Unit No: DT5X 33 DG4H 33. Unit Title: Mathematics for Engineering 1: Outcome 1. Additional Notes. Problem Solving involving Powers and Logs

Transcription

1 Unit No: DT5X 33 DG4H 33 Unit Title: Mathematics for Engineering : Outcome Additional Notes Problem Solving involving Powers and Logs (Evaluation using a scientific calculator,transposition and Indices) Engineering and Construction RD/PKN

2 25/8/2009 RD/PKN

3 Mathematics for Engineering Applying Algebraic Methods in Problem Solving Section - Revision It is assumed that you will have acquired some knowledge of indices and of logarithm and eponential epressions in previous mathematics courses. There follows a brief summary of the work covered previously. More information and eercises are available. Laws of Indices: Eample Multiplication a m % a n = a (m+n) 3 % 2 = 5 Division am a n = a (m n) 2 5 = 3 Powers (a m ) n = a mn ( 2 ) 3 = 6 Roots a m n = n a m 3 6 = 6 3 = 2 Reciprocals ka n = k a n 5 3 = 5 3 Zero inde a 0 = (2 3 ) 0 = TYPE A Some Casio f, Teas, Teet, early Sharp TYPE B Casio VPAM, Sharp DAL TYPE C Casio S-VPAM, All Graphics calculators y 2 +/ = y ( ( ) 2 ) = 0. ( 3) 4 3 +/ y 4 +/ = ( ( ) 3 ) y ( ) 4 = or SHIFT y 3 = = 2 Note: Your y key may be displayed as», y or ^ or a b. Try these ones a. 4-3 = b. ( 2 3 )4 = c. 8 4 = /8/2009 RD/PKN

4 2. The Constant ' e', e Like, is a naturally occurring constant and has a value of To check this, enter and use the e button, ie: e = (to 0 sf) Note, like it is a transcendental number and although it looks as though it might repeat it doesn t. What follows looks completely random. Functions containing e are used in 2 closely related ways: (i) as the base of Natural (or Naperian) logarithms (ii) in the laws of Natural growth and decay 3. Definition of a Logarithm If N is a positive number such that: N = a where a is positive, then is called the LOGARITHM of N to BASE a. It is written as log a N = N = a e log a N = Remember: NUMBER = BASE LOGARITHM Eample If 25 = 5 3 then log 5 25 = 3 4. Base for Logarithms Any positive number can be the base but the usual bases are: (i) Base 0 written as log 0 Use the LOG button on the calculator. (ii) Base e Logarithms to base e are known as Natural (or Naperian) logs. Written as log e or usually ln. (John Napier a.k.a. Lord Merchiston, , Edinburgh) Use the ln button on the calculator. y = log 0 transposes to = 0 y y = 0 transposes to = log 0 y y = ln transposes to = e y y = e transposes to = ln y 25/8/ RD/PKN

5 Eamples Check using the calculator: (all answers given correct to 4 sig fig). e.6 = e 0.3 = log 0 8 = ln 4.3 =.459 Note: ln = 0 log e e = log 0 = 0 log 0 0 = ln = log e ep he 5. Laws of Logarithms. log(a%b) = log A + log B 2. log A B = log A log B 3. log A n = n log A 4. log a y = log b y log b a Eamples Epress as a single log: Note that the base is not specified - they can t be evaluated (i). log 5 + log 4 = log(5%4) = log 20 Use Law (ii). log 32 log 8 + log 24 = log 32%34 8 = log 96 Use Laws & 2 Now using Law 3 then Law (iii) 2 log log = log log 3 = log 25 3 (iv) log 2 + k log = log 2 + log k = log((2)( k )) = log(2 k+ ) Eample (iv) used Indices Rule 25/8/ RD/PKN

