Differential calculus

Transcription

1 7.1 Kick off with CAS 7 7. Review of differentiation techniques Differential calculus 7.3 Applications of differentiation 7. Implicit and parametric differentiation 7.5 Second derivatives 7. Curve sketching 7.7 Derivatives of inverse trigonometric functions 7. Related rate problems 7.9 Review

2 7.1 Kick off with CAS Eploring differentiation with CAS In this topic we will be revising differentiation techniques and differentiating the inverse trigonometric functions. 1 Use CAS to determine: a d csin 1 a 3 b d b d csin 1 a 3 b d c d csin 1 a 5 b d. Can ou predict d csin 1 a a b d where a and b are positive real numbers? b Use CAS to determine: a d ctan 1 a 3 b d b d ctan 1 a 3 b d c d ctan 1 a 5 b d. Can ou predict d ctan 1 a a b d where a and b are positive real numbers? b 3 Use CAS to determine: a d ccos 1 a b d Can ou predict d ccos 1 a a b d? b d ccos 1 a 3 b d Please refer to the Resources tab in the Prelims section of our ebookplus for a comprehensive step-b-step guide on how to use our CAS technolog.

3 7. WORKED EXAMPLE 1 THINK Review of differentiation techniques Introduction In the Mathematical Methods course, the derivatives of functions and the standard rules for differentiation are covered. If = f() is the equation of a curve, then the gradient function is given b f( + h) f() = f () = lim. Using this formula to h 0 h obtain the gradient function or the first derivative is called using the method of first principles. Usuall the standard rules for differentiation are used to obtain the gradient function. In this section these fundamental techniques of differentiation are revised and etended. The table below shows the basic functions for = f() and the corresponding gradient functions for = f (), where k and n are constants and R. Function: = f() n sin(k) cos(k) e k log e () Gradient function: = f () n n 1 k cos(k) k sin(k) ke k The chain rule The chain rule is used to differentiate functions of functions. It states that if = f() = g(h()) = g(u), where u = h(), then = du du = g (u)h (). A quick proof is as follows. Let δ be the increment in and δ, and δu be the corresponding increments in and u. Provided that δu 0 and δ 0, δ δ = δ δu δu δ Now in the limit as δ 0, δu 0, δ δ, δ δu δu and du δ du, so that = du du. a If f () = " + 9, find the value of f ( 1). b Differentiate cos (3) with respect to. WRITE a 1 Write the equation in inde form. a f() = " = ( + 9) 1 Epress in terms of u and u in terms of. = u where u = MATHS QUEST 1 SPECIALIST MATHEMATICS VCE Units 3 and

4 3 Differentiate with respect to u and u with respect to. du = 1 1 u and du = = 1!u Find f () using the chain rule. f () = du du = 1!u 5 Substitute back for u and cancel factors. f () = " + 9 Evaluate at the indicated point. f ( 1) = 7 In this case we need to rationalise the denominator. State the final result. This represents the gradient of the curve at the indicated point. b 1 Write the equation, epressing it in inde notation.!( + 9) f ( 1) = f ( 1) =!13!13!13!13 The chain rule can also be used in conjunction with other mied tpes of functions. 13 b = cos (3) = (cos(3)) Epress in terms of u and u in terms of. = u where u = cos(3) 3 Differentiate with respect to u and u with respect to. Find using the chain rule. 5 Substitute back for u and state the final result. WOrKeD example think a 1 Write the equation. a If f () = sina b, find f a π b. du = u3 and du = 3 sin(3) b Differentiate e cos() with respect to. = du du = u3 3 sin(3) = cos3 (3)sin(3) WritE a f() = sina b Epress in terms of u and u in terms of. Let = sin(u) where u = = 1. 3 Differentiate with respect to u and u with respect to. du = cos(u) and du = = Topic 7 DIfferenTIaL CaLCuLus 305

5 Find f () using the chain rule. f () = du du = cos(u) = cos(u) 5 Substitute back for u. f () = cosa b Evaluate at the indicated point and simplif. f a π b = cos q a r b π π = π 3 cosaπ b 7 Substitute for the trigonometric values. f a π b = π 9 1 State the final result. f a π b = π 9 b 1 Write the equation. b Let = e cos(). Epress in terms of u and u in terms of. = e u where u = cos() 3 Differentiate with respect to u and u with respect to. Find using the chain rule. 5 Substitute back for u and state the final result. du = eu and du = sin() = du du = eu sin() = sin()ecos() The product rule The product rule states that if u = u() and v = v() are two differentiable functions of, and = u. v, then = u dv + v du. A quick proof is as follows. Let δ be the increment in and δu, δv and δ be the corresponding increments in u, v and. Then + δ = (u + δu)(v + δv) Epanding, + δ = uv + uδv + vδu + δuδv Since = u. v, δ = uδv + vδu + δuδv Divide each term b δ: δ δ = u δv δ + v δu δ + δu. δv δ Now in the limit as δ 0, δu 0, δv 0, δ δ, δv δ dv, δu δ du and and δu. δv 0. δ Hence, = u dv + v du. 30 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

6 WOrKeD example 3 a If f() = sin(), find f (). b Find d [5 e ]. think a 1 Write the equation. WritE a = f() = sin() = u. v State the functions u and v. u = and v = sin() 3 Differentiate u and v with respect to. Find f () using the product rule. Substitute for u, dv du, v and. du = 3 and dv = cos() dv f () = u + v du = cos() + 3 sin() 5 Simplif b taking out the common factors. f () = 3 ( cos() + sin()) b 1 Write the equation. b Let = 5 e = u. v. State the functions u and v. u = 5 and v = e 3 Differentiate u and v with respect to. Find using the product rule. Substitute for u, dv du, v and. 5 Simplif b taking out the common factors. Alternative notation The product rule can be used without eplicitl writing the u and v s. The setting out is similar, but an alternative notation is used. du = 5 and dv = e = u dv + v du = 5 e + e 5 = e (5 ) b 1 Write the equation. b d [5 e ] Use the product rule. = 5 d (e ) + e d (5 ) 3 Find the derivatives. = 5 e + e 5 Simplif b taking out the common factors and state the final result. d [5 e ] = e (5 ) The quotient rule The quotient rule states that if u = u() and v = v() are two differentiable functions of and = u v du, then v = u dv. A quick proof is as follows. v Topic 7 DIfferenTIaL CaLCuLus 307

7 Let = u v = u. v 1. Now b the product rule, = u d (v 1 ) + v 1 du. WOrKeD example think a 1 Write the equation. Using the chain rule, = u d dv (v 1 ) dv + du v 1 = uv dv + du v 1 v du u dv dz = v a Differentiate 3 sin() with respect to. b If f() = 3, find f (). " + 9 WritE a Let = 3 sin() = u v. State the functions u and v. u = 3 sin() and v = 3 Differentiate u and v with respect to. Find using the quotient rule. Substitute for u, dv du, v and. 5 Epand and simplif b taking out the common factors. Cancel the common factors and state the final result in simplest form. du dv = cos() and = 3 du v = u dv v = cos() 3 sin() 3 ( ) = 13 ( cos() sin()) = 3 cos() sin()) 5( 3 b 1 Write the equation. b Let f() = = u " + 9 v. State the functions u and v. u = 3 and v = " Differentiate u and v with respect to, using the chain rule to find dv. See Worked eample 1a for dv. du dv = 3 and = " MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

8 Find f () using the quotient rule. Substitute for u, dv du, v and. 5 Simplif b forming a common denominator in the numerator. Epand and simplif the terms in the numerator. Derivative of tan(k) The quotient rule can be used to find the derivative of tan(k). Let = tan(k) = sin(k) cos(k) = u, where u = sin(k) and v = cos(k). v Then du dv = k cos(k) and = k sin(k). Using the quotient rule, = v du u dv Hence, d (tan(k)) = k sec (k) = v f () = cos(k) k cos(k) sin(k) k sin(k) = (cos(k)) = k(cos (k) + sin (k)) since cos (k) + sin (k) = 1 cos (k) k = cos (k) = k sec (k) k cos (k). du v u dv v 3" f () = " + 9 (" + 9) f () = f () = 3( + 9) 1 " " Simplif using a = a b b 1 c. f () = 7 1 " c 7 = 3 ( + 9) Use inde laws to simplif and state the final result in simplest form. f () = 7 "( + 9) 3 Topic 7 Differential calculus 309

9 WOrKeD example 5 a Differentiate tan 5 a b with respect to. b If f() = tana 3 b, find f aπ b. think WritE a 1 Write the equation. a = tan 5 a b Epress in terms of u and u in terms of. = u 5 where u = tana b 3 Differentiate with respect to u and u with respect to using d (tan(k)) = k sec (k) with k = 1. Find using the chain rule. 5 Substitute back for u and state the final result. b 1 Write the equation. du = 10u and du = 1 sec a b = du du = 10u 1 sec a b = 5 tan a bsec a b b f() = tana 3 b = u. v State the functions u and v. u = and v = tana 3 b 3 Differentiate u and v with respect to. Find f () using the product rule. Substitute for u, dv du, v and. du dv = and = 3 sec a 3 b dv f () = u + v du f () = 3 sec a 3 b + tana 3 b 5 Substitute = π. f aπ b = 3 3 aπ b 1 π + tana cos a π 3 b 3 b π Substitute for the trigonometric values and simplif. 7 Epress the final result in simplest form b taking a common denominator and the common factor out in the numerator. f a π b = π a 1 b +!3π = π 3 +!3π f a π π(π + 3!3) b = MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

10 WOrKeD example Derivatives involving logarithms If = log e () = ln () where > 0, then = e so that = e. Since = 1 /, = 1 e = 1. In general, if a and b are constants and = log e (a + b), then = Furthermore, if = log e ( f()), then = f () f(). a a + b. This is the rule that we will now use to differentiate logarithmic functions. a Differentiate log e (cos(3)) with respect to. b Find d clog e a b d. think WritE a 1 Write the equation. a = log e (cos(3)) Use the general result. d = (cos(3)) cos(3) 3 State the derivative in the numerator. 3 sin(3) = cos(3) Simplif and state the final result. = 3 tan(3) b 1 Write the equation. b = log e a b Simplif using log laws. = log e (3 + 5) log e (3 5) 3 Differentiate each term. Form a common denominator. 5 Epand the numerator and denominator and simplif. State the final answer in simplest form. = (3 5) 3(3 + 5) = (3 + 5)(3 5) 9 15 (9 + 15) = 9 5 = c log e a b d = EErCisE 7. PraCtisE Review of differentiation techniques 1 WE1 a If f() =, find the value of f ( 1) b Differentiate 5 sin 3 () with respect to. Topic 7 DIfferenTIaL CaLCuLus 311

