UNIT 1: ANALYTICAL METHODS FOR ENGINEERS

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1 UNIT : ANALYTICAL METHODS FOR ENGINEERS Unit code: A/60/40 QCF Level: 4 Credit value: 5 OUTCOME 3 - CALCULUS TUTORIAL DIFFERENTIATION 3 Be able to analyse and model engineering situations and solve problems using calculus The calculus: the concept of the limit and continuity; definition of the derivative; derivatives of standard functions; notion of the derivative and rates of change; differentiation of functions using the product, quotient and function of a function rules; integral calculus as the calculation of area and the inverse of differentiation; the indefinite integral and the constant of integration; standard integrals and the application of algebraic and trigonometric functions for their solution; the definite integral and area under curves Further differentiation: second order and higher derivatives; logarithmic differentiation; differentiation of inverse trigonometric functions; differential coefficients of inverse hyperbolic functions Further integration: integration by parts; integration by substitution; integration using partial fractions Applications of the calculus: e.g. maima and minima; points of infleion; rates of change of temperature; distance and time; electrical capacitance; rms values; electrical circuit analysis; ac theory; electromagnetic fields; velocity and acceleration problems; comple stress and strain; engineering structures; simple harmonic motion; centroids; volumes of solids of revolution; second moments of area; moments of inertia; rules of Pappus; radius of gyration; thermonamic work and heat energy Engineering problems: e.g. stress and strain; torsion; motion; namic systems; oscillating systems; force systems; heat energy and thermonamic systems; fluid flow; ac theory; electrical signals; information systems; transmission systems; electrical machines; electronics You should judge your progress by completing the self assessment eercises. The basic calculus should have been covered at national level so it is puling why basic definitions are required by this syllabus. Most of this tutorial is the same as the tutorial for the national level module further mathematics for technicians. A brief cover of the basics is included here before going on to the more advanced work required for the higher level. D.J.Dunn

2 . REVISION OF BASIC DIFFERENTIATION LIMITS Consider the equation y. If we plot this function we get a curve stretching away to infinity at one end and ero at the other as shown. The smaller the value of, the larger the value of y. We can write L t meaning that in the limit as tends to ero, / approaches 0 infinity. Also note that the larger the value of the smaller the value of y. We can also write L t 0 meaning that in the limit as tends to infinity, / approaches ero. There are several techniques for finding limiting values. GRADIENTS AS LIMITS Consider the graph shown of y = f(). Suppose we want to find the gradient at point A. To do this choose a point at + Δ Δy f( Δ) f() Gradient Δ Δ If we make Δ smaller Δ becomes δ meaning a much smaller change. If we make it so small that the change approaches ero we use and and the gradient becomes eactly and this is called the differential coefficient. The ratio at the limit is called the differential coefficient and is denoted L t 0 f( Δ) f() Δ This method can be used to prove many types of differential coefficient. D.J.Dunn

3 . DIFFERENTIATION OF POLYNOMIALS Consider the polynomial case when y = f() = n It was shown in an earlier tutorial that: ( + a) n = a n + n C n- a n- + n C n- a n- + n C n-3 3 a n n If we replace a with Δ we get the epansion: ( + Δ) n = Δ n + n C n- Δ n- + n C n- Δ n- + n C n-3 3 Δ n n n n n n n n Δ C Δ C Δ n Δy ( Δ) n n Δ Δ Δ Δy n- n n n n Δ C Δ C Δ 3 n n.. Δ Δy Lt 0 Δ will always be n C n- = n n- For eample if n = 3 we are left with: Δy Δ... and the only term left will be the last one in the series containing (Δ) 0 = and this 3 3 Δ C Δ C Δ 0 Δ Δ Δ 0 In the limit as Δ0 Δy Δ 3 (3)() n- The differential coefficient of a polynomial is hence n The differential coefficient is the gradient of the function at a given point. This may also be defined as the RATE OF CHANGE with respect to the variable so is the rate of change of y or f() with. If the variable is time then we have the rate of change with time which is of course the velocity. 3. HIGHER ORDER DIFFERENTIALS Consider the function y = 3. The graph looks like this. (3) n n The gradient of the graph at any point is / =3. This may be evaluated for any value of. If we plot / against we get the following graph. This graph is also a curve. We may differentiate again to find the gradient at any point. This is the gradient of the gradient. We write it as follows. d d y 6 The graph is a straight line as shown with a gradient of 6 at all points. If we differentiate again we get d y d 3 d y 6 3 D.J.Dunn 3

