Solutions to Math 4 Second Exam November 5, 03. 5 points) Differentiate, using the method of your choice. a) fx) = cos 03 x arctan x + 4π) 5 points) If u = x arctan x + 4π then fx) = fu) = cos 03 u and by product rule du dx = arctan x + x x. So by chain rule + dfx) dx = dfu) du du dx = 03 cos 0 u sin u) ) ) du dx = 03 cos 0 x arctan x + 4π) sinx arctan x + 4π) arctan x + x ) x + b) gt) = log 3 sec 0 πt )) 5 points) By the base-change formula, gt) = lnsec 0 πt) ). Applying the chain rule repeatedly ln 3 we get dgt) d = ln 3 sec 0 πt ) sec 0 πt) = ln 3 sec 0 πt ) sec 0 πt) tan 0 πt) d 0πt = ln 3 sec 0 πt ) sec 0 πt) tan 0 πt) 0 πt ln 0 d πt = π ln 0 tan 0 πt) 0 πt ln 3 c) hz) = z ln z + arcsin z ) 5 points) Let fz) = z ln z and gz) = arcsinz ). Then using logarithmic differentiation and using chain rule ln fz) = lnz ln z ) = ln z)ln z) = ln z) f z) = ln z fz) z f z) = ln z z fz) = ln z z ln z = z ln z ln z z g z) = d z ) dz z ) = z z ) So h z) = f z) + g z) = z ln z ln z + z z )
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page of. points) Consider the curve with equation xy x + y =. a) Show that dy y ) 3. dx = x 4 points) There are several different ways to do this. First, let us note that we must have x, y 0 since if both x = y = 0, then the fraction is undefined as the denominator x + y = 0; while if one of x, y = 0, then the fraction is zero as the numerator xy = 0, in contradiction to the righthand side. Therefore, we may assume that x, y 0, i.e., the domain is {x, y) : x 0, y 0}. We now give three possible solutions: ) Differentiate implicitly using the quotient rule: ) d xy = d dx x + y dx ) ) x + y y + xy x+yy ) xy x +y x + y = 0 x + y ) y + xy ) xy x + yy ) x + y ) 3/ = 0 Clearly, this holds if and only if the numerator is zero, which we can expand as Canceling like terms, we obtain x y + x 3 y + y 3 + xy y x y xy y = 0. x 3 y + y 3 = 0, so solving for y yields y = y/x) 3 as claimed. ) Rewrite the equation as xy = x + y x y = x + y = y + x. Differentiating implicitly now gives d dx ) = d y + x ) dx 0 = y 3 y x 3, so y = x 3 y 3 = y/x) 3. 3) Use logarithmic differentiation: ) xy ln = ln x + y ln x + ln y ln x + y ) = 0 [ d ln x + ln y dx ln x + y )] = d dx 0) x + y y x + yy x + y = 0
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page 3 of Collecting like terms, we find that ) y y x + y y x = x + y x [ x + y ) ] y y = x x + y ) Solving for y then gives y = y/x) 3. y x + y ) x x + y ) ) x y = y y x b) Determine with reasoning the coordinates of all points on the curve, if any, where x = y; and for each such point, give the equation of the tangent line to the curve there. 4 points) Set y = x in the equation for the curve above to obtain x x =. Since x = x and x = x, this is equivalent to x x = x = since x 0 by a), so x =, i.e., x = ±. In other words, the only points on the curve with x = y are, ) and, ). The slope of the tangent line at each of these points is m = x/x) 3 = by a). Therefore, the tangent line to the curve through, ) has equation y = x ) or y = x +, and the tangent line to the curve through, ) has equation y + = x + ) or y = x. c) Determine with reasoning the coordinates of all points on the curve, if any, where the tangent line is horizontal or vertical; and for each such point, give the equation of the tangent line to the curve there. 4 points) By a), the slope of the tangent line to the curve at any point x, y) is m = y/x) 3. At a horizontal tangent, m = 0 so y = 0. But we concluded in a) that y = 0 is not in the domain. Therefore, there is no point on the curve with a horizontal tangent line. Similarly, at a vertical tangent, we must have /m = 0, so x = 0, which again is not in the domain. Therefore, there is no point on the curve with a vertical tangent line.
