Engineering Mathematics 2 26 February 2014 Limits of functions Consier the function 1 f() = 1. The omain of this function is R + \ {1}. The function is not efine at 1. What happens when is close to 1? f() f() 0.9 0.5132 1.1 0.4882 0.99 0.5013 1.01 0.4988 0.9999 0.5000 1.00001 0.5000 approaching 1 approaching 0.5 approaching 1 approaching 0.5 As gets closer to 1, either from the left or the right han sie, f() gets closer to 0.5. We say that the limit as approaches 1 of f() is 0.5 an write either f() 0.5 as 1 or lim 1 f() = 0.5. Eample: Calculate the limit 2 5+6 lim 3 3 Differentiation Given a function we are often intereste in how the values of the function f() are changing as we change. For eample, oes f() increase or ecrease when we increase? What is the rate of increase/ecrease? Eamples: The spee is the rate of change of istance travelle; The electric current is the rate of change of charge with respect to time. The average rate of change of f between 1 = a an 2 = a+h is f = f( 2) f( 1 ) = f(a+h) f(a) = f(a+h) f(a) 2 1 (a+h) a h We move to the instantaneous rate of change or erivative of f at by taking the limit of this as 2 approaches 1, that is as h 0. The erivative of the function f at the point a is f (a) = lim h 0 f(a+h) f(a) h if the limit eists. The value of the limit is then calle the rate of change or erivative of f at a an is enote by f (a), or by f (a). If the erivative of the function f eists at a point a, we say that f is ifferentiable at that point.
Engineering Mathematics 2 26 February 2014 Geometric Description Graphically, the average rate of change of f is the slope of the secant line joining two points on the graph of f As we move one of these closer to the other, the secant line becomes a tangent line an the erivative is the slope of the tangent line Figure 1: The erivative of a function Eample Recall the eample of the tile falling from a builing. The height of the tile at time t was h(t) = 40 4.9t 2 The tile reaches the groun when h(t) = 40 4.9t 2 = 0, which happens after 2.857 secons. The average spee of the tile for the last secon of its journey is h(2.857) h(1.857) 2.857 1.857 = 0 23.10 1 = 23.10 To get the instantaneous spee of the tile when it hits the groun we nee the limit of the average spee as t approaches the time of impact: h(2.857) h(t) lim = h (2.857) = 28 t 2.857 2.857 t The negative sign inicates that as time increases, the height ecreases. Eample: Let f be the function f() = 2+5. Determine whether f is ifferentiable at a an calculate its erivative at a. Eample: Let f() = 2. Let a R. Is f ifferentiable at a? If so, what is the erivative? The operation of fining the erivative of a function at a point is calle ifferentiation. By ifferentiation, we transform our function f into a new function enote by f. For eample, we saw that by ifferentiating the function f() = 2 we obtaine the function f() = 2. Note: The metho of calculating a erivative by using this limit formula is calle ifferentiation from first principles. We nee to introuce faster techniques for ifferentiating.
Engineering Mathematics 2 28 February 2014 Differentiation Rules The operation of fining the erivative of a function at a point is calle ifferentiation. By ifferentiation, we transform our function f() into a new function enote by f () or f. 1. Differentiating a constant function The erivative of a constant function is always equal to zero. For eample (3) = 0; 2. Differentiating a power function The rule for ifferentiating a power is ( 2) = 0, etc. (n ) = n n 1 (bring the power own an ecrease the power of by 1). For eample, ( 6 ) = 6 5 ; ( 100 ) = 100 99 This rule is also vali for negative or fractional powers. For eample, ( ) 1 5 = ( 5 ) = 5 5 1 = 5 6 ( ) ( ) = 1 2 = 1 2 1 2 1 = 1 2 1 2 3. Differentiating a function multiplie by a constant When ifferentiating a function multiplie by a constant, we only nee to ifferentiate the function while the constant stays the same. For eample, 4. Differentiating a sum of functions (Cf()) = C (f()) ( 4 7 ) = 4 ( 7 ) = 4 7 6 = 28 6 The erivative of a sum of functions is the sum of erivatives of the functions. For eample, (f() + g()) = (f()) + (g()) ( 9 + 1 ) 9 = ( 9 + 9) = 9 8 9 9 1 = 9 8 9 10
Engineering Mathematics 2 5 March 2014 Differentiation Rules 5. Differentiating eponential functions The erivative of the natural eponential function is equal to itself! That is, f() = e = f = e For eponential functions with ifferent bases the rule is slightly ifferent: (a ) = a ln(a) We also nee to know ( e C ) = C e C 6. Differentiating logarithmic functions The erivative of the natural logarithmic function is For bases other than e we have Also note that (ln()) = 1 (log a()) = 1 (ln(c)) = 1 1 ln(a) 7. Calculating the erivative of a function at a point To calculate the rate of change of a function at a point a, first ifferentiate the function an then let = a. For eample, the erivative of ln() at 2 is 1/2. Eamples: Differentiate (i) 6e 4 ; (ii) 4 ; (iii) 4 log() + 5; (iv)4 3e9 Eample: The voltage across the plates of a capacitor at any time t secons is given by v = V e t/cr, where V, C an R are constants. Given V = 200 volts, C = 0.10 10 6 faras an R = 2 10 6 ohms, fin (a) the initial rate of change of voltage, an (b) the rate of change of voltage after 0.2 s.
