Download Solutions to Exam 2 of Math 105a: Limits, Derivatives, and Applications and more Exams Calculus in PDF only on Docsity! Math 105a SolutionsExam 2 1. (a) dydx = 1q1 (x2)2 2x = 2xp1 x4 (b) To nd dydx given ey + ln y2 = xy + x, it is necessary to implicitly di erentiate both sides withrespect to x. ey dydx + 1y2 2y dydx = y + xdydx + 1dydx ey + 2y x = y + 1 dydx = y + 1ey + 2y x or simplify to get dydx = y 2 + yyey + 2 xy 2. (a) Apply l'Hôpital's rule since the limit leads to the indeterminate form (I.F.) 0=0. lim!0 tan() L0H= lim!0 1 cos2() 1 = lim!0 cos2() = cos2(0) = (b) Apply l'Hôpital's rule (twice) since the limit leads to the I.F. 1=1. limx!1 ex + xx2 L0H= limx!1 ex + 12x L0H= limx!1 ex2 =1 3. Is the following argument valid? Consider lim x!0+ 1x 1sinx . Because lim x!0+ 1x =1 and limx!0+ 1sinx =1it must surely be true that the limit in question has the value 1 1 = 0. The argument is not valid. Recall from our study of limits (Theorem 2.1) that limx!a (f(x) + g(x)) = limx!a f(x) + limx!a g(x) only if both limits on the right hand side of the equation exist. For our example, the individual limitslim x!0+ 1x and limx!0+ 1sinx are in nite, i.e., do not exist. Therefore, we cannot break up the (original)limit of a sum into the sum of individual limits. To nd a legitimate method to evaluate the limit startby rewriting 1x 1sinx using algebra, and then reevaluate the limit by applying l'Hôpital's rule. lim x!0+ 1x 1sinx = lim x!0+ sinx xx sinx L0H= limx!0+ cosx 1sinx+ x cosx L0H= limx!0+ sinx2 cosx sinx = 0. 4. (a) The local linearization of f(x) = arctan(x) near x = 1 is given by f(x) f(1) + f 0(1)(x 1). Note: f(1) = arctan 1 = 4 since tan(=4) = 1; and f 0(x) = 11 + x2 so f 0(1) = 12. Therefore, f(x) 4 + 12(x 1). 1 (b) To approximate arctan(1:01), let x = 1:01 then f(1:01) = arctan(1:01) 4 + 12(1:01 1) 0:7904. 5. A kite 100 feet above the ground moves horizontally at a speed of 8 ft/sec. At what rate is the anglebetween the string and the horizontal changing when 200 feet of string have been let out. Be sure toexpress your answer using the appropriate units. We are given dxdt = 8 ft/sec, but need to nd ddt .Using trigonometry we can establish a relationship between the base x of the right triangle and theangle . tan = 100xImplicitly di erentiating both sides with respect to t gives:1cos2 ddt = 100x2 dxdtWe can use the Pythagorean theorem to nd x and trigonometry to nd cos when z = 12. In partic- ular, if z = 200 ft then x =p2002 1002 = 100p3 and cos = p32 . Therefore, ddt = 100 cos2 x2 dxdt = 100 p3 2 2 100p32 = 150 radians/sec. In other words, the angle between the string and the horizontal is changing at a rate of 150 radians/sec. 6. Suppose the function f is continuous at the point P when x = c. The graph of f has a vertical tangent at P if lim x!c f 0(x) and lim x!c+ f 0(x) are either both +1 or both 1. a cusp at P if lim x!c f 0(x) and lim x!c+ f 0(x) are both in nite with opposite signs (one +1and the other 1). If f(x) = 3x2=3, then f 0(x) = 23px . To determine if the graph of f has a vertical tangent or cuspat x = 0 we need to look at the behavior of f 0(x) as x! 0 and x! 0+. In particular, 2
