Differential Calculus I - - : Fundamentals

Differential Calculus I - - : Fundamentals Assessment statements 6. Informal ideas of limits and convergence. Limit notation. Definition of derivative from first principles f9() 5 lim f ( h) f () h ( h ). Derivative interpreted as gradient function and as rate of change. Tangents and normals, and their equations. 6. Derivative of n (n 핈). The second derivative. 6.3 Local maimum and minimum points; testing for maimum or minimum. Points of infleion with zero and non-zero gradients. Graphical behaviour of functions including the relationship between f, f9 and f99. 6.6 Kinematic problems involving displacement s, velocit v, and acceleration a. Introduction Figure. Distance (metres) 5 4 3 Calculus is the branch of mathematics that was developed to analze and model change such as velocit and acceleration. We can also appl it to stud change in the contet of slope, area, volume and a wide range of other real-life phenomena. Although mathematical techniques that ou have studied previousl deal with man of these concepts, the abilit to model change was restricted. For eample, consider the curve in Figure.. This shows the motion of an object b indicating the distance ( metres) travelled after a certain amount of time (t seconds). Precalculus mathematics will onl allow us to compute the average velocit between two different times (Figure.). With calculus specificall, techniques of differential calculus we will be able to find the velocit of the object at a particular instant, known as its instantaneous velocit (Figure.3). The starting point for our stud of calculus is the idea of a limit. 3 4 5 6 7 8 9 t Time (seconds) 368

5 5 4 (, 4.5) 4 (7, 3.875) Distance (metres) 3 (4, ) Average velocit from t 4 to t seconds 4.5.5 m 4.375 m/s 6 s Distance (metres) 3 Instantaneous velocit at t 7 seconds.5 m/s 3 4 5 6 7 8 9 Time (seconds) t 3 4 5 6 7 8 9 t Time (seconds) Figure. Figure.3. Limits of functions A limit is one of the ideas that distinguish calculus from algebra, geometr and trigonometr. The notion of a limit is a fundamental concept of calculus. Limits are not new to us. We often use the idea of a limit in man non-mathematical situations. Mathematicall speaking, we have encountered limits on at least two occasions previousl in this book finding the sum of an infinite geometric series (Section 3.4) and computing the irrational number e (Section 4.3). Recall from Section 3.4 that we established that if the sequence of partial sums for an infinite series converges to a finite number L we sa that the infinite series has a sum of L. Further on in that section, we used limits to algebraicall confirm that the infinite series has a 4 8 sum of 4. As part of the algebra for this, we reasoned that as the value of n increases in the positive direction without bound (i.e. n `) the epression ( _ ) n converges to zero in other words, the limit of ( _ ) n as n goes to positive infinit is zero. We epress this result more efficientl using limit notation, as we did in Chapter 3, b writing lim _ n ` ( ) n 5. It is beond the requirements of this course to establish a precise formal definition of a limit, but a closer look at justifing this limit and a couple of others can lead us to a useful informal definition. Eample Evaluate lim _ n ` ( ) n b using our GDC to analze the behaviour of the function f () 5 ( _ ) for large positive values of. 369

Differential Calculus Ī - : Fundamentals Plot Plot Plot3 Y=(/)ˆX Y= Y3= Y4= Y5= Y6= Y7= WINDOW Xmin=- Xma=8 Xscl= Ymin=-. Yma= Yscl= Xres= TABLE SETUP TblStart= Tbl= Indpnt: Auto Ask Depend: Auto Ask 3 4 5 6 X X= Y.5.5.5.65.35.563 X 7 8 9 3 Y.78.39.95 9.8E-4 4.9E-4.4E-4.E-4 Y=.733E-4 The GDC screen images show the graph and table of values for 5 ( ). Clearl, the larger the value of, the closer that gets to zero. Although there is no value of that will produce a value of equal to zero, we can get as close to zero as we wish. For eample, if we wish to produce a value of within. of zero, then we could choose 5 and 5 ( ) 5 4. 976 56; and if we want a result within. of zero, then we could choose 5 4 and 5 ( ) 4 5. 59 65; and 67 77 6 so on. Therefore, we can conclude that lim _ n ` ( ) n 5. The line 5 c is a horizontal asmptote of the graph of a function 5 f () if either lim f () 5 c or lim f () 5 c. ` ` For eample, the line 5 (-ais) is a horizontal asmptote of the graph of 5 ( ) because lim n `( ) n 5. In calculus we are interested in limits of functions of real numbers. Although man of the limits of functions that we will encounter can onl be approached and not actuall reached (as in Eample ), this is not alwas the case. For eample, if asked to evaluate the limit of the function f () 5 as approaches 6, we simpl need to evaluate the function for 5 6. Since f (6) 5, then lim n 6 ( ) 5. However, it is more common that we are unable to evaluate the limit of f () as approaches some number c because f (c) does not eist. Hint: must be in radians because in calculus we are interested in functions of real numbers. Eample Evaluate lim sin. We are not able to evaluate this limit b direct substitution because when 5, sin 5 and is therefore undefined. Let s use our GDC again to analze the behaviour of the function f () 5 sin as approaches zero from the right side and the left side. 37

Although there is no point on the graph of 5 sin corresponding to 5, it is clear from the graph that as approaches zero (from either direction) the value of sin converges to one. We can get the value of sin arbitraril close to depending on our choice of. If we want sin to be within. of, we choose 5.5 giving sin.5.999 583 and.5.999 583 5. 47,.; and if we want sin to be within. of, then we choose 5. giving sin..999 999 3333. and.999 999 3333 5. 6667,. ; and so on. 4π π.5 sin π 4π Therefore, lim sin 5. Functions do not necessaril converge to a finite value at ever point it s possible for a limit not to eist. Eample 3 Find lim, if it eists. As approaches zero, the value of becomes increasingl large in the positive direction. The graph of the function (left) seems to indicate that we can make the values of 5 arbitraril large b choosing close enough to zero. Therefore, the values of 5 do not approach a finite number, so lim does not eist. Although we can describe the behaviour of the function 5 b writing lim 5 `, this does not mean that we consider ` to represent a number it does not. This notation is simpl a convenient wa to indicate in what manner the limit does not eist. Limit of a function If f () becomes arbitraril close to a unique finite number L as approaches c from either side, then the limit of f () as approaches c is L. The notation for indicating this is lim f () 5 L. c When a function f () becomes arbitraril close to a finite number L, we sa that f () converges to L. The line 5 c is a vertical asmptote of the graph of a function 5 f () if either lim f () 5 ` or lim f () 5 `. c c For eample, the line 5 ( -ais) is a vertical asmptote of the graph of 5 because lim 5 `. For our purposes in this course, it is also important to be able to appl some basic algebraic manipulation in order to evaluate the limits of some functions algebraicall, rather than b conjecturing from a graph or table. 37

