Math 100 Dec 08 - 2. U! December 2008 Math 100/180 Final...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2. U! December 2008 Math 100/180 Final Exam '2 117 — :L' "53—1 or determine that this limit (lees not exist. Evaluate limwal Evahiate hmma00 91‘" +0" or deternnne that this hnnt does not ex1st. :3 -:i:'3+-’l a; 3+c»-- - Find the derivative of Find the derivative of t3 cost. Find the derivative of Find f’(:r), if f(:z;) = arctan(m3). (Note: Another notation for arctan is ten”. ) ) l g) lf 1‘2 + my ~ 1/2 : 4, find (lg/dz); in terms of .‘L‘ and y. ) If f(:i:) = (cos IL)", find f'(:1:). ) Use a linear approximation to approximate V1002. ) Estimate the size of the error made in the linear a1_)proximati0n above. In other words, find an upper bound for the absolute value of the difference between V1002 and the answer to item above. (k) If f(:r) 2 :L' cos(:1:3), compute fm(0). Hint: Use Maclaurin series. (1) Find the absolute minimum value of = sin—1(ru/2) on the interval [—2, #1]. [Note: Another notation for sin—1 is arcsin. Newton’s Method is used to approximate a solution of the equation sin .‘L‘ : 1 — :L', starting with the initial approximation 1L1 : 7r/‘2. Find 1'2. (11) Given that f’(w) = 2:1: # (Ii/(1:4), :1: > 0, and f(1) = 3, find (in) A freshly brewed cup of coffee initially has temperature 95° C in a room that has a fixed temperature of 20° C. When the coffee temperature is 70° C, it is decreasing at a rate of 10 C per minute. When does this occur? Assume that the tenn)erature of the coffee in the cup satisfies Newton’s Law of Cooling. Gravel is being dumped from a conveyor belt at a rate of 2 ind/min,7 and its coarsness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 3 m high? Hint: The volume of .lTr’I'2/L. a right circular cone with base radius r and height h is d . Let f(:l:) : "f +12. :r—Z Find the domain of f(:r). Determine intervals where f(:L') is increasing or decreasing and the 3:; and ye coordinates of all local n'iaxinia or minima (if any). Determine intervals where f(:1:) is concave upwards or downards, and the :1:— coordinates of all inflection points (if any). You may use, without verifying it, the formula f”(1‘) = 32/(:L' - ‘2)3. Find and verify the equations of any asymptotes (horizontal, vertical or slant). Sketch the graph of y = f(:r), showing the features given in items (a) to ((1) above and giving the (egg) coordinates for all points occuring above and also all :r—intercepts (if any). . A cylindrical can without a top is made to contain 277r cm3 of liquid. Determine (with justification) the dimensions of the can that minimize the area of the metal used to make the can. Hint: The metal used consists of a circle (the bottom of the can), and a rectangle (the sides of the can). 1 1+5. No credit Will be Use the delimitiun of the dc'ri'uati'm: to determine f’(1‘), where 2 given for using derivative formulas. 7. The function is defined by (La:2+bru+c if.7;<0 f(1') = 2 if 1' = 0 2 +12 cosh—1) if a; > 0 Determine all the values of the constants (1., b, and c such that f is continuous at O, 01‘ determine that no such values exist. 8. Let £ be the tangent line to the. curve y = 1 + 6"" at the point P(a,b), where the point of tangency P is chosen so that the tangent line 1: passes through the origin 0(0, 0). (a) Find an equation that is satisfied by a, the :1:—coor(linate of the point P. (b) Prove that the equation in item (a) has exactly one real solution. 1. (a) December 2008 Math 100/180 Final Exam If we just tried evaluating directly, we’d be dividing by zero (12 — 1 = 0). But we can fortunately factor the numerator and denominator, then cancel some common factors to arrive at a value for the limit: 12—93 _ $(m-l) _ a: 1 =llm——————:hm :____ z—vl(:r+l)(x—1) z-+1:r+1 1+1 _ 1 ._ 2. We have a rational function and we are considering the limit at infinity. Since the degrees of the numerator and denominator are equal, the limit is just the ratio of the leading co- efficients of the numerator and denominator. More rigorously, we’ll divide the numerator and denominator by x3 and use some limit laws: , :63 + 5x . 