Further Calculus Implicit differentiation Parametric differentiation Related rates of change Small variations and linear approximations Stationary points Curve sketching - asymptotes Curve sketching the procedure Optimisation Area between curves Volumes of revolution Arc length of plane curves
. Main page To differentiate a function that is defined implicitly relating two variables, we differentiate each term of the equation with respect to one of the variables. Some terms in the equation may involve both x and y and so the product rule, quotient rule or chain rule may need to be used. For example: d( h( x) g( y)) d( h( x)) d( g( y)) g( y) + h( x) h '( x) g( y) + h( x) g '( y) Given f ( x, y) c we differentiate with respect to x so that d( f ( x, y)) 0. Implicit differentiation We will need to differentiate terms involving y with respect to x. We do this using the chain rule: d( g( y)) d( g( y)) g '( y)
. Main page To differentiate a function relating two variables that is defined parametrically we differentiate each function with respect to the parameter. 2 To find the second derivative d x 2 the chain rule again. we use 2 d x d dz 2 First let z Then using the chain rule dz z dz where z x Given x f ( t) and y g( t) we differentiate with respect to t so that f '( t) and g '( t) Parametric differentiation Using the chain rule: or. y If x and y then x
Main page Notation for rates with respect to time: x, y Using the chain rule to relate two rates e.g. and if y x or y x y y f ( x) f' ( x) x y so y xf '( x) or x f '( x) Similarly If dr r, S S so S ds f ( r) ds dr rf '( r) or r f' ( r) S f '( r) Related rates of change The formula relating the variables may be given or you may need to derive the formula from the situation. S r
Main page The small variation approximation is given by where the small changes in x and y are given respectively by So and if y x x and y. y x y f ( x) y f' ( x) x Percentage change To calculate a percentage change in y given a percentage change in x. If the % change in x is p% then y f '( x) x y f '( x) x y y x k x f '( x) x f ( x) So the % change in y is k x p%. The linear approximation to yf(x) at the point xx 0 is f ( x) f ( x0) + f '( x0 )( x x0) This formula can be used to approximate values of f(x) where a suitable x0 is to be chosen. Small variations and linear approximations
Main page Stationary points The derivative is zero Stationary points Nature of stationary points Second derivative test
Main page In a proper fraction the degree of the numerator is less than the degree of the denominator. Horizontal or Inclined Asymptotes These are the lines that the function approaches as x. If f(x) is a rational function then write f ( x ) p ( x ) + proper fraction where p(x) is a polynomials. The polynomial p(x) and the proper fraction can be found using algebraic long division or other methods. As x, the proper fraction 0 y f ( x) p( x). So we have the asymptote y p(x). If p(x) c we have a horizontal asymptote and if p(x) ax+b then we have an inclined asymptote. Vertical Asymptotes From values of x where the function is undefined 1 e.g. f ( x) has vertical asymptotes ( x 2)( x + 3) x 2 and x -3 Curve sketching: Asymptotes Approaching asymptotes We can consider the sign of A in A f ( x) p( x) + q(x) q( x) in order to find how the function approaches the asymptote. If A is positive as x or x - then q(x) f(x) approaches the asymptote from above, if negative then the approach is from below.
Back to Index page Geometry Curve sketching: Asymptotes Back to notes Find the assymptotes of f ( x) 2 2x + 5x 3 x + 2 Vertical asymptotes The denominator is zero when x + 2 0 x 2 So we have a vertical assymptote at x -2. Slanting or horizontal asymptotes. 5 f ( x) is an improper fraction so use algebraic long division to write f ( x) 2x + 1 x + 2 5 As x, 0 as it is a proper fraction so f ( x) 2x + 1 as x. x + 2 Thus the slanting asymptote is y 2x + 1. 5 When x > 0, > 0 so f ( x) < 2x + 1 (we are taking away a +ve number) x + 2 and so f ( x) 2x + 1 from below. 5 When x < 0, < 0 so f ( x) > 2x + 1 (we are taking away a -ve number) x + 2 and so f ( x) 2x + 1 from above.
Back to Index page Geometry Curve sketching: Asymptotes Back to notes 2) Find the asymptotes ( x 3)( x + 1) of f ( x) ( x + 2)( x 1) Vertical asymptotes The denominator is zero when ( x + 2)( x 1) 0 So we have a vertical assymptote at x -2 and at x 1. Slanting or horizontal asymptotes. x + 1 f ( x ) is an improper fraction so use algebraic long division to writ e f ( x ) 1 ( x + 2)( x 1) x + 1 As x, 0 as it is a proper fraction so f ( x) 1 as x. ( x + 2)( x 1) Thus th e horizontal asymptote is y 1. x + 1 When x is large and x > 0, > 0 so f ( x) < 1 (we are taking away a +ve number) ( x + 2)( x 1) and so f ( x) 1 from below. x + 1 When x is large and x < 0, < 0 so f ( x) > 1 (we are taking away a -ve number) ( x + 2)( x 1) and so f ( x) 1 from above.
Main page Intercepts on the axes x axis: y 0 f ( x) 0 Solve the equation to find the value(s) of x. Stationary points Find the stationary points using the methods on the Stationary points slide y axis: x 0 y f (0) Plot these points on the axes Asymptotes See previous slide. Draw in the vertical and horizontal or slanting asymptotes Further Calculus: Curve sketching: the procedure yf(x) Positive and negative regions The graph can only change from positive to negative at intercepts on the x axis and at vertical asymptotes. These values of x are the critical points. We need to find out whether yf(x) is positive or negative between these values. Use a table like the one below. x a b yf(x) + 0 - U D yf(x) is undefined (UD) at a vertical asymptote and zero at an intercept on the x axis. Identify these regions on the graph +
Main page Maximum or minimum The derivative is zero Deriving the formula for the variable to be optimised This formula must be in terms of 1 variable only: other variables must be eliminated using the constraints given Optimization Nature of stationary points Second derivative test
Further Calculus That s all folks! Main page