Differential Calculus Ī - : Fundamentals Eample 4 Evaluate each limit algebraicall. a) lim 5 3 ` b) lim p (3 4p p ) c) lim [( h) 6] ( 6) h h a) lim 5 3 ` 5 lim `( 5 3 ) Split the fraction into two terms and 5 lim 3_ ` ` evaluate the limit of each term separatel. 5 5 5 5 Therefore, lim 5 3 5 5. b) lim p (3 4p p ) 5 lim 3 lim p p 5 3 5 3 c) lim [( h) 6] ( 6) h h 5 lim h 5 lim h ` 4p lim p p h h 6 6 h h h h 5 lim h( h) h h 5 lim lim h h h 5 5 Therefore, lim [( h) 6] ( 6) 5. h h The limits in parts b) and c) of Eample 4 show that in some cases the limit of a function is itself a function. Eercise. In questions 4, evaluate each limit algebraicall and then confirm our result b means of a table or graph on our GDC. lim 4n n n ` lim (3 h h ) h 3 lim ( d ) d d 4 lim 9 3 3 37

In questions 5 7, investigate the limit of the epression (if it eists) as ` b evaluating the epression for the following values of :, 5,,, and. Hence, make a conjecture for the value of each limit. 5 lim 3 ` 3 6 lim 5 6 ` 5 7 lim 3 ` 3 8 Use the graphing or table capabilities of our GDC to investigate the values of the epression ( c ) c as c increases without bound (i.e. c `). Eplain the significance of the result. 9 If it is known that the line 5 3 is a horizontal asmptote for the function f (), state the value of each of the following two limits: lim f () and lim ` ` f (). If it is known that the line 5 a is a vertical asmptote for the function g() and g()., what conclusion can be made about lim a g ()? State the equations of all horizontal and vertical asmptotes for the following functions. a) f () 5 3 b) g () 5 ( ) c) g () 5 a b. The derivative of a function: definition and basic rules Tangent lines and the slope (gradient) of a curve In Section.6, we reviewed linear equations in two variables. And, later in Section., we established that an non-vertical line represents a function for which we tpicall assign the variables and for values in the domain and range of the function, respectivel. An linear function can be written in the form 5 m c. This is the slope-intercept form for a linear equation, where m is the slope (or gradient) of the graph and c is the -coordinate of the point at which the graph intersects the -ais (i.e. the -intercept). The value of the slope m, defined as m 5 5 vertical change, will be horizontal change the same for an pair of points, (, ) and (, ), on the line. An essential characteristic of the graph of a linear function is that it has a constant slope. This is not true for the graphs of non-linear functions. Consider a person walking up the side of a pitched roof as shown in Figure.4. At an point along the line segment PQ the person is eperiencing a slope of _ 3 4. Now consider someone walking up the curve shown in Figure.5, which passes through the three points A, B and C. As the person walks 373

Differential Calculus Ī - : Fundamentals 4 m 3 m along the curve from A to C, he/she will eperience a steadil increasing slope. The slope is continuall changing from one point to the net along the curve. Therefore, it is incorrect to sa that a non-linear function, whose graph is a curve, has a slope it has infinitel man slopes. We need a means to determine the slope of a non-linear function at a specific point on its graph. Figure.4 Figure.5 The slope (gradient) of a curve at a point is the slope of the line that is tangent to the curve at that point. Hint: The word curve can often mean the same as function, even if the function is linear. Imagine if the slope of the curve in Figure.5 stopped increasing (remained constant) C after point B. From that point on, a person walking up the curve would move along a line with a slope equal to the slope D B of the curve at point B. This A line containing point D in the diagram onl touches the curve once at B. Line (BD) is tangent to the curve at point B. Therefore, finding the slope of the line that is tangent to a curve at a certain point will give us the slope of the curve at that point. Finding the slope of a curve at a point or better finding a rule (function) that gives us the slope at an point on the curve is ver useful information in man applications. The slope of a line, or of a curve at a point, is a measure of how fast variable is changing as variable changes. The slope represents the rate of change of with respect to. To find the slope of a tangent line, we first need to clarif what it means to sa that a line is tangent to a curve at a point. Then we can establish a method to find the tangent line at a point. The three graphs in Figure.6 show different configurations of tangent lines. A tangent line ma cross or intersect the graph at one or more points. Figure.6 Figure.7 4 3 4 3 3 4 374 For man functions, the graph has a tangent at ever point. Informall, a function is said to be smooth if it has this propert. An linear function is certainl smooth, since the tangent at each point coincides with the original graph. However, some graphs are not smooth at ever point. Consider the point (, ) on the graph of the function 5 u u (Figure.7). Zooming in on (, ) will alwas produce a V-shape rather than smoothing out to appear more and more linear. Therefore, there is no tangent to the graph at this point.

Distance (metres) Figure.8 6 5 4 3 4 3 4 5 6 7 8 9 t Time (seconds) One wa to find the tangent line of a graph at a particular point is to make a visual estimate. Figure.8 reproduces the time-distance graph for an object s motion from the previous section (Figure.). The slope at an point (t, ) on the curve will give us the rate of change of the distance with respect to time t, in other words the object s instantaneous velocit at time t. In the figure, an estimate of the line tangent to the curve at (5, 3) has been drawn. Reading from the graph, the slope appears to be 4_ 6 5 _ 3. Or, in other words, the object has a velocit of approimatel.667 m/s at the instant when t 5 5 seconds. A more precise method of finding tangent lines makes use of a secant line and a limit process. Suppose that f is an smooth function, so the tangent to its graph eists at all points. A secant line (or chord) is drawn through the point for which we are tring to find a tangent to f and a second point on the graph of f, as shown in Figure.9a. If P is the point of tangenc with coordinates (, f ()), choose a point Q to be horizontall some h units awa. Hence, the coordinates of point Q are ( h, f ( h)). Then the slope of f ( h) f () the secant line (PQ) is m sec 5 5 f ( h) f (). ( h) h The right side of this equation is often referred to as a difference quotient. The numerator is the change in, and the denominator h is the change in. The limit process of achieving better and better approimations for the slope of the tangent at P consists of finding the slope of the secant (PQ) as Q moves ever closer to P, as shown in the graphs in Figure.9b and Figure.9c. In doing so, the value of h will approach zero. f() f() Q( h, f( h)) P(, f()) h f( h) f() Q( h, f( h)) P(, f()) h f( h) f() Figure.9a Figure.9b 375