11 Consolidate a Find d c d. b If f() =!cos(), find the value of f a π 1 b. 3 WE a If f() = cosa 3 b, find f a1 π b. b Differentiate e sin() with respect to. a Differentiate sin(!) with respect to. b If f() = e cos(), find the value of f a π b. 5 WE3 a If f() = 3 cos(), find f (). b Find d [ e 3 ]. a Differentiate e 3 cos() with respect to. b If f() = e, find f (). 7 WE a Differentiate 3 cos(3) with respect to. 3 b If f() =, find f ().! + 9 a If f() = 1 3e, find f (). b Find d c + 5 a3 3 5 b d. 9 WE5 a Differentiate tan a b with respect to. 3 b If f() = tana b, find f (π). 10 a If f() = tan(3), find f a π 9 b. b If f() = 5 tana b, find f a1 π b. 11 WE a Differentiate log e asina bb with respect to. b Find d clog e a b d. 1 a If f() = log e (" + 9), find f ( 1). b Differentiate cosalog e a bb with respect to. 13 Differentiate each of the following with respect to. a sin (3) b 5 cos 3 () c " + 9 d 1 Find for each of the following. a = 3 sin(5) b = cos() c = 3 sin(3) 15 Find f () for each of the following. 3 "3 + 5 d = cos() 3 a f() = e 1 b f() = e cos() c f() = cos(e ) d f() = e! 31 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

12 1 Find each of the following. a d [3 e ] b d [e 3 sin()] c d c e3 d d d c 1 3 e d 17 Find for each of the following. a = log e (5 + ) b = log e (" + 9) c = log e a b d = log e a b 1 a If f() = log e (sin(3)), find the eact value of f a π 1 b. b If f() = log e (tan()), find the eact value of f a π b. c If f() = cosa 3 b, find f a π b. d If f() = tana 3 b, find f a1 π b. 19 a Use the chain rule to show that d (sec(k)) = k sec(k)tan(k). b Use the chain rule to show that d (cos(k)) = k cosec(k)cot(k). c Use the quotient rule to show that d (cot(k)) = k cosec (k). d If = log e (tan(3) + sec(3)), show that = 3 sec(3). e If = log e acota b + coseca bb, show that = 1 coseca b. 0 If n, k, b and α are all constants, verif each of the following. a d [sin(n + α)] = n cos(n + α) b d [cos(n + α)] = n sin(n + α) c d [sinn (k)] = nk sin n 1 (k)cos(k) d d [cosn (k)] = nk cos n 1 (k)sin(k) e d [ek sin(b)] = e k (b cos(b) + k sin(b)) f d [ek cos(b)] = e k (k cos(b) b sin(b)) 1 If n and k are both constants, verif each of the following. a d [n sin(k)] = n 1 (n sin(k) + k cos(k)) b d [n cos(k)] = n 1 (n cos(k) k sin(k)) Topic 7 Differential calculus 313

13 Master 7.3 c d c sin(k) n d = 1 n+1(k cos(k) n sin(k)) d d c cos(k) n d = 1 n+1(k sin(k) n cos(k)) e d [n e k ] = n 1 e k (n + k) f d c ek n d = ek (k n) n+1 If a, b, c, d and n are all constants, verif each of the following. a d [log a e ("a + b)] = a + b b d clog e a a + b c + d b d = ad bc (a + b)(c + d) c d clog e a a + b c + d b d = (ad bc) (a + b)(c + d) d d [log e (sinn (b))] = nb tan(b) e d [log e (cosn (b))] = nb tan(b) f d [log e (tann (b))] = nb sin(b) 3 a If u(), v() and w() are all functions of and = u()v()w(), use the product rule to show that = v()w()du + u()w()dv + u()v()dw. b Hence, find for = 3 e cos(). sin(θ) Using the fundamental limit lim = 1, verif each of the following, using θ 0 θ the method of first principles where k is a constant and R. a d (sin(k)) = k cos(k) b c d (tan(k)) = k sec (k) e d (cosec(k)) = k cot(k)cosec(k) f Applications of differentiation d (cos(k)) = k sin(k) d d (sec(k)) = k tan(k)sec(k) d (cot(k)) = k cosec (k) Introduction There are man mathematical applications of differential calculus, including: finding tangents and normals to curves rates of change maima and minima problems curve sketching related rate problems kinematics. 31 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

14 WOrKeD example 7 think Some of these topics are revised in this section. Related rate problems are covered in more detail later in the topic. Maima and minima problems have alrea been covered in the Mathematical Methods course. Later topics include a more detailed coverage of the applications of differentiation to kinematics. Tangents and normal to curves Tangents to curves A tangent to a curve is a straight line that touches a curve at the point of contact. Furthermore, the gradient of the tangent is equal to the gradient of the curve at the point of contact. Find the equation of the tangent to the curve = at the point where = 3. Sketch the curve and the tangent. 1 Find the -coordinate corresponding to the given -value. Find the gradient of the curve at the given -value. This will be denoted b m T. 3 Find the equation of the tangent, that is, the straight line passing through the given point with the given gradient. Use 1 = m T ( 1 ). State the equation of the tangent and sketch the graph. WritE/draW = When = 3, = = The point is (3, ). = + 3 When = 3, ` = m T = 3 = 3 1 = 3, 1 = and m T = 3 = 3( 3) = = normals to curves The normal to a curve is a straight line perpendicular to the tangent to the curve. If two lines are perpendicular, the product of their gradients is 1. If m T is the gradient of the tangent and m N is the gradient of the normal, then m T. m N = 1. Topic 7 DIfferenTIaL CaLCuLus 315

15 WOrKeD example Determine the equation of the normal to the curve = 3 10 at the point where =. Sketch the curve and the normal. think 1 Find the -coordinate corresponding to the given -value. Find the gradient of the curve at the given -value. This is denoted b m T. 3 Find the gradient of the normal, which will be denoted b m N. Since the normal line is perpendicular to the tangent, the product of their gradients is 1. Find the equation of the normal, that is, the straight line passing through the given point with the given gradient. Use 1 = m N ( 1 ). 5 State the equation of the normal. To avoid fractions, give the result in the form a + b + k = 0. Sketch the graph. WritE/draW = 3 10 When =. = = The point is (, ). = 3 When =, ` = m T = 3 = 5 m N m T = 1 m T = 5 m N = 1 5 The normal has a gradient m N = =, 1 = and m N = = 1 ( ) = = 0 0 General results for fi nding tangents and normals to curves In general, to find the equation of the tangent to the curve = f() at the point where = a, the -value is = f(a), so the coordinates of the point are (a, f(a)). The gradient of the curve at this point is m T = f (a), so the equation of the tangent is given b f(a) = f (a)( a). The normal has a gradient of m N = 1, so the f (a) equation of the normal to the curve is given b f(a) = 1 ( a). f (a) 1 31 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

16 WOrKeD example 9 The tangent to the curve =! at a point is given b + c = 0. Find the value of c. think WritE 1 Find the gradient of the curve. =! = Then = 1 1 = 1 " Find the gradient of the tangent line. + c = 0, Rearrange to make the subject: = c 3 State the gradient of the tangent line. m T = 1 At the point of contact these gradients 1 are equal.! = 1 5 Solve the equation to find the -value at the point of contact. Find the coordinate at the point of contact. Substitute the -value into the curve.! = 3 = 9 1 = c When = 9, =! =!9 = 3. The point is (9, 3). 7 This point also lies on the tangent. + c = 0 c = = 9 1 State the answer. c = 9 WOrKeD example 10 think finding tangents and normals to other functions When finding the equation of the tangent and normal to trigonometric functions or other tpes of functions, we must use eact values and give eact answers. That is, give answers in terms of π or surds such as!3, and do not give answers involving decimals. Determine the equation of the tangent and normal to the curve = cos() at the point where = π. Sketch the tangent and the normal. 1 Find the -coordinate corresponding to the given -value. WritE/draW When = π, = cosa π b =! The point is a π,!b. Topic 7 DIfferenTIaL CaLCuLus 317

17 Find the gradient of the curve at the given -value. = sin() When = π, 3 Find the equation of the tangent, that is, the straight line passing through the given point with the given gradient. Use 1 = m T ( 1 ) ` = m T = π = sina π b =! 1 = π, 1 =! and m T =!! =!a π b State the equation of the tangent. =! +!π 5 Find the gradient of the normal, m N. Since the normal line is perpendicular to the tangent, the product of their gradients is 1. Find the equation of the normal, that is, the straight line passing through the given point with the given gradient. Use 1 = m N ( 1 ). 7 State the equation of the normal and sketch the graph, showing the tangent and normal. m N m T = 1 m T =! m N = 1!!! =! +! The normal has a gradient m N =! 1 = π, 1 =! and m N =!! =! a π b π =!!π +! π π 3π π. Rates of change The first derivative or gradient function,, is also a measure of the instantaneous rate of change of with respect to. 31 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

18 WOrKeD example 11 Find the rate of change of the area of a circle with respect to the radius. think 1 Define the variables and state the area of a circle. We require the rate of change of the area of a circle with respect to the radius. 3 State the required result. WOrKeD example 1 think WritE If the radius of the circle is r and the area is A, then A = πr. average rates of change The average rate of change is not to be confused with the instantaneous rate of change. The average rate of change of a function = f() over [a, b] is the gradient of the line segment f(b) f(a) joining the points:. b a The following worked eample illustrates these different ideas. da dr da dr = πr The tides in a certain ba can be modelled b D(t) = sina πt b where D is the depth of water in metres and t is the time in hours after midnight on a particular da. a What is the depth of water in the ba at am? b Sketch the graph of D(t) against t for 0 t. c Find when the depth of water is below 7.5 metres. d Find the rate of change of the depth at am. f(b) e Over the first hours, find the average rate of change of the depth. WritE/draW f(a) 0 a b a am corresponds to t =. Find D(). a D() = sina π b = 9 + 3!3 = metres Topic 7 DIfferenTIaL CaLCuLus 319

19 b 1 Find the period and amplitude of the graph. State the maimum and minimum values of the depth. Sketch the graph on the restricted domain. c 1 Solve an appropriate equation to find the times when the depth of water is below 7.5 metres. b n = π The period is π n or π = 1. π Over 0 t, there are two ccles. The maimum depth is = 1 metres and the minimum depth is 9 3 = metres. c D t D(t) = 7.5 = sina πt b sina πt b = 1 πt = 7π, 11π, 7π 11π + π, + π t = 7, 11, 19, 3 State when the depth is below 7.5 metres. The depth is below 7.5 metres between 7 am and 11 am and between 7 pm and 11 pm. d 1 Find the rate of change of depth with respect to time. Evaluate this rate at am, giving the correct units. e 1 Find the average rate of change over the first hours. d dd dt = π cosaπt b When t =, dd dt ` = π t = cosaπ 3 b = π m/h e t [0, ] D(0) = sin(0) = 9 D() = sina π b = 9 + 3! !3 9 D() D(0) = 0 0 State the required average rate, giving the correct units. The average rate is 3!3 m/h. 30 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

20 Eercise 7.3 PRactise Applications of differentiation 1 WE7 Find the equation of the tangent to the curve = at the point where = 3. Find the equation of the tangent to the curve =! + 5 at the point where = 1. 3 WE Determine the equation of the normal to the curve = at the point where =. Determine the equation of the normal to the curve = at the point where = WE9 The tangent to the curve = + 5 at a point is given b = + c. Find the value of c. The normal to the curve = at a point is given b + c = 0. Find the value of c. 7 WE10 Determine the equation of the tangent and normal to the curve = sina b at the point where = π 3. Determine the equation of the tangent and normal to the curve = 3e + at the point where it crosses the -ais. 9 WE11 For a sphere, find the rate of change of volume with respect to the radius. 10 For a cone, find the rate of change of volume with respect to: a the radius, assuming the height remains constant b the height, assuming the radius remain constant. 11 WE1 The tides in a certain ba can be modelled b D(t) = + cosa πt 1 b where D is the depth of water in metres and t is the time in hours after midnight on a particular da. a What is the depth of water in the ba at am? b Sketch the graph of D(t) against t for 0 t. c Find when the depth of water is below metres. d Find the rate of change of the depth at 3 am. e Over the first hours, find the average rate of change of the depth. 1 The population number, N(t), of a certain cit can be modelled b the equation N(t) = N 0 e kt, where N 0 and k are constants and t is the time in ears after the ear 000. In the ear 000, the population number was and in 010 the population had grown to a Find the values of N 0 and k. b What is the predicted population in 015? Topic 7 Differential calculus 31

