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# Math 100 Dec 02 Answers - Math 180 100 Final Exam December...

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Unformatted text preview: Math 180/ 100 Final Exam December 2002 Solutions 1.Short Answer Questions. (3.) Evaluate the following limit: 1” \$2+3x+2 I—l’n—ll 32—:5—2 Solution: Call this limit L. Then . (z+1)(:1:+2) . 2 l —— I ll . 1.1331 (:1:+1)(:r—2) 13195—2 3 a2+2 _ —1 (b) Determine an equation of the tangent line at (1, 5) to the curve y = 3:3 + 4. Solution: dy/da: : 3:122, so the slope at .7: = 1 is 3 ~ 12 = 3. Thus the equation of the line is y— 5 = 3(2: — 1). (c) Calculate the derivative of 9(3) : arctan{ln(s2 + 1)} Solution: We use the chain rule: g'(5) = 1 1 1+ln2(s2+1).32+1 - (23) ((1) Find the second derivative of Solution: We use the chain rule: f'(m) = 2cos(2m)esjn(2z) f"(\$) = (4 cos2(2:c) —- 4sin(2x))65in(2z) (e) Evaluate the following limit, if it exists _ sing: 11m r—~0 tan 3m Solution: Call this limit L. Then 3 L: lim c?” z 'sina: 1—*0 Sll'l 3m . 3:1: sina: 1 : hmcos3x~—.——-—-— :—’0 sm 3x a: 3 DJIH :1.1.1. (f) A bead is moving on a straight wire. It’s position a: is given as a function of time t below: \$(t) = t3 + t + 1 Find the value a > 0 so that the average velocity of the bead over the interval [0,a] is equal to 5. Solution: The average velocity is :1:(a) — 22(0) # a.3 + a a (l Therefore a : 2. (g)Suppose the function f is deﬁned as f(a:) : 0 if a: < 0 f(z)=Azz+Bif:1:_>_0 for constants A and B. For what value or values of A and B is f differentiable for all 2:? Solution: For f to be differentiable it must be continuous. Thus 0 = limxnm f(2:) = f(0) = B, so B = 0. For f to be differentiable we must have 0 : lim,,.0+ f’(a:) = f’(0) = A - 0, or A -0 = 0. This puts no restriction on A. Thus B : 0 and A can be any value. (h) Find the tangent line at (1,1) to the curve msin(zy - yz) : {1:2 — 1. Solution: We differentiate implicitly. dy dy - _ 2 _ 2 _ __ _ __ : sm(\$y y )+a:cos(\$y y ) (11+ xdz Zydx) 2:5 Therefore at (1, 1) we have dy dy 1 — —- -— : 2 + dx da: and thus dy/da: = —1 when :1: 2 1. Therefore the equation of the tangent line isy:1—(a:—1). (i) Indicate on the graph below where the estimates 2:1 and \$2 to the root of f (x) will be located if calculated using Newton’s method starting with initial estimate (guess) \$0. y! y: ICU) (j) Determine the slant asymptote of 31:3 +2.7:2 — 1 f(.'1:)= \$2~4\$+3' Solution: We use polynomial long division to determine that 472: — 43 =3 —-—. f(\$) 1+14+\$2_4\$+3 Therefore the slant asymptote is given by y = 33: + 14. (k) If f(l) : 2 and f’(1) = 3 use a linear approximation to estimate the value of f( 1.15). Solution: f(1.15) 2 2 + 3(0.15) : 2.45. (1) In the previous question, if it is known that If”(:1:)| < 2 for all :12, what is the maximum error in your approximation? 1 Solution: Error 3 5 - 2- (0.15)2 = (0.15)2. (m) Determine the ﬁrst three nonzero terms in the Taylor series based at :1: = 0 for cos(2:l:). Solution: We ﬁrst compute the series for cost: :1:2 3:4 :1— — — — cosm 2 + 4! Therefore 2 2 4 2 cos(2a:)=1-(:) +(Z) --... and the ﬁrst three nonzero terms are thus 1 — 2x3 + (2/3)::;4 — . .. (n) Evaluate . cos(2:1:) — 1 + 211:2 km _————. 1—»0 (1:4 Hint: The previous problem will be helpful here. Solution: Call this limit L. First we use from above that cos(2:1:)=1— 23:2 + (2/3):1:4 + . . .. Therefore cos(2n:) — 1 + 2m2 2 (2/3)a:4+ higher order terms. Therefore 4 5... L: m (2/3»: t“ ) 1—)0 :1: = 2/3+0 = 2/3. 