Differential Calculus Ī - : Fundamentals f() f() The word secant, as applied to a line, comes from the Latin word secare, meaning to cut. The word tangent comes from the Latin verb tangere, meaning to touch. Q( h, f( h)) P(, f()) Figure.9c h f( h) f() P(, f()) Figure.9d B evaluating a limit of the slope of the secant lines as h approaches zero, we can find the eact slope of the tangent line at P(, f ()). The slope (gradient) of a curve at a point The slope of the curve 5 f () at the point (, f ()) is equal to the slope of its tangent line at (, f ()), and is given b m tan 5 lim m sec 5 lim f ( h) f () h h h provided that this limit eists. Let s appl the definition of the slope of a curve at a point to find a rule, or function, for the slope of all of the tangent lines to a curve. Eample 5 Find a rule for the slopes of the tangent lines to the graph of f () 5. Use this rule to find the eact slope of the curve at the point where 5 and at the point where 5. Let (, f ()) represent an point on the graph of f. B definition, the slope of the tangent line at (, f ()) is: m 5 lim f ( h) f () 5 lim [( h) ] [ ] h h h h 5 lim [ h h ] [ ] h h 5 lim h h f() h h 5 4 h( h) 5 lim 3 h h 5 lim ( h) h 5 Therefore, the slope at an point (, f ()) on the graph of f is. 376

At the point where 5, the slope is () 5. This makes visual sense because the point (, ) is the verte of the parabola 5, and we epect that the tangent at this point is a horizontal line with a slope of zero. At the point where 5, the slope is () 5. This also makes visual sense because moving along the curve from (, ) to (, ) the slope is steadil increasing. In Eample 5, from the function f () 5 we used the limit process to derive another function with the rule. With this derived function we can compute the slope (gradient) of the graph of f () at a point from simpl inputting the -coordinate of the point. This derived function is called the derivative of f at. It is given the notation f 9(), which is commonl read as f prime of, or simpl, the derivative of f of. f() Tangent line at (, ) 5 4 3 Tangent line at (, ) The derivative and differentiation The derivative, f 9(), at a point in the domain of f is the slope (gradient) of the graph of f at (, f ()), and is given b f 9() 5 lim f ( h) f () h h provided that this limit eists. If the derivative eists at each point of the domain of f, we sa that f is smooth. The process of finding the derivative, f 9(), is called differentiation. If 5 f (), then f 9() is a formula for the instantaneous rate of change of with respect to. If finding the derivative of a function indicated with the function notation f (), then as shown alread the derivative is usuall denoted as f 9(). However, there are two other notations with which ou should be familiar. Commonl, if a function is given as d in terms of, then the derivative is denoted as 9, read as prime. The notation is also often used to indicate a derivative, and is read as the derivative d of d with respect to. Note: d is not a fraction. If, for eample, 5, the derivative can be denoted b writing d d ( ) 5. This is read as the derivative of with respect to is. Differentiating from first principles Depending on the particular purpose that ou have in differentiating a function, ou can consider the derivative as giving the slope of the graph of the function or the rate of change of the dependent variable (commonl ) with respect to the independent variable (commonl ). Both interpretations are useful and widel applied. Using the limit definition directl to find the derivative of a function (as we did in Eample 5) is often called differentiating from first principles. 377

Differential Calculus Ī - : Fundamentals slope 3 Eample 6 Differentiating from first principles, find the derivative of f () 5 3. f 9() 5 lim f ( h) f () 5 lim ( h) 3 3 h h h h 5 lim ( h)( h) 3 h h 5 lim ( h)( h h ) 3 h h 5 lim 3 3h 3h h 3 3 h h 5 lim h(3 3h h ) h h 5 lim h (3 3h h ) 5 3 Therefore, the derivative of f () 5 3 is f 9() 5 3. 3 (, ) (, ) As in Eample 5, the result for Eample 6 is a function that gives us the slope at an point on the graph of 5 3. For eample, the points (, ) and (, ) both lie on 5 3, and the slopes at these points are respectivel f 9() 5 3() 5 3 and f 9() 5 3() 5 3. Hence, the tangents at these points will be parallel, as shown in Figure.. Figure. f() g() 3 Tangent line at (, ) Tangent line at (, ) Figure. Let s eamine the relationship between the slopes of tangents to the curve f () 5 (Eample 5) and slopes of tangents to g () 5. Recall that we found the derivative of f () to be f 9() 5. It appears from the graphs of f and g, in Figure., that the slopes of tangents at points with the same -coordinate will be equal. For eample, the tangent to g at the point (, ) looks parallel to the tangent to f at (, ), as shown in Figure.. This implies that the derivatives of the two functions are equal. Rather than confirming this conjecture b finding the derivative of g () 5 b first principles (i.e. using the limit definition), let s use the graphical and computing power of our GDC. An GDC model is capable of computing the slope of a curve at a point either on the GDC s home screen, or its graphing screen. The screen images on page 379 show computing derivative values for 5 on the home screen. 378

MATH NUM CPX PRB : Frac : Dec 3:3 4:3 ( X 5: 6:fMin( 7 fma( This command finds the value of the derivative of in terms of, at the point. MATH NUM CPX PRB 4 3 ( X 5: 6:fMin( 7:fMa( 8:nDeriv( 9:fnInt( :Solver nderiv(x,x,) The eact command name and snta for computing the value of a derivative at a point ma var from one GDC model to another. nderiv(x,x,) 6 nderiv(x,x,) nderiv(x,x,-) - 4 nderiv(x nderiv(x,x,7),x,3) 34 6 nderiv(x,x,-9) -8 Eample 7 From first principles, find: a) 9 given 5 3 b) d d given 5 Our GDC results confirm our conjecture that the derivative of g() 5 is g 9() 5. We will appl the definition of the derivative, f 9() 5 lim f ( h) f (), h h in both a) and b). a) 95 lim [3( h) ( h)] (3 ) h h 5 lim (3 6h 3h h) (3 ) h h 5 lim 6h 3h h h h 5 lim (6 3h ) 9 5 6 h b) d d 5 d h h ( h) h 5 lim ( h) h h h ( d ) 5 lim h 5 lim h ( 5 lim ) ( h) h h h ( ( h) h ) 5 lim ( h h ) d ( d ) 5 or d d ( ) 5 379

