VIEWS: 1 PAGES: 6 POSTED ON: 10/2/2012 Public Domain
Chemistry 1S Calculus I Dr Paul May 4.12 Types of Stationary Point If xsp is the stationary point, then if we consider points either side of xsp, there are 4 types of behaviour of the gradient. x < xsp x = xsp x > xsp t.p.type (i) +ve zero -ve maximum (ii) -ve zero +ve minimum (iii) +ve zero +ve inflection (iv) -ve zero -ve inflection Stationary points like (iii) and (iv), where the gradient doesn't change sign produce S- shaped curves, and the stationary points are called points of inflection. Minimum Maximum +ve p.o.i -ve p.o.i 4.13 How to determine if a stationary point is a max, min or point of inflection The rate of change of the slope either side of a turning point reveals its type. But a dy rate of change is a differential. So all we need to do is differentiate the slope, dx with respect to x. In other words we need the 2nd differential, or d dy d2 y , more usually called (“dee-2-y by dee-x-squared”) dx d x dx 2 Examples 1. y(x) = 9x2 - 2 dy d2 y = 18x = 18 dx dx 2 27 Chemistry 1S Calculus I Dr Paul May 1 2. y(x) = 4x5 - x dy 1 d2 y 2 = 20x4 + 2 2 = 80x3 - 3 dx x dx x 3. p = 3q3 - 4q2 + 6 dp d2 p = 9q2 - 8q = 18q - 8 dq dq 2 4.14 Rules for stationary points d2 y i) At a local maximum, = -ve dx 2 d2 y ii) At a local minimum, = +ve dx 2 d2 y iii) At a point of inflection, = 0, dx 2 and we must examine the gradient either side of the turning point to find out if the curve is a +ve or -ve p.o.i. Examples 1. Taking the same example as we used before: y(x) = x3 - 3x + 1 (a cubic function – you should be expecting 2 t.p.s) dy = 3x2 - 3 stationary points at ( -1, 3) and (1, -1) dx d2 y = 6x dx 2 d2 y At stationary point (-1,3), x = -1, so = -6, so this point is a maximum. dx 2 d2 y At stationary point (1, -1), x = +1, so = +6, so this point is a minimum dx 2 So we can finally sketch the curve: 28 Chemistry 1S Calculus I Dr Paul May max at (-1,3) y 3 2 1 0 x -1 -2 min at (1,-1) -3 -3 -2 -1 0 1 2 3 2. y = x3 + 8 (a cubic function – you should be expecting 2 t.p.s) dy = 3x2, which is equal to zero at the turning point(s). dx If 3x2 = 0, x = 0, and so y = +8, so the t.p. is at (0,8). [The fact that we’ve found only one turning point for a cubic function (for which you would normally expect two) should tell you either you’ve done it wrong, or that the t.p. might be ‘unusual’]. d2 y Now = 6x dx 2 d2 y dy So, at the t.p. (0,8), i.e when x = 0, 2 = 0, so we have a point of inflection. But dx dx dy is +ve either side of this point (e.g. at x = +1, = +3 i.e. increasing, at x = -1, dx dy = +3, also increasing), so the curve has a positive point of inflection. dx 40 point of inflection at (0,8) 30 y 20 10 x 0 -10 -3 -2 -1 0 1 2 3 4 29 Chemistry 1S Calculus I Dr Paul May 3. Where are the turning point(s), and does it (or they) indicate a max or min in the function p(q) = 4 - 2q - 3q2 ? [How many t.p.s do we expect?] dp = -2 - 6q, which at the turning point = 0 dq so -2 - 6q = 0, 6q = -2, so q = - 1 , and p = 4 1 3 3 We have one t.p. at ( - 1 , 4 1 ). 3 3 d2 p = -6, which is –ve, so the t.p. is a maximum. dq 2 (This is consistent with what we said earlier, that for quadratics if the x2 term is -ve, we have a -shaped parabola). turning point at ( -1/3, 13/3 ) 6 p 4 2 q 0 -2 -4 -6 -3 -2 -1 0 1 2 4.15 More Complicated Differentiation Rules So far, we’ve only dealt with simple polynomial functions. But in chemistry we’re likely to meet a whole bunch of other, more complicated functions, which we also need to differentiate. For many of these, the ‘magic formula’ is insufficient, so we need to learn 3 more rules, and recognise when to apply them. These rules are: the Product Rule, the Quotient Rule, and the Function of a Function rule. 4.15A Product Rule Say we have functions like: y(x) = (3x2 – 2x +3)(4x4 +6x - 9); or x2 ln x ; or 3x sinx where the function y(x) can be written as two other functions multiplied together, e.g. in the second example we have x2×lnx. We name the first function u(x) and the second one v(x), so that 30 Chemistry 1S Calculus I Dr Paul May y(x) = u(x) v(x) then the differential of y(x) is given by the Product Rule: dy dv du = u. + v. dx dx dx In words: ‘The first function multiplied by the differential of the second, plus the second function multiplied by the differential of the first’. Examples 1. y(x) = (x2 + 2)(x + 1) du let u = x2 + 2, so that = 2x dx dv let v = x + 1, so that =1 dx dy dv du so = u. + v. = (x2 + 2).1 + (x + 1).2x dx dx dx = 3x2 + 2x + 2 (Note: in this case we can double check this is correct by expanding out the brackets) dy y(x) = x3 + x2 + 2x + 1, = 3x2 + 2x + 2 dx 1 2. y(x) = x3 (3 - + 3x2) x u v dy 1 1 = x3 ( 2 + 6x) + (3 - + 3x2).3x2 dx x x 1 1 3. () = (2 + 2)( + 2) u v dφ 1 2 1 1 = (2 + 2)( - 2 - 3 ) + ( + 2 )(2+2) dλ 31 Chemistry 1S Calculus I Dr Paul May 4.15B Quotient Rule Sometimes we come across rational functions, where one function is divided by another. Examples include: x 1 x2 log10 x y ( x) or y ( x) or y ( x) 3x 9 sin x x 1 Again, we call them u and v, making sure that u is on the top of the fraction, so that: u( x ) y(x) = , v( x) To differentiate this, we use the Quotient Rule: du dv v u = 2 dy dx dx dx v Example x2 1 du y(x) = so that u = x2 + 1 and = 2x 2x 3 dx dv and v = 2x + 3 and =2 dx dy (2 x 3)(2 x ) ( x 2 1)(2) = dx (2 x 3) 2 You may need to simplify/tidy up the answer: dy (4 x 2 6 x) (2 x 2 2) 2x2 6x 2 = dx (2 x 3) 2 (2 x 3) 2 There will be lots more examples of how to use both the Product Rule and the Quotient Rule in the tutorial/workshops. 32