6 6. Graphs of Eponential and Logarithm Functions (a) the natural growth curve (b) the eponential decay curve y = e y = e 0 e e We note the following points of importance in the 2 graphs. (i) each graph is continuous (ii) e and e are each positive for all values of (iii) one graph is the mirror image of the other in the y-ais (iv) the gradient of e is positive and increases rapidly as increases (v) the gradient of e is negative and decreases rapidly as increases (c) the natural logarithm curve y = log e 5 ln( ) e 0 ln( ) The graph, y = log e eists for > 0 only. The gradient is positive and decreases as increases. Note: the graph of y = log e is the mirror image of y = e in the line y = 25/8/ RD/PKN

7 An interesting fact which you will meet later in the calculus unit is: the function only mathematical function which does not change on differentiation, d That is d (e ) = e (gradient at any point is the same as original function value). e is the (d) y = e e In this graph, since e decreases as increases, the value of y tends to approach. As increases, e becomes smaller and smaller, and e gets closer and closer to. Revision Eercise. Evaluate correct to 4 significant figures: (i) 3.5e (ii) 5 e.5 (iii) 6( e 0.4 ) (iv) 3 ln.6 (v) 3 ln Change to eponential form: (i) log 3 p = q (ii) ln 2.6 = (iii) log 0 y = b (iv) 4 = log 6 3. Change to logarithmic form: (i) e =.2 (ii) 0 = 3 (iii) 0.3 = e (iv) p = a q 4. Epress as a single log: (i) log 2 + log 3 + log 4 (ii) 3 log 2 2 log 3 (iii) 2 log log 3 (iv) 2 log log 8 (v) log(a + 2) + log a (vi) a log + 3 log 25/8/ RD/PKN

8 Section 2 - Natural Laws of Growth and Decay The Natural laws of growth and decay are of the form y = Ae k or y = A( e k ) where A and k are constants which can be either positive or negative. They relate quantities in which the rate of increase of y is proportional to y itself for the growth law and the rate of decrease of y is proportional to y itself for the decay law. These laws occur frequently in Engineering and Science. Eamples include: (i) Linear Epansion A rod of length l at temperature 0 o C and having a positive coefficient of linear epansion of will become longer when heated. The natural growth law is: l = l ( is length of rod at 0 o 0 e t l 0 C, and t is the temperature) (ii) Atmospheric Pressure The pressure p at height h above ground level is given by p = p 0 e h c where p 0 is pressure at ground level and c is a constant. (iii) Tension in Belts T The relationship between the tension in a belt round a pulley wheel and its angle of lap is of the form T = T 0 e m where m is the coefficient of friction between belt and pulley and T and T 0 are the tension on the tight and slack sides of the belt respectively. Other Eamples Include: (iv) Change of Electrical Resistance with Temperature: R = R 0 e t (v) (vi) (vii) Discharge of a Capacitor Newton's Law of Cooling s = s o e kt Radioactive Decay N = N 0 e t q = Qe t CR 25/8/ RD/PKN

9 Eamples of Growth and Decay Curves Eample The decay of voltage, v volts, across a capacitor at time t seconds is given by: v = 250e t 3. The graph is drawn over the first 6 seconds. The natural decay curve is v = 250e t t 3 250e t Note: the rate of change can be calculated by finding the gradient at any point on the curve 6 Eample 2 The growth of current in the circuit is opposed by the emf induced in the coil of inductance L. The instantaneous current i amperes after a time t seconds is given by: i = E R e Rt L. This is the graph off the function i = E R e Rt L. When t is large, e Rt L is very small and which represents the final steady current. i = E R _ E R i t 25/8/ RD/PKN