Differential Calculus Ī - : Fundamentals Basic differentiation rules We have now established the following results: If f () 5, then f 9() 5. If f () 5, then f 9() 5. If f () 5 3, then f 9() 5 6. If f () 5 3, then f 9() 5 3. If f () 5, then f 9() 5. In addition, we know that if f () 5, then f 9() 5, since the line 5 has a constant slope equal to ; and that if f () 5, then f 9() 5 because the line 5 is horizontal and thus has a constant slope equal to. Furthermore, the graph of an function f () 5 c, where c is a constant, is a horizontal line, confirming that if f () 5 c, c [ 핉, then f 9() 5. In other words, the derivative of a constant is zero. This leads to our first basic rule of differentiation. The constant rule The derivative of a constant function is zero. That is, given c is a real number, and if f () 5 c, then f 9() 5. These following results: f () 5 f 9() 5 f () 5 5 f 9() 5 f () 5 5 f 9() 5 f () 5 f 9() 5 f () 5 3 f 9() 5 3 can be summarized in the single statement: if f () 5 n then f 9() 5 n n for n 5,,,, 3 In fact, this statement is true not just for these values but for an value of n that is a rational number (n [ 핈 ). This leads to our second basic rule of differentiation. Functions of the form f () 5 n are called power functions, so the differentiation rule d d ( n ) 5 n n gives the rule for differentiating power functions and is often referred to as the power rule. The derivative of n Given n is a rational number, and if f () 5 n, then f 9() 5 n n. Another basic rule of differentiation is suggested b our result that the derivative of f () 5 is f 9() 5. The derivative of a sum of a number of terms is obtained b differentiating each term separatel i.e. differentiating term-b-term. That is, d d ( ) 5 d d ( ) d d () 5 5. The sum and difference rule If f () 5 g() 6 h() then f 9() 5 g9() 6 h9(). A fourth basic rule of differentiation is illustrated b our result that the derivative of f () 5 3 is f 9() 5 6. Using the sum rule, f 9() 5 d d (3 ) 5 d d (3 ) d () 5 6. The fact that d d d (3 ) 5 6 suggests that 3 d d ( ) 5 3 5 6. In other words, the derivative of a function being multiplied b a constant is equal to the constant multipling the derivative of the function. 38

The constant multiple rule If f () 5 c g() then f 9() 5 c g9(). As mentioned before, and as ou have seen, there are different notations used for indicating a derivative or differentiation. These can be traced back to the fact that calculus was first developed b Isaac Newton (64 77) and Gottfried Leibniz (646 76) independentl of each other and hence, introduced different smbols for methods of calculus. The prime notations 9 and f 9() come from notations that Newton used for derivatives. The d notation is similar to that used b Leibniz for d indicating differentiation. Each has its advantages and disadvantages. For eample, it is often easier to write our four basic rules of differentiation using Leibniz notation as shown below. Constant rule: d (c) 5, c [ 핉 d Power rule: d d ( n ) 5 n n, n [ 핈 Sum and difference rule: d [g() h()] 5 d d d [g()] d d [h()] Constant multiple rule: d [c f ()] 5 c d d d [ f ()], c [ 핉 Eample 8 For each function: (i) find the derivative using the basic differentiation rules; (ii) find the slope of the graph of the function at the indicated points; and (iii) use our GDC to confirm our answer for (ii). Function Points a) f () 5 3 5 3 (3, 3), (3, 3) b) f () 5 ( 7) (, 9), (_ 7, ) c) f () 5 3 6 (4, ), (9, 3) d) f () 5 4 3 3 4 5 3 (5, 43), (, ) 4 a) (i) d d ( 3 5 3) 5 d d ( 3 ) d d ( ) 5 d d () d d (3) 5 3 () 5() 5 3 4 5 Therefore, the derivative of f () 5 3 5 3 is f 9() 5 3 4 5. (ii) Slope of curve at (3, 3) is f 9(3) 5 3(3) 4(3) 5 5 7 5 5. We should observe a horizontal tangent (slope 5 ) to the curve at (3, 3). Slope of curve at (3, 3) is f 9(3) 5 3(3) 4(3) 5 5 7 5 5 4. We should observe a ver steep tangent (slope 5 4) to the curve at (3, 3). 38

Differential Calculus Ī - : Fundamentals (iii) Not onl can we use the GDC to compute the value of the derivative at a particular value of on the home screen, but we can also do it on the graph screen. Plot Plot Plot3 Y= Xˆ3+X-5X- 3 Y= Y3= Y4= Y5= Y6= CALCULATE :value :zero 3:minimum 4:maimum 5:intersect 6:d/d 7: f()d WINDOW Xmin=-6 Xma=6 Xscl= Ymin=-4 Yma=4 Yscl=5 Xres= Y=Xˆ3+X -5X-3 X=-3 d/d=e - 6 turning point ( 3, 3) d/d=e - 6 CALCULATE :value :zero 3:minimum 4:maimum 5:intersect 6:d/d 7: f()d horizontal tangent X=3 The GDC computes a slope of E 6 at the point (3, 3). (E 6 5 3 6 5. ) Although the method the GDC uses is ver accurate, sometimes there is a small amount of error in its calculation. This most commonl occurs when performing calculus computations (e.g. the value of the derivative at a point). E 6 5. is ver close to zero which is the eact value of the derivative. Observe that the graph of 5 3 5 3 appears to have a turning point at (3, 3), confirming that a line tangent to the curve at that point would be horizontal. Let s check on our GDC that the slope of the curve is 4 at (3, 3). Again, the GDC ehibits a small amount of error in its result. Most GDCs are also capable of drawing a tangent at a point and displaing its equation as shown in the final screen image below. Y=Xˆ3+X -5X-3 d/d=4. X=3 =4.X+-85.3 The equation of the tangent line at (3, 3) is 5 4 85. We will look at finding the equations of tangent lines analticall in the last section of the chapter. b) (i) d d [( 7) ] 5 d [( 7)( 7)] differentiate term-b-term d after epanding 5 d d (4 8 49) 5 4 d d ( ) 8 d d () d d (49) 5 8 8 Therefore, the derivative of f () 5 ( 7) is f 9() 5 8 8. 38