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page 4 of ) 3. 9 points) Radiocarbon dating uses the equation T P ) = 8300 ln to predict the age, T in years), P of a specimen as a function of the proportion P of a carbon isotope that it contains. For example, if the proportion is measured to be P = 0., the predicted age of the specimen would be ) T 0.) = 8300 ln = 8300 ln 0 years. 0. a) Suppose the above measurement P = 0. has an experimental error uncertainty) of ±0.0. Use calculus to estimate the resulting error in the calculated age of the specimen. Simplify as much as possible but don t try to evaluate any logarithms involved in your answer). 4 points) To find the error in calculating T, it s most direct to use the method of differentials. To see how to use linearization instead, see the first portion of the solution to part )b) below.) We want to find dt, and we know that P = dp = 0.0. Since T P ) = 8300 ln, it follows P that dt/dp = 8300 P, so dt = dt P = 8300 dp P P. When P = 0. and dp = ±0.0, we find dt = ±830 years, so the error in calculating the age is ±830 years. b) Suppose it were true that P = 0.; then the predicted age of the specimen needs to be revised. Use the work of part a) to state an estimate for this revised age. Second, how does this estimate for the revised age compare with the function value T 0.) = 8300 ln 0. ) : is it larger or smaller than T 0.)? Explain your answer. 5 points) By the principle of linear approximation, T can be approximated near a point P 0 by its linearization at P 0 ; that is, T P ) T P o ) + T P 0 )P P 0 ) for P near P 0. When P = 0. and P 0 = 0., using the derivative that we found in the previous part, we get T P 0 ) = 83 0 3 years, so T 0.) 8300 ln0) 830 years. Notice that this method can be used to recover the error ±830 that we d found in part a). On the other hand, if you had used differentials to solve part a), you d have needed to be very careful to subtract not add) the error from the value 8300 ln0): the sign of T P ) at P = P 0 indicates that if P slightly increases, then T P ) should slightly decrease.) To determine whether this is an underestimate or an overestimate, we have to find the second derivative of T. Since T P ) = 8300, it follows that T is concave up as a function of P, so the P tangent line lies below the curve. Hence, the answer that we got above is smaller than the actual value T 0.).
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page 5 of 4. 4 points) a) A cook is making a pancake by pouring batter into a pan at a steady rate of cm 3 /s. Assuming that the pancake always keeps the shape of a circular disk with a thickness of 0.5 cm, determine the rate at which the radius of the pancake is changing when the circle has area 80 cm. Possibly useful formula: a circular disk of thickness h is just a cylinder, with volume πr h.) 7 points) Let V be the volume of the pancake, and r be the radius of the pancake. We are given dv dr =, and we wish to find when the area of the circle is 80 cm ; that is, when 80 πr = 80, or equivalently r = π. Now V = πr 0.5) = π r. Differentiate with respect to time: So at the moment when r = 80 π, dv = d π r) = πr dr = π 80 π dr = dr = π 80/π So the radius of the pancake is increasing at π 80/π = 80π = 5π cm/s.
