2.1 THE DEFINITION OF DERIVATIVE

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1 2.1 Te Derivative Contemporary Calculus 2.1 THE DEFINITION OF DERIVATIVE 1 Te grapical idea of a slope of a tangent line is very useful, but for some uses we need a more algebraic definition of te derivative of a function. We will use tis definition to calculate te derivatives of several functions and see tat te results from te definition agree wit our grapical understanding. We will also look at several different interpretations for te derivative, and derive a teorem wic will allow us to easily and quickly determine te derivative of any fixed power of x. In te last section we found te slope of te tangent line to te grap of te function f(x) = x 2 at an arbitrary point (x, f(x) ) by calculating te slope of te secant line troug te points (x, f(x) ) and (x+, f(x+) ), m sec = f(x+) f(x) (x+) (x), and ten by taking te limit of m sec as approaced 0 (Fig. 1). Tat approac to calculating slopes of tangent lines is te definition of te derivative of a function. Definition of te Derivative: Te derivative of a function f is a new function, f ' (pronounced "eff prime"), wose value at x is f '(x) f (x + ) # f (x) if te limit exists and is finite. Tis is te definition of differential calculus, and you must know it and understand wat it says. Te rest of tis capter and all of Capter 3 are built on tis definition as is muc of wat appears in later capters. It is remarkable tat suc a simple idea (te slope of a tangent line) and suc a simple definition (for te derivative f ' ) will lead to so many important ideas and applications. Notation: f '(x) Tere are tree commonly used notations for te derivative of y = f(x): empasizes tat te derivative is a function related to f D( f ) empasizes tat we perform an operation on f to get te derivative of f df dx f f(x+) f(x) empasizes tat te derivative is te limit of x =. We will use all tree notations so you can get used to working wit eac of tem.

2 2.1 Te Derivative Contemporary Calculus f '(x) represents te slope of te tangent line to te grap of y = f(x) at te point (x, f(x) ) or te instantaneous rate of cange of te function f at te point (x, f(x)). 2 If, in Fig. 2, we let x be te point a+, ten = x a. As 0, we see tat x a and lim f (a + ) # f (a) x "a f (x) # f (a) x # a so f '(a) f (a + ) # f (a) x "a f (x) # f (a) x # a We will use wicever of tese two forms is more convenient algebraically. Calculating Some Derivatives Using Te Definition. Fortunately, we will soon ave some quick and easy ways to calculate most derivatives, but first we will ave to use te definition to determine te derivatives of a few basic functions. In Section 2.2 we will use tose results and some properties of derivatives to calculate derivatives of combinations of te basic functions. Let's begin by using te graps and ten te definition to find a few derivatives. Example 1: Grap y = f(x) = 5 and estimate te slope of te tangent line at eac point on te grap. Ten use te definition of te derivative to calculate te exact slope of te tangent line at eac point. Your grapic estimate and te exact result from te definition sould agree. Solution: Te grap of y = f(x) = 5 is a orizontal line (Fig. 3) wic as slope 0 so we sould expect tat its tangent line will also ave slope 0. Using te definition: Since f(x) = 5, ten f(x+) = 5, so D( f(x) ) lim f (x + ) # f (x) 5 # 5 0 = 0. Using similar steps, it is easy to sow tat te derivative of any constant function is 0. Teorem: If f(x) = k, ten f '(x) = 0. Practice 1: Grap y = f(x) = 7x and estimate te slope of te tangent line at eac point on te grap. Ten use te definition of te derivative to calculate te exact slope of te tangent line at eac point. Example 2: Determine te derivative of y = f(x) = 5x 3 grapically and using te definition. Find te equation of te line tangent to y = 5x 3 at te point (1,5).