(ii) Slope of curve at (, 9) is f 9() 5 8() 8 5. (iii) c) (i) Slope of curve at ( 7, ) is f 9 ( 7 ) 5 8 ( 7 ) 8 5. Thus, we should observe a horizontal tangent to the curve at ( 7, ). Plot Plot Plot3 WINDOW Y=(X-7) Xmin=- Y= Xma=6 (,9) Y3= Xscl= Y4= Ymin=- Y5= Yma= Y6= Yscl= Y7= Xres= d/d=- There s no error this time in the GDC s computation of the slope at (, 9). The verte of the parabola is at ( 7, ), confirming that it has a horizontal tangent at that point. d d (3 6) 5 3 d d ( _ ) d d (6) 5 3 ( ) 5 3 ( _ ) 5 3 _ Therefore, the derivative of f () 53 6 is f 9() 5 3 or f 9() 5 3 _. (ii) Slope of curve at (4, ) is f 9(4) 5 3 4 5 3. 4 Slope of curve at (9, 3) is f 9(9) 5 3 9 5. Thus, because the slope at 5 9 is less than that at 5 4, we should observe the graph of the equation becoming less steep as we move along the curve from 5 4 to 5 9. (iii) Plot Plot Plot3 WINDOW Y=3 (X)-6 Y= 3 (X)-6 Xmin=- Y= Xma= Y3= Xscl= Y4= Ymin=-7 Y5= Yma=4 Y6= Yscl= Y7= Xres= X=4 Y= The slope of the graph of 5 3 6 appears to steadil decrease as increases. Let s check the results for (ii) b evaluating the derivative at a point on the home screen. The GDC confirms the slopes for the curve when 5 4 and 5 9, but again the GDC computations have incorporated a small amount of error. d) (i) d ( 4 3 3 d 4 5 3 4 ) 5 d 4 d ( 4 ) 3 d d ( 3 ) d d ( ) 5 d d () d ( 3 d 4 ) 5 (4 3 ) 3 (3 ) d () 5 4 d () 5 3 9 4 5 Y=(X-7) X=3.5 Y= nderiv(3 (X)-6,X,4).756 nderiv(3 (X)-6,X,9).59 383

Differential Calculus Ī - : Fundamentals Therefore, the derivative of f () 5 4 3 3 4 5 3 4 is f 9() 5 3 9 4 5. (ii) Slope of curve at (5, 43) is f 9(5) 5 5 3 9(5) 4(5) 5 5. Thus, there should be a horizontal tangent to the curve at (5, 43). Slope of curve at (, ) is f 9() 5 5. (iii) Your GDC is not capable of computing the derivative function onl the specific value of the derivative for a given value of. However, we can have the GDC graph the values of the derivative over a given interval of. We can then graph the derivative function found from differentiation rules (result from (i)) and see if the two graphs match. Plot Plot Plot3 Y= Xˆ4/4-(3Xˆ3) /-Xˆ+5X/+3/ 4 Y= Y3= Y4= Y5= WINDOW Xmin=-4 Xma=8 Xscl= Ymin=-5 Yma=5 Yscl= Xres= d/d=3.5e - 6 Horizontal tangent at (5, 43) The command nderiv(y, X, X) computes the value of the Plot Plot Plot3 Y= Xˆ4/4-(3Xˆ3) /-Xˆ+5X/+3/ 4 Y= nderiv(y,x, X) Y3= Xˆ3-(9X 4X+5/ )/- derivative of function Y in terms of for all. Values of the derivative of f () will be graphed as Y, and the derivative function, f 9() 5 3 9 4 5, determined b manual application of differentiation rules (part (i)), will be graphed as Y 3. Note that the graph of Y 3 will be in bold stle to distinguish it from Y, and that the equation Y has been turned off. Y 5 4 3 3 4 5 3 Y 4 5 nderiv(y, X, X ) Y 3 5 3 9 4 5 Since the two graphs match, this confirms that the derivative found in part (i) using differentiation rules is correct. 384

Eample 9 The curve 5 a 3 7 8 5 has a turning point at the point where 5. Determine the value of a. There must be a horizontal tangent, and a slope of zero, at the point where the graph has a turning point. d d 5 d d (a 3 7 8 5) 5 a d d ( 3 ) 7 d d ( ) 8 d d () d d (5) 5 3a 4 8 d d 5 when 5 : 3a () 4() 8 5 a 8 8 5 a 5 36 a 5 3 Recall that the derivative of a function is a formula for the instantaneous rate of change of the dependent variable (commonl ) with respect to the dependent variable (). In other words, as illustrated earlier in this section, the slope of the tangent at a point gives the slope, or rate of change, of the curve at that point. The slope of a secant line (that crosses the curve at two points) gives the average rate of change between the two points. Eample Boiling water is poured into a cup. The temperature of the water in degrees Celsius, C, after t minutes is given b C 5 9 8, for times t > minute. a) Find the average rate of change of the temperature from t 5 to t 5 6. b) Find the rate of change of the temperature at the instant that t 5 4. t 3_ a) 75 Temperature ( C) 5 5 75 5 5 (, 83.35) 8 C 9 m 3. (6, 3.38) t 3 3 4 5 6 7 8 9 3 4 5 Time (minutes) When t 5, C 83.35 and when t 5 6, C 3.38. The average rate of change from t 5 to t 5 6 is the slope of the line through the points (, 83.35) and (6, 3.38). Average rate of change 5 83.35 3.38 5 5.97 6 4 5.995. To an accurac of 3 significant figures, the average rate of change from 385

Differential Calculus Ī - : Fundamentals t 5 to t 5 6 is 3. C per minute. During that period of time the water is, on average, becoming 3 degrees cooler ever minute. b) Let s compute the derivative dc, i.e. the rate of change of degrees dt C with respect to time t, from which we can compute the rate the temperature is changing at the moment when t 5 4. dc dt 5 d dt ( 9 8 ) t 5 d (9 8 t 3_ ) 5 d (9) 8 d ( t 3_ ) 3_ dt dt dt 5 8 3 t 3_ 5 73 t 5 dc dt 5 73 5 73 t 5_ t 5 At t 5 4: dc dt 5 73 4 5 73 5 3 8.53 Therefore, the temperature s instantaneous rate of change at t 5 4 minutes is 8.53 C per minute. Temperature ( C) 75 5 5 75 5 5 m 8.53 (4, 4.75) 8 C 9 t 3 3 4 5 6 7 8 9 3 4 5 Time (minutes) Eercise. In questions 4, find the derivative of the function b appling the limit definition f 9() 5 lim f ( h) f (). h h f () 5 g() 5 3 3 h() 5 4 r () 5 5 Using our results from questions 4, find the slope of the graph of each function in 4 at the point where 5. Sketch each function and draw a line tangent to the graph at 5. 386

