Implicit Differentiation

Transcription

1 Revision Notes 2 Calculus 1270 Fall 2007 INSTRUCTOR: Peter Roper OFFICE: LCB 313 [ [email protected]] Standard Disclaimer These notes are not a complete review of the course thus far, and some subjects have been omitted here. These notes are meant solel as a guide and a summar of the harder and more important subjects in the course. The midterm will be drawn from the whole course, and ma include questions drawn from subjects not discussed in these notes. Implicit Differentiation [section 2.7 of book] Compare the two equations that describe familiar curves = (parabola) = 3 (circle) The first defines as an eplicit function of, because ever value of gives rise to a single value of. The second however defines as an implicit function of because each value of does not give a unique value for. How do we differentiate such implicit functions? For some we can re-arrange to find an implicit function, but this is not alwas possible and it is easier to use the chain rule. For eample, for the circle above and so d d (2 + 2 ) = d d 3 d d = 0 to find d/d( 2 ) use the chain rule, so that d d 2 = 2 d putting this back into the original equation we obtain 2 d + 2 = 0

2 and we can rearrange to give d = which gives us an epression for /d as we require. Infinitessimals/differentials and linear approimations [section 2.9 of book] Sometimes we need a quick and simple estimate of the change in a function f() that results when we change slightl. If we change b + then the resulting change in = f() is just = f( + ) f(). As shown in the figure tangent line + + =f() =f() However if the change is ver small then we can use the tangent line as an approimation to the curve and instead find as an approimation to. Therefore, if the slope of the tangent line is m = f (), and using the equation ( = m + c) for a straight line, we can write and in terms of infinitessimals, f () = f ()d

3 Optimization: finding maima, minima and critical points We ve introduced the idea of the differential of a curve at a point 0 as being the slope of the tangent line at that point. We now use this idea to eamine properties of curves. We note first that if a function, f() has a maima or a minima at some point, then the slope of the curve is zero AT THAT POINT because the tangent is flat. We call such a point an etrema of the function. However if there is an asmptote, the slope of the curve is infinite because it s pointing verticall up (or down) AT THAT POINT. So, to eamine the properties of the curve over some interval I, we compute it s differential, f () and determine whether or not the differential vanishes or goes to infinit at an point in the interval. However once we have found a point where the function has a tangent with a slope of zero, how do we tell if it s a maimum or a minimum? Two possible was are: The first derivative test Compute the slope of the function on either side of the etrema. If the function is increasing on the left side (/d > 0), and decreasing on the right (/d < 0) then we have a maima, if it is decreasing on the left and increasing on the right then we have a minima. The second derivative test If the second derivative of the function f () < 0, then we have a local maima. If f () > 0 then we have a local minima. Finding roots of equations numericall Often we need to find the roots, or zeros, of a function, b which I mean that we are tring to find where the function crosses the -ais. For some functions, such as quadratic and cubics, there eist eplicit formulae for computing their roots, but for most functions there is no general wa of finding them. Instead we turn to approimate, or numerical, techniques. We will talk about two similar techniques, one that alwas converges but can be ver slow, and one that is fast but doesn t alwas converge.

4 The bisection method Consider tring to find the roots of = f() The point of this technique is to find two points a and b, one that is above and one that is below the root 1, and then progressivel move each point, a and b, so that we narrow down the interval between them while keeping them on either side of the root. =f() a 1 m 1 b 1 We choose our first values for a and b, call them a 1 and b 1 b guessing at some numbers and then testing to see whether f(a 1 ) has the opposite sign to f(b 1 ) which means that the are on either side of the root. Note that if, sa, f(a 1 ) is positive, and f(b 1 ) is negative then their product is negative too. And so to test whether the both lie on either side of the root all we need to do is calculate f(a 1 ) f(b 1 ) and show that it is negative. Once we have found a suitable initial choice for a 1 and b 1 we can start to zoom in on the root. We first find the midpoint between a 1 and b 1 m 1 = a 1 + b 1 2 and we test whether f(a 1 ) f(m 1 ) or whether f(m 1 ) f(b 1 ) is negative, to determine whether the root lies between a 1 and m 1 or between m 1 and b 1. 1 Use the mean value theorem to show that the root lies between them