3 2.1 Te Derivative Contemporary Calculus 3 Solution: It appears tat te grap of y = f(x) = 5x 3 (Fig. 4) is increasing so te slopes of te tangent lines are positive except peraps at x = 0 were te grap seems to flatten out. Using te definition: Since f(x) = 5x 3, ten f(x+) = 5(x+) 3 = 5(x 3 + 3x 2 + 3x ) so f '(x) lim f (x + ) # f (x) te definition =lim 5(x 3 + 3x 2 + 3x ) # 5(x 3 ) eliminate 5x 3 5x 3 15x 2 +15x (15x 2 +15x ) =lim divide by (15x 2 +15x ) = 15x = 15x 2 so D( 5x 3 ) = 15x 2 wic is positive except wen x=0, and ten 15x 2 = 0. f '(x) = 15x 2 is te slope of te line tangent to te grap of f at te point ( x, f(x) ). At te point (1,5), te slope of te tangent line is f '(1) = 15(1) 2 = 15. From te point slope formula, te equation of te tangent line to f is y 5 = 15( x 1) or y = 15x 10. Practice 2: Use te definition to sow tat te derivative of y = x 3 is dy dx = 3x2. Find te equation of te line tangent to te grap of y = x 3 at te point ( 2, 8 ). If f as a derivative at x, we say tat f is differentiable at x. If we ave a point on te grap of a differentiable function and a slope (te derivative evaluated at te point), it is easy to write te equation of te tangent line. Tangent Line Formula If f is differentiable at a ten te equation of te tangent line to f at te point (a,f(a) ) is y = f(a) + f '(a)(x a). Proof: Te tangent line goes troug te point ( a, f(a) ) wit slope f '(a) so, using te point slope formula, y f(a) = f '(a) (x a) or y = f(a) + f '(a) (x a).

4 2.1 Te Derivative Contemporary Calculus Practice 3: Te derivatives D( x ) = 1, D( x 2 ) = 2x, D( x 3 ) = 3x 2 exibit te start of a pattern. Witout using te definition of te derivative, wat do you tink te following derivatives will be? D( x 4 ), D( x 5 ), D( x 43 ), D( x ) = D( x 1/2 ) and D( x π ). (Just make an intelligent "guess" based on te pattern of te previous examples. ) 4 Before going on to te "pattern" for te derivatives of powers of x and te general properties of derivatives, let's try te derivatives of two functions wic are not powers of x: sin(x) and x. Teorem: D( sin(x) ) = cos(x). Te grap of y = f(x) = sin(x) is well known (Fig. 5). Te grap as orizontal tangent lines (slope = 0) wen x = ± π 2 and x = ± 3π 2 and so on. If 0 < x < π 2, ten te slopes of te tangent lines to te grap of y = sin(x) are positive. Similarly, if π 2 < x < 3π 2, ten te slopes of te tangent lines are negative. Finally, since te grap of y = sin(x) is periodic, we expect tat te derivative of y = sin(x) will also be periodic. Proof of te teorem: Since f(x) = sin(x), f(x+) = sin(x+) = sin(x)cos() + cos(x)sin() so f '(x) lim f (x + ) # f (x) =lim {sin(x)cos() + cos(x)sin()} #{sin(x)} tis limit looks formidable, but if we just collect te terms containing sin(x) and ten tose containing cos(x) we get % & sin(x) # ' cos() $1 + cos(x) # sin() ( ) * now calculate te limits separately # = # $ lim % sin(x) & ' "0 ( ) # lim cos() *1& $ ' % "0 ( + # $ % lim cos(x) & ' "0 ( ) + $ lim + "0 % = sin(x). (0) + cos(x). (1) = cos (x). & sin() + ' te first and tird limits do not + ( depend on, and we calculated te second and fourt limits in Section 1.2 So D( sin(x) ) = cos(x), and te various properties we expected of te derivative of y = sin(x) by examining its grap are true of cos(x).

5 2.1 Te Derivative Contemporary Calculus 5 Practice 4: Use te definition to sow tat D( cos(x) ) = sin(x). (Tis is similar to te situation for f(x) = sin(x). You will need te formula cos(x+) = cos(x). cos() sin(x). sin(). Ten collect all te terms containing cos(x) and all te terms wit sin(x). At tat point you sould recognize and be able to evaluate te limits.) Example 3: For y = x find dy/dx. Solution: Te grap of y = f(x) = x (Fig. 6) is a "V" wit its vertex at te origin. Wen x > 0, te grap is just y = x = x wic is a line wit slope +1 so we sould expect te derivative of x to be +1. Wen x < 0, te grap is y = x = x wic is a line wit slope 1, so we expect te derivative of x to be 1. Wen x = 0, te grap as a corner, and we sould expect te derivative of x to be undefined at x = 0. Using te definition: It is easiest to consider 3 cases in te definition of x : x > 0, x < 0 and x = 0. If x > 0, ten, for small values of, x + > 0 so Df(x) lim x + # x = 1. If x < 0, ten, for small values of, we also know tat x + < 0 so Df(x) Wen x = 0, te situation is a bit more complicated and # = 1. Df(x) lim f (x + ) # f (x) 0 + # 0 wic is undefined since lim + = +1 and lim # = 1. D( x ) = +1 undefined if x > 0 if x = 0 1 if x < 0. Practice 5: Grap y = x 2 and y = 2x and use te graps to determine D( x 2 ) and D( 2x ). INTERPRETATIONS OF THE DERIVATIVE So far we ave empasized te derivative as te slope of te line tangent to a grap. Tat interpretation is very visual and useful wen examining te grap of a function, and we will continue to use it. Derivatives, owever, are used in a wide variety of fields and applications, and some of tese fields use oter interpretations. Te following are a few interpretations of te derivative wic are commonly used.