21 Consolidate c Find the rate of change of population in 015. d Find the average rate of growth over the first 30 ears. e Sketch the graph of N(t). 13 Find the equation of the tangent to each of the following curves at the point indicated. a = 9 at = b = 5 sin() at = π 3 c =! + 1 at = d = cos(3) at = π 1 1 Find the equation of the normal to each of the following curves at the point indicated. a = 1 at = 3 b = tan(3) at = π 9 3 c = log e ( 3) at = 1 d = ( 3) at = 3 15 a Find the value of c if = 1 + c is a tangent to the curve = 9. b Find the value of c if = + c is a tangent to the curve = ( 3). c Find the value of c if 3 + c = 0 is a normal to the curve = 3. d Find the value of c if + + c = 0 is a normal to the curve =! 3. 1 Given the function f : R R, f() = 5 3e : a show that the function is alwas increasing b sketch the graph, stating an ais intercepts and the equations of an asmptotes c find the equation of the tangent to the curve at = Given the function f : R R, f() = 3e : a show that the function is alwas decreasing b sketch the graph, stating an ais intercepts and the equations of an asmptotes c find the equation of the normal to the curve at = 1. 1 a Sketch the graph of = 3 log e ( 5), stating the ais intercepts, the equations of an asmptotes, and the domain and range. b Show that there are no stationar points. c Find the equation of the tangent to the curve = 3 log e ( 5) at the point where =. 19 a Sketch the graph of f() = log e (5 3), stating the ais intercepts, the equations of an asmptotes, and the domain and range. b Show that the function is a one-to-one decreasing function. c Find the equation of the normal to the curve = log e (5 3) at the point where = 1. 0 Given the function f : R R, f() = 3 e : a find the equations of the tangents to the curve = 3 e at the points where = 1 and = 0 b show that the tangents above are the onl two tangents to the curve which pass through the origin. 3 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

22 Master 1 a The tangent to the curve = 1 at the point where = a and a > 0 crosses the -ais at B and crosses the -ais at C. If O is the origin, find the area of the triangle OAB. b If the normal to the curve = 1 at the point where = a and a > 0 passes through the origin O, find the value of a and hence find the closest distance of the curve = 1 to the origin. c i The tangent to the curve = k, where k > 0, at the point where = a and a > 0 crosses the -ais at B and crosses the -ais at C. If O is the origin, find the area of the triangle OAB. ii The normal to the curve = k at the point where = a and a > 0 passes through the origin. Show that k = a + 1. The population number, P(t), of ants in a certain area is given b 50 P(t) = e 0.15t where t 0 is the time in months. a Find the initial population of the ants. b Find the rate at which the ant population is increasing with respect to time and evaluate the rate after 10 months. c Over the first 10 months find the average rate at which the ant population is increasing. 3 The current, i milliamps, in a circuit after a time t milliseconds is given b i = 10e 3t cos(10t) for t 0. a Find the rate of change of current with respect to time and evaluate it after 0.01 milliseconds, giving our answer correct to 3 decimal places. b Over the time interval from t = 0 to t = 0.0, find the average rate of change of current. The amount of a drug, D milligrams, in the bloodstream at a time t hours after it is administered is given b D(t) = 30te t3. a Find the average amount of the drug present in the bloodstream over the time from t = 1 to t = hours after it is administered. b Find the instantaneous amount of the drug after 1.5 hours. c Find the time when the amount of drug is a maimum and find the maimum amount of the drug in the bo. d For how long is the amount of the drug in the bo more than 10 milligrams? Topic 7 Differential calculus 33

23 7. AOS 3 Topic 1 Concept Implicit differentiation Concept summar Practice questions WOrKeD example 13 Implicit and parametric differentiation Introduction Up until now the relationship between and has alwas been eplicit: is the dependent variable and the independent variable, with in terms of written as = f(). It is said that depends upon and can be found directl. For eample: If = , then = +. If = 3 sin() + e, then = cos() e. If = "1, then =. "1 If = log e (5 + 3), then = There are times, however, when is not epressed eplicitl in terms of. In these cases there is a functional dependence or a so-called implicit relationship between and of the form f(, ) = c, where c is a constant, or g(, ) = 0. For eample: + = = 0 e + 3 sin( 3) = c These represent curves as an implicit relation and are not graphs of functions. In some cases it ma be possible to rearrange to make the subject, but in most cases this is simpl not possible. In this implicit form an epression for can still be obtained, but it will be in terms of both and. This can be obtained b differentiating each term in turn. For eample, d ( ) =, and in general, if n is a constant, then d (n ) = n n 1. Consider d ( ). To find this, use the chain rule and write d ( ) = d ( ) = ; furthermore, in general d (n ) = d (n ) = nn 1. This last result is known as implicit differentiation; it is reall just an application of the chain rule. Basicall, when it is necessar to differentiate a function of with respect to, differentiate with respect to and multipl b. Given + = 1, find an epression for think WritE/draW 1 Write the equation. + = 1 Take d ( ) of each term in turn. d ( ) + d ( ) = d (1) in terms of both and. 3 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

24 3 Use the results above together with d (c) = 0, since the derivative of a constant is zero. Transpose the equation to make the subject. 5 Cancel the common factor and divide b. State the final result, giving in terms of both and. Alternativel notice that in this particular eample it is possible to rearrange the equation to make the subject. 1 Write the equation. + = 1 + = 0 = = Rearrange to make the subject. = 1 3 There are two branches to the relation, which is a circle with centre at the origin and radius. Consider the branch or top half of the circle, which b itself is a function. Find in terms of, differentiating using the chain rule. 5 However, since = "1, epress in terms of both and as before. = ±"1 = 1 0 Consider = "1. = "1 = + = = Topic 7 Differential calculus 35

25 Further eamples involving implicit differentiation Sometimes it ma be necessar to use the product rule to obtain the required derivatives. WORKED EXAMPLE 1 THINK a Find d (3 ). b Given = 0, find an epression for in terms of both and. WRITE/DRAW a 1 Write the epression. a d (3) Use the product rule. d (3 ) = d (u. v) Let u = 3 and v =. 3 Find the derivatives. du dv = 3 and = d () = 1 = Use the product rule. d (u. v) = u dv + vdu 5 State the final result. d (3) = b 1 Write the equation. b = 0 Take d ( ) of each term in turn. d ( ) + d (3) d ( ) + d (5) d (7) + d () = 0 3 Substitute for the result from the second term from part a above and use implicit differentiation on each term. Transpose the equation to get all terms involving on one side of the equation. 5 Factor the terms involving on the right-hand side of the equation. Divide and state the final result, giving in terms of both and = = = (7 + 3) = MATHS QUEST 1 SPECIALIST MATHEMATICS VCE Units 3 and

26 WOrKeD example 15 Implicit differentiation with eponential or trigonometric functions Sometimes it ma be necessar to use the derivatives of eponential or trigonometric functions together with implicit differentiation techniques to obtain the required derivative. Given e + 3 sin( 3) = c, where c is a constant, find an epression for in terms of both and. think WritE 1 Write the equation. e + 3 sin( 3) = c Take d ( ) of each term in turn. d (e ) + d (3 sin( 3)) = d (c) d 3 Consider just the first term and use implicit differentiation. (e ) = d (e u ) where u = d du (e u ) du Use the product rule. e u d () = e u a + b 5 Substitute back for u. Consider just the second term and use implicit differentiation. d (e ) = a + be d (3 sin( 3)) = d (3 sin(v))dv dv where v = 3 so that dv = 3. d (3 sin( 3)) = 3 cos(v)a 3 b d 7 Substitute back for v. (3 sin( 3)) = 3 cos( 3)a 3 b Substitute for the first and a + second terms. be + 3 cos( 3)a 3 b = 0 9 Epand the brackets. e + cos( 3) 9 cos( 3) = 0 10 Transpose the equation to get all e + cos( 3) terms involving on one side of = e + 9 cos( 3) the equation. 11 Factor the terms involving on the e + cos( 3) right-hand side of the equation. = (e + 9 cos( 3)) cos( 3) e 1 Divide and state the final result, = giving e + 9 cos( 3) in terms of both and. Topic 7 DIfferenTIaL CaLCuLus 37

27 WOrKeD example 1 Parametric differentiation In the previous section, we saw that whether the variables and are given in the form = f() as an eplicit relationship or the form f(, ) = 0 as an implicit relationship, an epression for can be found. In this section, the two variables and are connected or related in terms of another variable, called a linking variable or a parameter. Often t or θ are used as parameters. A parameter is a variable that changes from case to case but in each particular case or instant it remains the same. For eample, an epression can be found for ; however, it will be in terms of the parameter b using the chain rule, since = dt dt, and noting that dt = 1. Alternativel, it ma also be possible to eliminate the parameter dt from the two parametric equations and obtain an implicit relationship between the two variables and. The following eamples will illustrate these concepts. a If = 3 cos(t) and = sin(t), find the gradient in terms of t. b For the parametric equations = 3 cos(t) and = sin(t), find an implicit relationship between and, and find the gradient. Hence, verif our answer to part a. think WritE a 1 Differentiate with respect to t. The dot a = 3 cos(t) notation is used for the derivative with respect to t. dt = # = 3 sin(t) Differentiate with respect to t. = sin(t) dt = # = cos(t) 3 Use the chain rule to find. = dt dt = # # Substitute for the derivatives. = cos(t) 3 sin(t) 5 State the gradient in simplest form. = 3 cot(t) b 1 Write the parametric equations. b = 3 cos(t) (1) = sin(t) () Epress the trigonometric ratios on their own. cos(t) = 3 sin(t) = (1) () 3 Eliminate the parameter to find the implicit relationship. Since cos (1) + sin (t) = 1, = 1 3 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

28 Use implicit differentiation on the implicit form. 5 Perform the implicit differentiation. Transpose the equation. 7 Make the subject. From the implicit differentiation, substitute for the parametric equations. 9 Simplif to verif the given result, as above. Eercise 7. PRactise Consolidate Implicit differentiation 1 WE13 Given = 7, find an epression for in terms of both and. Given! +! =, find an epression for in terms of both and. 3 WE1 a Find d ( ) b Given = 0, find an epression for in terms of both and. Find the gradient of the normal to the curve = 0 at the point where = in the first quadrant. 5 WE15 Given + e ( + ) = c, where c is a constant, find an epression for in terms of both and. If sin(3 + ) + + = c, where c is a constant, find in terms of both and. 7 WE1 a If = cos(t) and = sin(t), find the gradient in terms of t. b For the parametric equations = cos(t) and = sin(t), find an implicit relationship between and, and find the gradient. Hence, verif our answer to part a. a Given = t and = t t, find in terms of t. b For the parametric equations = t and = t t, epress in terms of and find. Hence, verif our answer to part a. 9 For each of the following implicitl defined relations, find an epression for in terms of and. a = 3 1 d 9 ( ) + 1 d 1 ( ) = d (1) = 0 9 = = cos(t) = 1 9 sin(t) b + 9 = 1 = 3 cot(t) c 1 9 = 1 d 3 + = 10 Topic 7 Differential calculus 39

