Calculus Lia Vas Te Derivative Te rate of cange Knowing and understanding te concept of derivative will enable you to answer te following questions. Let us consider a quantity wose size is described by a function.. How fast is te quantity canging at a given moment? 2. Is te quantity increasing or decreasing in size? 3. Wen will te size be maximal and wen will te size be minimal? 4. If te quantity is increasing, is te rate of te increase increasing or decreasing itself? 5. If anoter factor impacts te size of te quantity, ow does it rate of cange impacts te speed of te cange of te initial quantity? Rates of Cange. To introduce te concept of derivative, let us recall te definition of te average rate of cange of a function on an interval. Te average rate of cange of f(x) over te interval a x b is rise run f(b) f(a) Besides finding te rate of cange over an interval, it may be relevant to find te rate of cange at a specific point. Tis rate, called te instantaneous rate of cange or derivative of f at a can be computed from te above formula f(b) f(a) wen b approaces a. If we denote te difference b a by ten b a+ and te condition b a is equivalent to 0, te formula f(b) f(a) becomes f(a+) f(a). Hence, te can be computed as Te instantaneous rate of cange of f(x) at x a is lim 0 f(a+) f(a)
Geometric Interpretation. We ave seen tat te average rate of cange of f(x) from a to b represents te slope of te secant line. In te limiting case wen b a, te secant line becomes te tangent line as te previous figure illustrates. Tus, Te instantaneous rate of cange of f(x) at x a is te slope of te line tangent to te grap of f(x) at x a. f(a+) f(a) Tus, te formula lim 0 computes te slope m of te tangent line. Recall te pointslope equation of a line wit slope m passing point (x 0, y 0 ). y y 0 m(x x 0 ) Tis formula computes te equation of te tangent line to f(x) at x a for x 0 a, y 0 f(a) and m lim 0 f(a+) f(a). Example. Let f(x) x 2 + 4. (a) Find te average rate of cange of f(x) for x 2. Ten find te equation of te secant line of f(x) passing te grap of f(x) at x and x 2. (b) Find te instantaneous rate of cange of f(x) for x. Ten find te equation of te tangent line to f(x) at x. Solution. (a) Te average rate of cange of f(x) for x 2 can be computed as f(2) f() 2 22 + 4 ( 2 + 4) 4 + 4 4 3. Tis also computes te slope of te secant line. Te equation of te secant line can be obtained using te point slope equation wit m 3 and using eiter (, f()) or (2, f(2)) for point (x 0, y 0 ). For example, wit (, f()) (, 5) one gets te equation y 5 3(x ) y 3x + 2. (b) Te instantaneous rate of cange of f(x) at x can be computed as lim 0 f(+) f() ( + ) 2 + 4 ( 2 + 4) lim 0 lim 0 + 2 + 2 + 4 4 2 + 2 lim 0 lim 0 2 + 2. Tis also computes te slope of te tangent line. Te equation of te tangent line can be obtained using te point-slope equation wit m 2 and using (, f()) (, 5) for point (x 0, y 0 ). So, te tangent line is y 5 2(x ) y 2x + 3. An application. If f(x) computes te distance (in distance units) traveled x time units after te object started moving, ten te average rate of cange from x a to x b computes te average velocity between times a and b. Te instantaneous rate of cange at a computes te instantaneous velocity at time x a. 2
f(b) f(a) average velocity from x a to x b is f(a+) f(a) velocity at x a is lim 0 If [x] denotes te units of quantity x (tus [f(x)] denotes te units of f(x)), te units of te answer agree since [f(b) f(a)] [b a] distance units time units and [f(a + ) f(a)] lim 0 [] distance units time units. Example 2. Assume tat te distance traveled by a moving object x second after te object started moving can be computed by f(x) x 2 + 4 feet. (a) Determine te average velocity at wic te object was moving between te first and te second second. (b) Determine te velocity of te object one second after it started moving. Solution. Note tat te function describing te distance corresponds is te same function as in Example. Part (a) is asking for te average rate of cange of f(x) for x 2 wic we computed to be 3. Tus, te average velocity is 3 feet per second. Part (b) us asking for te instantaneous rate of cange of f(x) at x wic we computed to be 2 in