2. Theory Questions. (a) Evaluate the derivative of f (9:) : x/a: + 1 using the deﬁnition of the derivative, _ f (a? + h) - f(m) Solution: We will multiply by the conjugate. vm+h+ —\/m+1v—_____a:+h+1+\/m+1 :17 =11m——h— ﬂ) heo ¢m+h+1 +¢aT — —1im (\$+h+1)_ (14—1)) =lim h h—v0h(\/\$+h+1+\/IE+1hﬂoh(\/\$+h+1+\/\$T) And evaluating we have 1 I f (m) ‘ war (b) Consider functions f(z) deﬁned on the interval [0, 1] that have the following properties: (i)f is continuous in the interval [0, 1]. (ii) f is differentiable in the interval (0, 1). (iii) f(0) = 0 and f(1) = 1. Find the function deﬁned on the interval [0, 1] that has these properties that makes the maximum of ]f’(:z)] on the interval (0, 1) as small as possible. Hint: begin by sketching some graphs that satisfy the above properties. Solution: f(z) : x. Here’s the proof. If f(a) < a for any a then f’(c) > 1 for some c between a and 1, by the mean value theorem. If f (a) > a for any a then f'(c) > 1 for some c between 0 and a, also by the mean value theorem. Therefore only f (as) 2 x will minimize the absolute value of the derivative. Word Problems 3. A bead is constrained to move along a wire bent into the shape of the graph y = sin :12. At a certain instant in time the bead is at :1: = 1r/4 and dar/dt = 2. Find the rate of change of the distance between the origin and the bead at this instant in time. Solution: The distance to the origin of the point (z,y) is «11:2 + yz. So if the point is on the curve y = sinx then the distance is given by f(m) : V152 + sin2 2:. We are asked to ﬁnd df/dt and we use the chain rule. df df dx _:1:+sin:ccosm dz E 2 E 25 _ W i 3' At the instant cc 2 7r/4, dm/dt = 2, and thus df _ 1r/4+1/2 — _ —————— -2. dt 7r2/1S+1/2 4. Find the point or points on the graph y 2 x2 closest to (0, 3/2). Show that you have carefully checked all possibilities. Solution: Note that minimizing the square of the distance is equivalent. Thus we minimize f(a:) 2 2:2 + (:1:2 — 3/2)2. We compute f’(x) = 21: + 2(12 — 3/2) - 21: = 43¢»? — 1). Thus the critical points are :1: 2 0, a: 2 1 and a: 2 —1. We evaluate and ﬁnd that f(0) 2 9/4 and f(1) 2 f(—1) 2 5/4. Furthermore we note that lim f (:5) 2 lim f (as) 2 00. Therefore the minimum is obtained at the two I—‘m IH—m points (1, 1) and (—1, 1). 5. A 1 kg model boat is put into water at rest and its engines are turned on. Let v(t) denote the velocity of the boat in m/s, with t in seconds after the engines are turned on. It is known that v(t) solves the differential equation d1) 1) d_t_ _E‘ (this equation represents mass times acceleration equal to net force of the engines and friction). ' (a) Solve the differential equation for t. (b) Evaluate 31.120120). Solution: We note that Q 2 —1/10(v ~ 10). dt Let u 2 v -— 10. Then we note that du dz) — 2 — 2 —1 10 . dt dt / " This is an exponential growth differential equation and has solution u(t) 2 nos—”10. We also know that 110 2 0 and so no 2 —10. Thus v(t) 2 10 — 1064/10. Therefore tlim v(t) = 10. 6. Sketch the graph of Hi) = we‘r’ showing all of the following if they are present: (i) y-intercept (ii) 1 intercept (iii) critical points, local maxima and minima, intervals where f is increasing or decreasing (iv) inﬂection points and intervals where f is concave upward or downward (v) asymptotes (horizontal, vertical and slant) Give the (x,y) coordinates for all of the points listed above. Solution: f (0) = 0 is the only root and the only place where a: = 0. To ﬁnd the critical points we compute f’(:z:) = (1 — 2m2)e—I2. Therefore the critical points are where (1 / ﬂ, 6'1/2 / ﬂ) and (—1/\/§, —6_1/2/\/§). When a: < —1/\/§, f is decreasing. When —1/\/§ < :r < l/ﬁ, f is increasing. Lastly we note that when :r > l/x/i, f is decreasing. Therefore (1 / ﬂ, 6—1/2 / ﬂ) is a local max and (—~1/\/§,—e”1/2/x/§) is a local min. Now we inspect the second derivative, f”(a:) : e"Iz(—6m + 413) Therefore the possible inflection points are :1: = 0, a: 2 J37? and a: = ~m. We note that for at < —\/§—/_2, f is concave down. For —\/IV§ < a: < 0, f is concave up. For 0 < a: < \/3/—2, f is concave down. And for :1: > m, f is concave up. Therefore the inflection points are (M,°\/3/—26~3/2) and (—\/3’/—,—\/2We—3/2). Finally we note that lim f(\$) : lim f(:1:) = 0. (ta—00 1—900 Therefore there is a horizontal asymptote of y = 0. Thus the graph should look like this: —‘/2 (—L)"L'e > {OCctl WMX The End ...
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Math 100 Dec 02 Answers - Math 180 100 Final Exam December...

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