10 EVALUATION OF EXPRESSIONS We have seen how eponential functions, and hence logarithmic functions form the basis for rules for growth, decay, cooling, atmospheric pressure to name but a few. We need to be able to evaluate such epressions. Eamples Evaluate to 3 decimal places:. ln( p p 2 ) when p = 7000 and p 2 = 8000 ln( p p 2 ) = ln( ) = i 0 ( e 20t ) when i 0 = 2 and t = 0.06 i 0 ( e 20t ) = 2( e ( 20%0.06) ) =.398 Eercise Evaluate to 3 decimal places:. ln( p p 2 ) when p = 2500 and p 2 = i 0 ( e 5t ) when i 0 = 3.5 and t = p 0 e 0.5h when p 0 = and h = 5 4. N 0 e lt when N 0 = 500, l = 0.25 and t = A ln( c c 2 ) when c = 50, c 2 = 38 and A = KRT log 0 A when K = 5, R = 0.4, T = 56 and A = 35 25/8/ RD/PKN

11 SECTION 3: INDICIAL, LOGARITHMIC AND EXPONENTIAL EQUATIONS When solving Indicial, Eponential and Logarithm equations it is usually necessary to re-arrange the equation by changing its form (i.e. Carry out trasposition). Certain recognised methods can be used. Simple eamples of these are as follows: (a) Type k = e Working: k = e (Remember NUMBER = BASE LOG ) ln k = or log e k = (Change to log form to "undo" from the inde position) (b) Type log e = q ln = q You might see either form Working: log e = q ln = q (Change to eponential form to "undo" from log position) = e q = e q (c) Type log a p + log a q = log a Working: log a p + log a q = log a log a (p%q) = log a (Use first log law) pq = (d) Type p = q (ie, with base other than 0 or e) Working: p = q (Taking log of both sides use base 0 or base e) log 0 p = log 0 q log 0 p = log 0 q Log law 3 = log 0 q log 0 p /8/ RD/PKN

12 Worked Eamples (a) Type k = e Solve for : (i) 5e 0.3 = 4. (i) 5e 0.3 = 4. e 0.3 = = ln 4. 5 = ln ( 4. ) = divide by 5 log e both sides divide by (ii) 220( e 0.5 ) (ii) 220( e 0.5 ) = 48 e 0.5 = = e 0.5 ln = 0.5 ln( 48 ) 220 = = 4 sf divide by 220 add e 0.5 to both sides log e both sides divide by 0.5 Try the following eamples: Eercise 2 Solve for (to 4 significant figures): (i) e 2 = 5.6 (ii) 5e 0.5 = 4. (iii) e = 0.6 (iv) 00( e 0.3 ) = 80 4 sf 25/8/ RD/PKN

13 (b) Type log e = q log 0 = q Solve for : (i) log 0 ( + ) = 2.3 (i) log 0 ( + ) = = = = 98.5 (ii) 2 ln + ln 5.3 = both sides (ii) 2 ln + ln 5.3 = 4.6 ln 2 + ln 5.3 = 4.6 ln 2 = 4.6 ln (4.6 ln 5.3) = e (4.6 ln 5.3) = e = (iii) 4 ln 6.8 = 3.2 e both sides only + ve sol n. for real answer 4 ln 6.8 = 3.2 ln 6.8 = 0.8 (iii) 6.8 = e 0.8 is under the line! 6.8 = e 0.8 % 6.8 e 0.8 = = Eercise 3 Solve for (i) log 0 = 3 (ii) log 8 = 4 (iii) log 0 (3 + ) = 0.8 (iv) 3 ln ln 5 = 2.4 (v) 6 log = /8/2009 RD/PKN

14 (c) Type log a p + log a q = log a Eample Solve for : (i) log( 3) + log = log 4 Working: log(( 3)) = log 4 ( 3) = = 0 Use first log law This is a quadratic equation which can be solved either by (i) factorisation or (ii) quadratic formula = 0 (i) Factorisation Method ( + )( 4) = 0 + = 0 4 = 0 = = 4 But log(-) does not eist so = 4 (ii) Quadratic Formula For a 2 + b + c = 0 formula is = b! b2 4ac 2a For equation = 0 a = b = -3 c = -4 = ( 3)! (( 3) 2 4%%( 4)) 2% 3! 25 = 2 = 3!5 2 = 4 or = Only feasible answer is = 4 Eercise 4 Solve for (3 decimal places): (i) log( 5) + log = log 6 (ii) 2 log(3) log = log.6 25/8/ RD/PKN

