Single Variable Calculus Warnings

Transcription

1 Single Variable Calculus Warnings These notes highlight number of common, but serious, first year calculus errors. Warning. The formula g(x) = g(x) is vali only uner the hypothesis g(x). Discussion. In some cases, when g(x) =, the it g(x) oes not exist at all. But in other cases it oes exist an when it oes, it can take any value at all. Here are two simple examples. In both examples g(x) = x a, which has the property that g(x) =. In the first example, g(x) oes not exist at all (because g(x) is huge an positive for x a small an positive an huge an negative for x a small an negative). In the secon it oes exist an takes the value = g(x) = x a g(x) oes not exist = 23.45(x a) g(x) = x a g(x) = 23.45(x a) x a = = Warning 2. Let a an b be two real numbers with a < b. Then ac < bc only if c >. When c < we have ac > bc an when c =, we have ac = bc. Discussion. Here is an example with a = an b = 2. For c = 3, ac = ()(3) = 3 < bc = (2)(3) = 6. For c = 3, ac = ()( 3) = 3 > bc = (2)( 3) = 6. For c =, ac = ()() = = bc = (2)() =. A relate inequality is that a 2 < b 2 if < a < b. If a < b <, then a 2 > b 2. Warning 3. L Hôpital s rule, g(x) = f (x) g (x) is not vali if the it g(x) exists an is nonzero. c Joel Felman. 2. All rights reserve. August 29, 2 Single Variable Calculus Warnings

2 Discussion. Here is an example with a =. If = 3x an g(x) = 4+5x, then x g(x) = x 3x 4+5x = = f (x) x g (x) = 3 x 5 = 3 5 Warning 4. x [f(a)] is not the same as f (a). Discussion. The expression x [f(a)] is, by efinition, the erivative of the constant function g(x) = f(a). In other wors, [f(a)] means x first evaluate at x = a, to get a constant, f(a). Then ifferentiate the constant to get zero. On the other han, f (a) means first fin the erivative f (x). Then evaluate the erivative at x = a. Typically, this is not zero. Here is an example. Let = x + an a =. Then f (x) = for all x. In particular, f () =. But if we first evaluate at x =, we get f() = 2. The erivative of the constant function g(x) = 2 is g (x) =. y y = = x+ y = g(x) = f() = 2 x The two functions an f() (which I am also calling g(x) here) both take the value when x =, but they have ifferent slopes an hence have ifferent erivatives: f () = but x [f()] = x =. Warning 5. The usual formula for the erivative of sinx, namely x sinx = cosx, is vali only if x is measure in raians. c Joel Felman. 2. All rights reserve. August 29, 2 Single Variable Calculus Warnings 2

3 Discussion. Any erivative measures the instantaneous rate of change of one quantity per unit change of a secon quantity. If you change the units of either quantity, the numerical value of the erivative changes. For example, consier the average rate of change of sin x as x moves from to a right angle. If x is measure in raians, average rate of change = change in sinx change in x = π 2 = 2 π because sinx changes from to as x changes from raians to π 2 raians. But, if x is measure in egrees, average rate of change = change in sinx change in x = 9 = 9 The two answers are ifferent even though x move from to a right angle in both cases. Here is a sample problem in which one gets the wrong answer by expressing an angle in the wrong units. Suppose that you wish to compute, approximately, the sin of two egrees, but that you can t just use your calculator. (Preten that you are the person esigning the software use by the calculator.) Recall that f(x )+f (x )(x x ) gives an approximate value for for all x close to x. Apply this approximation to = sin(x) with x =. Then = sinx f(x ) = f() = f (x) = cosx f (x ) = f () = so that sinx = f(x )+f (x )(x x ) = +(x ) = x. If we apply this with x = 2 we get sin2 2 which is both wrong an riiculous, as sinx is never bigger than one in magnitue. The problem arose because we use sinx = cosx in the computation x f(x )+f (x )(x x ) = +(x ) = x c Joel Felman. 2. All rights reserve. August 29, 2 Single Variable Calculus Warnings 3

4 thereby implicitly assuming that x is in raians. We shoul have first rewritten 2 = 2 π 8 = raians an then applie sin(2 ) = sin(.34966ra) [f(x )+f (x )(x x )] x= = x x= = For comparison purposes, the exact value of sin of 2 egrees is to 7 ecimal places. Our rather simple mine approximation scheme gave an error of =.7, which is a relative error of % =.2%. Warning 6. x ax is not the same as x xa. Discussion. x is the rate of change of per unit change in x. In x ax, the exponent changes as x changes, with the base, a, remaining fixe. In x xa, the base changes as x changes, with the exponent, a, remaining fixe. As a concrete example, suppose a = 4. As x changes from 4 to 5, a x = 4 x changes from 4 4 = 256 to 4 5 = 24 the average rate of change of a x = 4 x from x = 4 to x = 5 is = 768 x a = x 4 changes from 4 4 = 256 to 5 4 = 625 the average rate of change of x a = x 4 from x = 4 to x = 5 is = 369 The correct formulae for the instantaneous rates of change of a x an x a are x ax = lna a x x xa = ax a Warning 7. The slope of the curve y = at x = a is not f (x). It is f (a). Discussion. Consier the problem Fin the tangent line to the curve y = x 2 at the point (,). The tangent line is a straight line. Any straight line has only one slope. In this case the slope of the tangent line is the same as the slope of y = x 2 at x =, which is x x2 x= = 2x x= = 2. So the tangent line is y = 2(x ). If you neglect to evaluate x x2 = 2x at x =, you get y = (2x)(x ) instea. This is not a straight line at all, because as x changes the quantity you put in for the slope (2x) also changes. As you move along a straight line, the slope is not allowe to change. c Joel Felman. 2. All rights reserve. August 29, 2 Single Variable Calculus Warnings 4

