Math 100 Dec 94 Answers - Math 100/120/153 Solutions/Dec....

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Unformatted text preview: Math 100/120/153 Solutions/Dec. 1994 Math 100/120/153 Solutions Dec. 1994 1. This question tests your ability to take derivatives using the Product Rule and Chain Rule. (11) f (2:) = (21: + 3)e' (b) y=ln(a:+\/:c§‘-—l =ln(z+(z2—1)§) iv- dz- dy dz f'(z) = 2e’+(2z+3)e' = (2.'t:+5)ez f’(1) = (2+5)e1=7e 1 ;+— ,/:2-:-i(1+%(2'1)_§(2$) l a: x+Vz§—l(1+ 1:2-1) 1 z+\/:z:2—l :c+\/12—1. «22—1 1 22—1 (c) Slope of y = ‘/arctan(z2) at 2: = l Marctan (2:2) 1 + $4 1 2 (arctan (2:2)) 'é .‘B l Atx=l,y=‘/arctan =f§=azfi Ly dz 1:! l Varctanl . 1 [Oh—- 1 2:1: ) Page 1 (d) To find the points where y = 22' -— 2’ has a horizontal asymptote, solve dy _ - 21 _. 2 1 d1: — 0. Notice that 2 —- (2 ) - y = (2’)’ — 2' 3% = (22V ln(2’) -2’ 1:12 22‘-2ln2-2’ln2 (22“! — 2') ln2 = 0 Thus, since In 2 ;£ 0, 2224-1 _22 = 0 2214-1 _ 2:: To solve the equation, we equate the exponents. 22: + 1 = z z = —1 Hence y has a horizontal tangent at :1: = —-1. 2. The definition of the derivative of f (x) at a: = a is, I _ - f(z)-f(a) f(a)— Inn—r;- 2-H! Now apply it to f(1:) = Vx’ + 5. I _ . f(r)-f(2) “2) - lama? _ 1i ¢x2+5—\/22+5 — 1:312 3—2 ¢z2+5—3 ¢z2+5+3 lm——— 'Jz2+5+3 2—»2 :z—2 (, Math 100/120/153 Solutions/Dec. 1994 Page 2 _ 2:2 + 5 — 9 3 dzy = lim——————— —+3+9———+2-3—3 = 0 =42 (:r—2)(\/z2+5+3) 2 ( )dri:2 — lim ___i'—4——— 12% — g = o 2—»2 (1-2)(‘/32+5+3) :5 (fig 5 = “m ___(3=+_2)(£i 2:2 = fl 1H2 (a: — 2) (V32 + 5 + 3) _ =42 2:2 + 5 + 3 (c) The piece of information f (3) = 1 tells you that the curve passes through _ 4 _ 2 the point (3,1). Part a) tells you that the slope of the tangent line to the - é " 5 curve at that point is —%. So the tangent line has a slope of —-%. Additionally, . 1 _ . I I _ part b) indicates that the curve is concave-up at (3.1), so our tangent line 'Il'y checking the answer by finding f'(z) usmg the differentiation rules and lies b flow the curve_ substituting in a: = 2. Y 3. (a) Usmg impliCit differentiation y 1”,” m dy dy 2_ 2___ = 33] dz + 213g + 3 dz 0 I my"! “1‘: concave upward slopez- —L- Now when x = 3, y is equal to 1, so ’ dy dy 2_ 2_ = 3(1) dz: 3:3 + 2(3)(1)+(3) dz 2:3 0 Rewrite th's as ‘ :12 _ :1 _ _1 d: 3:3 ‘ 3 + 9 ” 2 3 ‘ (b) To find f”(2:) we implicitly differentiate again the first line of the pre- vious calculation: 4. (a) dy 2 ,d’y d9 ‘13! 2J2}! 6y(—) +3y—+2y+2z—+2z—+z——-=0 t c dz dz” d3 dz d-“r’ lim M(t) = lim 4 .6— = lim 4" = 9 = o 2—4—00 t-b-oo 1 + e“ et l—i—oo e' + 3 3 When 1: = 3, we know that y = f(3)=1 and £- = f’(3) = mg. lim M(t)= lim 4 =—4—=4 2 t—boo t—roo l + '1 1+ 0 6(1) “mg +2(1)+ 2(3) +2(3) + (202% = o e Math 100/ 120/153 Solutions/Dec. 1994 (b) Now w d 1 _ __ —t - dt _ dt4(l+3e) = 4(—1) (1 + 3e")_2 (—3e-') __ 128“ (1+3e")2 14 1 4 4 MMTM) ‘ Z'1+3e-¢(4_1+3e-t) 1 4 + 12¢" — 4 1+3e“ ( 1+3e“ ) 12e“ (1 + :se—t)2 Since expressions (1) and (2) are identical, we have proven that 1 ZM(4 — M). (c) The maximal rate of growth of M means the maximum of find this by solving dM F. We J‘M dM dtz — 0. We already know —d-t- from part b). 2’14. 1 dt’ "dt l2e“ (1+ 33")2 Use the quotient. rule to take the derivative. (PM dt2 —l2e"(1 + 3e")2 — 12e“(2) - (l + 3e“)(—3e") (1 + 3e")4 -—12e"(l + 3c" — 6e") (1+ 3e“)3 —12e"(1 - 3e") (1 + 3e“)3 dM (1) (2) Page 3 d2 Now we set (it)? = 0 and solve for t. We know that —12e" ¢ 0 so we must have 1 — 3e"‘ = 0 1 -1 _ _ e ' 3 —t — ln(3‘l) t —- ln3 . . . dM To find the maxxmum value, we substitute t = ln3 Into our formula for 71?. dM 12e-'n3 12(1) 1 71? mm, = (1+ 3e-ln3)2 6. Start off by (implicitly) differentiating the given equation with respect to t: _L§ _ _ifl _ LE C2 dt R2 dt 52 dt 2?. - 52 1.19 _ i151 dt _ 02 dt R2 dt dS - dR . We want to know Et— when R = 250 Q, S = 1000 Q, I = 100 Q/minute and % = 10 Q/minute. We have everything we need to fill in all the unknowns in the above equation, except C. We get this value by plugging our R and S values into the original equation. 