15 d. Type p = q Solve for : (i) 3 = = 5.6 log 0 3 = log log 0 3 = log = log 0 3 log =.568 Using log 0 and 3rd log law (ii)(ii) 5 ( ) = 3. Use natural logs this time (for a change). (You could use base 0 logs instead). 5 ( ) = 3. ln(5 ( ) ) = ln(3. ) ( ) ln(5) = ln(3.) ln(5) ln(5) = ln(3.) (ln(5) ln(3.)) = ln(5) = ln(5) (ln(5) ln(3.)) = Log law Eercise 5 Solve for (to 3 decimal places): (i) 3 = 2 (ii) 2.4 = 3.6 ( 2) (iii).8 (+3) = 6.52 (iv) 3.5 = (v).3 (2 ) = 6.3 (+) /8/ RD/PKN

16 SECTION 4: CHANGING THE SUBJECT OF FORMULAE Sometimes it is necessary to change the subject of a formula. If the formula contains logarithm, eponential or indicial epressions, certain procedures are used. A. Formula Containing Several Logarithms Eample : Re-arrange the formula lna - lnb = c to make a the subject ln a ln b = c ln a b = c a b = ec a = be c Second log law, undo ln with e Eample 2: Re-arrange ln (y + ) = ln ( + 2) + lna to make y the subject ln(y + ) = ln( + 2) + ln A = ln(a( + 2)) y + = A( + 2) y = A( + 2) First log law Eample 3: Re-arrange 2 ln y = 5 + ln3y for y 3 ln y = 5 + ln(3y) ln y 3 = 5 + ln(3y) ln y 3 ln(3y) = 5 ln y3 3y = 5 y 3 3y = e(5) yyy/ = e (5) 3y/ yy 3 = e(5) y 2 = 3e (5) y = (3e (5) ) Log law 3 then 2 and some algebra 25/8/ RD/PKN

17 B. Formula Containing a Logarithm of More Than One Variable Eample: Re-arrange the formulae L = k log 0 d 2 d a. for d 2 and b. for d Working for a: L = k log 0 d 2 d L k = log 0 0 L k = d 2 d d 0 L k = d 2 d 2 d continuing for b: d 0 L k = d 2 d = d 2 0 L k C. Formulae Containing Variable in Inde Form Eample: Re-arrange the formula l = l e µq to make µ the subject 0 Working: log e I = I 0 e ( q) I I 0 = e ( q) I I0 I I 0 = q log e q = log e is the same as ln 25/8/ RD/PKN

18 Eample: change the subject of i = E R e t CR to t i = E R e t CR ir = E e t CR ir E = e t CR e t CR + ir E = e t CR = ir E CR t = ln ir E This is a difficult but common problem t = CR ln ir E t = CR ln ir E Eample: Change the subject of PV n = C to n Working: pv n = c v n = c p log 0 (v n ) = log 0 c p (Note: base is not e or 0). Log law 3 n log 0 (v) = log 0 c p n = log 0 ( c p ) log 0 (v) /8/ RD/PKN

19 Eercise 6: Changing the Subject of Formulae Re-arrange the following to give a formula in the letter indicated in brackets. (i) a = log 0 log 0 y [] (ii) 2 log 0 a + 3 log 0 b = c [b] (iii) T = ln + ln y [y] (iv) 2 ln ln y = z [] (v) p = 3 (ln ln y) [] (vi) ln(y + ) = sin + ln A [y] (vii) ln(y 3) = ln( + 2) + ln a [y] (viii) 2 ln y = 3 + ln(4y) [y] (i) Q = k log 0 ( p 2 p ) [p 2 ] (ii) p = A log e ( v 2 v ) [v ] (iii) m = ln T T 2 [T ] (iv) a = 2 ln( y ) [y] (i) v = we t [t] (ii) T T 2 = e s [s] (iii) = o e kt [t] (iv) i = Ie kt [t] (v) i = E ( R e kt ) [t] (vi) P = P o e h c [c] 25/8/ RD/PKN

