Further Differentiation and Applications

Transcription

1 Avance Higher Notes (Unit ) Prerequisites: Inverse function property; prouct, quotient an chain rules; inflexion points. Maths Applications: Concavity; ifferentiability. Real-Worl Applications: Particle motion; optimisation. Derivative of Inverse Functions Given a function f, the erivative of its inverse f can be foun. Theorem: Given a function f, the erivative of f is given by, D (f ) ( Df ) f In Leibniz notation, this formula takes on a much more memorable form, remembering that if y f (x) then x f (y), An example will be given (using both formulae) to show how the theorem is use. Example Obtain the erivative of the function efine by y f (x) x. Taking the positive root allows us to efine an inverse function given by, f x. Note that Df x. Hence, D (f ) x (x) M Patel (April 0) St. Machar Acaemy

2 Avance Higher Notes (Unit ) Note that ifferentiating the inverse irectly gives the same answer. Using the Leibniz formula instea, recalling that x f (y) y, x y y This is the same answer as above (the inverse function use here has variable y, so interchanging x an y gives the same result). On the other han, x y Derivatives of Inverse Trigonometric Functions The erivatives of arcsine, arccosine an arctangent can be foun using the above formula. sin x x cos x x tan x + x M Patel (April 0) St. Machar Acaemy

3 Avance Higher Notes (Unit ) Example Differentiate y cos (x). By the chain rule, ( x ) (x) 9x Example Differentiate y sin ( x ). ( x ) ( x ) x 6 x Example 4 Differentiate y tan + x. + ( + x ) + x + x + x ( + x ) + x Often in this course, answers to erivatives will be freakishly complex looking, as in Example 4. Don t let this convince you that the answer is wrong. Just follow the rules an get an answer. M Patel (April 0) St. Machar Acaemy

4 Avance Higher Notes (Unit ) Implicit Differentiation Implicit an Explicit Functions Definition: A function f is given explicitly (f is an explicit function) if the output value y is given in terms of the input value. An explicit function is recognise when y is given as a function of x. Definition: A function f is given implicitly (f is an implicit function) if the output value y is not given in terms of the input value. An implicit function is usually ientifie when the variables x an y are mixe up in a higgle-piggle manner. An implicitly efine function may or may not be solve for y. Example 5 xy efines y implicitly as a function of x. However, for x 0, it can be written explicitly as y x. Example 6 The function y xy 8 sin y + x 5 efines y implicitly as a function of x ; the expression on the LHS is so higgle-piggle in x an y, that y cannot be written explicitly as a function of x. First Derivatives of Implicit Functions To fin for an implicitly efine function, ifferentiate both sies of the equality with respect to x an solve the resulting equation for. M Patel (April 0) 4 St. Machar Acaemy

5 Avance Higher Notes (Unit ) The chain rule will most likely be use an sometimes the prouct or quotient rule too. Just remember that y is a function of x. Example 7 Fin for the function efine implicitly by the equation x + 5. xy First, write own what the intent is, ( x + xy ) 5 When there is a lonely constant on one sie of the equation, it can be very easy to forget to ifferentiate it; so ifferentiate it first. Next, remember that y is a function of x, so the secon term on the LHS nees an application of the prouct rule, x + y + xy 0 Solving this equation for the erivative gives, ( ) x + y xy Example 8 Fin when 6y + xy x. (6 y + xy x ) y + y + x x 0 (y + x) x y M Patel (April 0) 5 St. Machar Acaemy

6 Avance Higher Notes (Unit ) x y y + x x y 6 y + x Functions efine implicitly by equations that look a little more menacing are still no match for implicit ifferentiation. Example 9 Fin when y xy 8 sin y + x 4. ( y xy 8 sin y + x ) 4 4y y + x 8 cos y + x 0 (4y x 8 cos y) y x y x 4 y x 8 cos y Example 0 Fin when sin x ln (xy) + x y 7. (sin x ln (xy) + x y) 7 sin x y + x cos x ln (xy) xy x sin x x sin x + x y xy cos x ln (xy) x y y sin x M Patel (April 0) 6 St. Machar Acaemy

7 Avance Higher Notes (Unit ) (x sin x + x y) xy cos x ln (xy) x y y sin x ( xy cos x ln ( xy ) + x y + y sin x ) x sin x + x y Secon Derivatives of Implicit Functions The secon erivative can also be foun using implicit ifferentiation. The iea is to ifferentiate a line of working that is use in fining the first erivative. Example Fin an y when xy x 4. Differentiating implicitly we get, y + x 0 From this we get, y x Differentiate implicitly the line of working to get, + y + x y x 0 y x y x M Patel (April 0) 7 St. Machar Acaemy