1 _ 1+1 0 “ R s .1. _ L+__L C ” 250 1000 l _ 4+1 C ‘ 1000 C = ESQ—0:200!) Math 100/120/153 Solutions/Dec. 1994 Now put it all into the. differential equation. d5 2 1 1 Et— _ 1000 (—2002(10)———2502(100)) 200252(10) _ 250242000) 2002 2502 25-10—16-100 250 - 1600 = -1350 Q/minute Thus 5 is decreasing by 1350 fl/minute. 6. Let a: be the height of the box, y be the depth of the box, and 2 be the width of the box (see the following diagram). We can write down a formula for the volume of the box: V = zyz 5ft 8ft However, we have the following constraints. 22+2y=8 y=4—a: 2z+z=5 2:5—22: We can make our formula for volume into a function of a: only by substituting in for y and z. V(a:) = 2(4 —- z)(5 - 22:) Page 4 I This is the function that we want to maximize (with respect to 1:). (‘11—: = (4 — x)(5 — 21:) - 1(5 - 2x) - 2:0(4 — 2:) = 62:2 — 262 + 20 = (6x — 20)(:r: — 1) = 0 Thus 0;: = 0 for z = (3—0 = g or a: = 1. When 2: = 13—0, our other values are y = g aid 2 = -—§. But 2 cannot be negative! So instead try :2: = 1. Then y = 3 and z = 3. The volume in this case is 9 ft’. This is the maximum volume. 7. (a) Use f’(a:) to find intervals over which f(z) increases, decreases, or has local maxima or local minima. , __ 2(1 — 1:2) __ 2(1— z)(1 +2) “3) ‘ (1+z2)2 ‘ (1+z2)’ Since (1 + :22)2 > 0, we need to consider only the sign of (1 — 1:)(1 + z). l -— a: 1 + 2: f’(z) — negative —1 < a: < 1 + positive 1 < a: —- + negative So, f has a local minimum at (—1,0), and a local maximum at (1,2). f(2:) is increasing on (—1,1) and decreasing on (-oo, —1) and (1,00). (b) We do the same thing we did for part a), except this time we use f”(2:) instead of f'(x). 2(—2:c)(1 + 1:2)2 — 2(1—- 2:2)(2)(2:c)(1 + 2:2) (1 + zz)‘ 2(—21:)(1 + z2)2 — 2(1— $2)(2)(2x)(1 + x2) (1 + 1:”)4 2(—21:)(1 + 1:2) — 2(1 — $2)(2)(2:c) (1 + :52)3 f"(z) Math 100/ 120/ 153 Solutions/Dec. 1994 —4a:(1 + 2:2) - 83(1 — 2:2) (c) See the following diagram. (1 + z’)3 -4a:(l + 2:2 + 2 - 22:2) (1 + 11:2)3 —4a:(3 — 1:2) (1 + 22)3 Now we know that f”(2:) = 0 when a: = 0 or a: = is/fi. The denominator of f ” (x) is always positive, so we only need to worry about the sign of the numerator. z :c—s/fi a:+\/§ f"(:c) negative —\/5 < a: < 0 - - + positive 8. (a) 0 < 1: < x/i + - + negative “m “13 z 9 J5 < z + + + positive 2—H sin(1r:1:) 0 Using L’Hépital’s theorem, .1. lim .ln1 = lim —5——— 2-H sm(1rz) z—u 1rcos(7r:1:) _ 1 ' ’— “(-1) f (:c) is concave up on (—\/§,0) and (Jim). 1 f (z) is concave down on (—00, -—\/§) and (0, J5). 7r Hence f (2:) has points of inflection at a: = 0,:l:\/§. (these are the points where the concavity changes sign.) ('3) Page 5 Math 100/ 120/ 153 Solutions/Dec. 1994 Again, use L’Hopital's theorem. 1. e2! _ 1 — 2e2t e336 te2t _ 151(1) 6"" + 2te” 2 = —— = 2 1 + 0 (c) The In function is continuous, so lnlimz = limln 2:. 1“ [Hm (“exp/2,] = lim [ln(seczrl’z] 2—90 1—)!) ln(sec z) 0 z- lim 2—90 272 0 1 —— secz tan 2; lim flit—— using L’Hopital’s theorem. z-w 2a: ll .5: = % using L’Hopital’s theorem. ll '5: 9. (a) The given limit is finite. Notice that the denominator goes to zero. lim 2:2 -— 4 = 0 242 In order for the limit to be finite, the numerator also must go to zero. limf(:c)—5=0=>li_.n12f(2)=5 2—52 (h) The definition of the derivative of f (3) at a: = 2 is 1—92 1—2 We have to massage this to get it into a form where we can use the limit that we were already given. Note that since f(a:) is continuous at a: = 2, we can Page 6 « say that f(2) = 5. nn—quz+2 I _ . f (2) _ :1: — 2 :1: + 2 = mgflnzwu+m 2—02 2: —-4 = limf(:)-5-limz+2 :42 z — 4 :42 == 3 - 4 = 12 . dy dv . . . 