Engineering Mathematics 2 6 March 2014 Differentiation Rules 8. Differentiating trigonometric functions We have the following rules: We will also nee (sin()) = cos() (cos()) = sin() (tan()) = 1 cos 2 () = 1 + tan2 () (sin(a + b)) = a cos(a + b) (cos(a + b)) = a sin(a + b) In orer to ifferentiate the trigonometric functions, we nee to introuce the raian measure of angles. In other wors, all our angles have to be measure in raians instea of egrees. One raian is the angle subtene at the center of a circle by an arc that is equal in length to the raius of the circle. The conversion between raians an egrees is given by the following formula angle in egrees = angle in raians 180 π Eample: Convert the following angles from egrees into raians: 180 ; 120 ; 60 ; 45 ; 270 ; 1 Eample: Calculate the following values (all angles are in raians) sin(0.1) ; 0.1 sin(0.01 0.01 ; sin(0.001) 0.001 It looks like we have sin() lim 0 In other wors, when an angle is very small, the value of the sine is approimately equal to the value of the angle itself (in raians). This property is not true if we re working with egrees instea. Eample: Differentiate the following: f() = 5sin(100 0.40); f() = 2cos(5 + 0.20) Eample: An alternating voltage is given by: v = 100 sin(200t) volts, where t is the time in secons. Calculate the rate of change of voltage when (a) t = 0.005 s an (b) t = 0.01 s
Engineering Mathematics 2 19 March 2014 9. Higher orer ifferentiation The secon orer erivative of a function is efine as the erivative of the erivative of that function an enote by 2 f 2 = ( ) f For eample, 2 (45 + 6) = (204 + 6) = 80 3 By successive ifferentiation, further higher erivatives can be obtaine. f() = 4 we have f = 43, 2 f 2 = 122, 3 f 3 = 24, 4 f 4 = 24, 5 f 5 = 0 an all successive erivatives after the fifth one are zero as well. For eample, if Eample: the acceleration of a moving object is efine as the secon erivative of its position function. So, if X(t) is the istance travelle as a function of time we have V (t) = X t = spee; V A(t) = t = 2 X t 2 = acceleration Recall the eample of the falling tile whose height was a function of time h(t) = 40 4.9t 2 The istance travelle is (t) = 4.9t 2 so the spee is v(t) = t is a(t) = v t = 9.8. = 9.8t hence the acceleration Eercise: Fin the first 3 erivatives of the following functions: f() = 8 3 + 6; f() = sin(3); f() = 5 ln(). Applications of ifferentiation Maimum an Minimum problems Suppose we have a quaratic function f() = a 2 + b + c. If a is positive then we know that the graph is -shape. When is f() a minimum? We can see that the minimum occurs at the point where the tangent line is horizontal, i.e. the point such that f = 0. (The slope of a horizontal line is zero!) Eample: For what value of oes the quaratic f() = 2 2 3 take its minimum value? What is this minimum value? If the coefficient of a of 2 in the quaratic f() = a 2 + b + c is negative, the graph of f is -shape an so it will have a maimum value which occurs at the point which satisfies f = 0 What we have sai about maima an minima is not restricte to quaratic polynomials. If f is any ifferentiable function then we can look for a maimum or minimum of f by fining zeros of the erivative function. This iea can be use to solve many everyay problems.
Engineering Mathematics 2 20 March 2014 Applications of ifferentiation Maimum an Minimum problems 1. The current I (in amperes) in an alternating current circuit is given by I(t) = 2 sin(t) + 2 cos(t) where t is time (secons). circuit? What is the peak current (maimum amplitue) for this 2. A rectangular fiel has imensions by 100/. Fin any etreme values (maimum or minimum values) of the perimeter, P. 3. The height of a ball moving vertically is H(t) = 1 2 gt2 + v 0 t + s 0 If the ball is launche from an initial height of s 0 = 2m with an initial spee of v 0 = 10m/s, fin the maimum height it climbs to. 4. A rectangular bo has imensions, 10 2 an 16 2 where is between 0 an 5m. Fin the etreme values of V, the volume of the bo.