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page 6 of b) Note: this part is independent of part a).) A chunk of butter is placed on the hot pancake after cooking; it has a box shape with a square base of side length cm, and a height of cm. As the butter melts, it maintains the same volume and the same square-base box shape, but the dimensions change. When the side length of the square is 4 cm, it is growing at a rate of 0. cm/s. Determine the rate at which the height of the butter is changing at this instant. 7 points) Let V be the volume of the butter, x be its side length and h its height. We wish to find dh dx when x = 4 and = 0.. Because the butter keeps a square base, V = x h. Differentiate with respect to time: dv = d x h ) = x dx dh h + x, The volume of the butter is constant, so dv = 0. So by the product rule. x dh = xdx dh h = = h dx x. The starting volume of the butter is )) = 4 cm 3, and this is constant, so when x = 4, h = V x = 4 4 = dx dh. Plugging the values of h, x, into our expression for 4 gives dh = /4) 4 0 = 80. So the height of the butter is decreasing, with a rate of change of 80 cm/s. Alternate solution: Let x be the side length of the butter and h its height. The starting volume of the butter is )) = 4 cm 3, and this is constant, so x h = 4, i.e. h = 4x. Differentiate with respect to time: dh = 4 x 3 ) dx = 4 )4) 3 ) = 0 80, so the height of the butter is decreasing, with a rate of change of 80 cm/s.
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page 7 of 5. 8 points) Consider the function gx) = x + x. a) State the domain of g. points) The domain of g is, 0) 0, ) = {x R: x 0}. b) Determine, with complete reasoning, whether g has any asymptotes horizontal or vertical), and give their equations. Compute both one-sided limits for any vertical asymptotes. 4 points) Vertical: Since g is continuous on its domain, 0) 0, ), the only possibility for a vertical asymptote is x = 0. To show that x = 0 really is a vertical asymptote, we compute the left-hand and right-hand limits: lim gx) = lim x + ) = x 0 x 0 x lim gx) = lim x + ) = +, x 0+ x 0+ x where we have used the fact that lim = and lim x 0 x x 0+ x = +. Thus, x = 0 is a vertical asymptote. Horizontal: To find horizontal asymptotes, we take the limit as x + and x : lim gx) = lim x + ) = + x x x lim gx) = lim x + ) = +. x x x Since neither of these limits are finite numbers, we conclude that g has no horizontal asymptotes.
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page 8 of For easy reference, gx) = x + x. c) On what intervals) is g increasing? decreasing? Explain completely. 4 points) To find the intervals on which g is increasing and decreasing, we examine the signs of g x). We first compute g x). Since gx) = x + x = x + x, we have g x) = x x = x3 x x = x3 x = x3 x. Setting g x) = 0, we get x 3 =, so x = is the only critical point of g. Note that x = 0 is technically not a critical point of g because it is not in the domain of g. Despite this, x = 0 is still a number worth considering in our sign analysis., 0): Here, x 3 < 0 and x > 0. Thus, g x) < 0, so g is decreasing on, 0). 0, ): Here, x 3 < 0 and x > 0. Thus, g x) < 0, so g is decreasing on 0, )., ): Here, x 3 > 0 and x > 0. Thus, g x) > 0, so g is increasing on, ). d) On what intervals) is g concave up? concave down? Explain completely. 4 points) To find the intervals on which g is concave up and concave down, we examine the signs of g x). We first compute g x). Since g x) = x x, we have g x) = + 4x 3 = + 4 x 3 = x3 + 4 x 3 = x3 + x 3. Setting g x) = 0, we get x 3 =, so x = 3 = 3 is the only critical point of g. Note that x = 0 is technically not a critical point of g because it is not in the domain of g. Despite this, x = 0 is still a number worth considering in our sign analysis., 3 ): x 3 + < 0 and x 3 < 0. Thus, g x) > 0, so g is concave up on, 3 ). 3, 0): x 3 + > 0 and x 3 < 0. Thus, g x) < 0, so g is concave down on 3, 0). 0, ): x 3 + > 0 and x 3 > 0. Thus, g x) > 0, so g is concave up 0, ).