5 Warning 8. Suppose that you solve = approximately by some numerical proceure, like Newton s metho. Suppose further that you want the root accurate to three ecimal places. So you iterate until the first three (or four or five or six, ) ecimal places stabilize. While this is strongly suggestive that you have three ecimal place accuracy, it is not conclusive. Also, if you substitute your guess into f an fin that the result is zero to three (or four or five or six, ) ecimal places, you still cannot conclue efinitively that your root is accurate to three ecimal places. Discussion. It is perfectly possible for Newton s metho to behave very nicely for, for example, fifty iterations an look like it is converging to some value an then suenly go crazy on the fifth iteration. For example, if the graph of f looks like an you start with x < (or even x positive but not too big), Newton s metho will start off looking like it is going to converge to zero, even though there is no root near x =. The above example also shows that it is possible for f() to be extremely small even if f has no root near x =. Even if oes have a root r near your guess x n, you cannot tell how big x n r, the error in your guess, is just from how close f(x n ) is to zero. If the graph of f looks like f(x n ) x n r then f(x n ) is a lot smaller than x n r. If the slope of the graph is small enough it is perfectly possible to have f(x n ) = with x n r =. Assuming that is continuous, a sure way to test if x n =.234 is accurate to four ecimal places is to check if f(.2335) an f(.2345) are of opposite sign. If so f has to be zero somewhere between x =.2335 an x = c Joel Felman. 2. All rights reserve. August 29, 2 Single Variable Calculus Warnings 5

6 Warning 9. If c is a constant b a cx = cx b = c(b a) not c2 b a 2 = b2 a 2. a 2 Discussion.Theintegral b cxrepresents theareabetween y = an y = c witha x b. a Becausetheheightcisconstant, thisisarectangleofwithb aanheightcanconsequently of area c(b a). Warning. The square root cos 2 x is cosx not cosx. Discussion. Suppose that you are walking along a path an that at time t you are at x(t) = sin 3 t an y(t) = cos 3 t an that t runs from to 2π. Then your velocity in the x irection is x (t) = 3costsin 2 t an your velocity in the y irection is y (t) = 3sintcos 2 t. Your spee is x (t) 2 +y (t) 2 = 9cos 2 tsin 4 t+9cos 4 tsin 2 t = 3 cos 2 tsin 2 t(sin 2 t+cos 2 t) = 3 cos 2 tsin 2 t Spee is istance travele per unit time. It is always positive. In particular the refers to the positive square root. The total istance travele is 2π 2π 2π speet = 3 cos 2 tsin 2 tt = 3 costsint 2π t = 3 sin(2t) t 2 If we ha forgotten the absolute value signs, we woul now have that the total istance travele is 3 2π 2 sin(2t)t =, which is riiculous! The real answer is 2π π/2 3 2 sin(2t) t = sin(2t) t (Graph the integran to see this!) π/2 ] π/2 = 6 sin(2t)t = 6 [ 2 cos(2t) = 6 Warning. The computation is wrong. x x = x2 = = 2 Discussion. In fact, the answer is riiculous. The integran x 2 >, so the integral has to be positive. The flaw in the argument is that the Funamental Theorem of Calculus, which c Joel Felman. 2. All rights reserve. August 29, 2 Single Variable Calculus Warnings 6

7 says that if F (x) = then b a x = F(b) F(a) is applicable only when F (x) exists an equals for all a x b. In this case F (x) = x 2 oes not exist for x =. The given integral is improper. The correct computation is t x = x2 t + = t + x+ x2 ] [ /t t + t + t + x = x2 [ /t ] t + x + x t + [ 2 ] t 2 = t = t + t Warning 2. The inefinite integral 5 3x x is NOT ln 5 3x +C. Discussion. Any time you think you know an inefinite integral, you can always check your guess by ifferentiating it. In this case, when 5 3x > so that ln 5 3x = ln(5 3x), x ln(5 3x) = 5 3x x (5 3x) = 35 3x 5 3x by the chain rule. We can get the correct answer by substituting y = 5 3x, y = 3x, x = y 3 5 3x x = y y 3 = 3 y y = 3 ln y +C = 3 ln 5 3x +C Warning 3. The inefinite integral x /2 +x 3/2 x is NOT x/2 /2 + x /2 /2 +C. Discussion. The incorrect computation leaing to the answer x/2 /2 + x /2 /2 [ x = x /2 +x 3/2] x = x /2 +x3/2 x /2 x+ +C is x 3/2 x = x/2 /2 + x /2 /2 +C The mistake occurre in the very first step: x /2 +x 3/2 is not equal to x /2 + x 3/2. For example, when x = 4, x /2 = 2 an x 3/2 = 8, so saying = + is the same as saying x /2 +x 3/2 x /2 x 3/2 2+8 = 2 + 8, which is absur. c Joel Felman. 2. All rights reserve. August 29, 2 Single Variable Calculus Warnings 7