20 SECTION 5: PROBLEMS INVOLVING EXPONENTIAL AND LOGARITHM FUNCTIONS. The pressure p at height h above ground level is given by P = P o e h c, where P o is the pressure at ground level and c is a constant. Whe P o is 0.3%0 3 pascals and the pressure at a height of 570 metres is 98.7%0 3 pascals, determine the value of c. Working: Use your solution to Eercise 6, 3 (vi) and substitute in the numbers c = The temperature q o 2 C of an electrical conductor at time t seconds is given by: q 2 = q ( e t T ), where q is the initial temperature and T seconds is a constant. Determine: Working: a. q when q 2 = 50 o C, t = 30 seconds and T = 80s b. the time t for q 2 to fall to half the value of q if T remains at 80s a. b. q 2 = q ( e t T ) q 2 ( e t T ) = q 50 = e q = 59.9 o C q 2 = 2 q 2 q = q ( e t T ) 2 = e t T e t T = 2 = 2 t T = ln 2 t = T ln 2 = 80 ln 2 = It takes seconds for q 2 to fall to ½ value of q 25/8/ RD/PKN

21 Eample 3: Calculate T from the Formula = 0.5, T 2 =25, q = 3.3 q = log e T 2 T when q = log e T 2 T q = log e T 2 T e (q ) = T 2 T T e (q ) = T 2 T = T 2 e (q ) = 25 e (3.3%0.5) = = 7.00 Problems can often be solved graphically as well as theoretically. Eample 4 The formula i = 2( e 0t ) gives the relationship between the instantaneous current, i amperes, and the time, t seconds, in an inductive circuit. Plot a graph of i against t, taking values of t from 0 to 0.3 seconds at intervals of 0.05 seconds. Hence find: a. the initial rate of growth of the current when t = 0, and; b. the time taken for the current to increase from to.6 amperes Verify these results using theoretical methods. Solution by Graphical Method A table of values of for i calculated. t from 0 to 0.3 seconds is drawn up and the corresponding values t ( e 0t ) A graph of the results is shown below. 25/8/ RD/PKN

22 M O N a. When t = 0, the initial rate of growth will be given by the gradient of the tangent at 0. The tangent at 0 is the line OM and its gradient may be found by using a MN suitable right angled triangle MNO and finding the ratio. Initial rate of growth of: i = MN ON = amperes 0.05 seconds = 20 amperes per second ON b. Point P on the curve corresponds to a current of.0 amperes and the time at which this occurs is read from the t scale and is 0.07 seconds. Similarly point Q corresponds to current of.6 amperes at a time of 0.6 seconds. Hence time between P and Q is = 0.09 seconds, ie the time for the current to increase from to.6 amperes is 0.09 seconds continued on net page 25/8/ RD/PKN

23 Solution by Theoretical Methods a. The rate of growth of current can be found by using calculus. [Students who have already completed HN Unit Calculus will understand the solution]. This method of solution is not a required part of the present unit. For interest only: a. Differentiating to find the rate of change of current with time: i = 2( e ( 0t) ) = 2 2e ( 0t) di dt = 20e( 0t) When t = 0 di dt = 20e( 0t) = 20 Hence when t = 0 seconds, the rate of change of current is 20 amp/second. b. Wheni =.6 A When i = 2( e ( 0t) ) i 2 = e( 0t) e ( 0t) = i 2 0t = ln 2 i t = ln ( i ) 2 0 t = ln (.6 ) 2 0 t = = 60.9%0 3 seconds i =.0 A t = ln (.0 2 ) 0 = t = 69.3%0 3 seconds Time taken for current to increase from.0 A to.6 A is seconds (92 ms). 25/8/ RD/PKN