1 + (SQ/(:19) 11m —— = hm -—-——-—-—— Iqw 2x3 _ 1.2 + 4 “.00 2 _ ($2)/($3) + 4/(m3) limIaw1+5/(m2) _ 1+0 _ 1 “lilnzsoo2—1/x+4/(m3) 2—0+0'2' In general, as long as everything is defined, (f/g)/ = (f’g — g’f)/gg. If f = 1' then f’ :: 1. And if g = 3 + 61 then 9’ 2 er. So the derivative is: (1)(3 + 62) — (ez)(z) _. 3 + eI — max (3—1—61)2 _ (3+ez)2 . In general, as long as everything is defined, (fg)’ = f’g + g’f. If f = t3 then f’ = 3152. And if g 2 cost then 9' : ~sint. So the derivative is simply: (3t2)(cos t) — (sin t)(t3) = 3132 cost — t3 sin t. In general, as long as everything is defined, (f o g)’ = (f’ o g)g’. Here f(a:) = 6‘” so that f’(z-) 2 er, and g($) = fl so that g’(rr) : The derivative is therefore: _ 2fi. We have another composition as above. This time, = arctan(:r) so that f’(2:) = 1/(1 + 3:2), and 9(a) : m3 with g'(a:) = 3m? The derivative we get is: (69(1))(1/(2fin (1/(1+(9($))2)(31‘2) = (1/(1+($3)2)(3$2) : 13:19 Throu hout this solution we will use ' ’ to denote 51—1” for shorter notation. g y 111: To start, We will differentiate with respect to m, but we will be using the fact that y is a function of :L' (y = and therefore we will have to use the chain rule: d 2 2 d — 'e 2—420. da:( +$y y) dm Now let’s work more with the leftmost term to get: ($2)' + (551/), - (2/2)’ = 206 + (1)04) + (1M1) - (2y)(y’) = 2x + y + my’ — 2113/4 If the above is zero, then we can solve for y’ : d1: m—Zy- (2$+y)+($—2y)y'=0 => y (h) We’ll need logarithmic differentiation. We take the ln of both sides and differentiate with respect to m : lnf(:c)=ln((cos:r)x)=xln(cos:c) => (fiXf’wD:(1)(ln(cosx))+( 1 Note that we used the chain rule in computing the derivative of xln(cos We now multiply both sides of that latter equation by f and simplify: )(—sinx)(m). COS 1? sin a; f’fx) — (f(~”€))(1n(00S 96) — 1 C08$) = (cos w)x(ln(cos m) — :ctan As a linear approximation, f(a + h) = f(a) + hf’(a) + 0(h2), provided f is at least twice differentiable at a, and h is near 0. Let = fl with f’(;r) = We know that \/ 100 = 10, and from here we will approximate v 100.2 2 \/ 100 + 0.2. Using our formula: 1 «100.- : «100 + (0.2)(9 100) + 0022) = 10.01 + 009). So our linear approximation is just 10.01. We use Taylor’s theorem on the above: 2 h // f(a + h) = 1(a) + ma) 2 (c) where c E (a, a + 11). Our error is therefore |fl;f”(c)| for some 0 that we don’t know. To get an upper bound, we’ll find the largest possible value for |f”| on (100,100.2). For = fl, f”(m) -%:L'_3/2, which is increasing and negative. So an upper bound on |f”| on our interval is |f”(100)| = | — §100-3/2| = @105. The upper bound on the error is then Lgf”(100) 0.22 1 : (2)(4000) ’ 200000: We’ll most happily follow the advice of the hint — otherwise this would be a horrendous mess. cosy : Z;O(—1)jy2j/(2j)l and it’s of course 1 at y = 0. If we let y = $2, then: Z<—1)jm4j+1/<2j>! ' 0 J: flit) = I C08(962) = 56 Z(~1)j($2)2j/(21)! = Now we need to think carefully. We’re expanding in powers of 3:. When evaluating the 71’“ derivative at 0, all terms of order n — 1 or smaller will be wiped out by differentiation. And all terms of order (n + 1) or larger in x will be zero because we’re evaluating at zero and they will all have a factor of x. So we only care about the term of the series with $9, This corresponds to j : 2. Then the derivative at 0 will be 9! (due to the repeated differentiations) times the coefficient of 3:9 : f<9>(0) = (9!)