29 10 For each of the following implicitl defined relations, find an epression for in terms of and. a = 0 b = 0 c = 0 d = 0 11 For each of the following implicitl defined relations, find an epression for in terms of and. a 3 + = b = c e + + cos() = 0 d e + cos() + = 0 1 For each of the following implicitl defined relations, find an epression for in terms of and. a log e ( + 3) + 5 = 10 b log e (3) = 0 c = 0 d 3 sin(3) sec() + = 0 13 a Find the gradient of the tangent to the curve = 3 ( ) at the point (1, 1). b Find the equation of the tangent to the curve = 0 at the point (1, ). c For the circle = 0, find the gradient of the tangent to the circle at the point (, 1). d A certain ellipse has the equation = 0. Find the gradient of the normal to the ellipse at the point (1, ). 1 a Find the equation of the tangent to the circle + = 5 at the point in the fourth quadrant where = 3. b Find the gradient of the tangent to the ellipse + = 1 at the point in the first quadrant where = 1. 9 c Find the gradient of the normal to the hperbola 1 = 1 at the point in the fourth quadrant where = 5. 9 d Find the gradient of the normal to the curve ( + ) = 5( ) at the point (3, 1). 15 For each of the following, find in terms of the parameter. a The parabola = t and = t b The ellipse = 3 sin(t) and = cos(t) c The rectangular hperbola = 5t and = 5 t d The hperbola = 3 sec(t) and = tan(t) 1 In each of the following, a and b are constants. Find in terms of the parameter. a = at and = at b = a cos(t) and = b sin(t) c = at and = a t d = a sec(t) and = b tan(t) 17 Check our answers to question 1a d b eliminating the parameter and determining b another method. 330 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

30 Master a A curve called the folium of Descartes has the equation 3 3a + 3 = 0. Its graph is shown at right. Find for this curve, given that a is a constant. Rene Descartes ( ) was a French mathematician and philosopher who lived in the Dutch republic. He is noted for introducing the Cartesian coordinate sstem in two dimensions and is credited as the father of analtical geometr, the link between algebra and geometr. 0 b Show that the folium of Descartes can be represented b the parametric equations = 3at 3at and = t 1 + t3, and find the gradient of the curve at the point t. 19 a A curve called a lemniscate has the implicit equation ( + ) = ( ). Its graph is shown at right. Find an epression for in terms of and. 0 b Show that the curve can be represented b the parametric! cos(t)! sin(t)cos(t) equations = and =, sin (t) + 1 sin (t) + 1 and find in terms of t. 0 A curve known as the Cissoid of Diocles has the equation = 3. Its graph is shown at right. a Find for this curve, given that a is constant. a Show that a parametric representation of the Cissoid curve is given b the equations = at at3 and = 1 + t 1 + t, and find in terms of t. b Show that another alternative parametric representation of the Cissoid curve is given b the equations = a sin (t) and = a sin3 (t), and find in terms of t. cos(t) Second derivatives 0 a Introduction If = f() is the equation of the curve, then the first derivative is = f () and it is the gradient of the curve. It is also the rate of change of with respect to, often abbreviated to wrt. This is still a function of, so if the first derivative is differentiated again, the second derivative or the rate of change of the gradient is obtained. This is denoted b d a b = d = f (). Notice the position of the s in this notation and that the two dashes after f indicate the second derivative with respect to. Topic 7 Differential calculus 331

31 AOS 3 Topic 1 Concept Second derivatives Concept summar Practice questions WOrKeD example 17 think Similarl, the third derivative is denoted b d3 3 = f (). In general, the n th derivative is denoted b dn = f (n) (). n For eample, consider the general cubic equation = f() = a 3 + b + c + d where a, b, c and d are constants. The first derivative or gradient function is = f () = 3a + b + c. The second derivative or rate of change of gradient function is d = a + b. The third derivative is d3 derivatives are zero. If f() = " " 5, find f (9). 3 1 Epress the function in simplified form using inde laws. Find the first derivative, using the basic laws for differentiation. 3 Find the second derivative, using the basic laws for differentiation, b differentiating the first derivative again. WritE f() = "5 3 5 = 3 = 3 5 = 1 3 f () = = 3 1 f () = = 1 3" 3 Substitute in the value for. f (9) = 1 3"9 3 = State the final result. f (9) = 1 1 WOrKeD example 1 = a, and all further 3 using product and quotient rules Often when we differentiate we ma need to use rules such as the product and quotient rules. Find d for = log e (3 + 5). think WritE 1 Write the equation. Let = log e (3 + 5) 33 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

32 Use the product rule. 3 State the result for. Differentiate with respect to again. 5 Use the quotient rule on the first term and the product rule again on the second term. Perform the required derivatives. 7 Simplif the numerator in the first epression and state the final result. Add the first and last terms b forming a common denominator. 9 State the final result in simplest form. WOrKeD example 19 think = d (log e (3 + 5)) + d ( ) log e (3 + 5) = log e (3 + 5) d = d a b + d ( log e (3 + 5)) d = d (3 )(3 + 5) d (3 + 5)(3 ) (3 + 5) + d ()(log d e (3 + 5)) + (log e (3 + 5)) d = (3 + 5) 3(3 ) + log (3 + 5) e (3 + 5) d = (3 + 5) + log e (3 + 5) d = log e (3 + 5) (3 + 5) (3 + 5) d = log e (3 + 5) (3 + 5) d = log 3(9 + 0) e (3 + 5) + (3 + 5) finding second derivatives using implicit differentiation Implicit differentiation techniques can be used to find relationships between the first and second derivatives. Equations involving the function and its first and second derivatives are called differential equations. The are eplored in greater depth in later topics. If a and b are constants and = "a " + b, show that d + a b = a + b. WritE 1 Write the equation and square both sides. = "a + b = (a + b ) Epand the brackets. = a + b 3 Take d ( ) of each term in turn. d ( ) = d (a ) + d (b ) Use implicit differentiation to find the first derivative. = a + b3 Topic 7 DIfferenTIaL CaLCuLus 333

33 5 Take d ( ) of each term in turn again. d a b = d (a) + d 1b3 Use the product rule on the first term. d a b + d () = a + 1b 7 Use implicit differentiation. d + a b = a + 1b Divide each term b and state the final result. d + a b = a + b WOrKeD example 0 think using parametric differentiation to find second derivatives When finding d with parametric differentiation, often our first thought might be to use the chain rule as d = d dt d t. However this rule is incorrect and cannot be used, as dt and d t are not equivalent; furthermore, d 1 / d. To correctl obtain d, we must use implicit differentiation in conjunction with the parametric differentiation. Because is itself a function of t, we obtain the second derivative using d = d a b = d dt a b dt. If = 3 cos(t) and = sin(t), find d in terms of t. 1 Differentiate with respect to t. The dot notation is used for the derivative with respect to t. WritE = 3 cos(t) dt = # = 3 sin(t) Differentiate with respect to t. = sin(t) dt = # = cos(t) 3 Use the chain rule to find. Substitute for the derivatives. 5 State the gradient in simplest form. = dt dt = # # = cos(t) 3 sin(t) = 3 cot(t) Take d ( ) of each term. d = d a b = d a 3 cot(t)b 33 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

34 7 Use implicit differentiation and cot(t) = cos(t) sin(t). Use the quotient rule. d = d dt d = ± a cos(t) 3 sin(t) b dt d dt ( cos(t))(3 sin(t)) d (3 sin(t))( cos(t)) dt dt (3 sin(t)) 9 Perform the derivatives in the numerator. 10 Simplif using sin (t) + cos (t) = 1 11 Substitute for dt = 1 / dt 1 State the final result. Eercise 7.5 PRactise Consolidate Second derivatives 1 WE17 If f() = "3, find f (). 3 If f() = cosa b, find f aπ 3 b. 3 WE1 Find d for = 3 log e ( + 5). Find d for = e 3. 5 WE19 If a and b are constants and = a + b, show that d + a b = b. If a and b are constants and 3 = (a + b 3 ), show that d + a b = b. 7 WE0 If = cos(t) and = sin(t), find d in terms of t. Given = t and = t t, find the rate of change of gradient and evaluate when t =. 9 a If f() = 3!, find f (). b If f() =, find f (1). 3 5 c If f() = log e ( 3), find f (3). d = a 1 sin (t) 1 cos (t) b dt (3 sin(t)) d = 1 dt 9 sin (t) d = 3 sin (t) 1 3 sin(t) d = 9 sin 3 (t) = 9 cosec3 (t) d If f() = e, find f (1). e If f() = tana b, find f aπ 3 b. f If f() = ecos(), find f a π b. Topic 7 Differential calculus 335

35 10 Find d if: a = log e ( ) b e 3 cos() c = 3 e d cos(3) e = log e ( + 7) f = log e ( + " + 1). 11 If a and b are constants, find d for each of the following. a = a b a b = 1 c = a 1 In each of the following, a and b are constants. Find d in terms of the parameter. a = at and = at b = at and = a t c = a cos(t) and = b sin(t) 13 Let a, b and k be constants. a If = a + b, prove that d + a b = 0. b Given that =!a + b, verif that d + a b = a + 3b. a cos(k) + b sin(k) c If =, show that d + + k = 0. 1 If a, b and n are constants, show that: a d [(a + b)n ] = a n(n 1)(a + b) n [log e (a + b) n ] = a n (a + b) b d c d [log e (a + b) n ] = an(a b) (a + b) d d [(a + b) n ] = an(a + b) n (a(n 1) + b). 15 A circle of radius a with centre at the origin has the equation + = a. a Show that d = a 3. b The radius of curvature ρ of a plane curve is given b ρ = c1 + a b d Show that the radius of curvature of a circle has a magnitude of a. 1 An involute of a circle has the parametric equations = cos(θ) + θ sin(θ) and = sin(θ) θ cos(θ). Show that the radius of curvature is θ. d 3 33 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

36 17 A ccloid is the curve obtained b a point on a circle of radius a rolling along the -ais. Its graph is shown below. a Master 0 π π π It has the parametric equations = a(θ sin(θ)) and = a(1 cos(θ)). Show that: a = cotaθ b b d = 1 a sin a θ b c the radius of curvature of the ccloid is a sina θ b. 1 a Given the implicit equation for an astroid = a 3, where a is a constant, epress in terms of both and ȧ a a 0 b Show that the astroid can be epressed in parametric form as = a cos 3 (θ) and = a sin 3 (θ), and find and d in terms of θ. 19 a If = e sin(3), find d. b If = u()v() where u() and v() are differentiable functions, show that d = v u()d + du dv + u v()d. Use this result to verif d for = e sin(3). 0 a If and are both epressed in terms of a parameter t, that is, = f(t) and = g(t), show that d = # $ # $ #. 3 b Show that d 3 d = a b. a Topic 7 Differential calculus 337