In questions 6, a) find the derivative of the function, and b) compute the slope of the graph of the function at the indicated point. Use a GDC to confirm our results. 6 5 3 4 point (, ) 7 = 6 point (3, ) 8 5 3 point (, ) 9 5 5 3 point (, ) 5 ( )( 6) point (, 6) 5 3 3 point (, ) 5 3 point (, ) 3 The slope of the curve 5 a b at the point (, 4) is. Find the value of a and the value of b. In questions 4 7, find the coordinates of an points on the graph of the function where the slope is equal to the given value. 4 5 3 slope 5 3 5 5 3 slope 5 6 5 5 slope 5 7 5 3 slope 5 8 Use the graph of f to answer each question. C f B D E F A a) Between which two consecutive points is the average rate of change of the function greatest? b) At what points is the instantaneous rate of change of f positive, negative and zero? c) For which two pairs of points is the average rate of change approimatel equal? 9 The slope of the curve 5 4 6 at the point (3, 3) is equal to the slope of the curve 5 8 3 at (a, b). Find the value of a and the value of b. The graph of the equation 5 a 3 7 has a slope of 3 at the point where 5. Find the value of a. 387

Differential Calculus Ī - : Fundamentals Find the coordinates of the point on the graph of 5 at which the tangent is parallel to the line 5 5. Let f () 5 3. f ( h) f () a) Evaluate for h 5.. h f ( h) f () b) What number does approach as h approaches zero? h 3 From first principles, find the derivative for the general quadratic function, f () 5 a b c. Confirm our result b checking that it produces: (i) the derivative of when a 5, b 5, c 5 (ii) the derivative of 3 4 when a 5 3, b 5 4, c 5. 4 A car is parked with the windows and doors closed for five hours. The temperature inside the car in degrees Celsius, C, is given b C 5 t 3 7 with t representing the number of hours since the car was first parked. a) Find the average rate of change of the temperature from t 5 to t 5 4. b) Find the function that gives the instantaneous rate of change of the temperature for an time t,, t, 5. c) Find the time t at which the instantaneous rate of change of the temperature is equal to the average rate of change from t 5 to t 5 4. If the graph of a function is smooth at a particular point, the function is considered to be differentiable at this point. In other words, a tangent line eists at this point. All functions that will be differentiated in this course will be differentiable at all values in the function s domain. Figure. verte 388 horizontal tangent.3 Maima and minima first and second derivatives The relationship between a function and its derivative The derivative, written in Newton notation as f 9() or in Leibniz notation as d, is a function derived from a function f that gives the slope of the d graph of f at an in the function s domain (given that the curve is smooth at the value of ). The derivative is a slope, or rate of change, function. Knowing the slope of a function at different values in its domain tells us about properties of the function and the shape of its graph. In the previous section, we observed that if a graph turns at a particular point (for eample, at the verte of a parabola), then it has a horizontal tangent (slope 5 ) at the point. Hence, the derivative will equal zero at a turning point. In Section.5, we found the verte of the graph of a quadratic function b using the technique of completing the square to write its equation in verte form. We can also find the verte b means of differentiation. As we look at the graph of a parabola moving from left to right (i.e. domain values increasing), it either turns from going down to going up (decreasing to increasing), or from going up to going down (increasing to decreasing) (Figure.).

Eample Using differentiation, find the verte of the parabola with the equation 5 8 4. Find the value of for which the derivative, d, is zero. d d d 5 d d ( 8 4) 5 8 5 5 4 Thus, the -coordinate of the verte is 4. To find the -coordinate of the verte, we substitute 5 4 into the equation, giving 5 4 8(4) 4 5. Therefore, the verte has coordinates (4, ). 5 4 4 decreases as increases increases as increases Figure.3 increases from left to right along the -ais. 5 8 4 3 4 5 6 (4, ) 7 8 We know that the parabola in Eample will open up because the coefficient of the quadratic term,, is positive. The parabola has a negative slope (decreasing) to the left of the verte and a positive slope (increasing) to the right of the verte (Figure.3). As the values of increase, the derivative of 5 8 4 will change from negative to zero to positive, accordingl. d d 5 8 d d, for, 4 and d d 5 for 5 4 and d d. for. 4 In other words, the function f () 5 8 4 is decreasing for all, 4; it is neither decreasing nor increasing at 5 4; and it is increasing for all. 4. A point at which a function is neither increasing nor decreasing (i.e. there is a horizontal tangent) is called a stationar point. A convenient wa to demonstrate where a function is increasing or decreasing and the location of an stationar points is with a sign chart for the function and its derivative, as shown in Figure.4 for f () 5 8 4. The derivative f 9() 5 8 is zero onl at 5 4, thereb dividing the domain of f (i.e. 핉 ) into two intervals:, 4 and. 4. f 9() 5 8 is a continuous function (i.e. no gaps in the domain) so it is onl necessar to test one point in each interval in order to determine the sign of all the values of the derivative in that interval. f 9() can onl change sign at 5 4. For eample, the fact that f 9(3) 5 (3) 8 5, means that f 9(), for all when, 4. Therefore, f is decreasing for all in the open interval (`, 4). f() 8 4 f () 8 4 f () f() Figure.4 Sign chart for f 9() and f (). 389