((—1)2)(1/(2(2))!) = 91/41: 362,880/24 = 15,120 Let’s compute the derivative of sin—1(x/2). We’ll need the chain rule. Here’s what we get: 1 (w/1+(a:/2)2 The above is always positive. So f is increasing. Thus it will achieve its absolute min— imum on [-2, —1] at the left endpoint. The minimum value is f(—2) : sin—1(—2/2) : sin—1(—1) : —7r/2. f’(:r) )( )- loll—t f($) (m) If x" is a guess for the root of f(z) then Newton’s Method tells us mn+1 = xn — For sincr =1— x, 2 sins: — 1+ :5. The derivative of f is cosa: + 1. Given $1 = 7r/2 we find 232 : $22$1_ f(:cl) :W/2_ sm(7r/2)—1+7r/2 :W/2_ 1-1+7r/2 :0 f’(a:1) cos(7r/2) + 1 O + 1 (n) f’(z) = 2x — E3; = 2m1 — 355—4. We will find the antiderivative of both sides both sides. Noting that in general, the derivative of act" for constants a and n with respect to a: is naxn'l, we know that the antiderivative of am" is :‘hx'H'l, with the special case that the antiderivative of g is aln When taking antiderivatives we also wind up with an arbitrary constant. When we apply this to finding the antiderivative of f’ (3:) we get: m1+1 x—4+1 — 3 1+ 1 —4 +1 f(1)=3 =>12+1-3+c=3 => C=1andthusf(a:):1:2+;1§+1. f(m)=2 +C=z2+x—3+C. 2. Let T be the temperature of the coffee, To be the ambient temperature, and k be the propor- tionality constant. Then, Newton’s Law of Cooling states: dT — = k — T dt (T 0) Let’s change variables (although we can solve without doing this if we wanted) to the variable 5 : T — To. Then % = 9% and k(T — To) 2 k8. We have a simpler differential equation now: dS _ E; _ for which we know the general solution is S = S(0)ek‘. It remains to find It and So so that we can find the temperature at any time. We’re told it; = % : —1° C/min when T = 70 which is when S = 70 — 20 = 50. Therefore, M50) : —1 => k = —1/50. The coffee at time t = O has temperature T := 95° C so that S = 95 — 20 = 75. Therefore 5(0) 2 75. Therefore, S(t) = 75e-t/50 and by adding 20 to both sides, T(t) = 20 + 7551/50. We’re asked to find when the coffee is at T = 70° C. kS ‘7 -l 70 = 90 + 75e—‘/5° => 50 = 75e—‘/5° => — : «it/50 :> ln( 3 )= 4/50 oolio 2 => t: —501n(§) = 501n(§)min. .4 3. Let V be the volume of the cone, r be its radius, and h be its height. Then the base diameter (which is TWICE the base radius), 27" is equal to the height h. So h : 27" and r : h/2. Let’s look at our formula for volume before we go any farther, V = grzh. If we differentiated that with respect to time, we’d have the derivative of volume, plus derivatives of 7‘ and h. But we know how r and h relate, so we only have to worry about one of them. Which one? The problem is asking us to find how fast the height is increasing at a certain point, so this would suggest eliminating r and replacing it with r = h/2. Then V = %(h/2)2h = 1"—2h3. Now let’s differentiate: VI ll 7r 7r —3h2h' = —h2h’ 12 4 We know V’ = 2 from the problem. We’re asked to find h’ when h = 3. With simple algebra: 8 2: 33214:: h’ = ——. 4 97r 4. (b) (e) f is a rational function and therefore is defined everywhere except when the denominator is zero. The denominator is only zero at a: = 2 so the domain of f is: lR\{2} i.e. all real numbers except 2. We need to find the derivative for that. And we’ll use the quotient rule. Using the same skeleton as in question 1 part c, with f = 2:2 + 12, and g = x — 2, we have: (:r+2)($—2)—(1)(.r2+12):x2—4m—12 : (m-6)(:r+2)‘ / :1} 2 f ( ) (:12 — 2)2 (z — 2)2 (x — 2)2 Notice how we wound up putting it in factored form. This makes the next step easier. We can easily see the derivative is 0 at x = 6 and :1: = —2, and it is undefined at a: = 2. The factors that give roots of zero, (1 — 6) and (1+2) both have multiplicty 1 so the derivative will change sign at these points. However, the singularity term, (z — 2)2 has multiplicity 2, an even number, so the derivative will not change sign as one passes through the singularity. Our intervals of interest are: (—00,—2), (—2,2), (2,6), and (600). If you let x tend to +00 in f’(m) you should be able to verify that it approaches 1 and hence the derivative is positive on (6,00). Then it must be negative on (2,6) and also on (—2,2) and positive again on (—00, —2). : (10—6)(10+2) __ Let’s check that by picking some representative numbers. (10_2)2 Egg : 3/16 > 0. It is positive on (6, f’(3) = (3—_6)(—3+2—) = (—41%(51 = —15 < 0. It is negative on (2,6. (3~2)2 ) f’(0) = W = ((4%?) = «3 < 0. It is negative on (—2,2). f’(—3) : (‘3(‘_§>_‘;f’+2) = = 9/25 > 0. 