37 7. Curve sketching Curve sketching and derivatives We can sketch a curve from its main characteristic or critical points and aial intercepts. We will use the derivatives as an aid to sketching curves. Because the first derivative, = f (), represents the slope or gradient of a curve at an -value, we sa that a curve is increasing when its gradient is positive, { : f () > 0}; that is, the graph slopes upwards to the right. A curve is decreasing when its gradient is negative, { : f () < 0}; that is, the graph slopes upwards to the left. The second derivative, d = d a b, is the rate of change of the gradient function or the gradient of the first derivative and is also useful in sketching graphs. Consider the graph of = f() =. + The graph has a minimum turning point at = 0, + that is at the origin, (0, 0). This is also where the + derivative function f() = + fʹ() = = f () intersects the -ais Note that f () = and f () =. When < 0, to the left of the minimum turning point, the gradient is negative; when > 0, to the right of the minimum turning point, the gradient is positive. So the gradient changes sign from negative to zero to positive, and the rate of change of the gradient is positive. Consider the graph of = f() =. The graph has a maimum turning point at = 0, that fʹ() = is at the origin, (0, 0). This is also where the derivative function = f () intersects the -ais. Note that f() = + + f () = and f () =. When < 0, to the left + of the maimum turning point, the gradient is positive; when > 0, to the right of the maimum turning point, the gradient is negative. So the gradient changes sign from positive to zero to negative, and the rate of change of the gradient is negative. Consider the graph of = f() = 3. 9 For this function, f () = 3 + and f () =. The origin, (0, 0), is a special point on the graph. This is where the fʹ() = derivative function + = f () intersects the -ais. When < 0, to the left of the origin, the gradient is positive; + f() = 3 9 when > 0, to the right of the origin, the gradient is + positive. At the origin f (0) = 0 and f (0) = 0. So the gradient changes from positive to zero to positive, and the rate of change of the gradient is zero at this point. In this case the point at the origin is a horizontal stationar point of inflection. The tangent to the curve = 3 at the origin is the -ais, and the tangent crosses the curve. 33 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

38 Stationar points Although we have used the above graphs to make some important observations about the first and second derivatives, the results are true in general and can be used as an aid to sketching graphs. A stationar point on a curve is defined as a point on the curve at which the slope of the tangent to the curve is zero; that is, where the gradient = f () = 0. All turning points are stationar points. There are two tpes of turning points: maimum turning points and minimum turning points. A maimum turning point is a point on the curve at which the -coordinate has its highest value within a certain interval. At such a point the slope of the curve changes from positive to zero to negative as increases. The rate of change of the gradient is negative at such a point. If there are higher -values outside the immediate neighbourhood of this maimum, it is called a local maimum. If it is the highest -value on the whole domain, it is called an absolute maimum. A maimum turning point is like the top of a hill. At = a, f (a) = 0 and f (a) < 0. If < a, then f () > 0. If > a, then f () < 0. A minimum turning point is a point on the curve at which the -coordinate has its lowest value within a certain interval. At such a point the slope of the curve changes from negative to zero to positive as increases. The rate of change of the gradient is positive at such a point. If there are lower -values outside the immediate neighbourhood of this minimum, it is called a local minimum. If it is the lowest -value on the whole domain, it is called an absolute minimum. A minimum turning point is like at the bottom of a valle. At = a, f (a) = 0 and f (a) > 0. If < a, then f () < 0. If > a, then f () > 0. Note that for both a maimum turning point and a minimum turning point, the first derivative is zero. To distinguish between a maimum and a minimum, we use the sign test or the second derivative test. On a continuous curve, between a local maimum and a local minimum there is another critical point called an inflection point. At such a point the curve changes from being concave to conve or vice versa. The tangent to the curve at such a point crosses the graph. The rate of change of the gradient is zero at a point of inflection. There are two tpes of inflection points = a = a Topic 7 Differential calculus 339

39 Horizontal stationar point of inflection = a < a, f (a) > 0 < a, f (a) > 0 = a, f (a) = 0 = a, f (a) = 0 > a, f (a) > 0 > a, f (a) < 0 f (a) = 0 f (a) = 0 Inflection point At = a, f (a) 0 f (a) = 0 0 Concavit A non-stationar point of inflection is where the tangent to the curve moves from being above the curve to below the curve or vice versa. Curves are also defined as being concave up or concave down. When a tangent is above the curve at each point, then the derivative function is decreasing and f () < 0. = a 0 = a If the tangent to the curve is below the curve at each point, then the derivative function is increasing and f () > 0. = a = a 30 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

40 At a point of inflection, the tangent will pass through the curve, and the curve will change from concave down to concave up or vice versa on either side of the point of inflection. Concave up Concave up WOrKeD example 1 Concave down 0 Concave down 0 sketching the graphs of cubic functions A general cubic function is of the form = a 3 + b + c + d where a 0. A cubic function is a polnomial of degree 3. When sketching the graphs of cubic functions, we need to find where the graph crosses the -ais. To solve the cubic equation a 3 + b + c + d = 0, we factorise or use the factor theorem. To find the stationar points we solve when the first derivative = 3a + b + c = 0. To find the inflection point we solve when the second derivative d = a + b = 0. We can use the second derivative to determine the nature of the stationar points. The values for obtained for these equations are not necessaril integers, and there are several different possibilities for the shapes of the graphs. For eample, the cubic ma cross the -ais at three distinct points and ma have both a maimum and a minimum turning point as well as a non-horizontal point of inflection. Sketch the graph of the function f : R R, f() = b finding the coordinates of all ais intercepts and stationar points and establishing their nature. Also find the coordinates of the point of inflection, and find and draw the tangent to the curve at the point of inflection. think WritE/draW 1 Factorise the function. f() = = ( + + 9) = ( + 3) Determine and find the ais The graph cross the -ais when = 0, at = 0 or = 3. intercepts. The intercepts are (0, 0) and ( 3, 0). 3 Find the first derivative and factorise. Find the stationar points b equating the first derivative to zero. f() = f () = = 3( + + 3) = 3( + 3)( + 1) f () = 3( + 3)( + 1) = 0 when = 1 or = 3 Topic 7 DIfferenTIaL CaLCuLus 31

41 5 Find the -values at the stationar points. Find the second derivative. f () = f () = + 1 = ( + ) 7 Test the two stationar points b using the second derivative to determine their nature. Find the point of inflection b equating the second derivative to zero and find the -value. 9 Determine the gradient at the point of inflection. 10 Determine the equation of the tangent at the point of inflection. 11 Sketch the graph and the tangent using an appropriate scale. We see that the tangent crosses the curve at the point of inflection. f() = ( + 3) f( 1) = 1( 1 + 3) = f( 3) = 0 The stationar points are ( 3, 0) and ( 1, ). When = 3, f ( 3) = < 0. The point ( 3, 0) is a local maimum. When = 1, f ( 1) = > 0. The point ( 1, ) is a local minimum. f () = ( + ) = 0 when =. f( ) = ( + 3) = The point (, ) is the point of inflection. f ( ) = 3( ) = 3 Use 1 = m( 1 ): P (, ), m = 3 + = 3( + ) + = 3 = 10 = ( 3, 0) (, ) ( 1, ) 10 Sketching the graphs of quartic functions A general quartic function is of the form = a + b 3 + c + + e where a 0. A quartic function is a polnomial of degree. When sketching the graphs of quartic functions, we need to find where the graph crosses the -ais. To solve the quartic equation a + b 3 + c + + e = 0, we factorise or use the factor theorem. To find the stationar points, we solve when the first derivative = a3 + 3b + c + d = 0, and to find the points of inflection, we solve when the second derivative d = 1a + b + c = 0. The graph of a quartic function will cross the -ais four times at most, and will have at most three stationar points and possibl two inflection points. 3 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

42 WOrKeD example Sketch the graph of = b finding the coordinates of all ais intercepts, stationar points and points of inflection, and establishing their nature. think WritE/draW 1 Factorise. = = ( ) = (! )(! + ) Determine and find the ais intercepts. 3 Find the first derivative and factorise. Find the stationar points b equating the first derivative to zero. 5 Find the -values at the stationar points. Find the second derivative and factorise. 7 Test the stationar points using the second derivative to determine their nature. The graph crosses the -ais when = 0, at = 0 or = ±!. The -intercepts are (0, 0), (!, 0), (!, 0).!.5 = = 1 3 = (3 ) = (!3 )(!3 + ) = (!3 )(!3 + ) = 0 = 0 or = ±!3. When =!3, = (!3) (!3) = 9. When =!3, = (!3) (!3) = 9. The stationar points are (0, 0), (!3, 9) and (!3, 9). = 1 3 d = 1 1 = 1(1 ) = 1(1 )(1 + ) When = 0, d = 1 > 0. The point (0, 0) is a local minimum. There are -values lower than zero in the range of the function. When =!3, d = < 0. The point (!3, 9) is an absolute maimum. When =!3, d = < 0. The point (!3, 9) is an absolute maimum. Topic 7 DIfferenTIaL CaLCuLus 33

43 Find the points of inflection b equating the second derivative to zero and find the -value. 9 List an other features about the graph. 10 Sketch the graph, showing all the critical points using an appropriate scale. sketching the graphs of reciprocal functions using calculus 1 In this section, we sketch graphs of the form = a + b + c. If b ac > 0, the graph will have two vertical asmptotes and one turning point. It can be shown that the graph will not have an inflection points. We can deduce the graph of the function f() = 1 from the graph of the function g() b noting that: g() the -intercepts of g() become the equations for the vertical asmptotes of f() the reciprocal of a positive number is positive, so the parts of the graph where g() is above the -ais remain above the -ais for f() the reciprocal of a negative number is negative, so the parts of the graph where g() is below the -ais remain below the -ais for f() if g() crosses the -ais at (0, a), then f() will cross the -ais at a0, 1 a b if g(), then f() 0 + (from above) if g(), then f() 0 (from below) if g() has a local maimum at (p, q), then f() will have a local minimum at ap, 1 q b if g() has a local minimum at (p, q), then f() will have a local maimum at ap, 1 q b if g() = 1, 1 g() = 1. d = 1(1 )(1 + ) = 0 = ±1 When = ±1, = 5. The points ( 1, 5) and (1, 5) are both points of inflection. Let f() =, then f( ) = f(), so the graph is an even function and the graph is smmetrical about the -ais. ( 3, 9) 10 ( 3, 9) ( 1, 5) (1, 5) (, 0) (0, 0) (, 0) WOrKeD example 3 Sketch the graph of =. State the equations of an asmptotes. + Find the coordinates of an ais intercepts and an turning points. State the maimal domain and range. 3 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

44 THINK WritE/draW 1 Factorise the denominator. = f() = + = ( + )( ) Vertical asmptotes occur when the denominator is zero. 3 Determine ais intercepts. First find the -intercepts. The lines = and = are both vertical asmptotes. The graph does not cross the -ais, as the numerator is never zero. Find the -intercepts. The graph crosses the -ais when = 0, f(0) = 1 at (0, 1) 5 Find the equations of an other asmptotes. Use the chain rule to differentiate the function. 7 Stationar points occur when the gradient is zero. Equate the gradient function to zero and solve for. Determine the -value of the turning point. The second derivative will be complicated; however, we can determine the nature of the turning point. 9 Using all of the above information, we can sketch the graph using an appropriate scale. Draw the asmptotes as dotted lines and label the graph with all the important features. As ±, 0 +. The plus indicates that the graph approaches from above the asmptote. The line = 0 or the -ais is a horizontal asmptote. f() = + = ( + ) 1 f () = ( + )( + ) ( + ) f () = ( + ) ( + ) f () = ( + ) = 0 + = 0 = 1. Substitute = 1: f( 1) = = 1 5 Since the graph of = + has a local minimum at = 1, the graph of = + has a 1, b as a local maimum. = = = 10 ( 1, ) (0, 1) Topic 7 Differential calculus 35