6 2.1 Te Derivative Contemporary Calculus General Rate of Cange f '(x) is te rate of cange of te function at x. If te units for x are years and te units for df people f(x) are people, ten te units for dx are year, a rate of cange in population. Grapical Slope f '(x) is te slope of te line tangent to te grap of f at te point ( x, f(x) ). Pysical Velocity If f(x) is te position of an object at time x, ten f '(x) is te velocity of te object at time x. If te units for x are ours and f(x) is distance measured in miles, ten te units for f '(x) = df miles dx are our, miles per our, wic is a measure of velocity. 6 Acceleration If f(x) is te velocity of an object at time x, ten f '(x) is te acceleration of te object at time x. If te units are for x are ours and f(x) as te units miles our, ten te units for te acceleration f '(x) = df dx are miles/our our = miles 2, miles per our per our. our Magnification f '(x) is te magnification factor of te function f for points wic are close to x. If a and b are two points very close to x, ten te distance between f(a) and f(b) will be close to f '(x) times te original distance between a and b: f(b) f(a) f '(x) ( b a ). Business Marginal Cost If f(x) is te total cost of x objects, ten f '(x) is te marginal cost, at a production level of x. Tis marginal cost is approximately te additional cost of making one more object once we ave already made x objects. If te units for x are bicycles and te units for f(x) are dollars, ten te units for f '(x) = df dollars dx are bicycle, te cost per bicycle. Marginal Profit If f(x) is te total profit from producing and selling x objects, ten f '(x) is te marginal profit, te profit to be made from producing and selling one more object. If te units for x are bicycles and te units for f(x) are dollars, ten te units for f '(x) = df dollars dx are bicycle, dollars per bicycle, wic is te profit per bicycle. In business contexts, te word "marginal" usually means te derivative or rate of cange of some quantity. One of te strengts of calculus is tat it provides a unity and economy of ideas among diverse applications. Te vocabulary and problems may be different, but te ideas and even te notations of calculus are still useful.

7 2.1 Te Derivative Contemporary Calculus 7 Example 4: A small cork is bobbing up and down, and at time t seconds it is (t) = sin(t) feet above te mean water level (Fig. 7). Find te eigt, velocity and acceleration of te cork wen t = 2 seconds. (Include te proper units for eac answer.) Solution: (t) = sin(t) represents te eigt of te cork at any time t, so te eigt of te cork wen t = 2 is (2) = sin(2) 0.91 feet. Te velocity is te derivative of te position, so v(t) = d (t) dt = d sin(t) dt = cos(t). Te derivative of position is te limit of ( )/( t), so te units are (feet)/(seconds). After 2 seconds te velocity is v(2) = cos(2) 0.42 feet per second = 0.42 ft/s. Te acceleration is te derivative of te velocity, so a(t) = d v(t) dt = d cos(t) dt = sin(t). Te derivative of velocity is te limit of ( v)/( t), so te units are (feet/second) / (seconds) or feet/second 2. After 2 seconds te acceleration is a(2) = sin(2) 0.91 ft/s 2. Practice 6: Find te eigt, velocity and acceleration of te cork in te previous example after 1 second? A MOST USEFUL FORMULA: D( x n ) Functions wic include powers of x are very common (every polynomial is a sum of terms wic include powers of x), and, fortunately, it is easy to calculate te derivatives of suc powers. Te "pattern" emerging from te first few examples in tis section is, in fact, true for all powers of x. We will only state and prove te "pattern" ere for positive integer powers of x, but it is also true for oter powers as we will prove later. Teorem: If n is a positive integer, ten D( x n ) = n. x n 1. Tis teorem is an example of te power of generality and proof in matematics. Rater tan resorting to te definition wen we encounter a new power of x (imagine using te definition to calculate te derivative of x 307 ), we can justify te pattern for all positive integer exponents n, and ten simply apply te result for watever exponent we ave. We know, from te first examples in tis section, tat te teorem is true for n= 1, 2 and 3, but no number of examples would guarantee tat te pattern is true for all exponents. We need a proof tat wat we tink is true really is true.

