Math 11 Final Examination May 11, 2011 Answer Key

Math 11 Final Examination May 11, 2011 Answer Key 1. [20 Points] Evaluate each of the following limits. Please justify your answers. Be clear if the limit equals a value, +∞ or −∞, or Does Not Exist. 0 x2 − 1 = lim = 0 by DSP 2 (x + 1) − 1 x→1 4 − 1 (a) lim x→1 (b) lim x→3− x2 − 8x + 15 , where g(x) = x2 + 7. 1 − 8x + g(x + 1) x2 − 8x + 15 x2 − 8x + 15 = lim 1 − 8x + g(x + 1) x→3− 1 − 8x + (x + 1)2 + 7 lim x→3− = lim x→3− = lim x→3− x2 − 8x + 15 x2 − 8x + 15 = lim 1 − 8x + (x2 + 2x + 1 + 7) x→3− 1 − 8x + (x2 + 2x + 8) (x − 5)(x − 3) x−5 −2 x2 − 8x + 15 = lim = lim = − = +∞ 2 − − x − 6x + 9 (x − 3)(x − 3) x→3 x − 3 0 x→3 √ √ 8−x (8 − x)( x + 1 + 3) 8−x x+1+3 = lim √ ·√ = lim (c) lim √ x→8 (x + 1) − 9 x + 1 − 3 x→8 x + 1 − 3 x + 1 + 3 x→8 √ √ √ −(x − 8)( x + 1 + 3) = lim −( x + 1 + 3) = −( 9 + 3) = −6 = lim x→8 x→8 x−8 x2 − 5x − 14 DNE since RHL6=LHL. see below. x→7 |7 − x|  7−x if x ≤ 7, that is 7 − x ≥ 0 Note: |7 − x| = −(7 − x) if x > 7, that is 7 − x < 0 (d) lim RHL: lim x→7+ LHL: lim x→7− x2 − 5x − 14 x2 − 5x − 14 (x − 7)(x + 2) = lim = lim = lim x + 2 = 9 x→7+ x→7+ x→7+ |7 − x| −(7 − x) x−7 x2 − 5x − 14 (x − 7)(x + 2) x2 − 5x − 14 = lim = lim = lim −(x + 2) = −9 x→7− x→7− x→7− |7 − x| 7−x 7−x 2. [30 Points] Compute each of the following derivatives. Simplify numerical answers. Do not simplify your algebraically complicated answers. (a) f ′ π , where f (x) = sec2 (2x) + sin(4x). 12 f ′ (x) = 2 sec(2x) sec(2x) tan(2x)2 + 4 cos(4x) 1         π π  π  π  2π 2π 2π 4π 2 = 4 sec sec tan + 4 cos = 4 sec tan + 4 cos 12 12 12 12 12 6 6 3  2   16 2 1 1 = √ +2 =4· √ · √ +4 2 3 3 3 3 f′ Note: sec x = 1 π 1 and sec = π = cos x 6 cos 6 1 2 √ !=√ 3 3 2 ! 3 i  √ 3 (x2 + 1) 7 etan x d h  2 d √ ln (x + 1) 7 + ln etan x − ln 1 + cos x ln = (b) dx dx 1 + cos x       1 1 3 1 1 d 3 2 2 · 2x + sec x − · (− sin x) ln(x + 1) + tan x − ln(1 + cos x) = = dx 7 2 7 x2 + 1 2 1 + cos x = 6x sin x + sec2 x + + 1) 2(1 + cos x) 7(x2 s   5 (c) where g(x) = 1 + x        1 5 −5 6 5 · 7 cos g ′ (x) = s · − sin ·   x x x2 5 2 1 + cos7 x g ′ (x), (d) dy , dx if cos7 3 exy + sin3 x = ln(xy) + sin(e9 ). 