45 10 From the graph we can state the domain and range. The domain is R\{, } and the range is a, R (0, ). 5 sketching rational functions A rational function is defined as f() = P() where both P() and Q() are Q() polnomials. In this section we will sketch the graphs of simple rational functions where P() is a linear, quadratic or cubic function and Q() is a simple linear or quadratic function. To sketch the graphs of these tpes of rational functions, we consider the following main points. ais intercepts of rational functions The curve ma cross the -ais where = 0, that is, at values of where the numerator P() = 0. The curve ma cross the -ais where = 0, that is P(0) Q(0). asmptotic behaviour of rational functions A function is not defined when the denominator is zero. We thus have a vertical asmptote for each value of where Q() = 0. A vertical asmptote is never crossed. To obtain the equations of other asmptotes, if the degree of P() Q(), we divide the denominator into the numerator to obtain an epression of the form S() + R() Q(), so that R() is of a lower degree than Q(). Now as, then S(), so = S() is the equation of an asmptote. S() ma be of the form = c (a constant), so we get a horizontal line as an asmptote, or it ma be of the form = a + b, in which case we have an oblique asmptote. It is also possible that S() = a, so we can even get quadratics as asmptotes. stationar points of rational functions It is easiest to find the gradient function from the divided form of these tpes of functions. Equate the gradient function to zero and solve for. Note that these stationar points ma be local minima or local maima. We are not concerned with finding the inflection points of these tpes of graphs. However, we can find the second derivative and use it to determine the nature of the stationar points. Using this information we can sketch the curve. Sometimes graphs of these tpes ma not cross the - or -aes, or the ma not have an turning points. WOrKeD example Sketch the graph of = 3 5. State the equations of an asmptotes. Find 9 the coordinates of an ais intercepts and an stationar points, and establish their nature. 3 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

46 THINK 1 Determine ais intercepts. First, find the -intercepts. Vertical asmptotes occur when the denominator is zero. 3 Simplif the epression b dividing the denominator into the numerator. Find the equations of an other asmptotes. 5 Find the first derivative b differentiating the divided form. Stationar points occur when the gradient is zero. Equate the gradient function to zero and solve for. WritE/draW The graph crosses the -ais when the numerator is zero. Solve 3 5 = 0. 3 = 5 = " The graph crosses the -ais at (" 3 5, 0) or (3.7, 0). = The line = 0 or the -ais is a vertical asmptote. = = = 9 As, from below. 9 As, from above. 9 The quadratic = is an asmptote. 9 = 9 1 = 9 + = 9 + = 9 + = 0 9 = 3 = 7 = " 3 7 = 3 7 Determine the -value of the When = 3, = turning point. Find the second derivative. = Determine the sign of the second derivative to determine the nature of the turning point. d = = = 3. When = 3, d = = 3 > 0. The point ( 3, 3) is a local minimum turning point. Topic 7 Differential calculus 37

47 10 Using all of the above information, we can sketch the graph using an appropriate scale. Draw the asmptotes as dotted lines, and label the graph with all the important features. Eercise 7. PRactise Curve sketching ( 3, 3) = 0 10 = 9 10 (3.7, 0) 1 WE1 Sketch the graph of = 3 + b finding the coordinates of all ais intercepts and stationar points, and establishing their nature. Find the coordinates of the point of inflection. Find and draw the tangent to the curve at the point of inflection. Show that the graph of = 3 a + a, where a R\{0}, crosses the -ais at (a, 0) and (0, 0), has turning points at (a, 0) and a a 3, a3 b, and has an inflection 7 point at a a 3, a3 b. Show that the equation of the tangent to the curve at the point 7 of inflection is given b = a3 7 a 3. 3 WE Sketch the graph of = + 0 b finding the coordinates of all ais intercepts and stationar points, and establishing their nature. Find the coordinates of the point of inflection. The graph of = a + b + c has one stationar point at (0, ) and points of inflection at (±!, 1). Determine the values of a, b and c. Find and determine the nature of an other stationar points. 5 WE3 Sketch the graph of = 1 1. State the equations of an asmptotes. Find the coordinates of an ais intercepts and an turning points. State the maimal domain and range. 1 Sketch the graph of = b finding the equations of all straight line 1 asmptotes. Find the coordinates of an ais intercepts and turning points. State the maimal domain and range. 1 7 WE 3 Sketch the graph of =. State the equations of an asmptotes. Find the coordinates of an ais intercepts and an stationar points, and establish their nature. Sketch the graph of = + 9. State the equations of an asmptotes. Find the coordinates of an ais intercepts and an stationar points, and establish their nature. 3 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

48 Consolidate 9 Sketch the graphs of each of the following b finding the coordinates of all ais intercepts and an stationar points, and establishing their nature. Also find the coordinates of the point(s) of inflection. a = 3 7 b = 9 3 c = d = Sketch the graphs of each of the following b finding the coordinates of all ais intercepts and an stationar points, and establishing their nature. Also find the coordinates of the point(s) of inflection. a = b = c = d = Sketch the graphs of each of the following b finding the coordinates of all ais intercepts and an stationar points, and establishing their nature. Also find the coordinates of the point(s) of inflection. a = 3 b = c = = ( )( + ) 3 d = + 3 = ( 1) 3 ( + 3) 1 Sketch the graphs of each of the following. State the equations of an asmptotes. Find the coordinates of an ais intercepts and an turning points. State the domain and range. a = 1 b = c = d = Sketch the graphs of each of the following. State the equations of an asmptotes. Find the coordinates of an ais intercepts and an stationar points, and establish their nature. a = + b = c = 3 d = Sketch the graphs of each of the following. State the equations of an asmptotes. Find the coordinates of an ais intercepts and an stationar points, and establish their nature. a = 3 3 b = c = d = (3 + ) 3 15 Sketch the graphs of each of the following. State the equations of an asmptotes. Find the coordinates of an ais intercepts and an stationar points, and establish their nature. a = 1 b = c = d = ( + 1) 1 a The function f() = 3 + b + c + d has a stationar point of inflection at (1, ). Find the values of b, c and d. b A cubic polnomial = 3 + b + c + d crosses the -ais at = 5 and has a point of inflection at (1, 1). Find the equation of the tangent at the point of inflection. Topic 7 Differential calculus 39

49 Master c The function f() = 3 + b + c + d crosses the -ais at = 3 and has a point of inflection at (, ). Find the values of b, c and d. d A cubic polnomial = a 3 + b 3 + c has a point of inflection at =. The tangent at the point of inflection has the equation = 1 +. Find the values of a, b and c. A 17 a The curve = + b + 7 has a local maimum at a, b. Determine the 3 values of A and b. State the equations of all straight-line asmptotes and the domain and range. A b The curve = has a local minimum at (3, ) and a vertical b + c asmptote at =. Determine the values of A, b and c. State the equations of all straight-line asmptotes and the domain and range. 1 c State conditions on k such that the graph of = k 5 has: i two vertical asmptotes ii onl one vertical asmptote iii no vertical asmptotes. d Sketch the graph of = 1 b finding the coordinates of an stationar + 9 points and establishing their nature. Also find the coordinates of the points of inflection, and find the equation of the tangent to the curve at the point of inflection where > 0. 1 a Show that the graph of = 3 a, where a R\{0}, crosses the -ais at (±a, 0) and (0, 0) and has turning points at a!3a 3,!3 a3 b and 9 a!3a 3, a3!3 b. Show that (0, 0) is also an inflection point. 9 b Show that the graph of = ( a) ( b), where a, b R\{0}, has turning points at (a, 0) and a a + b (a b)3, b and an inflection 3 7 point at a a + b (a b)3, b. 3 7 c Show that the graph of = ( a) 3 ( b), where a, b R\50, has a stationar point of inflection at (a, 0), a turning point at a a + 3b 7(a b), b 5 and an inflection point at a a + b (a b), b a Consider the function f() = a + b where a, b R\{0}. State the equations of all the asmptotes and show that if ab > 0, then the graph has two turning b points at a Å a,!abb and a b,!abb and does not cross the -ais. Åa However, if ab < 0, then there are no turning points, but the graph crosses the -ais at = ± Å b a. 350 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

50 AOS 3 Topic Concept 1 Derivatives of inverse circular functions Concept summar Practice questions b Consider the function f() = a3 + b where a, b R\{0}. State the equations of all the asmptotes and show that the graph has a turning point at 3 b a Å a, 3"3 a b b and an -intercept at = b Å3 a. c Consider the function f() = a3 + b where a, b R\{0}. State the equations of all the asmptotes and show that the graph has a turning point at 3 b a Å a, 3"3 ab b and an -intercept at = b Å3 a. d Consider the function f() = a + b where a, b R\{0}. State the equations of all the asmptotes and show that if ab > 0, the graph has two turning points b at a±,!abb and does not cross the -ais. However, if ab < 0, then there Å a are no turning points, but the graph crosses the -ais at = ± b Å a. 0 a The amount of a certain drug, A milligrams, in the bloodstream at a time t hours after it is administered is given b A(t) = 30te t3, for 0 t 1. i Find the time when the amount of the drug, A, in the bo is a maimum, and find the maimum amount. ii Find the point of inflection on the graph of A versus t. b The amount of a drug, B milligrams, in the bloodstream at a time t hours after it is administered is given b B(t) = 15t e t, for 0 t 1. i Find the time when the amount of the drug, B, in the bo is a maimum, and find the maimum amount. ii Find the points of inflection correct to decimal places on the graph of B versus t. c Sketch the graphs of A(t) and B(t) on one set of aes and determine the percentage of the time when the amount of drug B is greater than drug A in the bloodstream. Derivatives of inverse trigonometric functions Introduction In this section, the derivatives of the inverse trigonometric functions are determined. These functions have alrea been studied in earlier topics; recall the definitions and alternative notations. The inverse function = sin 1 () or = arcsin() has a domain of [ 1, 1] and a range of c π, π d. It is equivalent to = sin() and sin(sin 1 ()) = if [ 1, 1], and sin 1 (sin()) = if c π, π d. Topic 7 Differential calculus 351

51 The inverse function = cos 1 () or = arccos() has a domain of [ 1, 1] and a range of [0, π]. It is equivalent to = cos() and cos(cos 1 ()) = if [ 1, 1], and cos 1 (cos()) = if [0, π]. The inverse function = tan 1 () or = arctan() has a domain of R and a range of a π, π b. It is equivalent to = tan() and tan(tan 1 ()) if R, and tan 1 (tan()) = if a π, π b. WOrKeD example 5 think The derivative of sin 1 () First we must determine the derivative of sin 1 (). Let = sin 1 (). From the definition of the inverse function, = sin(), so = cos() and = 1 cos(). However, we need to epress the result back in terms of. Using sin() =, draw a right-angled triangle. Label the angle 1 1 with the opposite side length being, so the hpotenuse has a length of 1 unit. From Pthagoras theorem, the adjacent side length is "1. 1 Thus, cos() = "1. It follows that d 1 (sin 1 ()) =. "1 A better result is if = sin 1 a b, where a is a positive real constant; then a = 1. Note however that the maimal domain of = sin 1 a b is a, "a a whereas the domain of the derivative is < a. Although the function is defined at the endpoints, the gradient is not defined at the endpoints; for this reason the domain of the derivative is required. Determine for each of the following, stating the maimal domain for which the derivative is defined. a = sin 1 a b a 1 State the function and the maimal domain. Use the result d asin 1 a a bb = 1 "a with a =. b 1 State the function and the maimal domain. WritE b = sin 1 () a = sin 1 a b is defined for ` ` 1; that is, [, ] = 1 " for (, ) b = sin 1 () is defined for 1; that is, c 1, 1 d 35 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