part (b) of Example. Tus, te object as (instantaneous) velocity of 2 feet per second second after it started moving. Derivative. Te instantaneous rate of cange of f(x) at x a is te derivative of f(x) at x a. Te notation f (a) is used to denote te derivative of f(x) at x a. Tus, te following concepts are all equivalent. () Te derivative f (a) of f(x) at x a; (2) Te instantaneous rate of cange of f(x) at x a; (3) Te slope of te tangent line to f(x) at x a; (4) f (a) lim 0 f(a+) f(a) Alternative formula. If we denote te canging quantity a + by x so tat x a, te quotient f(a+) f(a) can be written as f(x) f(a). Wen 0, x a so te derivative f (a) can also x a be found as follows. f (a) lim x a f(x) f(a) x a 3
Example 3. Find te derivative of f(x) x 2 + 4 at x. Solution. Recall tat we ave found te instantaneous rate of cange of tis function at x to be 2 in Example. Tus f () 2. Example 4. Find te derivative of f(x) at x 2. x Solution. Te problem is asking you for f f(2+) f(2) (2) lim 0 lim 0 2+ 2 2 (2+) 2(2+) lim 0 lim 0 2(2 + ) lim 0 2(2 + ) 4. Example 5. Find te derivative of f(x) x 2 + 4 at x a. Solution. Te problem is similar to Examples (b) and 3 except tat te x value is a instead of. f (a) lim 0 f(a+) f(a) (a + ) 2 + 4 (a 2 + 4) lim 0 lim 0 a 2 + 2a + 2 + 4 a 2 4 lim 0 2a + 2 lim 0 2a + 2a. Derivative as a function. Te previous example illustrates tat derivative at x a can be considered as a function of a. By using more familiar x instead of a to denote te independent variable, we obtain tat te derivative f (x) can be considered to be a function of x since at every value of x it computes te slope of te tangent at te point (x, f(x)). f (x) lim 0 f(x+) f(x) Example 6. Find derivative of te line f(x) mx + b. Solution. f f(x+) f(x) m(x+)+b (mx+b) mx+m+b mx b (x) lim 0 lim 0 lim 0 m lim 0 m. Tus, te derivative of mx + b is m. Alternatively, you can simply argue tat, since te tangent line to a line is te same line, te slope of te tangent line is m at every point x. Example 7. Find derivative of f(x) x 2. Solution. f (x) lim 0 f(x+) f(x) lim 0 (x+) 2 x 2 lim 0 x 2 +2x+ 2 x 2 lim 0 2x+ 2 lim 0 2x + 2x. Tus, te derivative of x 2 is 2x. Notation for derivative. If te function f(x) is denoted by y, sometimes y is used to denote f (x). Tere are oter notations for derivative besides f (x) and y. Note tat te quotient f(x+) f(x) measures te quotient of te cange in y over cange of x. Tese two canges are denoted by y and x and te limit wen x 0 is denoted by dy. Tis notation is known as te Leibniz notation. In tis notation, te formula computing te derivative can be written as follows. dy lim y x 0 x In some cases, te notations d df d f(x) or are also used. In tis case, te term as a function tat maps y f(x) to its derivative y f (x). In tis case, te term d differentiation operator. Keep in mind tat te notations d. To summarize, te following denote te derivative of y f(x). and dy 4 is considered is refer to as df f(x) or are equivalent to f (x), y
f (x), y, dy, df, d f(x) If te derivative y of a function y f(x) is evaluated at x a, te following notation is also used besides f (a) dy xa Te sign of derivative. Recall tat te derivative at a point is te slope of te tangent line at tat point. Te slope of te tangent reflect te fact if te function is increasing, decreasing or constant. Tus we ave te following. f(x) increasing around x a f (a) positive f(x) decreasing around x a f (a) negative f(x) constant around x a f (a) equal to zero Example 8. Recall tat we determined te derivative of f(x) x 2 to be f (x) 2x. From te grap of x 2 we see tat it is an increasing function for x positive and decreasing function for x negative. Tis fact is also reflected in te sign of te derivative f (x) 2x : for x > 0, 2x is positive and for x < 0, 2x is negative. Units of te derivative. If [x] denotes te units of quantity x and [f(x)] denotes te units of f(x), te units of derivative are determined as follows. Units of derivative f (x) [f (x)] lim 0 [f(x + ) f(x)] [] units of f(x) units of x [f(x)]. [x] In Example 2 we ave seen tat te te velocity is in feet per second if te distance is in feet and time is in seconds. Determining rates of cange from a table or a grap. If a function is given by a table, te formula f(b) f(a) can be used to compute te average rate of cange. Te instantaneous rate of cange cannot be computed exactly but can be approximated by te average rate of cange of consecutive points, or wit te average of te two average rates of canges at consecutive points as follows. f (x n ) average of f(xn) f(x n ) x n x n and f(x n+) f(x n) x n+ x n 5
