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2Derivative as a Rate of ChangeIf y = f(x) and if x changes from the value x1 to x2, then y changes fromf(x1) to f(x2). So, the change in y, which we denote by ∆y, is f(x2) - f(x1)when the change in x is ∆x = x2 – x1. The average rate of change of y withrespect to x, over the interval [x1, x2] , is thenThis can also be interpreted as the slope of the secant line.2 12 1( ) ( )f x f x yx x xΔΔ−=−To illustrate, average velocity=where the interval is from t = a to t = a + h.hafhaftimentdisplaceme )()( −+=
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xy Tangent lineSecant liney = f(x)PQf(x2)-f(x1)x2 – x1Average rate of change is the slope of the secant line.Instantaneous rate of change is the slope of the tangent line at P.
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Definition of Instantaneous Rate of ChangeInstantaneous Rate of Change=f’(x1) is the instantaneous rate of change at x1.Note that a positive rate means a quantity increases with respect to theother quantity, that is y increases with x. If it is negative, then thequantity decreases with respect to other quantity.Note that heat, velocity, density, current, temperature, pressure, molarconcentration, fluid flow, bacterial growth, reaction rate, blood flowand cost are just some of the few quantities that maybe analyzedthrough derivatives.We consider the average rate of change over a smaller and smallerintervals by letting x2 approaches x1 and therefore letting ∆x approach 0.The limit of this average rate of change is called the (instantaneous) rateof change of y with respect to x at x = x1, which is interpreted as theslope of the tangent line to the curve y = f(x) at x = x1.2 12 102 1( ) ( )x x xf x f x yLim Limx x xΔΔΔ→ →−=−
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Given a set of data, we may approximate instantaneous rate of change invalues using average values or the graph representing the set of data.For example:Page 147 number 24The population P (in thousands) of Canada from 1994 to 2002 is shown inthe table. (Midyear estimates are given).a) Find the average rate of growthi. From 1996 to 2000ii. From 1998 to 2000iii. From 2000 to 2002b) Estimate the instantaneous rate of growth in 2000 by taking theaverage of two average rates of change.Year 1994 1996 1998 2000 2002P 29036 29672 30248 30791 31414
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p.147 #28If a cylindrical tank holds 100,000 liters of water, which can be drainedfrom the bottom of the tank in an hour, then Torricelli’s law gives thevolume V of water remaining in the tank after t minutes asa)Find the rate of change at which the water is flowing out of the tank (theinstantaneous rate of change of V with respect to t) as a function of t.b)For times t = 0, 20, 40, and 60,find the flow rate and the amount ofwater remaining in the tank.2601000,100)( −=ttV
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1. Using Ohm’s Law where V volts is the electromotive force, R ohms isthe resistance and I amperes is the current in an electric circuit, find therate of change in I with respect to R and find the instantaneous rate ofchange of I with respect to R in an electric circuit having 120 volts whenthe resistance is 20 ohms. (Take that V is constant)Other ExamplesSolution:Ohm’s Law states that V = IR. Thus I = V/R = VR-1. So, I= 120R-1.We have 22R120R120dRdI −=−= −And so when R = 20 ohms, then = -0.30 ampere/ohmThe negative signs implies that current is decreasing at this conditions.2)20(120dRdI −=
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2. A solid consists of a right circular cylinder and a hemisphere on eachend, and the length of the cylinder is twice its radius. Let r units be theradius of the cylinder and the two hemispheres, and V(r) cubic units bethe 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, thevolume of the cylinder is V1 = πr2h = πr2(2r) = 2πr3.Since two hemispheres are equal to a sphere, the volume isV2 = 4/3 πr3.So the volume of the solid is V(r) = 2πr3+ 4/3 πr3= 10/3 πr3. Theinstantaneous rate of change isV ’(r) = 10πr2.Note that the rate of change of the volume with respect to r can beobtained when r is given.
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3.Sand is being dropped onto a conical pile in such a way that the height ofthe pile is always twice the base radius. Find the rate of change of thevolume 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, thevolume of the cone isV = 1/3πr2hV = 1/3 πr2(2r)So,V(r) = 2/3 πr3.The rate of change of the volume with respect to r isV ’(r) = 2 πr2.When a) h = 4m, r = 2 then v’(2) = 2 π (2)2= 8π m2.b) h = 8m, r = 4, then v’(4) = 2 π (4)2= 32π m2.