4 WORKED EXAMPLE No. The distance moved by a bo (in metres) with uniform acceleration is given by s = 5t. Find the distance moved, velocity and acceleration after seconds. distance s 5t 70 m ds velocity v 0t 0 m/s acceleration dv d s 0 m/s 4. RULES IF INTEGRATION 4. DIFFERENTIATION OF A SUM This is straight forwards, each term is differentiated separately so if for eample y = u + v + w +... / = du/ + dv/ + dw/ +... WORKED EXAMPLE No. Find the gradient of the curve y = at the point = / = and at point, / = 3() + 6() - = 3 4. FUNCTION OF A FUNCTION If a variable y depends on a second variable u which in turn depends on a third variable, then du and this is the chain rule. du du If we have y = f(u) and u = f () then we find that: f(u) f() f (u) and f() du Put another way, we can substitute a function into another function to simplify the differentiation process. WORKED EXAMPLE No. 3 Given that y = ( ) find at the point = Substitute u = f() = ( ) so y = f(u) = u du f () 3 f (u) u du f(u) f() u( 3) = ( )( + 3) At = (4 6 )(4 3) 54 D.J.Dunn 4

5 4.3 DIFFERENTIATION OF A PRODUCT When it is difficult to multiply out an epression we can differentiate with the following rule. dv du y = u v u v WORKED EXAMPLE No. 4 Find the gradient of the curve y = ( 3 + )( + ) at the point = Let u = ( 3 + )and v = ( + ) 3 dv du 3 u v () (3 Put = ()() (3)(3) 3 ) 4.4 DIFFERENTIATION OF A QUOTIENT v u This rule helps us differentiate a function of the form y = u/v v WORKED EXAMPLE No. 5 Find the gradient of the curve u v 4 3 du v u v dv 4 at the point = 3 y du 4 dv Put = 3 ( 4) ( 4) D.J.Dunn 5

6 5. DIFFERENTIATION OF EXPONENTIALS AND LOGARITHMS 5. EXPONENTIAL In outcome (tutorial ) it was shown that the gradient of the function f() = e has a gradient at all points the same as the function itself, so it follows that if y = f() = e then Consider the equation y = e where = a d d ae ae a d e and a d e 5. NATURAL LOGARITHMS Consider the equation y = ln() from which = e y and Inverting we have 5.3 OTHER LOGARITHMS To integrate base 0 logarithms we use the conversion log() = {ln()}log(e) = ln() Hence if y = log() then 6. DIFFERENTIATION OF CIRCULAR FUNCTIONS y = sin(θ) show cos θ e The diagram shows an arc of length Rθ and a chord. As θ becomes small the length of the chord and arc become the same. Also the chord becomes the opposite side of a right angle triangle so as θ0 RθRsin(θ) hence sin(θ) θ (radian) y d sin( θ) To find we use f(θ Δθ) f( θ) Lt Δθ 0 L t 0 sin(θ Δθ) sin( θ) Δθ Using the trigonometric identity sin A sin B = sin {½ (A B)} cos {½ (A + B) } A becomes (θ + Δθ) and B becomes θ sin θ Δθ θ cos θ Δθ θ L t 0 Δθ sinδθ/ cos θ Δθ/ L t 0 ( Δθ)/ As Δθ/0 sin(θ/) θ/ (radian) as eplained above. Δθ/ cosθ Δθ/ L t Lt cosθ Δθ/ cos θ ( Δθ)/ 0 0 D.J.Dunn 6