10. First, we note that d— = v and E = a. Now 1t 15 Simply a matter of integrating and calculating constants of integration. (1—!) _ 6t if 0 < t 5 10 dt _ —9.8 if10<t<T Um _ 3t2+C if0<tglo “ —9.8t+D if10<t<T To calculate the constants C and D we are told that the rocket starts at rest, so 11(0) = 0. Thus C = 0. When the fuel runs out (after 10 seconds) the velocity function switches from one formula to the other. The velocity must be continuous, so we can stipulate, [33”!=10 = [—9.8t+D]|t=m 3(10)2 = —9.8(10)+D D = 300+98 D = 398 So now we have the velocity function. v(t)_ 3:2 if0<t510 ‘ -9.8t+398 if10<t<T Now we integrate v(t) to get y(t). g __ 3t2 if0<t310 dt ‘ —-9.8t+398 if10<t<T (t) __ t3+E if0<t510 3’ " —4.9t2+398t+F if10<t<T Math 100/120/ 153 Solutions/Dec. 1994 Again, we use the information we are given to calculate the constants of integration, E and F. The rocket starts from the ground at t = 0, so y(0) = 0 => E = 0. Also, the height of the rocket is continuous, so, [t3]|t=10 = [—4.9t2+398t+F]|l=10 103 = -4.9(10)2+398(10)+F F = 1000+490—3980 F = —2490 Thus, the height function for the rocket is (t); t3 if0<t$10 y ' -4.9t’+398t-2490 if 10<t<T 11. (a) We can centre the tangent line approximation at a: = 2 since we know f(2) and f’(2). The linear approximation function is fit) z 14(3) = f(2) + f'(2)(17 - 2) = 4-1(2:—2) = 6—2: L(3) = 6—3=3 (b) To find the smallest interval that we are sure contains f (3), we must calculate a bound on the error of our estimate of f (3). We know that Im) — Len = gm“: — 2)’ for some c between a: and 2. We don’t know what that c is, so instead we choose a c that makes our error estimate as large as possible (so that we don’t underestimate the error). man—Len s §crgllzagllf”(c)l(x-2)2 lf(3)-L(3)l s gflgllmcms-zf Page 7 It is given to us that i S f "(3) 5 -1- for all a: > 0. Thus, we are guaranteed 2a: a: that I f”(:c)| will be no bigger than % on the interval [2,3] (try substituting 1 l :1: = 2 and :1: = 3 into —z and E to try to get an idea of what :c-value will give you a maximum bound for f"(:r)). |f(3) — Lam < 1 max moms — 2)” = g (1)2 = § — 2 «[2,3] The smallest interval that we can be sure contains the value for f (3) is 1 l 11 13 [3‘2'3+zi - [7’ Ii 12. (a) We are looking for solutions to the equation f (z) = a: for ze[a, b]. Define the function g(z) = f (z) — a: and try to show that g(c) = 0 for some ce[a, b]. Notice that, since a S f (1:) 5 b Vze[a, b] (“for all a: in [a,b]”), we can say 9(a) = f(a) -a 20 and 9(5) = f(b) —bS 0 Thus, since 9 is continuous, by the Intermediate Value Theorem, we know that there exists some c in [a, b] such that g(c) = 0. Therefore, g(c)=0=f(c)-c = f(c)=c as required. (b) We use the method of contradiction for this proof. Assume that there are two :r—values that make I (1:) = 1:, and call them an and 2:2. Then by the Mean Value Theorem, there is a c between 2:; and :2 such that no): “Ill-“332) ___ 251-552 :1 121-12 $1—12 But we were given that |f'(z)| < 1 for a S :1: 5 b. The above result contra- dicts this constraint, so we know that there cannot be two (or more) roots in [a, b]. This fact, together with part a) proves that there is exactly one root to the equation f (1:) = 2:. ...
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This note was uploaded on 01/30/2011 for the course MATH 100 taught by Professor Lamb during the Fall '08 term at The University of British Columbia.

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Math 100 Dec 94 Answers - Math 100/120/153 Solutions/Dec....

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