Differential Calculus Ī - : Fundamentals Geometricall speaking, a function is continuous if there is no break in its graph; and a function is differentiable (i.e. a derivative eists) at an points where it is smooth. Increasing and decreasing functions and stationar points If f 9(). for a,, b, then f () is increasing on the interval a,, b. If f 9(), for a,, b, then f () is decreasing on the interval a,, b. If f 9() 5 for a,, b, then f () is constant on the interval a,, b. If f 9() 5 for a single value 5 c on some interval a, c, b, then f () has a stationar point at 5 c. The corresponding point (c, f (c)) on the graph of f is called a stationar point. It is at stationar points, or endpoints of the domain if the domain is not all real numbers, where a function ma have a maimum or minimum value. These points at which etreme values of a function ma occur are often referred to as critical points. Whether a function is increasing or decreasing on either side of a stationar point will indicate whether the stationar point is a maimum, minimum or neither. Eample Consider the function f () 5 3 3 4, [ 핉. a) Find an stationar points of f. b) Using the derivative of f, classif an stationar points as a maimum or minimum. f () f () f () 3 3 4 f () 6( )( ) increasing stationar decreasing f() stationar increasing a) f 9() 5 6 6 5 6( ) 5 6( )( ) 5 5 or 5 With a domain of all real numbers there are no domain endpoints that ma be an etreme value. Thus, f has two critical points: one at 5 and the other at 5. When 5 : 5 () 3 3() () 4 5 6 f has a stationar point at (, 6). When 5 : 5 () 3 3() () 4 5 f has a stationar point at (, ). b) Construct a sign chart for f 9() and f () (right) to show where f is increasing or decreasing. The derivative f 9() has two zeros, at 5 and 5, thereb dividing the domain of f into three intervals that need to be tested. Since f 9(3) 5 6()(4) 5 4., then f 9(). for all,. Likewise, since f 9() 5 6(4)() 5 4., then f 9(). for all.. Thus, f is increasing on the open intervals (`, ) and (, `). Since f 9() 5,, then f 9(), for all such that,,. Thus, f is decreasing on the open interval (, ), i.e.,,. From this information, we can visualize for increasing values of that the graph of f is going up for all,, then turning down at 5, then going down for values of from to, then turning up at 5, and then going up for all.. The basic shape of the graph of f will look something like the rough sketch shown right. Clearl, the stationar point (, 6) is a maimum and the stationar point (, ) is a minimum. 39

The graph of f () 5 3 3 4 from Eample (Figure.5) visuall confirms the results acquired from analzing the derivative of f. (, 6) 5 5 Figure.5 4 3 3 4 5 3 3 4 (, ) 5 For Eample, we can epress the result for part b) most clearl b saing that f () has a relative maimum value of 6 at 5, and f () has a relative minimum value of at 5. The reason that these etreme values are described as relative (sometimes described as local ) is because the are a maimum or minimum for the function in the immediate vicinit of the point, but not for the entire domain of the function. A point that is a maimum/minimum for the entire domain is called an absolute, or global, maimum/minimum. The first derivative test From Eample, we can see that a function f has a maimum at some 5 c if f 9(c) 5 and f is increasing immediatel to the left of 5 c and decreasing immediatel to the right of 5 c. Similarl, f has a minimum at some 5 c if f 9(c) 5 and f is decreasing immediatel to the left of 5 c and increasing immediatel to the right of 5 c. It is important to understand, however, that not all stationar points are either a maimum or minimum. The plural of maimum is maima, and the plural of minimum is minima. Maima and minima are collectivel referred to as etrema the plural of etremum (etreme value). Etrema of a function that do not occur at domain endpoints will be turning points of the graph of the function. Eample 3 For the function f () 5 4 3, find all stationar points and describe them completel. f 9() 5 d d ( 4 3 ) 5 4 3 6 5 ( 3) 5 5 or 5 3 The implied domain is all real numbers, so 5 and 5 3 are the critical points of f. When 5, 5 f () 5. When 5 3, 5 f ( 3 ) 5 ( 3 ) 4 ( 3 ) 3 5 8 6 54 8 5 7 6. Therefore, f has stationar points at (, ) and ( 3, 7 6 ). 39

Differential Calculus Ī - : Fundamentals Because f has two stationar points, there are three intervals for which to test the sign of the derivative. We could use some form of a sign chart as shown previousl, or we can use a more detailed table that summarizes the testing of the three intervals and the two critical points as shown below. Interval/point, 5,, 3 5 3. 3 6 5 4 3 4 3 3 7 (, 6) 3 3 Test value 5 5 5 Sign of f 9() f 9() 5, f 9() 5, f 9() 5 8. Conclusion f decreasing none f decreasing abs. min. f increasing On either side of 5, f does not change from either decreasing to increasing or from increasing to decreasing. Although there is a horizontal tangent at (, ), it is not an etreme value (turning point). The function steadil decreases as approaches zero, then at 5 the function has a rate of change (slope) of zero for an instant and then continues on decreasing. As approaches 3, f is decreasing and then switches to increasing at 5 3. Therefore, the stationar point (, ) is neither a maimum nor a minimum; and the stationar point ( 3, 7 6 ) is an absolute minimum. Or, in other words, f has an absolute (global) minimum value of 7 6 at 5 3. The reason that an absolute, rather than a relative, minimum value occurs at 5 3 is because for all, 3 the function f is either decreasing or constant (at 5 ) and for all, 3 f is increasing. First derivative test for maima and minima of a function Suppose that 5 c is a critical point of a continuous and smooth function f. That is, f (c) 5 and 5 c is a stationar point or 5 c is an endpoint of the domain. I. At a stationar point 5 c:. If f 9() changes sign from positive to negative as increases through 5 c, then f has a relative maimum at 5 c. relative maimum f () f (). If f 9() changes sign from negative to positive as increases through 5 c, then f has a relative minimum at 5 c. c relative minimum f () f () c 39

3. If f 9() does not change sign as increases through 5 c, then f has neither a relative maimum nor a relative minimum at 5 c. no etreme f () f () c II. At a domain endpoint 5 c: If 5 c is an endpoint of the domain, then 5 c will be a relative maimum or minimum of f if the sign of f 9() is alwas positive or alwas negative for. c (at a left endpoint), or for, c (at a right endpoint), as illustrated below. relative maimum relative maimum f () relative minimum f () f () relative minimum f () c c c c If it is possible to show that a relative maimum/minimum at 5 c is the greatest/least value for the entire domain of f, then it is classified as an absolute maimum/minimum. Eample 4 Appl the first derivative test to find an local etreme values for f (). Identif an absolute etrema. f () 5 4 3 9 5 f 9() 5 d d (4 3 9 5) 5 8 f 9() 5 8 5 6( 3 ) 5 6( 5)( 4) 5 Thus, f has stationar points at 5 5 and 5 4. To classif the stationar point at 5 5, we need to choose test points on either side of 5, for eample, 5 3 (left) and 5 (right). Then we have f 9(3) 5 6()(7) 5 4. f 9() 5 6(5)(4) 5, 393