52 Epress in terms of u. = sin 1 (u) where u = 3 Differentiate with respect to u and u with respect to. Find using the chain rule. 5 Substitute back for u and state the final result. WOrKeD example think Derivative of cos 1 a a b The derivative of cos 1 a b where a is a real positive constant is required. a Let = cos 1 a a b. From the definition of the inverse function, a = cos(), so = a cos() and = a sin(), and = 1 a sin(). We need to epress the result back in terms of. Using cos() =, draw a right-angled triangle. Label the a angle, with the adjacent side length being and the length of the hpotenuse being a. From Pthagoras theorem, the opposite side length is "a. Therefore, sin() = "a. a It follows that d acos 1 a a bb = 1, "a and if a = 1, we obtain d 1 (cos 1 ()) =. "1 du = 1 "1 u and du = = du du = "1 u = "1 for a 1, 1 b Determine for each of the following, stating the maimal domain for which the derivative is defined. a = cos 1 a 3 b b = cos 1 a 5 b a 1 State the function and the maimal domain. WritE a a a = cos 1 a 3 b is defined for ` ` 1; that is, or c 3, 3 d Epress in terms of u. = cos 1 a u b where u = 3 Topic 7 DIfferenTIaL CaLCuLus 353

53 3 Differentiate with respect to u using d du acos 1 a u a bb = 1 with a = 3, "a u and differentiate u with respect to. Find using the chain rule. 5 Substitute back for u and state the final result. b 1 State the function and solve the inequalit to find the maimal domain of the function. Derivative of tan 1 a a b We also need to determine the derivative of tan 1 a b, where a is a real positive a constant. du = 1 and du "9 u = = du du = "9 u = for a 3 "9, 3 b b = cos 1 a 5 5 b is defined for ` ` or 5 5 c, 5 5 d Epress in terms of u. = cos 1 a u b where u = 5 3 Differentiate with respect to u and u with respect to. Find using the chain rule. 5 Substitute back for u. Simplif the denominator using the difference of two squares. 7 State the final result. du = 1 and du "3 u = 5 = du du = 5 "3 u = 5 "3 (5 ) = 5!( + (5 ))( (5 )) = 5!( + 5)( 5) for a 5, 5 b Let = tan 1 a a b. From the definition of the inverse function, a = tan(), so = a tan(), = a sec (), and = 1 a sec (). 35 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

54 WOrKeD example 7 think a 1 State the function. However, we need to epress the result in terms of. Using tan() =, draw a right-angled triangle. Label the angle, a with the opposite side length being and the adjacent side length being a. From Pthagoras theorem, the length of the hpotenuse is "a +. We know that sec 1 a () =, and from above, cos() =. cos () "a + It follows that = cos () = 1 a a q a "a + r. Thus, d atan 1 a a bb = a a +, and if a = 1, then d (tan 1 ()) = The domain is defined for all R. Find for each of the following. a = tan 1 a 3 b b = tan 1 a 5 + b 7 WritE a = tan 1 a 3 b Epress in terms of u. = tan 1 a u b where u = 3 3 Differentiate with respect to u and u with respect to, using d du atan 1 a u a bb = a with a =. a + u Find using the chain rule. 5 Substitute back for u and state the final result. du = du and 1 + u = 3 = du du = u = b 1 State the function. b = tan 1 a 5 + b 7 Epress in terms of u. = tan 1 a u b where u = a + a 3 Differentiate with respect to u and u with respect to, using d du atan 1 a u a bb = a with a = 7. a + u Find using the chain rule. du = 7 du and 9 + u = 5 = du du = u Topic 7 DIfferenTIaL CaLCuLus 355

55 5 Substitute back for u. Epand the denominator, and simplif and take out common factors. 7 State the final result. WOrKeD example think = (5 + ) = = = 5( ) = finding second derivatives Recall that the second derivative d = d a b is also the rate of change of the gradient function. Find d if = sin 1 a 3 5 b. WritE 1 State the function and the maimal domain. = sin 1 a 3 5 b is defined for ` 3 ` 1; that is or c 5 3, 5 3 d. Epress in terms of u. = sin 1 a u b where u = Differentiate with respect to u and u with respect to u. du = 1 and du "5 u = 3 Find using the chain rule. 5 Substitute back for u. = du du = 3 "5 u = 3 "5 9 Write in inde notation. = 3(5 9 ) 7 Differentiate again using the chain rule. Simplif the terms. 9 State the final result in simplest form and the domain. d = 3 1 1(5 9 ) d = 7 (5 9 ) 3 1 d = 7 for a 5 "(5 9 ) 3 3, 5 3 b 3 35 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

56 further eamples Because sec() = 1 cos(), cosec() = 1 1 and cot() =, it follows that sin() tan() sec 1 () = cos 1 a 1 b, cosec 1 () = sin 1 a 1 b and cot 1 () = tan 1 a 1 b. WOrKeD example 9 think Find is defined. if = cos 1 a 3 b, stating the maimal domain for which the derivative WritE 1 State the function and the maimal domain. = cos 1 a 3 b for ` 3 ` 1 Solve the inequalit to find the maimal domain of the function is equivalent to Epress in terms of u. = cos 1 (u) where u = 3 = 3 1 Differentiate with respect to u and u with respect to. 5 Find using the chain rule. Substitute back for u. 7 Simplif b taking a common denominator in the square root in the denominator. Simplif, noting that " =. 9 State the final result and the domain. du = 1 and du "1 u = 3 = 3 = 3 "1 u = 3 Å 1 9 = 3 9 Å = 3 " 9 = 3 for < 3 or > 3. " 9 EErCisE 7.7 PraCtisE Derivatives of inverse trigonometric functions 1 WE5 Determine for each of the following, stating the maimal domain for which the derivative is defined. a = sin 1 a 5 b b = sin 1 (5) Topic 7 DIfferenTIaL CaLCuLus 357

57 Consolidate Given f() = sin 1 (), find f a 1 b. 3 WE Determine for each of the following, stating the maimal domain for which the derivative is defined. a = cos 1 a 3 b b = cos 1 a 3 b 5 Determine the gradient of the curve = cos 1 a 3 b at the point where = WE7 Find for each of the following. a = tan 1 () b = tan 1 a 3 b 5 Find the gradient of the curve = tan 1 a 3 5 b at the point where =. 7 7 WE Find d if = cos 1 a 3 b. Find d if = tan 1 a 3 b. 9 WE9 Find if = sin 1 a b, stating the maimal domain for which the derivative is defined. 10 Find if = tan 1 a b, stating the maimal domain for which the derivative! is defined. 11 Determine the derivative of each of the following, stating the maimal domain. a sin 1 a 3 b b sin 1 (3) c sin 1 a 3 b d sin 1 a + 3 b 5 1 Determine the derivative of each of the following, stating the maimal domain. a cos 1 a b b cos 1 () c cos 1 a 3 b d cos 1 a b 7 13 Determine the derivative of each of the following. a tan 1 a b b tan 1 () c tan 1 a 5 b d tan 1 a + 5 b 1 Determine the derivative of each of the following, stating the maimal domain. a sin 1 a! 3 b b sin 1 a 3 b c cos 1 a e b d cos 1 a 3 b 15 Find d if: e tan 1 a 3 b f tan 1 a 5 b a = sin 1 a 5 b b = cos 1 a 5 b c = tan 1 a 7 b. 35 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

58 Master 1 For each of the following: i sketch the graph of the function, and state the domain and the range ii find the eact equation of the tangent to the curve at the point indicated. a = sin 1 a 3 b, = 3 b = cos 1 a 5! b, = 5 c = tan 1 a 5 b, = 5 17 a If f() = sin 1 a b, find f (). b If f() = cos 1 a b, find f (3). c If f() = 3 tan 1 a b, find f (1).!3 1 Consider the function T() = tan 1 a 3 b + tan 1 a 3 b. a Find the values of T(3) and T( 3). b Find T (). What can ou deduce about the function T()? c State the domain and range of T(), and sketch its graph. 19 If a and b are positive constants, verif the following. a d csin 1 a b a b d = b b d "a b csin 1 a b a b d = b 3 "(a b ) 3 0 If a and b are positive constants, verif the following. a d ccos 1 a b a b d = b b d "a b ccos 1 a b a b d = b 3 "(a b ) 3 1 If a and b are positive constants, verif the following. a d ctan 1 a b a b d = ab a + b b d If a and b are positive constants, verif the following. a d csin 1 a a b b d = a "b a b d ccos 1 a a b b d = a "b a c d ctan 1 a a b b d = ab a + b ctan 1 a b a b d = ab3 (a + b ) Topic 7 Differential calculus 359

59 7. AOS 3 Topic 1 Concept 3 Related rates of change Concept summar Practice questions WOrKeD example 30 Related rate problems Introduction When two or more quantities var with time and are related b some condition, their rates of change are also related. The steps involved in solving these related rate problems are listed below. 1. Define the variables. Write down the rate that is provided in the question. 3. Establish the relationship between the variables.. Write down the rate that needs to found. 5. Use a chain rule or implicit differentiation to relate the variables. A circular metal plate is being heated and its radius is increasing at a rate of mm/s. Find the rate at which the area of the plate is increasing when the radius is 30 millimetres. think WritE 1 Define the variables. Let r mm be the radius of the metal at time t seconds. Let A mm be the area of the plate at time t seconds. The radius is increasing at a rate of mm/s. Note that the units also help to find the given rate. 3 The plate is circular. Write the formula for the area of a circle. Find the rate at which the area of the plate is increasing. 5 Form a chain rule for the required rate in terms of the given variables. Since one rate, dr, is known, we need dt to find da dr. 7 Substitute the given rates into the required equations. Evaluate the required rate when r = State the final result with the required units, leaving the answer in terms of π. dr dt = mm/s. A = πr. da =? and evaluate this rate when r = 30. dt da dt = da. dr dr dt A = πr da dr = πr da dt = da dr. dr dt = πr da = πr when r = 30 dt da dr ` = 10π mm /s r = 30 The area is increasing at 10π mm /s. 30 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

60 Decreasing rates If a quantit has a rate decreasing with respect to time, then the required rate is given as a negative quantit. WOrKeD example 31 A spherical balloon has a hole in it, and the balloon s volume is decreasing at a rate of cm 3 /s. At what rate is the radius changing when the radius of the balloon is cm? think WritE 1 Define the variables. Let r cm be the radius of the balloon at time t seconds. Let V cm 3 be the volume of the balloon at time t seconds. dv The volume is decreasing at a rate of cm 3 /s. Notice that the units also help find dt = cm3 /s the given rate. 3 The ballon is spherical. Write the formula for the volume of a sphere. We need to find the rate at which the radius changing. 5 Form a chain rule for the required rate in terms of the given variables. Since one rate, dv, is known, we need dt to find dr dv. 7 Substitute the given rates into the required equations. V = 3 πr3 dr =? and evaluate this rate when r =. dt dr dt = dr. dv dv dt Since V = 3 πr3 dv dr = πr and dr dv = 1 / dv dr = 1 πr dr dt = dr dv. dv dt = 1 πr = 1 πr Evaluate the required rate when r =. When r =, dr dt ` = 1 r = 3π. 9 State the final result in the required units, The radius is decreasing at a rate of 1 leaving the answer in terms of π. 3π cm/s. relating the variables It is often necessar to epress a required epression in terms of onl one variable instead of two. This can be achieved b finding relationships between the variables. Topic 7 DIfferenTIaL CaLCuLus 31