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page 9 of e) Using the information you ve found, sketch the graph y = gx). Label and provide x, y) coordinates of any local extrema and inflection points. 4 points)
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page 0 of 6. 0 points) Bruce, a legendary music producer of the 970s, has determined that the profit P generated by a hit song depends on the decibel level, x, of cowbell sound featured in the song, as follows with amounts here given in thousands of dollars): } gross revenue = x 3/ + 6x / + 8 production cost = 9 x = profit P x) = x 3/ + 6x / + 8) 9 x If Bruce s eight-track mixing board permits him to set the cowbell sound level x to any non-negative value less than or equal to 6 decibels, what should x be in order to maximize profit P x)? Give a complete mathematical justification for your answer. We want to maximize the function P x) = x 3 + 6x + 8) 9 x on the domain 0 x 6. Because the domain is closed we will use the Closed Interval Method. We first find the derivative of P x): P x) = 3 x + 6 x 9 = 3 x x + 3 x) = 3 x x ) x ) P x) is not defined for x = 0 and P x) = 0 for x =, 4. These are the critical points. The endpoints of the interval are x = 0, 6. We evaluate P x) at these points: P 0) = 8 P ) = + 6 + 8 9 = 0.5 P 4) = 4 3 + 6 4 + 8) 9 4 = 0 P 6) = 6 3 9 + 6 6 + 8) 6 = 4 By the Closed Interval Method, P 6) = 4 will be the maximum of P x) on the interval [0, 6]. So Bruce needs to set the cowbell sound level to 6 decibels in order to maximize profit.
Math 4, Autumn 03 Solutions to Second Exam November 5, 03 Page of 7. 0 points) There s been a zombie outbreak on the xy-plane! Our hero is located at the origin at time t = 0, and begins to escape by running straight east positive x-direction) at a constant speed of 0 ft/s. At the same time, a zombie initially located at the point 50 ft, 50 ft) starts to move south negative y-direction) at 5 ft/s. What is the closest distance that the zombie gets to our hero? Justify your answer completely. Let xt) be the distance run by the hero at time t and let yt) be the distance run by the zombie at time t. Because the velocity of the hero is 0 ft/s, xt) = 0 t. The velocity of the zombie is 5 ft/s, so yt) = 5 t. The position of the human at time t will be 0t, 0) and the position of the zombie will be 50, 50 5t). The distance between the two can now be written as: Dt) = 0t 50) + 5t 50) = 00t 000t + 500 + 5t 500t + 500 = 5t 500t + 5000 = 5 5 t t + 40 The problem asks to find the minimum of the distance, over the domain t 0. We compute the derivative of Dt): D t) = 5 5 t t + 40 t ) = 5 5t 6) t t + 40 Notice that the denominator is always larger than 0. D t) = 0 for t = 6. For t < 6, D t) < 0 and for t > 6 we have that D t) > 0. This implies, by the First Derivative Test for Absolute Extrema, that Dt) has an absolute minimum at t = 6. The minimum distance between the hero and the zombie will be D6) = 5 5 6 6 + 40 = 5 5 4 = 0 5 ft. Alternate solution: The following alternate solution does not use the fact that displacement equals time times constant) velocity, only the fact that velocity is the derivative of position. Let xt) be the distance run by the hero at time t and let yt) be the distance run by the zombie at time t. Because the velocity of the hero is double the velocity of the zombie, xt) = yt). The position of the human at time t will be yt), 0) and the position of the zombie will be 50, 50 yt)). The distance between the two can now be written as Dt) = yt) 50) + yt) 50) = 4y 00y + 500 + y 00y + 500 = 5y 300y + 5000 = 5 y 60y + 000 We now want to find the maximum of Dt) so we look for points were D t) = 0: D t) = 5 y 60y + 000 yy 60y ) = 5y 30) 5 y 60y + 000 We used the fact that y = 5. Notice that D t) is always defined and is 0 only when y = 30. Because the velocity is positive, y will be an increasing function such that y0) = 0. Notice that for y < 30 D t) < 0 and that for y > 30 we have that D t) > 0. This will imply, by the First Derivative Test for Absolute Extrema, that Dy) will have an absolute min at y = 30. So the distance between the hero and the zombie will be the smallest when y = 30: D = 5 30 60 30 + 000 = 5 00 = 0 5 ft