Differential Calculus Ī - : Fundamentals So f has a relative maimum at 5 5. f ( 5 ) 5 4 ( 5 ) 3 9 ( 5 ) ( 5 ) 5 5 6.5 Therefore, f has a relative maimum value of 6.5 at 5 5. To classif the stationar point at 5 4, we need to choose test points on either side of 4, for eample, 5 (left) and 5 5 (right). Then we have f 9() 5, f 9(5) 5 6(5)() 5 9. So f has a relative minimum at 5 4. f (4) 5 4(4) 3 9(4) (4) 5 5 343 Therefore, f has a relative minimum value of 343 at 5 4. Change in displacement and velocit Consider the motion of an object such that we know its position s relative to a reference point or line as a function of time t given b s(t). The displacement of the object over the time interval from t to t is: change in s 5 displacement 5 s(t ) s(t ) The average velocit of the object over the time interval is: v avg 5 displacement change in time 5 s(t ) s(t ) t t The object s instantaneous velocit at a particular time, t, is the value of the derivative of the position function, s, with respect to time at t. velocit 5 ds 5 s9(t) dt Eample 5 A rocket is launched upwards into the air. Its vertical position, s metres, above the ground at t seconds is given b s(t) 5 5t 8t. a) Find the average velocit over the time interval from t 5 second to t 5 seconds. b) Find the instantaneous velocit at t 5 second. c) Find the maimum height reached b the rocket and the time at which this occurs. 394

s() s() a) v avg 5 5 [5() 8() ] [5 8 ] 5 3 metres per second (or m s ) b) s9(t) 5 t 8 s9() 5 8 5 8 m s c) s9(t) 5 t 8 5 t 5.8 Thus, s has a stationar point at t 5.8. t must be positive and ranges from time of launch (t 5 ) to when the rocket hits the ground, i.e. h 5. s(t) 5 5t 86 8 4(5)() 8t 5 t 5 (5) t.547 or t 3.655 So, the rocket hits the ground about 3.66 seconds after the time of launch. Hence, the domain for the position (s) and velocit (v) functions is < t < 3.66. Therefore, the function s has three critical points: t 5, t 5.8 and t 3.66. The maimum of the function, i.e. the maimum height, most likel occurs at the critical point t 5.8. Let s confirm this. Appling the first derivative test, we determine the sign of the derivative, s9(t), for values on either side of t 5.8, for eample, t 5 and t 5. s9() 5 8. and s9() 5,. Neither of the domain endpoints, t 5 and t 3.66, are at a maimum or minimum because the function is not constantl increasing or constantl decreasing before or after the endpoint. Since the function changes from increasing to decreasing at t 5.8 and s(.8) 5 5(.8) 8(.8) 5 7., then the rocket reaches a maimum height of 7. metres.8 seconds after it was launched. The relationship between a function and its second derivative You ma have wondered wh the strateg we are appling to locate and classif etrema for a function focuses on using the first derivative of the function. This implies that we are interested in using some other tpe of derivative, namel the second derivative. There is another useful test for the purpose of analzing the stationar point of a function that makes use of the derivative of the derivative, i.e. the second derivative, of the function. When we differentiate a function 5 f (), we obtain the first derivative f 9() ( also denoted as d d ). Often this is a function that can also be differentiated. The result of doing so is the derivative of f 9(), which is denoted in Newton notation as f () or in Leibniz notation as d d and called the second derivative of f with respect to. For eample, if f () 5 3, then f 9() 5 3 and f () 5 6 ( or d d 5 6 ). 395

Differential Calculus Ī - : Fundamentals s 5 5 5 5 v 5 5 5 5 a 5 5 5 5 Figure.6 Position function: s(t) 5t 8t 3 4 Velocit function: v(t) s (t) t 8 3 4 Acceleration function: a(t) v (t) s (t) 3 4 It would be incorrect to graph a function and its first and/ or second derivative on the same aes. For eample, the position s(t), velocit v (t) and acceleration a(t) functions graphed on separate aes in Figure.6 will have different units on each vertical ais: metres for s(t), metres per second for v(t) and metres per second per second for a(t). t t t Second derivatives, like first derivatives, occur often in methods of appling calculus. In Eample 5, the function s(t) gave the position, in metres above the ground, of a projectile (to rocket) where t, in seconds, is the time since the projectile was launched. The function s 9(t), the first derivative of the position function, then gives the rate of change of the object s position, i.e. its velocit, in metres per second (m s ). Differentiation of this function gives the rate of change of the object s velocit, i.e. its acceleration, measured in metres per second per second (m s ). The graphs of the position, velocit and acceleration functions for Eample 5 aligned verticall (Figure.6) nicel illustrate the relationships between a function, its first derivative and its second derivative. The slope of the graph of s (t) is initiall a large positive value (graph is steep), but steadil decreases until it is zero (horizontal tangent) at t 5.8 and then continues to decrease, becoming a large negative value (again, steep, but in the other direction). This corresponds to the real-life situation in which the rocket is launched with a high initial velocit (v() 5 8 m s ) and then its velocit decreases steadil due to gravit. The rocket s velocit is zero for just an instant when it reaches its maimum height at t 5.8 and then its velocit becomes more and more negative because it has changed direction and is moving back (negative direction) to the ground. The rate of change of the velocit, v 9(t), is constant and it is negative because the velocit is decreasing from positive values to zero to negative values. This is clear from the fact that the graph of the velocit function, v (t), is a straight line with a negative slope. It follows then that the acceleration function the rate of change of velocit is a negative constant, a 5 in this case, and its graph is a horizontal line. In Eample 5, it is not possible to have a negative function value for s (t) because the rocket s position is alwas above, or at, ground level. In man motion problems in calculus, we consider a simplified version b limiting an object s motion to a line with its position given as its displacement from a fied point (usuall the origin). At a position left of the fied point, the object s displacement is negative, and at a position right of the fied point, the displacement is positive. Velocit can also be positive or negative depending on the direction of travel (i.e. the sign of the rate of change of the object s displacement). Likewise, acceleration is positive if velocit is increasing (i.e. rate of change of velocit is positive) and negative if velocit is decreasing. A common misconception is that acceleration is positive for motion in the positive direction (usuall right or up ) and negative for motion in the negative direction (usuall left or down ). Acceleration indicates how velocit is changing. Even though an object ma be moving in a positive direction (e.g. to the right) if it is slowing down, then its acceleration is acting in the opposite direction and would be negative. In Eample 5, the rocket was alwas accelerating in the negative direction, m s, due to the force of gravit. Note: A more accurate value for the acceleration of a free-falling object due to gravit is 9.8 m s. 396