61 WOrKeD example 3 A conical funnel has a height of 5 centimetres and a radius of 0 centimetres. It is positioned so that its ais is vertical and its verte is downwards. Initiall it is filled with oil. The oil leaks out through an opening in the verte at a rate of cubic centimetres per second. Find the rate at which the oil level is falling when the height of the oil in the funnel is 5 centimetres. (Ignore the clindrical section of the funnel.) think WritE/draW 1 Define the variables. Let r cm be the radius of the oil in the funnel. Let h cm the height of the oil in the funnel. dv Write the rate given in the question. dt = cm3 /s 3 Write which rate is required to be found. dh dt =? when h = 5 Make up a chain rule for the required rate, in dh terms of the given variables. dt = dh dv dv dt 5 Find the relationship between the variables. The variables r and h change; however, the height and radius of the funnel are constant, and this can be used to find a relationship between h and r. We need to epress the volume in terms of h onl. 7 Find the rate to substitute into the chain rule. Substitute for the required rates. 9 Evaluate the required rate when h = State the final result with the required units, leaving the answer in terms of π. 0 5 h Filled with oil Using similar triangles, 0 5 = r h r = h 5 Substitute into V = 1 3 πr h: r V = 1 3 πr h V = 1 3 πah 5 b h = 1πh3 75 dv dh = 1πh 5 dh dv = 5 1πh dh dt = dh dv dv dt = 5 1πh dh dt ` = 5 h = 5 1π5 = 1 π The height is decreasing at a rate of 1 π cm/s. 3 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

62 WOrKeD example 33 think Method 1: Using the chain rule Determining the required variables An alternative method to solving related rate problems is to use implicit differentiation. A ladder 3 metres long has its top end resting against a vertical wall and its lower end on horizontal ground. The top end of the ladder is slipping down at a constant speed of 0.1 metres per second. Find the rate at which the lower end is moving awa from the wall when the lower end is 1 metre from the wall. WritE/draW 1 Define the variables. Let metres be the distance of the base of the ladder from the wall, and let metres be the distance of the top of the ladder from the ground. The top end of the ladder is slipping down at a constant speed of 0.1 m/s. = 0.1 m/s dt 3 Appl Pthagoras theorem. + = 3 = 9 We need to find the rate at which the =? when = 1. lower end is moving awa from the wall dt when the lower end is 1 m from the wall. 5 Construct a chain rule for the required rate in terms of the given variables. dt = dt Epress in terms of. + = 9 = 9 = ±"9 7 Since one rate,, is known, we dt need to find. Substitute the given rates into the required equation. = "9 since > 0 = "9 = "9 dt = dt "9 = Topic 7 DIfferenTIaL CaLCuLus 33

63 9 Evaluate the required rate when = State the final result with the required units, giving an eact answer. Method : Using implicit differentiation The first steps are identical to method 1 above. From this point, we use implicit differentiation to find the required rate. EErCisE 7. PraCtisE Related rate problems dt ` =! = 1 10 =! 10 dt ` =! = 1 5 m/s The lower end is moving awa from the wall at a rate of! 5 m/s. 5 Take d dt ( ) of each term in turn. d dt ( ) + d dt ( ) = d dt (9) Use implicit differentiation. dt + dt = 0 7 Rearrange to make the required rate the subject. dt = dt Find the appropriate values. When = 1, =! and dt = 0.1 = Substitute in the appropriate values of and. 10 State the final result as before. dt ` =! = 1 10 =! 10 dt ` =! = 1 5 m/s The lower end is moving awa from the wall at a rate m/s. of! 5 1 WE30 A circular oil slick is epanding so that its radius increases at a rate of 0.5 m/s. Find the rate at which the area of the oil slick is increasing when the radius of the slick is 0 metres. A circular disc is epanding so that its area increases at a rate of 0π cm /s. Find the rate at which the radius of the disc is increasing when the radius is 10 cm. 3 WE31 A spherical basketball has a hole in it and its volume is decreasing at a rate of cm 3 /s. At what rate is the radius changing when the radius of the basketball is cm? A metal ball is dissolving in an acid bath. Its radius is decreasing at a rate of 3 cm/s. At what rate is the ball s surface area changing when the radius of the ball is cm? 5 WE3 A conical vase has a height of 0 cm and a radius of cm. The ais of the vase is vertical and its verte is downwards. Initiall it is filled with water which leaks out through a small hole in the verte at a rate of cm 3 /s. Find the rate at which the water level is falling when the height of the water is 1 cm. 3 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and

64 Consolidate A cone is such that its radius is alwas equal to half its height. If the radius is decreasing at a rate of cm/s, find the rate at which the volume of the cone is decreasing when the radius is cm. 7 WE33 A ladder 5 metres long has its top end resting against a vertical wall and its lower end on horizontal ground. The bottom end of the ladder is pushed closer to the wall at a speed of 0.3 metres per second. Find the rate at which the top end of the ladder is moving up the wall when the lower end is 3 metres from the wall. A kite is 30 metres above the ground and is moving horizontall awa at a speed of metres per second from the bo who is fling it. When the length of the string is 50 metres, at what rate is the string being released? 9 a A square has its sides increasing at a rate of centimetres per second. Find the rate at which the area is increasing when the sides are centimetres long. b A stone is dropped into a lake, sending out concentric circular ripples. The area of the disturbed water region increases at a rate of m /s. Find the rate at which the radius of the outermost ripple is increasing when its radius is metres. 10 A spherical bubble is blown so that its radius is increasing at a constant rate of millimetres per second. When its radius is 10 millimetres, find the rate at which: a its volume is increasing b its surface area is increasing. 11 A mothball has its radius decreasing at a constant rate of 0. millimetres per week. Assume it remains spherical. a Show that the volume is decreasing at a rate that is proportional to its surface area. b If its initial radius is 30 millimetres, find how long it takes to disappear. 1 The sides of an equilateral triangle are increasing at a rate of centimetres per second. When the sides are!3 centimetres, find the rate at which: a its area is increasing b its height is increasing. 13 a Show that the formula for the volume, V, of a right circular cone with height h is given b V = 1 3 πh3 tan (α), where α is the semi-verte angle. b Falling sand forms a heap in the shape of a right circular cone whose semiverte angle is If its height is increasing at centimetres per second when the heap is 5 centimetres high, find the rate at which its volume is increasing. Topic 7 Differential calculus 35

65 1 a The volume, V cm 3, of water in a hemispherical bowl of radius r cm when the depth of the water is h cm is given b V = 1 3 πh (3r h). A hemispherical bowl of radius 10 centimetres is being filled with water at a constant rate of 3 cubic centimetres per second. At what rate is the depth of the water increasing when the depth is 5 cm? b A drinking glass is in the shape of a truncated right circular cone. When the glass is filled to a depth of h cm, the volume of liquid in the glass, V cm 3, is given b V = π 3 (h3 + 10h + 3h). Lemonade is leaking out from the glass at a rate of 7 cm 3 /s, Find the rate at which the depth of the lemonade is falling when the depth is cm. 15 a A rubber flotation device is being pulled into a wharf b a rope at a speed of metres per minute. The rope is attached to a point on the wharf 1 metre verticall above the ding. At what rate is the rope being drawn in when the ding is 10 metres from the wharf? b A car approaches the ground level of a 30-metre-tall building at a speed of 5 kilometres per hour. Find the rate of change of the distance from the car to the top of the building when it is 0 metres from the foot of the building. 1 The distance, q cm, between the image of an object and a certain lens in terms of p cm, the distance of the object from the lens, is given b q = 10p p 10. a Show that the rate of change of distance that an image is from the lens with respect to the distance of the object from the lens is given b dq dp = 100 (p 10). b If the object distance is increasing at a rate 0. cm/s, how fast is the image distance changing, when the distance from the object is 1 cm? 17 a When a gas epands without a change of temperature, the pressure P and volume V are given b the relationship PV 1. = C, where C is a constant. At a certain instant, the pressure is pascals and the volume is cubic metres. The volume is increasing at a rate of cubic metres per second. Find the rate at which the pressure is changing at this instant. b The pressure P and volume V of a certain fied mass of gas during an adiabatic epansion are connected b the law PV n = C, where n and C are constants. Show that the time rate of change of volume satisfies dv dt = V dp np Dt. 1 a A jet aircraft is fling horizontall at a speed of 300 km/h at an altitude of 1 km. It passes directl over a radar tracking station located at ground level. Find the rate in degrees per second at which the radar beam to the aircraft is turning when the jet is at a horizontal distance of 30 kilometres from the station. 3 Maths Quest 1 SPECIALIST MATHEMATICS VCE Units 3 and

66 Master b A helicopter is fling horizontall at a constant height of 300 metres. It passes directl over a light source located at ground level. The light source is alwas directed at the helicopter. If the helicopter is fling at 10 kilometres per hour, find the rate in degrees per second at which the light source to the helicopter is turning when the helicopter is at a horizontal distance of 0 kilometres from the light source. 19 A man metres tall is walking at 1.5 m/s. He passes under a light source metres above the ground. Find: a the rate at which his shadow is lengthening b the speed at which the end point of his shadow is increasing c the rate at which his head is receding from the light source when he is metres from the light. 0 a A train leaves a point O and travels south at a constant speed of 30 km/h. A car that was initiall 90 km east of O travels west at a constant speed of 0 km/h. Let s be the distance in km of the car from the train after a time of t hours. i Show that s = 10"5t 7t + 1. ii Find when the train and the car are closest, and find the closest distance in kilometres between the car and the train. iii Find the rate at which the distance between the train and the car is changing after 1 hour. b Two railwa tracks intersect at 0. One train is 100 km from the junction and moves towards it at 0 km/h, while another train is 10 km from the junction and moves towards the junction at 90 km/h. i Show that the trains do not collide. ii Find the rate at which the trains are approaching each other after 1 hour. Topic 7 Differential calculus 37

67 ONLINE ONLY 7.9 Review the Maths Quest review is available in a customisable format for ou to demonstrate our knowledge of this topic. the review contains: short-answer questions providing ou with the opportunit to demonstrate the skills ou have developed to efficientl answer questions using the most appropriate methods Multiple-choice questions providing ou with the opportunit to practise answering questions using CAS technolog stuon is an interactive and highl visual online tool that helps ou to clearl identif strengths and weaknesses prior to our eams. You can then confidentl target areas of greatest need, enabling ou to achieve our best results. Etended-response questions providing ou with the opportunit to practise eam-stle questions. a summar of the ke points covered in this topic is also available as a digital document. REVIEW QUESTIONS Download the Review questions document from the links found in the Resources section of our ebookplus. Units 3 & Differential calculus Sit topic test 3 MaThs QuesT 1 specialist MaTheMaTICs VCe units 3 and