d  xy3 d e + sin3 x = ln(xy) + sin(e9 ) Implicit Differentiation dx dx     dy dy 1 3 3 2 2 xy x3y e x + y + 3 sin x cos x = +y +0 dx xy dx 3xy 2 exy 3 dy 1 dy 1 3 + y 3 exy + 3 sin2 x cos x = + dx y dx x 1 dy 1 dy 3 − = − y 3 exy − 3 sin2 x cos x dx y dx x   1 1 dy 3 3 = − y 3 exy − 3 sin2 x cos x 3xy 2 exy − y dx x 3xy 2 exy 3 1 3 − y 3 exy − 3 sin2 x cos x dy x = 1 dx 3xy 2 exy3 − y 2 (e) g ′′ (x), where g(x) = Z 2011 √ ln t + ln √ t dt. x Z 2011 √ Z x √ √ √ √ √  d d = ln t + ln t dt = − ln t + ln t dt = − ln x + ln x ( FTC Part I ) dx x dx 2011        1 1 1 1 1 1 ′′ √ √ √ g (x) = − +√ = − + x 2 x 2 ln x x 2x ln x 2x g ′ (x) d cos x x dx We can solve this two ways: first try Logarithmic Differentiation and using the properties of logs, (f) Let y = xcos x , so that ln y = ln(xcos x ) = cos x ln x Next use implicit differentation to differentiate both sides w.r.t x. d d (ln y) = (cos x ln x) dx dx   1 dy 1 Then = cos x + (ln x)(− sin x). y dx x  cos x  dy =y − sin x ln x . As a result, dx x   dy cos x Finally, = xcos x − sin x ln x . dx x cos x The second option is to rewrite y = xcos x = eln(x ) = ecos x ln x .     d  cos x ln x  1 d cos x cos x ln x (x )= e + ln x(− sin x) =e cos x Then differentiate, dx dx x  cos x  = xcos x − ln x sin x . x 3. [25 Points] (a) Z π 9 tan(3x) dx = π 18 1 =− 3 Here Compute each of the following integrals. ln 1 √2 3 2 !! Z π 9 π 18 1 =− 3 1 sin(3x) dx = − cos(3x) 3 Z 1 2 √ 3 2 1 1 du = − ln |u| u 3 1 2 √ 3 2 1 =− 3   1 − ln ln 2 √ !! 3 2 √    √  √  1 1 ln 3 or ln 3 1 = − ln 1 − ln 3 = − 0 − ln 3 = = ln √ 3 3 3 6 3 u = cos(3x) du = −3 sin(3x)dx − 31 du = sin(3x)dx and x= x= π 18 π 9 √ 3 2

1 3π π =⇒ u = cos 9 = cos 3 = 2 =⇒ u = cos 3 3π 18
= cos π 6
= (b) Z (c) Z e4 e Here (d) Z or =  dx = Z 3 √ dx = 3 x ln x Z x u = ln x du = x1 dx ex ex 2 +ln x+1 5 x5 + 2x 2 + 1 dx = x 4 1 1 √ du = 3 u Z 4 u − 21 Z 3 x4 + 2x 2 + √ du = 6 u 1 1 x5 4 5 dx = + x 2 + ln |x| + C x 5 5 4 =6 1 √ 4− √  1 = 6(2 − 1) = 6 x = e =⇒ u = ln e = 1 x = e4 =⇒ u = ln e4 = 4 Z Z Z e 2 e e x2 ln x x2 dx = e e e dx = e xe dx = eu du = eu + C = ex + C 2 2 2 and 2 +1 2 Here 1 2 4. 2 5 x2 + 1 +C u = x2 du = 2x dx du = x dx [10 Points] Give an ε–δ proof that lim 6 − 5x = −4. x→2 Scratchwork: we want |f (x) − L| = |(6 − 5x) − (−4)| < ε |f (x) − L| = |(6 − 5x) − (−4)| = |6 − 5x + 4| = |10 − 5x| = | − 5(x − 2)| = | − 5||x − 2| = 5|x − 2| (want < ε) ε 5|x − 2| < ε means |x − 2| < 5 ε ε So choose δ = to restrict 0 < |x − 2| < δ. That is 0 < |x − 2| < . 5 5 ε Proof: Let ε > 0 be given. Choose δ = . Given x such that 0 < |x − 2| < δ, then as desired 5 |f (x) − L| = |(6 − 5x) − (−4)| = | − 5x + 10| = | − 5(x − 2)| = | − 5||x − 2| = 5|x − 2| < 5 · 5. [10 Points] Let f (x) = x+2 . x−3 ε = ε. 5 Calculate f ′ (x), using the limit definition of the derivative. (x + h) + 2 x + 2 − f (x + h) − f (x) (x + h) − 3 x − 3 f ′ (x) = lim = lim h→0 h→0 h h   (x + h + 2)(x − 3) − (x + 2)(x + h − 3) (x + h − 3)(x − 3) = lim h→0 h 4 x2 + xh + 2x − 3x − 3h − 6 − (x2 + xh − 3x + 2x + 2h − 6) h→0 h(x + h − 3)(x − 3) = lim x2 + xh + 2x − 3x − 3h − 6 − x2 − xh + 3x − 2x − 2h + 6 h→0 h(x + h − 3)(x − 3) = lim −5h −5 −3h − 2h = lim = lim h→0 h(x + h − 3)(x − 3) h→0 (x + h − 3)(x − 3) h→0 h(x + h − 3)(x − 3) = lim −5 (x − 3)2 = Free double check for yourself using the Quotient Rule: f ′ (x) = 6. x−3−x−2 −5 (x − 3)(1) − (x + 2)(1) = = 2 2 (x − 3) (x − 3) (x − 3)2 [15 Points] Compute Z Match!! 