Example 9. Te temperature in degrees Fareneit is being measured several times during te day and te data is collected in te table below. time 6 am 9 am noon 3 pm 6 pm 9 pm temperature 68 72 75 74 73 70 (a) Find te average rate at wic te temperature canged from 6 am to noon and from noon to 6 pm. (b) Estimate te instantaneous rate of cange at 9 am. Solution. For convenience, let us represent 6 am wit 0 ours and let x denote te number of ours after 6 am. (a) 6 am and noon correspond to x 0 and x 6. Te rate is 75 68.67 degrees Fareneit 6 0 per our. 6 pm and noon correspond to x 2 and x 6. Te rate is 73 75 0.333. So, te 6 0 temperature decreased by 0.33 degrees Fareneit per our on average from noon to 6 pm. Te negative sign represents te fact tat te temperature decreased on average from noon to 6 pm. (b) Compute te average rates of cange from 9 am to noon to be 75 72 and from noon to 3 6 3 pm to be 74 75 + 3 0.333. Ten average te two rates and get 0.333 degrees per 9 6 3 2 3 our. Tus, at noon te temperature is increasing by about 0.33 degrees Fareneit per our. If a function is given by a grap, te slope of te secant line computes te average rate of cange and te slope of te tangent line computes te instantaneous rate of cange. Example 0. Te grap below represents te distance traveled by te object in miles as a function of time in ours. Use te grap for te following. (a) Estimate te initial velocity (velocity at point A). (b) Compare te velocities at points A and B. (c) Estimate te velocity at point E. (d) Find te average velocity between points A and E. (e) Explain wat appens to te object between points C and D. Solution. (a) Te tangent at (0,0) is a line tat seem to be passing te point (0.5,3) so its slope can be computed as 3 0 6. Tus, te object as te initial velocity of 6 miles per our. 0.5 0 (b) At point B te slope of te tangent line is less steep tan at point A. So, te velocity at B is smaller tan at point A. (c) Te tangent at (5,3) is a line tat seem to be passing te point (5.5,3.5) (or point (4.5, 2.5)) so its slope can be computed as 5.5 5 0.5. Tus, te object as velocity of mile per our at 3.5 3 0.5 point E. (d) Te point A is (0,0) and te point E is (5,3). Te average velocity is 5 0.67 miles per 3 0 our. 6
(e) Te distance is constant so te velocity is zero. You can also argue tat te velocity is zero since te slope of te tangent line remains zero between points C and D. Practice problems.. Find te average rate of cange of te following functions over te given interval. (a) y x+2, [0, 2] (b) y x + 3, [, 5] 2. Find te instantaneous rate of cange of te following functions at te given point. (a) f(x) x 2 3x, x 2 (b) f(x) x+2, x 3. Find an equation of te tangent line to te curve at te given point. (a) f(x) x 2 3x, x 2 (b) f(x) 5x x 2 3, x 2 4. Find te derivative of te following functions at a given point using te definition of derivative at a point. (a) f(x) x 2 3x, x 2 (b) f(x) x+2, x 5. Find te derivative of te following functions using te definition of derivative. (a) f(x) x 2 3x (b) f(x) 5x 3 (c) f(x) x 6. If we approximate te gravitational acceleration g by 9.8 meters per seconds squared, te distance from te initial eigt of an object dropped from it to te ground can be described as s(t) g 2 t2 4.9t 2. (a) Find te average velocity of te object in te first tree seconds. (b) Find te velocity of te object tree seconds into te fall. (c) Find te formula computing te velocity at any point t. (d) If te eigt is 300 meters, find te velocity at te time of te impact wit te ground. You may use part (c). 7. Te distance in miles a traveling car passes after it started moving is represented in te following table as a function of time in ours. time (ours) 0 0.5.5 2 2.5 distance (miles) 0 30 52 52 76 04 (a) Find te average velocity of te veicle between te first and te second our. (b) Estimate te velocity two ours after te veicle started moving. (c) Based on te information given, estimate te initial velocity of te veicle. (d) Based on te information given, wat can you say about te movement of te veicle between te first our and te first our and a alf? 8. Use te given grap to approximate te following. 7