24 Eercise 7: Practical Problems. In an eperiment involving Newton's law of cooling the temperature s( o C) is given by: =. Find the volume of constant when = 56.6 o C, = 6.5 o o e ( k t) k o C and t = 83.0 seconds. 2. The length l metres of a metal bar at temperature T o Ct o C is given by: l = l 0 e T, where and are constants. Determine: l 0 a. the value of when l =.993m, l 0 =.894m and T = 250 o C and; b. the value of l 0 when l = 2.46, T = 30 o C and = 68.2% Given W = nb, find the value of when W = 4300, B = 900 and n =.296% The equation i = 2.4e ( 6t) gives the relationship between the instantaneous current, i ma, and the time, t seconds. Plot a graph of i against t for values of t from 0 to 0.6 seconds at 0. second intervals. Use the curve obtained to find: a. The rate at which the current is decreasing when t = 0.2 seconds and; b. The time when the current is 0.54 ma. Verify your answer to (b) by a calculation method. 5. N = N 0 e epresses the decay of radioactive material used in a nuclear moisture/density meter for soil compaction control. If N = 2 N 0, calculate the value of. 6. The atmospheric pressure P at a height h (in km) above sea level obeys the rule: P = P 0 e 0.5h where P 0 is the pressure at sea level and equals pascals. a. Find the pressure at a height of 2 km. b. At what height is the pressure equal to half the value at sea level? c. At what height is the pressure equal to one-tenth of the value at sea level? 25/8/ RD/PKN

25 7. The current i in a circuit is described by the law: i = i 0 ( e 20t ) where i 0 is equilibrium current and t is the time in seconds. Find the value of t when i = 2 amperes if i 0 = 2.5 amperes. 8. The work done in the isothermal epansion of a gas from pressure p to p 2 is given by the formula: w = w 0 ln p p 2 If the initial pressure p is 7000 pascals find the final pressure p 2 if w = 3w If f = p 4 calculate the value of when = 600, = 0.56, t 2 log D 3 e d + D p t d =.4 and f = The relation between pressure p and volume v of a gas is pv n = c. Determine the value of n given that p = 200 when v = 0.8 and c, the constant is Quantities p and q are related by the equation p = 7.43 e kq t, where k and t are constants. Find t when k = 37%0 3, q = 72.8 and p = /8/ RD/PKN

26 Answers Revision eercise (i) (ii) (iii).978 (iv).40 (v) (i) p = 3 q (ii) 2.6 = e (iii) y = 0 b (iv) 4 = 6 3(i) = ln.2 (ii) = log 0 3 (iii) = ln 0.3 (iv) log a p = q 4(i) log 24 (ii) log( 8 ) 9 (ii) log 2 3 (iv) log = 0 (v) log(a(a + 2)) ) (vi) log( (a+3) ) Eercise (4 sf) Eercise 2 (i) (ii) (iii) (iv) Eercise 3 (i) 000 (ii) 3 (iii) 3.30 (iv) (v) Eercise 4 (i) 6 (ii) Eercise 5 (i) (ii) (iii) (iv) (v) /8/ RD/PKN

27 Eercise 6 (i) = 0 3 y (ii) b = 3 0 (2c) a (iii) y = y = et (iv) = ye z (v) = y e 3p (vi) y = Ae (sin ) (viii) y = a( + 2) + 3 (viii) y = 4e 3 2(i) p 2 = 0 Q k p (ii) v = v 2 e p A (iii) T = T 2 e m (iv) y = e ( 2a) 3(i) t = log e ( V ) W (ii) s = ln T T 2 (iii) k = t ln( 0 ) (iv) t = ( k ) ln( i I ) (v) t = ( k ) ln( ( Ri )) E (vi) c = h ln p po Eercise % = 203.8%0 6, l 0 = mas, pascals, 4.62 km, 5.35 km % /8/ RD/PKN