8 2.1 Te Derivative Contemporary Calculus Proof of te teorem: Since f(x) = x n, ten f(x+) = (x+) n, and in order to simplify f(x+) f(x) = (x+) n x n, we will need to expand (x+) n. However, we really only need to know te first two terms of te expansion and to know tat all of te oter terms of te expansion contain a power of of at least 2. Te Binomial Teorem from algebra says (for n > 3) tat 8 (x+) n = x n + n. x n 1 + a. x n b. x n n were a and b represent numerical coefficients. (Expand (x+) n for at least a few different values of n to convince yourself of tis result.) Ten D( f(x) ) lim f (x + ) # f (x) (x + ) n # x n ten expand to get { x n + n # x n$1 + a# x n$2 2 + b # x n$ n } $ x n { n # x n$1 + a# x n$2 2 + b# x n$ n } #{ n # x n$1 + a# x n$2 + b# x n$ n$1 } eliminate x n x n factor out of te numerator divide by te factor { n. x n 1 + a. x n 2 + b. x n n 1 } = n. x n 1 + lim { a. x n 2 + b. x n n 1 } separate te limits eac term as a factor of, and 0 = n. x n = n. x n 1 so D( x n ) = n. x n 1. Practice 7: Use te teorem to calculate D( x 5 ), d dx ( x2 ), D( x 100 ), d dt ( t31 ), and D( x 0 ). We will occasionally use te result of te teorem for te derivatives of all constant powers of x even toug it as only been proven for positive integer powers, so far. Te result for all constant powers of x is proved in Section 2.9 Example 5: Find D( 1/x ) and d dx ( x ). Solution: D( 1 x ) = D(x 1 ) = 1x ( 1) 1 = 1x 2 = 1 x 2. d dx ( x ) = D( x1/2 ) = (1/2)x 1/2 = 1 2 x. Tese results can be obtained by using te definition of te derivative, but te algebra is sligtly awkward. Practice 8: Use te pattern of te teorem to find D( x 3/2 ), d dx ( x1/3 ), D( 1 x ) and d dt ( tπ ).

9 2.1 Te Derivative Contemporary Calculus 9 Example 6: It costs x undred dollars to run a training program for x employees. (a) How muc does it cost to train 100 employees? 101 employees? If you already need to train 100 employees, ow muc additional will it cost to add 1 more employee to tose being trained? (b) For f(x) = x, calculate f '(x) and evaluate f ' at x = 100. How does f '(100) compare wit te last answer in part (a)? Solution: (a) Put f(x) = x = x 1/2 undred dollars, te cost to train x employees. Ten f(100) = $1000 and f(101) = $ , so it costs $4.99 additional to train te 101 st employee. (b) f '(x) = 1 2 x 1/2 = 1 2 x so f '(100) = = 1 20 undred dollars = $5.00. Clearly f '(100) is very close to te actual additional cost of training te 101 st employee. IMPORTANT DEFINITIONS AND RESULTS Definition of Derivative: f '(x) lim f (x + ) # f (x) if te limit exists and is finite. Notations For Te Derivative: f '(x), Df(x), d f(x) dx Tangent Line Equation: Te line y = f(a) + f '(a). ( x a ) is tangent to te grap of f at ( a, f(a) ). Formulas: D( constant ) = 0 D( x n ) = n. x n-1 (proven for n = positive integer: true for all constants n) D( sin(x) ) = cos(x) and D( cos(x) ) = sin(x) D( x ) = +1 if x > 0 undefined if x = 0 1 if x < 0. Interpretations of f '(x): Slope of a line tangent to a grap Instantaneous rate of cange of a function at a point Velocity or acceleration Magnification factor Marginal cange