5 If f(a 1 ) f(m 1 ) < 0 then the root lies between a 1 and m 1 and so we make a new interval b setting a 2 = a 1 and b 2 = m 1 and find a new midpoint m 2 = a 2 + b 2 2 Similarl if f(m 1 ) f(b 1 ) is negative then we set a 2 = m 1 and b 2 = b 1 and again find the new midpoint. We then test m 2 to find which of f(a 2 ) f(m 2 ) or f(m 2 ) f(b 2 ) is negative. We then continue b setting a 3 and b 3 equal to the resulting boundaries, and so on until we have reached a desired degree of accurac. So, the net question is how do we decide how accurate our answer is at each iteration? This is actuall prett straightforward: at the n th iteration we ve shown that the root lies between a n and b n. This means the midpoint m n is within half of the distance between a n and b n of the true root. Therefore, the quantit h n = b n a n 2 tells us how close m n is to the correct answer. This bisection method alwas converges to the correct answer, although it can be etremel slow. Newton s method Newton s method rests upon the idea that when ou are close to a curve, the tangent line is a good approimation to the curve. This time we start with a single guess at the root, call it a 1. We find the net approimation, a 2 to the root b finding the intersection with the ais of the tangent line to f() at the point a 1, as shown in the figure. Therefore if the slope of the tangent line is f () then the new guess is a 2 = a 1 f(a 1) f (a 1 ) We can then continue the iteration b finding a 3 from the tangent line at a 2 until we have reached the desired degree of accurac. Newton s method does not alwas converge to the correct answer, but it can be significantl faster than the bisection method.

6 tangent line a 2 a 1 =f() Differential Equations We frequentl encounter problems of the form d = f() where we have a differentiated quantit on the LHS and some function of on the right. Such problems are called differential equations. The problem then becomes: how do we invert the differentiation to find as a function of? Simple, first order, ordinar differential equations (ODE s) can be solved b integration, eg: ds dt = 10 is an ODE for the speed of a vehicle travelling at 10 mph, where s is the distance. Integrate to find the total distance gone from time t = 0 to t = 1 hour 1 0 ds = 10 dt ds = dt = 10t] 1 0 = 10 miles

7 Eponential growth and deca ODE s that have the form dn dt = kn describe eponential growth (if k > 0) or deca (k < 0). Again solve b simple integration 1 dn = k dt N 1 N dn = k dt ln(n) = kt + C N(t) = ep(kt + C) = ep(kt) ep(c) = K ep(kt) where, since C is a constant, K = ep(c) is also a constant. What actuall is K? Think about setting t = 0, ie when the sstem first starts off. Then we have: N(t = 0) N(0) = K ep(k 0) = K and so K = N(0) is the initial value for the problem. [In fact we can use the initial value of an ODE to find the constants of integration.] Separable differential equations An ODE that can be manipulated into the form where all s and d s are one one side of the equation, while all s and s are on the other is called a separable differential equation, and can be integrated directl. Eamples d = 1 2 ( + 1) + 1 = 2 d d = ( + 1) 1 = d + 1 d = 2 sin() 1 = sin() d 2 and in the same wa as for eponential growth, use the initial values (if given) to find numerical values for the constant of integration.

8 Integration For the midterm I m going to assume that ou understand that integration is the inverse of differentiation, and I m more concerned with our being able to practicall solve integrals. With this in mind, some common errors... for which ou would be penalized for... are: Forgetting infinitessimals If one side of an equation is multiplied b an infinitessimal, then the other side must also be multiplied b an infinitessimal too, for eample: = 7 sin() d and so the integral becomes and so = 7 sin() d = 7 cos() + C Constants of integration When indefinite integrals (i.e. those that do not have limits) are evaluated, the solution must include a constant of integration: f()d = F () + C but definite integrals (i.e. those that have limits) do not have a constant of integration b Integrating f(3), f(4), etc a f()d = F (b) F (a) This is a special case of integration b substitution. To integrate (e.g.) sin(5) make the substitution u = 5, so: du = 5 d or d = 1 5 du

9 and therefore sin(5) d = sin(u) 1 5 du = = 1 5 cos(u) = 1 5 cos(5) Integration b substitution Some integrals ma be transformed to a simpler integral b substitution. transforming an integral When Step 1 transform the variable, e.g. u = sin() Step 2 transform the infinitessimal, e.g. du = cos()d Step 3 (for definite integrals) transform the limits, e.g. u 1 = sin( 1 ), u 2 = sin( 2 )