8 x − 3 dx 0 using each of the following three different methods: (a) Area interpretations of the definite integral, (b) Fundamental Theorem of Calculus, (c) Riemann Sums and the limit definition of the definite integral ∗ ∗ ∗ . n X ***Recall i=1 3 |{z} 8  i= n(n + 1) 2 and n X 1=n i=1 5 5 1 25 1 (base) (height) = (5)(5) = 2 2 2 1 9 1 Area Below x-axis = (base) (height) = (3)(3) = 2 2 2 Z 8 16 25 9 − = = 8 x − 3 dx = Then 2 2 2 0 (a) Area Above x-axis = (b) Z 8 0 x2 x − 3 dx = − 3x 2 (c) Here a = 0, b = 8, ∆x = 8 = 0   64 − 24 − (0 − 0) = 32 − 24 = 8 2 8−0 8 8i 8i = and xi = a + i∆x = 0 + = . n n n n 5 Z 8 0 x − 3 dx = lim n→∞ n X i=1   n X 8i 8 f (xi )∆x = lim f n→∞ n n i=1 = lim n→∞ n  X 8i i=1  8 −3 n n ! n n 8 X 8i 8 X − 3 n n n = lim n→∞ i=1 i=1 n 64 X 8 i − (3n) 2 n n = lim n→∞ i=1 ! = lim   64 n(n + 1) − 24 n2 2 = lim  64  n  2 n = lim   n→∞ n→∞ n→∞  n+1 n 1 32(1) 1 + n    − 24  − 24 = 32 − 24 = 8 [10 Points] Find the equation of the tangent line to y = cos(ln(x + 1)) + ln(cos x) + esin x + sin(ex − 1) at the point where x = 0. 7.  1 1 (− sin x) + esin x cos x + cos(ex − 1)ex + = − sin(ln(x + 1)) x+1 cos x   1 1 ′ y (0) = − sin(ln(0 + 1)) (− sin 0) + esin 0 cos 0 + cos(e0 − 1)e0 + 0+1 cos 0  y′ = 0 + 0 + 1 + 1 = 2 ←− Slope Point (0, y(0)) = (0, 2) because y(0) = cos(ln(0 + 1)) + ln(cos 0) + esin 0 + sin(e0 − 1) = cos 0 + ln 1 + e0 + sin 0 =1+0+1+0=2 Point-Slope Form y − 2 = 2(x − 0) y = 2x + 2 8. [20 Points] Let f (x) = x ex = xe−x . For this function, discuss domain, vertical and horizon6 tal asymptote(s), interval(s) of increase or decrease, local extreme value(s), concavity, and inflection point(s). Then use this information to present a detailed and labelled sketch of the curve. Take my word that lim f (x) = 0 and lim f (x) = −∞. x→∞ x→−∞ • f (x) has domain (−∞, ∞) so No Vertical Asymptotes. • Vertical asymptotes: none • Horizontal asymptotes: at y = 0 towards ∞, since lim f (x) = 0. x→∞ • First Derivative Information: f ′ (x) = xe−x (−1) + e−x = e−x (−x + 1) The critical points occur where f ′ is undefined (never here) or zero. Also note that the exponential function is always non-zero, which implies that −x + 1 = 0 As a result, x = 1 is the critical number. Using sign testing/analysis for f ′ , f′  ⊕ ⊖ t - 1 ց local max Therefore, f is increasing on (−∞, 1) and decreasing on (1, ∞) with local max at (1, f (1)) = (1, e−1 ). f ր • Second Derivative Information f ′′ (x) = e−x (−1) + (−x + 1)e−x (−1) = e−x (−1 + x − 1) = e−x (x − 2) Possible inflection points occur when f ′′ is undefined (never here) or zero (x = 2) (again note the exponential piece is non-zero). Using sign testing/analysis for f ′′ , f ′′  ⊖ ⊕ t - 2 S infl. point Therefore, f is concave down on (−∞, 2), whereas f is concave up on (2, ∞) with I.P. at (2, f (2)) = (2, 2e−2 ). f T • Piece the first and second derivative information together  t 1 f f ր T  local max t - 2 ց T  infl. 7 pt. ց S  9. [15 Points] A conical tank, 14 feet across the entire top and 12 feet deep, is leaking water. The radius of the water level is decreasing at the rate of 2 feet per minute. How fast is the water leaking out of the tank when the radius of the water level is 2 feet? **Recall the volume of the cone is given by V = 13 πr2 h The cross section (with water level drawn in) looks like: 7  6 B r  wB - 12 6 aB t B h e B r B ? ? B • Diagram • Variables Let r = radius of the water level at time t Let h = height of the water level at time t Let V = volume of the water in the tank at time t dV =? when r = 2 feet Find dt dr ft and = −2 dt min • Equation relating the variables: 1 Volume= V = πr2 h 3 • Extra solvable