10 2.1 Te Derivative Contemporary Calculus PROBLEMS Matc te graps of te tree functions in Fig Fig. 9 sows six graps, tree of wic are wit te graps of teir derivatives. derivatives of te oter tree. Matc te functions wit teir derivatives. In problems 3 6, find te slope m sec of te secant line troug te two given points and ten calculate m tan = lim m sec. 3. f(x) = x 2 (a) ( 2,4), ( 2+, ( 2+) 2 ) (b) (0.5, 0.25), (0.5+, (0.5+) 2 ) 4. f(x) = 3 + x 2 (a) (1,4), (1+, 3+(1+) 2 ) (b) (x, 3 + x 2 ), (x+, 3 + (x+) 2 ) 5. f(x) = 7x x 2 (a) (1, 6), (1+, 7(1+) (1+) 2 ) (b) (x, 7x x 2 ), (x+, 7(x+) (x+) 2 ) 6. f(x) = x 3 + 4x (a) (1, 5), (1+, (1+) 3 + 4(1+) ) (b) (x, x 3 + 4x), (x+, (x+) 3 + 4(x+) ) 7. Use te grap in Fig. 10 to estimate te values of tese limits. (It elps to recognize wat te limit represents.) (a) lim f (0 + ) # f (0) (b) lim f (1+ ) # f (1) (c) lim f (2 + ) #1 (d) lim w"0 f (3 + w) # f (3) w (e) lim f (4 + ) # f (4) (f) lim s"0 f (5 + s) # f (5) s 8. Use te grap in Fig. 11 to estimate te values of tese limits. (a) lim g() # g(0) (b) lim g(1+ ) # g(1) (c) lim g(2 + ) # 2 (d) lim w"0 g(3+ w) # g(3) w (e) lim g(4 + ) # g(4) (f) lim s"0 g(5 + s) # g(5) s

11 2.1 Te Derivative Contemporary Calculus In problems 9 12, use te Definition of te derivative to calculate f '(x) and ten evaluate f '(3). 9. f(x) = x f(x) = 5x 2 2x 11. f(x) = 2x 3 5x 12. f(x) = 7x 3 + x Grap f(x) = x 2, g(x) = x and (x) = x 2 5. Calculate te derivatives of f, g, and. 14. Grap f(x) = 5x, g(x) = 5x + 2 and (x) = 5x 7. Calculate te derivatives of f, g, and. In problems 15 18, find te slopes and equations of te lines tangent to y = f(x) at te given points. 15. f(x) = x at (1,9) and ( 2,12) 16. f(x) = 5x 2 2x at (2, 16) and (0,0) 17. f(x) = sin(x) at (π, 0) and (π/2,1) 18. f(x) = x + 3 at (0,3) and ( 3,0) 19. (a) Find te equation of te line tangent to te grap of y = x at te point (2,5). (b) Find te equation of te line perpendicular to te grap of y = x at (2,5). (c) Were is te tangent to te grap of y = x orizontal? (d) Find te equation of te line tangent to te grap of y = x at te point (p,q). (e) Find te point(s) (p,q) on te grap of y = x so te tangent line to te curve at (p,q) goes troug te point (1, 7). 20. (a) Find te equation of te line tangent to te grap of y = x 3 at te point (2,8). (b) Were, if ever, is te tangent to te grap of y = x 3 orizontal? (c) Find te equation of te line tangent to te grap of y = x 3 at te point (p,q). (d) Find te point(s) (p,q) on te grap of y = x 3 so te tangent line to te curve at (p,q) goes troug te point (16,0). 21. (a) Find te angle tat te tangent line to y = x 2 at (1,1) makes wit te x axis. (b) Find te angle tat te tangent line to y = x 3 at (1,1) makes wit te x axis. (c) Te curves y = x 2 and y = x 3 intersect at te point (1,1). Find te angle of intersection of te two curves (actually te angle between teir tangent lines) at te point (1,1). 22. Fig. 12 sows te grap of y = f(x). Sketc te grap of y = f '(x). 23. Fig. 13 sows te grap of te eigt of an object at time t. Sketc te grap of te object's upward velocity. Wat are te units for eac axis on te velocity grap?