7 y = cos(θ) show sin θ d cos( θ) To find we use L Using the trigonometric identity t 0 f(θ Δθ) f( θ) cos(θ Δθ) cos( θ) L t Δθ 0 Δθ cos A cos B = sin {½ (A + B)} sin {½ (A B)} A becomes (θ + Δθ) and B becomes θ - sin θ Δθ θ sin θ Δθ θ L t 0 Δθ -sinθ Δθ/ sin Δθ/ L t 0 ( Δθ)/ As Δθ/0 sin(θ/) θ/ (radian) as eplained above. -sinθ Δθ/ Δθ/ L t Lt -sinθ Δθ/ sinθ 0 ( Δθ)/ 0 y = tan (θ) Show cos ( θ) tan(θ) = sin(θ)/cos(θ) so using the preceding proofs for sin and cos we can use the quotient rule to complete this. Let u = sin(θ) and v = cos(θ) cos(θ) and sin(θ) and note that cos ( θ) sin ( θ) v u cos(θ) cos(θ) sin(θ)sin(θ) cos (θ) sin (θ) v cos (θ) cos (θ) cos (θ) sinθ y = sec (θ) Show cos ( θ) sec(θ) = /cos(θ) so we can use the quotient rule to complete this. Let u = and v = cos(θ) v u cos(θ)(0) sin(θ) sin (θ) 0 and sin(θ) v cos (θ) cos (θ) y = cosec (θ) Show tan( θ)sin( θ) cosec(θ) = /sin(θ) so we can use the quotient rule to complete this. Let u = and v = sin(θ) 0 and cos(θ) v u sin(θ)(0) cos(θ) cos(θ) cos(θ) v sin (θ) sin (θ) sin(θ)sin(θ) tan( θ)sin( θ) y = cotan (θ) Show sin (θ) cotan(θ) = /tan(θ) so we can use the quotient rule to complete this. Let u = and v = tan(θ) du v u dv tan(θ)(0) cos (θ) 0 cos ( θ) tan (θ) = sin (θ)/cos (θ) v tan (θ) cos (θ) cos (θ)sin (θ) sin (θ) D.J.Dunn 7

8 WORKED EXAMPLE No. 6 The distance moved by an oscillating bo is related to time by the function: =. sin(t) mm. Find the distance moved, velocity and acceleration after 0.3 seconds. distance.sin(t).sin (0.6) (.)(.5646) mm velocity v ()(.)cos(t) (.4)cos(0.6).98 mm/s dv d acceleration ()()(.)sin(t) - 4.8sin(0.6) -.7mm/s WORKED EXAMPLE No. 7 Given that y = 4 cos(5 - ) find the equation for Substitute u = f() = (5-) so y = f(u) = 4 cos(u) du f () 5 f (u) -4sin(u) du f(u) f() -4sin(u)(5) = -0 sin (5-) SELF ASSESSMENT EXERCISE No.. Evaluate the first and second derivative of the function p = 8e -0.t when t =. (Answers and 0.5). The motion of a mechanism is described by the equation = 50 Cos(0.5t) mm. Calculate the distance, velocity and acceleration after 0.3 seconds. (Answers mm, mm/s and -.36 mm/s ) 3. Evaluate the first and second derivatives of the function = when = 4 (Answers 658 and 456) 4. The motion of a bo is described the equation = Asin(t) where is the distance moved and t is the variable time. Show by successive differentiation and a substitution that the acceleration is given by a = -. D.J.Dunn 8

9 5. Given y = ( ) 4 find the equation for 6. Given y = sin(θ ) find the equation for 4( ) 3 ( -4) (θ cos θ) 7. Given y = 5cos(θ 3 ) find the equation for (-5θ sin 3 θ) 8. Given V = 3sin(θ) + cos(θ) evaluate dv/ at θ = 30 o (.598) 9. Given y = e -0.t + ln(t) evaluate / at t = 0. s (9.80) 0. Given y = 5 + e evaluate / at = (4.778). Given y = ( + 3)( - ) find / ( ). Given F = ( )(e ) find df/ e ( + ) 3. Given y = sin(t) cos(4t) find / cos(t) cos(4t) 4 sin(t) sin(4t) 4. Given y find ( +) - 5. Given y find -( -) Given y find at = 43/6 e 7. Given y find at = Given y (cos ) find - cos()sin() 3 9. Given y ( 5 3)(6cos ) find (8 + 30) cos () - ( ) cos() sin() 0. Given y e 4 3 find 4e 4 6e D.J.Dunn 9

