Introduction to the Calculus, Fall 2011 Final Exam Solutions

Math 111, Introduction to the Calculus, Fall 2011 Final Exam Solutions 1. (5 points each) Calculate each of the following limits. You should say if the limit is equal to a particular value, is ±∞ or if it does not exist. |x| x→0 x2 Solution: For x > 0, |x|/x2 = x/x2 = 1/x, so the limit from above is +∞. For x < 0, |x|/x2 = −x/x2 = −1/x, so the limit from below is also +∞. Therefore the overall limit is +∞ . √ 3− x (b) lim x→9 x − 9 Solution: We have √ √ √ (3 − x)(3 + x) 9−x −1 3− x √ √ = √ . = = x−9 (x − 9)(3 + x) (x − 9)(x + 3 x) 3+ x (a) lim Therefore the limit is equal to −1 −1 √ = x→9 3 + 6 x lim x 4x2 + 1 √ Solution: We multiply numerator and denominator by 1/ x2 but note that for √ x < 0 we have 1/ x2 = −1/x so, for x < 0: (c) lim √ x→−∞ √ x x.(−1/x) −1 =p =p 2 2 2 4x + 1 (1/x )(4x + 1) 4 + 1/x2 √ −1 Taking the limit as x → −∞ we have −1/ 4 which is 2 (d) lim+ [ln(x2 − 1) − ln(x2 + x − 2)] x→1 Solution: We can rewrite 2 2 ln(x −1)−ln(x +x−2) = ln  x2 − 1 x2 + x − 2   = ln (x − 1)(x + 1) (x − 1)(x + 2)   = ln x+1 x+2  . This is now continuous at x = 1, so when we take the limit we get ln(2/3) 2. (a) (2 points) What does it mean to say that a function f (x) is continuous at x = a? Solution: It means that f is defined at x = a, and that lim f (x) = f (a). x→a (b) (3 points) Draw the graph of a function f (x) such that limx→2 f (x) exists, but f (x) is not continuous at x = 2. Solution: The function should either be not defined at x = a, or it should be equal to some value other than the limit. For example, just take the graph of y = x and remove the point (2, 2). (c) (10 points) Use the precise (/δ) definition of limit to prove that f (x) = 1 − 2x is continuous at x = 3. Solution: We have f (3) = −5 so we have to prove that limx→3 (1 − 2x) = −5. For any  > 0, let δ = /2. Then, if 0 < |x − 3| < δ, we have |x − 3| < /2 2|x − 3| <  2|3 − x| <  |6 − 2x| <  |(1 − 2x) − (−5)| <  which proves that the limit is −5 as desired. 3. (10 points) Use the limit definition of derivative to calculate f 0 (1) where 1 f (x) = 2 + . x Solution: It’s much easier to go for f 0 (1) directly, rather than to find f 0 (x) and then substitute in x = 1: f (1 + h) − f (1) h→0 h

1 2 + 1+h − 2 + 11 = lim h→0 h 1 1 − = lim 1+h 1 h→0 h 1 − (1 + h) = lim h→0 h(1 + h) −1 = lim h→0 1 + h −1 = 1 f 0 (1) = lim So we have f 0 (1) = −1 4. (5 points each) Calculate each of the following derivatives: 1 + t2 √ t Solution: We can rewrite f as t−1/2 + t3/2 so the derivative is (a) f 0 (t) where f (t) = f 0 (t) = − (b) f 0 π  , where f (x) = 3 Solution: We have t−3/2 3t1/2 + 2 2 π  π  g(x) , and it is known that g = 1 and g 0 =0 sin(x) 3 3 f 0 (x) = sin(x)g 0 (x) − g(x) cos(x) sin2 (x) and so f 0 (π/3) = (c) sin(π/2)(0) − (1) cos(π/3) −1/2 −2 = = √ 2 3 sin (π/3) ( 3/2)2 dy as a function of x and y, where yex+y = x dx Solution: Differentiating the given equation we get y 0 ex+y + y(1 + y 0 )ex+y = 1 and so y 0 [ex+y + yex+y ] = 1 − yex+y and y0 = 1 − yex+y ex+y + yex+y (d) the second derivative g 00 (x), where g(x) = x ln x Solution: g 0 (x) = ln x + x x1 = ln x + 1 so g 00 (x) = 0 Z (e) F (ln π), where F (x) = 1 x x cos(et ) dt 0 Solution: By the Fundamental Theorem of Calculus F 0 (x) = cos(ex ) and so F 0 (ln π) = cos(eln π ) = cos(π) = −1 5. (5 points each) Calculate each of the following integrals: Z (a) x2 (1 − x) dx R Solution: This is x2 − x3 dx = x3 /3 − x4 /4 + c for a constant c. √ Z 2 t +e t √ (b) dt t √ Solution: We can substitute for u = t in which case 1 dt 1= √ 2 t du so √ dt = 2 t. du Therefore the integral is Z √ 2 2(u4 + eu ) du = 2u5 /5 + 2eu + c = t5/2 + 2e t + c 5 for a constant c. Z e sin(ln(x)) (c) dx x 1 dx Solution: Substituting u = ln x we have du = x and so the integral becomes Z x=e Z u=1 sin(u) du = sin(u) du = [− cos(u)]u=1 u=0 = − cos(1) + cos(0) x=1 u=0 which is 1 − cos(1) Z 3 (d) |x − 1| dx 0 Solution: It’s easiest just to draw the graph and figure out the relevant area. This consists of two triangles: one is base 1, height 1, so area is 1/2; the other is base 5 2, height 2, so the areas is 2. The total area, and hence the integral is 2 6. (20 points) Let f (x) = 10x + 9x2 + 3x3 . Take my word for it that (x + 1)3 f 0 (x) = −2(x − 5) , (x + 1)4 and f 00 (x) = 6(x − 7) . (x + 1)5 Sketch the graph of f (x), clearly indicating horizontal and vertical asymptotes, local maxima and minima, inflection points, and intervals on which f is increasing, decreasing, concave