12 2.1 Te Derivative Contemporary Calculus 24. Fill in te table wit te appropriate units for f '(x). units for x units for f(x) units for f '(x) ours miles people automobiles dollars pancakes days trout seconds miles per second seconds gallons study ours test points A rock dropped into a deep ole will drop d(x) = 16x 2 feet in x seconds. (a) How far into te ole will te rock be after 4 seconds? 5 seconds? (b) How fast will it be falling at exactly 4 seconds? 5 seconds? x seconds? 26. It takes T(x) = x 2 ours to weave x small rugs. Wat is te marginal production time to weave a rug? (Be sure to include te units wit your answer.) 27. It costs C(x) = x dollars to produce x golf balls. Wat is te marginal production cost to make a golf ball? Wat is te marginal production cost wen x = 25? wen x= 100? (Include units.) 28. Define A(x) to be te area bounded by te x and y axes, te line y = 5, and a vertical line at x (Fig. 14). (a) Evaluate A(0), A(1), A(2) and A(3). (b) Find a formula for A(x) for x 0: A(x) =? (c) Determine d A(x) dx. (d) Wat does d A(x) dx represent? 29. Define A(x) to be te area bounded by te x axis, te line y = x, and a vertical line at x (Fig. 15). (a) Evaluate A(0), A(1), A(2) and A(3). (b) Find a formula wic represents A(x) for all x 0: A(x) =? (c) Determine d A(x) dx. (d) Wat does d A(x) dx represent? 30. Find (a) D( x 12 ) (b) d dx ( 7 x ) (c) D( 1 x 3 ) (d) d xe dx (e) D( x 2 ) 31. Find (a) D( x 9 ) (b) d x2/3 dx (c) D( 1 x 4 ) (d) D( x π ) (e) d x+5 dx

13 2.1 Te Derivative Contemporary Calculus In problems 32 37, find a function f wic as te given derivative. (Eac problem as several correct answers, just find one of tem.) f '(x) = 4x f '(x) = 3x 2 + 8x 34. D( f(x) ) = 12x d f(t) dt = 5 cos(t) d f(x) 36. dx = 2x sin(x) 37. D( f(x) ) = x + x2 Section 2.1 PRACTICE Answers Practice 1: Te grap of f(x) = 7x is a line troug te origin. Te slope of te line is 7. For all x, m tan f (x + ) # f (x) 7(x + ) # 7x 7 7 = 7. Practice 2: f(x) = x 3 so f(x + ) = (x + ) 3 = x 3 + 3x 2 + 3x dy dx f (x + ) # f (x) { x 3 + 3x 2 + 3x } # x n 3x 2 + 3x (3x 2 + 3x + 2 ) = 3x 2. At te point (2,8), te slope of te tangent line is 3(2) 2 = 12 so te equation of te tangent line is y 8 = 12(x 2) or y = 12x 16. Practice 3: D( x 4 ) = 4x 3, D( x 5 ) = 5x 4, D( x 43 ) = 43x 42, D( x 1/2 ) = 1 2 x 1/2, D( x π ) = πx π 1 Practice 4: D( cos(x) ) cos(x + ) # cos(x) cos(x)cos() # sin(x)sin() # cos(x) cos(x) cos() 1 sin(x) sin() cos(x). (0) sin(x). (1) = sin(x). Practice 5: See Fig. 16 for te graps of y = x 2 and y = 2x. D( x 2 ) = +1 if x > 2 undefined if x = 2 1 if x < 2 D( 2x ) = +2 if x > 0 undefined if x = 0 2 if x < 0

14 2.1 Te Derivative Contemporary Calculus Practice 6: (t) = sin(t) so (1) = sin(1) 0.84 feet, v(t) = cos(t) so v(1) =cos(1) 0.54 feet/second. a(t) = sin(t) so a(1) = sin(1) 0.84 feet/second Practice 7: D( x 5 ) = 5x 4, d x 2 dx = 2x1 = 2x, d x 100 dx = 100x99, d t 31 dt = 31t30, D( x 0 ) = 0x 1 = 0 or D( x 0 ) = D( 1 ) = 0. Practice 8: D( x 3/2 ) = 3 2 x1/2, d x 1/3 dx = 1 3 x 2/3, D( x 1/2 ) = 1 2 x 3/2, d t π dt = π tπ 1.