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Derivative as a Rate of Change

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.

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 .

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

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

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.

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.

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.

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 .

4. If water is being drained from a swimming pool and V(t) liters is the volume of water in the pool t minutes after draining starts, where V(t) = 250(1600 – 80t + t 2 ). Find The average rate at which the water leaves the pool during the first 5 minutes. How fast the water is flowing out of the pool 5 min after the draining starts? Solution: b) The rate of change of the volume with respect to t is V ’(t) = 250(-80 + 2t). So, at t = 5mins, the rate of change of the volume is V ’(5) = 250[-80 + 2(5)] = -17,500 liters/min. We find V(0) = 250(1600) = 400,000 and V(5) = 250(1600 – 80(5) + 5 2 ) = 306,250. The average rate at which the water leaves the pool during the first 5 minutes is

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.

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.

Other Problems: 1. A wave produced by a simple sound has the equation P(t) = 0.003 sin 1800  t where P(t) dynes/cm 2 is the difference between the atmospheric pressure and the air pressure at the eardrum at t seconds. Find the instantaneous rate of change of P(t) with respect to t at (a) 1/9 sec and (b) 1/8 sec. A Cepheid variable star is a star whose brightness alternately increases and decreases. The most easily visible such star is Delta Cephei, for which the interval between times of maximum brightness is 5.4 days. The average brightness of this star is 4.0 and changes by  0.35. Its brightness B is modeled by where t is measured in days. Find the rate of change of the brightness after 1day. (page 229, #61)

3. A company estimates that in t years, the number of its employees will be N(t) = 1000 (0.8) t/2 (a) How many employees do the company expect to have in 4 years. (b) At what rate is the number of employees expected to be changing in 4 years? Consider a blood vessel with radius 0.01 cm, length 3 cm, pressure difference 3000 dyne/cm 2 and viscosity  = 0.027. Using the law of laminar flow: (a) Find the velocity of the blood along the center line, at r = 0.005 cm and at the wall. (b) Find the velocity gradient (instantaneous rate of change of v with respect to r) at r = 0, r = 0.005 and r = 0.01 cm. (page 212, #25) If R denotes the reaction of the body to some stimulus of strength x, the sensitivity S is defined to be the rate of change of reaction with respect to x. A particular example is that when the brightness x of a light source is increased, the eye reacts by decreasing the area R of the pupil. The experimental formula has been used to model the dependence of R on x, when R is measured in mm 2 and x measured in brightness, find the sensitivity. (page 212, #30)

Another application of derivatives is rectilinear motion . Rectilinear Motion is classified as horizontal motion or vertical motion (free fall) For Horizontal Motion: We let x = f(t) be the distance function. Now, velocity is v = dx/dt and acceleration is a = dv/dt = d 2 x/dt 2 . The rate of change in acceleration is called jerk and this is j = da/dt = d 2 v/dt 2 = d 3 x/dt 3 . In some cases, s is used instead of x. Example 1: If a ball is given a push so that it has an initial velocity of 5m/s down a certain inclined plane, then the distance it has rolled after t seconds is s = 5t + 3t 2 . Find the velocity after 2 sec. How long does it take for the velocity to reach 35m/s? (page 210, number 6)

Example 3. The motion of a spring that is subject to a frictional force or a damping force is often modeled by the following function: where s is measured in cm and t in seconds. Find the velocity after t seconds. (page 229, #63) Example 2: A particle moves according to a law of motion s = t 3 – 12t 2 + 36t, t  0 where t is measured in seconds and s in meters. Find the velocity at time t. What is the velocity after 3 sec? When is the particle at rest? When is the particle moving forward? Find the total distance travelled during the first 8 sec. Find the acceleration at time t and after 3 sec. (page 210, #1)

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

Exercises: 1. A stone is dropped from a building 256 ft high. a. Write the equation of motion of the stone. b. Find the instantaneous velocity of the stone at 1 sec and 2 sec. c. Find how long it takes the stone to reach the ground. d. What is the speed of the stone when it reaches the ground? 2. A ball is thrown vertically upward from the ground with an initial velocity of 32 ft/sec. Estimate how high the ball will go and how long it takes the ball to reach the highest point? Find the instantaneous velocity of the ball at t = 0.75 sec and t = 1.25 sec. Find speed of the ball at t = 0.75 sec and t = 1.25 sec. Find the speed of the ball when it reaches the ground. 3. In an opera house, the base of a chandelier is 160 ft above the lobby floor. Suppose the phantom of the opera dislocates the chandelier and is able to give the chandelier an initial velocity of 48 ft/sec and causes it to fall and crash on the floor below. a. Write the equation of motion of the chandelier. b. Find the instantaneous velocity of the chandelier at 1 sec. c. Find how long it takes the chandelier to hit the floor. d. What is the speed of the chandelier when it hits the floor?

Rectilinear motion is a motion of a particle or object following a straight line. We will discuss two types of motion: Horizontal and Vertical motion . 1. Horizontal Motion Suppose that a particle moves along a horizontal line, with its location x at a time t given by its position function x = f(t) . Thus, we make the line of motion a coordinate axis with an origin and a positive direction; f(t) is merely the x-coordinate of a moving particle at a time t. x=0 x = f(t)

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.

