) = -16t 2 + v 0 t + y 0 (or simply y = -16t 2 + v 0 t + y 0 ) The book uses s = -16t 2 + v 0 t + s 0 If a particle is projected straight upward from an initial height y 0 (ft) above the ground at time t = 0 (sec) and with initial velocity v 0 (ft/sec) and if air resistance is negligible, then its height y (in feet above the ground) at time t is given by a formula known from Physics, y(t) = where g denotes the acceleration due to the force of gravity. At the surface of the earth, g  32 ft/sec 2 . The velocity of the particle at time t is V(t) = = -32t + v 0 , and the acceleration of the particle is a = =-g.

30. y is increasing y is decreasing v > 0 v < 0 x y y=y(t) Ground level y = 0

33. c) Find the speed of the ball when it reaches the ground. First, we find the time where y = 0. So, 0 = -16t 2 + 32t, -16t(t – 2) = 0 where t = 0 (initial time, before the motion takes place) and t = 2 (final time, where the ball reaches the ground). Thus, if t = 2 sec. v = -32t + 32 v = -32(2) + 32 v = -32 ft/sec. It follows that the speed of the ball when it reaches the ground is  v  = 32 ft/sec .

34. Related Rates

37. 2. A man 6 ft tall is walking toward a building at the rate of 5 ft/sec. If there is a light on the ground 50 ft from the building, how fast is the man’s shadow on the building growing shorter when he is 30 ft from the building? Differentiate both sides wrt t, Let y – shadow, and x = distance of man from light. 6 y 50 ft x By similar triangles, and so

38. 3. A bacterial cell is spherical in shape. If the radius of the cell is increasing at the rate of 0.01  m/day when it is 1.5  m , what is the rate of increase in volume of the cell at that time? Solution: Let V – volume,and r – radius. Given: dr/dt = 0.01  m/day Unknown: dV/dt when r = 1.5  m

39. 4. A water tank in the form of an inverted cone is being emptied at the rate of 6 m 3 /min. The height of the cone is 24 m and the radius is 12 m. Find how fast the water level is lowering when the water is 10 m deep? Solution: Let V – volume, and r – radius and h – height of the water. Given: dV/dt = - 6 ft 3 /min, Unknown: dh/dt when h = 10m By similar triangles 24m 12m 12m h 24m r

40. 5. A trough is 12 ft long and its ends are in the form of an isosceles triangles having an altitude of 3ft and a base of 3ft. Water is flowing into the trough at the rate of 2ft 3 /min. Find how fast is the water level rising when the water is 1 ft deep. Equation: V = ½ xh(12). So, V = 6xh = 6h 2 . Diff. both sides wrt t, So, 2 = 12(1) dh/dt Solution: Let V – volume, h = height, and x be the base. Given: Unknown: when h = 1 ft. 3 x 3 h By similar  s, So, x = h

41. 6. A horizontal trough is 16m long and its ends are isosceles trapezoids with an altitude of 4m, a lower base of 4m and an upper base of 6m. Water is being drained from the trough at a rate of 10 m 3 /min. How fast is the water level lowering when the water is 2m, deep?

44. 7. A girl is using straw to drink coke from a right cylindrical glass at the rate of 6 cm 3 /sec . If the height of the glass is 12 cm and the diameter is 6 cm , how fast is the level of coke falling at a constant rate? Note that r will not change with time, so we may directly substitute its value in the equation of volume. Thus, we have V =  (3) 2 h =9  h. Differentiate both sides wrt t, r h cm/sec