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4. If water is being drained from a swimming pool and V(t) liters is thevolume of water in the pool t minutes after draining starts, where V(t)= 250(1600 – 80t + t2). Finda. The average rate at which the water leaves the pool during the first 5minutes.b. How fast the water is flowing out of the pool 5 min after the drainingstarts?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 isV ’(5) = 250[-80 + 2(5)] = -17,500 liters/min.a) We find V(0) = 250(1600) = 400,000 and V(5) = 250(1600 – 80(5) + 52)= 306,250. The average rate at which the water leaves the pool duringthe first 5 minutes ismin/liters187505000,400250,30605)0(V)5(VtV−=−=−−=∆∆
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11In manufacturing companies, the costs of producing their products is amajor concern. The following terms are useful in dealing with problemsinvolving costs.Total Cost Function, C(x) – expression giving the total amount needed toproduce x units of a certain product.Marginal Cost Function, C’(x) – Rate of change in cost when x units ofproduct is produced.In a similar sense, we may consider the revenue of the company. Thus, wehaveTotal Revenue Function, R(x) – expression giving the total amount earnedin the sales of x units of a certain product.Marginal Revenue Function, R’(x) – Rate of change in revenue when xunits of product is sold.
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Solution:a. Since C(x) = 1500 + 3x + x2then C’(x) = 3 + 2xb. When x = 40, C’(x) = 3 + 2(40) = 83 dollars/watch.5. The number of dollars in the total cost of manufacturing x watches in acertain plant is given by C(x) = 1500 + 3x + x2. Find (a) the marginalcost function and (b) the marginal cost when x = 40.
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Other Problems:1. A wave produced by a simple sound has the equation P(t) = 0.003 sin1800πt where P(t) dynes/cm2is the difference between theatmospheric pressure and the air pressure at the eardrum at tseconds. Find the instantaneous rate of change of P(t) with respect to tat (a) 1/9 sec and (b) 1/8 sec.2. A Cepheid variable star is a star whose brightness alternately increasesand decreases. The most easily visible such star is Delta Cephei, forwhich 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. Itsbrightness B is modeled bywhere t is measured in days. Find the rate of change of the brightnessafter 1day. (page 229, #61)2( ) 4.0 0.35sin5.4tB tπ = + ÷
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3. A company estimates that in t years, the number of its employees will beN(t) = 1000 (0.8)t/2(a) How many employees do the company expect tohave in 4 years. (b) At what rate is the number of employees expected tobe changing in 4 years?4. Consider a blood vessel with radius 0.01 cm, length 3 cm, pressuredifference 3000 dyne/cm2and viscosity η = 0.027. Using the law oflaminar flow:(a) Find the velocity of the blood along the center line, at r = 0.005 cm andat the wall. (b) Find the velocity gradient (instantaneous rate of change ofv with respect to r) at r = 0, r = 0.005 and r = 0.01 cm. (page 212, #25)( )2 24Pv R rlη= −5. If R denotes the reaction of the body to some stimulus of strength x, thesensitivity S is defined to be the rate of change of reaction with respect tox. A particular example is that when the brightness x of a light source isincreased, the eye reacts by decreasing the area R of the pupil. Theexperimental formulahas been used to model the dependence of R on x, when R is measured inmm2and x measured in brightness, find the sensitivity. (page 212, #30)0.40.440 241 4xRx+=+
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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 = d2x/dt2.The rate of change in acceleration is called jerk and this isj = da/dt = d2v/dt2= d3x/dt3.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 acertain inclined plane, then the distance it has rolled after t seconds iss = 5t + 3t2.(a) Find the velocity after 2 sec.(b) How long does it take for the velocity to reach 35m/s? (page 210,number 6)
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Example 3.The motion of a spring that is subject to a frictional force or a dampingforce is often modeled by the following function:where s is measured in cm and t in seconds. Find the velocity after tseconds. (page 229, #63)1.5( ) 2 sin2ts t e tπ−=Example 2:A particle moves according to a law of motion s = t3– 12t2+ 36t, t ≥ 0where t is measured in seconds and s in meters.(a) Find the velocity at time t.(b) What is the velocity after 3 sec?(c) When is the particle at rest?(d) When is the particle moving forward?(e) Find the total distance travelled during the first 8 sec.(f) Find the acceleration at time t and after 3 sec.(page 210, #1)
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17Vertical MotionHere, 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 y0 (ft) above the ground attime t = 0 (sec) and with initial velocity v0 (ft/sec) and if air resistance is negligible, thenits height y = f(t) (in feet above the ground) at time t is given by a formula known fromPhysics,y = f(t) = ½ gt2+ v0t + y0.where g denotes the acceleration due to the force of gravity. At the surface of the earth,g ≈ -32 ft/s2(or -9.8 m/s2). Thus, y= f(t) = -16t2+ v0t + y0. (or y= f(t) = -4.9t2+ v0t + y0)y is increasing y is decreasingv > 0, decreasing v < 0, increasingy=f(t)Ground levely = 0
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18Exercises: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.a. Estimate how high the ball will go and how long it takes the ball to reach thehighest point?b. Find the instantaneous velocity of the ball at t = 0.75 sec and t = 1.25 sec.c. Find speed of the ball at t = 0.75 sec and t = 1.25 sec.d. 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 thephantom of the opera dislocates the chandelier and is able to give the chandelier an initialvelocity 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?
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