(a) Compare te values of derivative at points x.5 and x 2.5. (b) Arrange te following in increasing order: f( 2) f( ) f(0), f() and f(.5). (c) Arrange te following in increasing order: f ( 2), f ( ), f (0), f () and f (.5). (d) Estimate te following: f ( 2) f (0) and f (). Solutions.. (a) f(0) 2 and f(2) 4 so te average rate is f(2) f(0) 2 0 4 2 2 8. (b) f() 4 2 and f(5) 8 so te average rate is f(5) f() 5 2. (a) f f(2+) f(2) (2+) (2) lim 0 lim 2 3(2+) (2 2 3(2)) 0 + lim 2 0 lim 0 +. (b) f f(+) f() () lim 0 lim 0 lim 0 lim 3(3+) 0. 3(3+) 9 ++2 +2 lim 0 +3 3 8 2 4 0.207. lim 0 4+4+ 2 6 3 4+6 lim 0 3 (+3) 3(+3) 3. (a) From part (a) of te previous problem we ave tat f (2) so te slope is. Since f(2) as well, te function passes te point (2,). Tus te tangent line is y (x 2) y x. (b) Calculate te slope to be f f(2+) f(2) 5(2+) (2+) (2) lim 0 lim 2 3 (5(2) 2 2 3) 0 0+5 4 4 lim 2 3 0+4+3 0 lim 2 0 lim 0. Since f(2) 0 4 3 3, te function passes (2,7). Tus te tangent line is y 7 (x 2) y x + 5. 4. (a) As in 2 (a), f (2). (b) As in 2 (b) f () 9. 5. (a) f (x) lim 0 f(x+) f(x) lim 0 2x 3+ 2 lim 0 2x 3 + 2x 3. (b) f (x) lim 0 f(x+) f(x) lim 0 (x+) 2 3(x+) (x 2 3x) lim 0 x 2 +2x+ 2 3x 3 x 2 +3x lim 0 5(x+) 3 (5x 3) lim 0 5x+5 3 5x+3 lim 0 5 5. Alternatively, you can simply say tat, since te tangent line to a line is te initial line, te slope of te tangent line is 5 at every point x. (c) f f(x+) f(x) (x) lim 0 lim 0 lim 0. x(x+) x 2 x+ x lim 0 x (x+) x(x+) lim 0 x(x+) 6. (a) Te average velocity in te first tree seconds is s(3) s(0) 3 4.9(9) 3 4.7 meters per second. (b) v(3) s 4.9(3+) (3) lim 2 4.9(3) 2 4.9[9+6+ 0 lim 2 9] 4.9[9+6+ 0 lim 2 9] 0 lim 0 4.9(6 + ) 4.9(6) 29.4 meters per second. 8
(c) Te velocity at time t can be found as v(t) s (t) lim 0 4.9(t+) 2 4.9t 2 4.9[t lim 2 +2t+ 2 t 2 ] 4.9[2t+ 0 lim 2 ] 0 lim 0 4.9(2t + ) 9.8t meters per second. (d) If te eigt is 300 meters, ten te object its te ground wen s(t) 300 4.9t 2 300 t 2 6.22 t ±7.82. Since we are looking for positive solution, we conclude tat te object its te ground approximately 7.82 seconds after it is dropped. Plugging t 7.82 in te formula tat computes te velocity v(t) 9.8t found in part (c), we obtain tat te velocity at te time of te impact is v(7.82) 76.68 meters per second. 7. Let s(t) denote te distance as a function of time. (a) Te average velocity of te veicle between t and t 2 is s(2) s() 2 per our. 76 52 2 24 miles (b) Te velocity two ours after te veicle started moving can be estimated to be te average of s(2) s(.5) 76 52 48 and s(2.5) s(2) 04 76 56. Tis average is 48+56 52 miles per 2.5 0.5 2.5 2 0.5 2 our. (c) Based on te information given, te initial velocity can be approximated by te average velocity in te first alf our wic is s(0.5) s(0) 30 60 miles per our. 0.5 0 0.5 (d) Based on te information given, te veicle seem to be not moving between te first our and te first our and a alf. 8. (a) Te derivative is negative bot at x.5 and x 2.5. Te downwards slope at 2.5 is steeper tan at.5 so te derivative at 2.5 is a larger negative number tan at.5. (b) Te function is maximal at f() so tis is te largest value. f(.5) is te next largest value, te values f( 2) and f(0) are bot equal to zero. Te value f( ) is negative and te smallest of te five given. (c) f (0) is te largest since tis is te only positive value of te five. f ( ) f () 0 since te slopes of tangent lines bot at x and x are zero. Te downwards slope at -2 is steeper tan at.5 so f (.5) is less negative tan f ( 2). (d) Te tangent line at -2 passes (-2,0) and, approximately (-2.5, 2) (or (-.5, -2)). Tus te 2 0 slope is 2 4. Tus f ( 2) 4. Te tangent line at 0 passes (0,0) and, 2.5 ( 2) 0.5 approximately (0.5, 2) (or (-0.5, -2)). Tus te slope is 2 0 4. Tus f (0) 4. In part (c) 0.5 0 we discussed ow f () 0. 9