up or concave down. You do not need to indicate locations where the graph crosses the axes, or y-coordinates of maxima, minima or inflection points. Do not try to draw the graph exactly to scale, but do make sure it clearly shows all the above information. Solution: There is a vertical asymptote at x = −1. To find the horizontal asymptote, we look at the limit as x → ±∞: 3 + x9 + x102 10x + 9x2 + 3x3 10x + 9x2 + 3x3 lim = lim = lim 3 x→±∞ x→±∞ 1 + 3 + 32 + x→±∞ x + 3x2 + 3x + 1 (x + 1)3 x x 1 x3 =3 and so y = 3 is a horizontal asymptote. The first derivative is positive for x < −1 and −1 < x < 5, and negative for x > 5, so f is increasing on (−∞, −1) and (−1, 5), and decreasing on (5, ∞). There is a local maximum at x = 5. The second derivative is positive for x < −1 and x > 7, and negative for −1 < x < 7, so f is concave up on (−∞, −1) and (7, ∞), and concave down on (−1, 7). There is an inflection point at x = 7. (x = −1 is not an inflection point because the function is not defined there.) Altogether the graph looks something like this: 7. The function P has domain [−1, 1] and is given by P (s) = s3 + 6s2 − 1. (a) (2 points) Explain how you know that P has an absolute maximum in this case. (State the properties of P that are relevant.) Solution: P is a continuous function on a closed interval, so it has an absolute maximum and minimum by the Extreme Value Theorem. (b) (8 points) At what value of s does P (s) reach its absolute maximum? Solution: The critical points occur where P 0 (s) = 0. We have P 0 (s) = 3s2 + 12s = 3s(s + 4) so either s = 0 or s = −4. But s = −4 is not in the domain of the function so is not relevant. At s = 0, we have P (s) = P (0) = −1. We also have to check the endpoints: s = −1 and s = 1. At s = −1, P (s) = P (−1) = 4 and at s = 1, P (s) = P (1) = 6. There are no points where P is not differentiable, so the absolute maximum occurs at s = 1 . √ 8. (10 points) A bowl is formed by rotating the graph of the function y = x for 0 ≤ x ≤ 4 around the x-axis. (Distances along the x and y-axes are measured in centimeters.) Water is then poured into the bowl at a rate of 10 centimeters cubed per second. How long does it take to fill the whole bowl with water? Solution: The volume of the bowl is given by Z 4 πf (x)2 dx 0 √ where f (x) = x. So the integral becomes Z 4  x=4 πx dx = π x2 /2 x=0 = 8π. 0 So the volume is 8π centimeters cubed. If the bowl is filled at a rate of 10 centimeters cubed per second, then the time taken to fill is 8π 4π = seconds . 10 5 9. (10 points) During a tennis match, the ball is hit so that it travels across the net (and perpendicular to it) at a speed of 20 meters per second. What is the maximum rate (in radians per second) at which a spectator has to turn her head in order to follow the ball if she is level with the net and 10 meters away from the point where the ball crosses? Solution: If the variable x measures the distance travelled by the ball along its path (with x = 0 at the point when it crosses the net) and θ measures the angle at which the spectator has to turn to view it (with θ = 0 when the ball is crossing the net) then we have x tan θ = . 10 Differentiating with respect to time we get 1 dθ 1 dx = . 2 cos θ dt 10 dt We know that dx dt = 20 so we get dθ = 2 cos2 θ. dt We want the maximum rate which is when cos θ = 1 so the maximum rate is 2 radians/second . 10. (10 points) A farmer has $1000 to purchase fencing for a new pen. He wants to build a rectangular pen whose perimeter is constructed with reinforced barbed wire fence. He also wants to divide this pen into two equal rectangular regions, using wood fencing. If barbed wire fencing costs $10 per meter, and wood fencing costs $5 per meter, find the maximum area that can be enclosed by such a pen. Solution: Suppose that the dimensions of the pen are x meters and y meters with the extra length of fence down the middle being x meters. Then there is 2x + 2y meters of barbed wire fencing and x meters of wood fencing, for a total cost of 10(2x + 2y) + 5x. This will be equal to 1000 so we have 25x + 20y = 1000. We want to maximize the area, that is A = xy = x((1000 − 25x)/20) = 50x − 45 x2 . The range of possible values of x is 0 ≤ x ≤ 40. The maximum will occur at either a critical point or endpoint. At the endpoints x = 0, 40 we have A = 0. To find the critical point: 5 0 = A0 (x) = 50 − x 2 so x = 20. Then A(20) = 500 so this is the maximum. The maximum area is therefore 500 square meters .