Examples: The position function of a particle moving in a horizontal line is , where x is in feet and t is in seconds. a. Find its location x when its velocity v is zero. b. Find t, x, and v when a = 0. Solution: Position function: Velocity function: v = Acceleration function: a =

a. When v = 0, . It follows that (2t +1)(t – 2) = 0 , and so t = -1/2 and t = 2, the object is, therefore, at rest during these times.  If t = -0.5 sec, x = 2/3(-1/2) 3 -3/2(-1/2) 2 -2(-1/2) + 4 x = 109/24 ft or 4.54 ft. This means that the object (which is already moving before t = 0, prior to the point of reference) is moving from the left at t = 0. It is 4.54 ft from the right of the point of reference x = 0, when at that moment v = 0 or when the object momentarily stopped.  If t = 2 sec, x = 2/3(2) 3 -3/2(2) 2 -2(2) + 4 x = -2/3 ft or -0.67 ft. This means that the object is 0.67 ft from the letf at 0 when the object momentarily stopped. b. When a = 0, 4t – 3 = 0. So, t = ¾ when the velocity becomes constant. If t = 3/4, then x = 2/3(3/4) 3 – ½(3/4) 2 – 2(3/4) + 4 = 2.6 ft If t = 3/4, then v = 2(3/4) 2 -3(3/4) – 2 = -3.13 ft/sec

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 =

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:

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

Other Examples: (page 154) Suppose a sprinter running in a 100-meter race is s meters from the finish line t seconds after the start of the race where s = 100 – ¼ (t 2 + 33t) Find the sprinter’s speed At the start of the race and When the sprinter crosses the finish line. Solution: The velocity function is v = -1/4 (2t + 33). At the start of the race, t = 0. So, v = -1/4[2(0) + 33] = -8.25 m/sec. Thus, the speed is 8.25 m/sec. b) When the sprinter crosses the finish line, s = 0. So, solve for t when s = 0. We get, 0 = 100 – ¼ (t 2 + 33t), t 2 + 33t – 400 = 0. Solving for t using quadratic formula, t = 9.43. So the velocity is v = -1/4 [2(9.43) + 33]= -12.96m/sec. Thus, the speed is 12.96m/sec.

If a ball is given a push so that it has an initial velocity of 24 ft/sec, down a certain inclined plane, then s = 24t + t 2 , where s feet is the distance of the ball from the starting point at t seconds and the positive direction is down the inclined plane. a) What is the instantaneous velocity of the ball at 2 seconds? b) How long does it take for the velocity to increase to 48 ft/sec? Solution: The velocity function is v = 24 + 2t. At t 1 sec, the instantaneous velocity is v = 24 + 2t 1 . b) If v = 48ft/sec, 48 = 24 + 2t. And so, t = 6/5 sec. Assign # : p. 154 Numbers 27 and 35, to be submitted on Tues.

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.

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

Examples: 1. A ball is thrown vertically upward from the ground with an initial velocity of 32 ft/sec. a) Estimate how high the ball will go and how long it takes the ball to reach the highest point? b) Find the instantaneous velocity of the ball at 1.25 sec. c) Find the speed of the ball when it reaches the ground. Solution: Since v 0 = 32 ft/sec (positive because the motion is upward) and y 0 = 0, the equation of motion is y = -16t 2 + 32t . The velocity function is v = -32t + 3 2.

a) Estimate how high the ball will go and how long it takes the ball to reach the highest point? To determine the highest point where the ball can go is to find the y when v = 0. So, -32t + 32 = 0. Solving for t gives us t = 1 sec . This is the length of time where the ball will reach its highest point. Since t is already known, the value of y when t = 1 sec is y = -16(1) 2 + 32(1) = 16 ft. b) Find the instantaneous velocity of the ball at 1.25 sec. At t = 1.25 sec, the instantaneous velocity is v = -32(1.25) + 32 = -8 ft/sec . This means that the ball is already on its way down.

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 .

A problem in related rates is one involving rates of change of related variables. These variables have specific relationship for values of t , where t is a measure of time. The following steps should be taken in solving problems involving rate of change wrt time t for two or more related variables: Select a letter to represent each variable. Identify the constant rates of change that are given and the rate of change that must be found. Find an equation that expresses the relation between the variables. Differentiate both sides of the equation wrt t (applying implicit diff.). Replace the given rates of change with their constant values and the variables by their values at the particular time of interest. Solve the resulting equation for the unknown rate of change.

A spherical balloon is being inflated so that its volume is increasing at the rate of 5 m 3 /min. At what rate is the diameter increasing when the diameter is 12 m? Solution: Let V – volume, D – diameter. Given : Unknown: when D = 12 m Equation:

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

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

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

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

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?

6. The adiabatic law (no gain or loss of heat) for the expansion of air is PV 1.4 = C , where P is in psi, V in in 3 and C is a constant. At a specific instant, the pressure is 40 psi and is increasing at a rate of 8 psi/sec . If C = 5/16 , what is the rate of change of volume at this instant? Solution: Let P – pressure, and V – volume. Given: P = 40 psi and dP/dt = 8 psi/sec , and C = 5/16 Unknown: dV/dt

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

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