8 Avance Higher Notes (Unit ) y ( y ) x Logarithmic Differentiation Another technique to a to our arsenal of ifferentiating functions involves taking natural logarithms of both sies of an equation before ifferentiating. This technique is known as logarithmic ifferentiation. Logarithmic ifferentiation shoul be use when any one of the following inicators are present. Brackete terms with fractional powers. Variable is in the power. Prouct or quotient of more than functions. Example Differentiate y ( x ) ( x ) (4 x + 7) 4. Taking natural logarithms of both sies an using the power comes own rule for logarithms gives, ln y ln (x ) + ln ( x) 4 ln (4x + 7) This is much frienlier to ifferentiate. Differentiating an simplifying each term on the RHS gives, y x x 4 x + 7 y x x 4 x + 7 Remembering what y actually is gives, M Patel (April 0) 8 St. Machar Acaemy

9 Avance Higher Notes (Unit ) ( x ) ( x ) (4 x + 7) 4 x x 4 x + 7 Example x Differentiate y x e cos x. Taking natural logarithms gives, x ln y ln x + ln ( e ) + ln (cos x) ln y ln x x + ln (cos x) Differentiating gives (notice that term on the LHS cropping up again), y x + ( sin x ) cos x y x tan x x x e cos x x tan x x e cos x x x e cos x x x e sin x Example 4 Differentiate y sin x x. Taking natural logarithms gives, ln y sin x. ln x Differentiating (using the prouct rule) gives, y cos x. ln x + sin x. x M Patel (April 0) 9 St. Machar Acaemy

10 Avance Higher Notes (Unit ) y (cos x. ln x + x sin x) sin x x (cos x. ln x + x sin x) Parametric Differentiation It is possible to write a function y f (x) ifferently as functions of a common variable. This sometimes makes it easier to work out. Parametric Functions Definition: A curve is efine parametrically if it can be escribe by functions x an y, calle parametric functions (aka parametric equations of the curve), of a common variable known as the parameter. Normally, the parameter is enote by t or θ, an the parametric functions x an y are thus written as x (t) an y (t) or x (θ) an y (θ). It often helps to think of the parameter as time; as the value of t changes, a curve is generate in the x y plane. Think of an ant walking about in the x y plane; for each value of t, it has a certain position (i.e. coorinate) in the plane given by (x (t), y (t) ). Sometimes the curve is a function, but it can be any type of shape in the plane (for example, a circle). Many shapes are often easier to escribe implicitly. Example 5 Show that the point (0, ) lies on the curve escribe by the parametric equations x cos θ, y sin θ (θ [0, π]), stating the value of θ. Note that x + y, i.e. the parametric equations efine a circle with centre (0, 0) an raius. The point (0, ) clearly lies on this circle an the angle is obviously π/. Another way to see this is to put x 0 an y into the parametric equations an show that there is only one value of θ that satisfies both equations. Inee, 0 cos θ leas to M Patel (April 0) 0 St. Machar Acaemy

11 Avance Higher Notes (Unit ) θ π/ or π/, whereas sin θ leas to θ π/. So, there is a common value that satisfies both equations an it is θ π/. Example 6 Show that the point (, 5) oes not lie on the curve efine parametrically by x t, y 4t. If the given point i lie on the curve, then t t. However, t gives y 4, not 5. Thus, (, 5) oes not lie on the curve. First Derivatives of Parametric Functions Theorem: Given parametric functions x (t) an y (t), the erivative of y with respect to x is given by, In Newton Notation, this is written as, y x Example 7 Fin when x t, y ln t. The erivatives with respect to t are x 4t an y t. Hence, y x / t 4t M Patel (April 0) St. Machar Acaemy

12 Avance Higher Notes (Unit ) 4t Example 8 Fin the equation of the tangent line to the curve at t efine t parametrically by the equations x, y t The erivatives are, + t. t + t ( t ), 4 t ( t ) ( t ) Hence, 4 t ( t ) + t At t, x, y 5 an tangent line is, 4 5. Hence, the equation of the y x + After a wee bit of simplifying, this becomes, 7x + 5y M Patel (April 0) St. Machar Acaemy

13 Avance Higher Notes (Unit ) Secon Derivatives of Parametric Functions Theorem: Given parametric functions x (t) an y (t), the secon erivative of y with respect to x is given by, Evaluating this leas to, y y x x y x y y x x Example 9 Fin y when x + t an y 4 4t. If using the secon form, calculate the relevant quantities first. So, x, ẋ 0, y 8t an ẏ 8. Then use the formula, y x y y x x y.( 8) ( 8 t ).0 y 4 7 y 8 9 Example 0 Show that the curve efine parametrically by the equations x t an y t + has no point of inflexion. t t M Patel (April 0) St. Machar Acaemy