10 6. DIFFERENTIATION OF INVERSE CIRCULAR FUNCTIONS This is easily confused with /sin, /cos and /tan which are the reciprocal rather than the inverse of the function. This is further confused because sometimes the inverse functions are written as sin -, cos - and tan -. The modern way of writing inverse functions is shown below. An inverse trigonometric function is the function that returns the angle given the sine, cosine or tangent. These are written as follows. If = sin(θ) then θ = arcsin() If = cos(θ) then θ = arccos() If = tan(θ) then θ = arctan() From our triangle we have b/c = sin(θ). For simplicity let b/c = = sin(θ) θ = arcsin() cos(θ) and hence cos(θ ) d arcsin( θ) cos( θ) sin (θ) Clearly this can only apply to < without having to consider the root of a negative number. In the same way we can show:- d arccos( θ) and d arctan( θ) 7. DIFFERENTIATION OF HYPERBOLIC FUNCTIONS These were defined in an earlier tutorial as: e e e e sinh () e e sinh () and cosh () and tanh () cosh () e e d y y = sinh(θ) show that cosh(θ) θ θ θ θ e e dsinh ( θ) e e sinh ( θ) cosh(θ) d y y = cosh(θ) show that sinh(θ) θ θ θ θ e e dcosh ( θ) e e cosh ( θ) sinh(θ) d y y = tanh(θ) show that cosh (θ) sinh ( θ) tanh ( θ) Let u = sinh(θ) and v = cosh(θ) cosh ( θ) cosh( θ) sinh( θ) v u anh ( θ) cosh (θ) sinh (θ) v cosh (θ) cosh (θ) D.J.Dunn 0

11 WORKED EXAMPLE No.8 Differentiate y = sinh( + ). Let + = d d d y = sinh() cosh() d cosh() cosh( ) 8. DIFFERENTIATION OF INVERSE HYPERBOLIC FUNCTIONS An inverse hyperbolic function is the function that returns the value given the sinh, cosh or tanh. These are written as follows. If = sinh() then = arcsinh() If = cosh() then = arccosh() If = tanh() then = arctanh() also written as = sinh - () also written as = cosh - () also written as = tanh - () Let = sinh() = arcsinh() d cosh() d cosh(θ ) Since cosh () sinh () = then d cosh( ) d arcsinh( ) sinh () In the same way we can show:- d arccosh( ) and d arctanh( ) SELF ASSESSMENT EXERCISE No. d arcsinh( ). Prove that. Differentiate sinh () with respect to Answer sinh() cosh() - cosh() 3. Differentiate with respect to Answer 3 sinh () sinh () 4. Differentiate e 4 sinh() with respect to Answer 4e 4 sinh()+e 4 cosh() 5. Differentiate arcsinh(4) with respect to Answer 4( + 6 ) -½ 6. Differentiate arctanh( ) with respect to Answer 4 - D.J.Dunn

12 TABLE OF DIFFERENTIAL COEFFICIENTS ae k Eponential and Logarithmic n(a) log() then Trigonometric function sin(a) acos(a) k ake Differential Coefficient Differential Coefficient cos(a) tan() sec() cosec() cot() arcsin() arccos() arctan() sinh() cosh() tanh() sech() cosech() coth() arcsinh() arcchosh() arctanh() Hyperbolic function -asin(a) cos - () or sec () or tan () sec() tan() -cosec() cot() -sin - () or cosec () Differential Coefficient cosh() sinh() or sech () cosh () -sech() tanh() -cosech() coth() -cosech () D.J.Dunn