information: Note that h is not mentioned in the problem’s info. But there is a relationship, via similar triangles, between r and h. We must have r h = 7 12 =⇒ h = 8 12r 7 After substituting into our previous equation, we get:   4 12r 1 = πr3 V = πr2 3 7 7 • Differentiate both sides w.r.t. time t.   d d 4 3 4 dr dV (V ) = πr = π · 3r2 · =⇒ dt dt 7 dt 7 dt =⇒ dV 12 dr = πr2 dt 7 dt • Substitute Key Moment Information (now and not before now!!!): 12 dV = π(2)2 (−2) dt 7 • Solve for the desired quantity: 96π ft3 dV =− dt 7 min 96π • Answer the question that was asked: The water is leaking out of the tank at a rate of cubic 7 feet every minute. 10. [15 Points] √ Let R be the region inside the top semicircle of radius one, centered at the origin, given by y = 1 − x2 . Find the area of the largest rectangle that can be inscribed in this region R. Two vertices of the rectangle lie on the x–axis. Its other two vertices lie on the semicircle. √ y = 1 − x2  6 '$ (x, y) r " 1 " - ? are computing extreme values for.) (Remember to state the domain of the function you • Diagram: We already have a diagram. • Variables: Let x = x-coordinate of point (x, y). Let y =-coordinate of point (x, y). Let A =area of inscribed rectangle. • Equation: √ Then the area A = 2xy = 2x 1 − x2 must be maximized. The (common-sense-bounds)domain of A is {x : 0 ≤ x ≤ 1} . p −8x2 + 4 (−2x) −4x2 + 4(1 − x2 ) √ = √ . • Maximize: Next A′ = (2x) √ + 1 − x2 (2) = 2 1 − x2 2 1 − x2 2 1 − x2 Setting A′ = 0 we solve for x2 = talking distance.) 1 2 or x = √1 . 2 (We take the positive square root here because we’re Sign-testing the critical number does indeed yield a maximum for the area function. A′ ⊕ A ⊖ 1 ր√2ց MAX 9 r q q  2 then y = 1 − √12 = 1 − 21 = 12 =    √1 area that occurs is A = 2xy = 2 √12 = 1 square unit. 2 • Answer: Since x = √1 2 √1 . 2 As a result, the largest 11. [15 Points] Consider the region in the first quadrant bounded by y = ex + 1, y = 4, and the y–axis. (a) Draw a picture of the region. See me for a sketch. Note that the curves intersect when 1 + ex = 4 which is when ex = 3 which implies x = ln 3. (b) Compute the area of the region. Z Z ln 3 Z ln 3 4 − (ex + 1) dx = top − bottom dx = Area= 0 0 = (3 ln 3 − eln 3 ) − (0 − e0 ) ln 3 0 ln 3 3 − ex dx = 3x − ex 0 = 3 ln 3 − 3 + 1 = ln 27 − 2 (c) Compute the volume of the three-dimensional object obtained by rotating the region about the horizontal line y = −2 Z ln 3 Z ln 3 2 2 π[62 − (3 + ex )2 ] dx π[(outer radius) − (inner radius) ] dx = Volume= 0 0 = Z ln 3 0 π[36 − (9 + 6ex + e2x )] dx = = π[(27 ln 3 − 6eln 3 = π[27 ln 3 − 18 − − 9 2 1 2 ln 3 ) 2e − (0 − Z ln 3 0 6e0 1 π[27 − 6ex − e2x ] dx = π[27x − 6ex − e2x ] 2 − 1 0 2 e )] = π[27 ln 3 − 6(3) − 1 ln(32 ) 2e +6+ ln 3 0 1 2] + 6 + 12 ] = π[27 ln 3 − 12 − 28 ] = π[27 ln 3 − 12 − 4] = π[27 ln 3 − 16] 12. [15 Points] Consider an object moving on the number line such that its velocity at time t seconds is v(t) = 4 − t2 feet per second. Also assume that the position of the object at one 5 second is . 3 (a) Compute the acceleration function a(t) and the position function s(t). a(t) = −2t Z t3 s(t) = 4 − t2 dt = 4t − + C 3 5 Use the initial condition s(1) = 3 1 set 5 s(1) = 4 − + C = ⇒ C = −2 3 3 Finally, s(t) = 4t − t3 −2 3 (b) Compute the total distance travelled for 0 ≤ t ≤ 3. 10 Total Distance = Z 3 0 2 |4 − t | dt = Z 2 0 2 4 − t dt + t3 = 4t − 3 Z 3 2 −(4 − t2 ) dt   t3 + −4t + 3 0 2 3 2     8 8 − (0 − 0) + (−12 + 9) − −8 + = 8− 3 3 =8− 8 8 −3+8− 3 3 = 13 − = = 16 3 39 16 − 3 3 23 3 11