14 Avance Higher Notes (Unit ) First note that t 0. If there is a P of I, then either y oes not exist or it is obtaine by solving y 0. To obtain the secon erivative, use the first form (it s easier) to get, y 4t ( t + ) 4 6 Solving this for t gives t 0. But this contraicts the fact that t 0. Alternatively, since t 0, the equation y 0 has no solutions for t. Either way, there is no P of I. Planar Motion Motion in a plane is often best escribe by parametric equations. The following efinitions exten the D efinitions from Unit to D. Definition: Planar motion is motion in a plane an is escribe by functions of time x (t) an y (t). Definition: The isplacement of a particle at time t in a plane is escribe by the isplacement vector, s (t) ef (x (t), y (t) ) x (t) i + y (t) j The magnitue of isplacement, aka istance (from the origin), at time t is, s (t) ef x + y The irection measure from the x axis (aka irection of isplacement), θ, at any instant of time t is, M Patel (April 0) 4 St. Machar Acaemy

15 Avance Higher Notes (Unit ) tan θ y x where θ is the angle between x i an s. Definition: The velocity of a particle at time t in a plane is escribe by the velocity vector, v (t) ef s (x (t),y (t ) x (t) i + y (t) j The magnitue of velocity, aka spee, at time t, is, v (t) ef x + y The irection of motion (aka irection of velocity), η, at any instant of time t is, tan η y x where η is the angle between x i an v. Definition: The acceleration of a particle at time t in a plane is escribe by the acceleration vector, a (t) ef v (ẋ (t),ẏ (t ) ẋ (t) i + ẏ (t) j The magnitue of acceleration, at time t, is, a (t) ef x + y The irection of acceleration, ψ, at any instant of time t, is, M Patel (April 0) 5 St. Machar Acaemy

16 Avance Higher Notes (Unit ) tan ψ y x where ψ is the angle between ẋ i an a. Example Fin the magnitue an irection of the isplacement, velocity an acceleration at time t of the particle whose equations of motion are x t t an y t + 4t. We eal first with the isplacement. x () 4 an y (). Hence, the magnitue of isplacement is metres. The irection is foun by solving tan θ, which gives 7 6. The velocity components are given by x t an y t + 4. So, x () 0 an y () 8. Hence, the magnitue of velocity is metres per secon. The irection is foun by solving tan η 4/5, which gives 8 7. The acceleration components are given by ẋ 6t an ẏ. So, ẋ () an ẏ (). Hence, the magnitue of acceleration is metres per secon square. The irection is foun by solving tan ψ /6, which gives 9 5. Relate Rates of Change Definition: A relate rate of change refers to an equation involving erivatives, in particular, erivatives arising from y as a function of x an both x an y as functions of a thir variable u. Problems involving relate rates of change arise naturally in real-life. To solve them requires use of the Chain Rule an occasionally the Leibniz relation. M Patel (April 0) 6 St. Machar Acaemy

17 Avance Higher Notes (Unit ) Example If a spherical balloon is inflate at a constant rate of 40 cubic centimetres per secon, fin the rate at which the raius is increasing when the iameter is 6 cm. Something about a sphere, it s iameter (an hence raius), rate of change of volume with respect to time an rate of change of raius with respect to time. Souns like the volume of a sphere woul be a goo start. V 4 π r Remember that V an r are secretly functions of time, t. Differentiate both sies w.r.t. t to get, V 4πr r Rearranging this gives, r 4πr V If the iameter is 6 cm, then the raius is 8 cm. Hence, r 4π(64). 40 r r 40 56π 5 6π cm/s Example In a capacitive circuit, the formula for the total capacitance C of capacitors in series is, M Patel (April 0) 7 St. Machar Acaemy

18 Avance Higher Notes (Unit ) C C + C If C is increasing at a rate of 0 5 faras per minute an C is ecreasing at a rate of 0 9 faras per minute, at what rate is C changing (to. p.) when C faras an C fara? Remember that C, C an C are secretly functions of time. Firstly, when C faras an C fara, C / fara (easy aing fractions; on t forget to reciprocalise!). Next, ifferentiate the given equation w.r.t. t to get, C C C C C C With a little algebraic jiggery-pokery, this becomes, C C C C + C C C C List the quantities an their values. C /, C, C, 0 C 5 an 0 9 (it s negative because it s ecreasing). Now it s just number crunching, C. (0 5) +. ( 0 9) C 8 5 C 0 faras (. p.) M Patel (April 0) 8 St. Machar Acaemy