1: is positive on (—00, —2). The concavity can change when the second derivative is zero or singular. We’re given f”(a:) = which has only a singularity at 2. We can check easily that f” is positive on (2, 00) and therefore concave up, and negative on (—oo,2) and therefore concave down. f/l(5) : 32 (5-2)3 f”(0) = Frizz? 2 Ti? 2 —4 < 0, It is negative on (—00,2). : 33 = 3—3 > 0. It is positive on (2,00). The vertical asymptotes can be found when we have our rational function reduced as much as possible (any common factors in the numerator and denominator canceled out). Fortunately, it already is. Then those vertical asymptotes will occur at the lines as = c where c is a point where the rational function is singular. Therefore we anticipate :r = 2 is a vertical asymptote. Checking, limz_,2+ 1:32 : 7+/0+7 : +00 and limzdg— ii? = ’+/0‘7 = —00. Indeed we have avertical asymptote. Dividing — 2) into (302 + 011," + 12) yields + 2) with remainder 16. Hence our function : (x + 2) + 771%. As :7: goes out to either positive or negative infinity, that term With the 16 will go to zero and we’ll be left with (m + 2). Thus we anticipate y = a: + 2 is our slant asymptote. m2+12 m—2 limzszmo( — (:r + 2)) = limxsioo(;1—_6—2) = 0. So there is certainly a slant asymptote, y=m+2. See sketch provided on the last page. 5. We’ll start off with our constraint that V = 7rr2h = 277r cm3 so that r2h = 27. Now we want to minimize the area, which is the sum of a circle area at the bottom (area WW) and the area of the sides (area 27r7'h — the circumference times the height). So our total area is A : 7W2 + 27rrh. Let’s take the equation we got from volume and solve for either r or h. lt’s nicer if we solve for it because then we don’t need to take a square root: h = 27/r2. If we substitute this into our equation for A : A 2 71'1"2 + 27rr(27/T2) = 71'7"2 + 547T/T To find local maxima or minima we will find when A’(r) is either zero or undefined. 2r3 — 54 2 A'(r)=97rr—547r/r =7r T2 lf A’(r) : 0 then 27‘3 54 — 0 —:> r3 — 27 —> 7' — 3 where we took the real root of the equation for obvious physical reasons (we can’t have a radius with an imaginary component to it). We see that A’(r) is undefined when 1' = 0, because then T2, the denominator, is 0. We want the cylinder to have a nonzero radius so we’ll exclude this solution here. To see that we do have a minimum at r : 37 we could either substite r = 3 into A”(r) = 2n + (3)(54)/r4 which is clearly positive there, from which we conclude we have a minimum; or, we could see that A’ is negative for r < 3 and positive for r > 3 which tells us that we have a minimum. The minimizing dimensions are: r = 3 cm and h = 27/7"2 2 3 cm. . We shall use the definition: m) z m 3 m and we will clear those fractions by multiplying the numerator and denominator by + h) + 5)(z + 5) yielding: lim(m+5)~((m+h)+5) hm —h _hm -1 h—io h((x+h+5)(m+5) _/Hoh((:n+h+5)(as+5) h—»0((x+h)+5)($+5) by canceling the h—s, and now noting how + h) + 5 goes to :r + 5 as h goes to 0 we are left with: , __ 1 f(:c)_ (1+5)2‘ . Okay, we’re told that x = ‘2 at a: = 0. So no matter how our constants work out, if we want f to be continuous, we’d better have that limz_,o flat) 2 2. We’ll start off by claiming that we don’t care about the function on the right of zero because lim$_,0+ 2 + $2 cosh—1) = limxfi0+ 2 + limx_,0+ x2 cos(x‘1)= 2 + 0 = 2. We just need to show limmdo+ :52 cos($“1) = 0. Let’s start by noting that for any real argument, cos will take on a value between 1 and 1, so we can write: ~ 2_ <- 2 —1 < - 2 xgrg+$( 1)_i£n)x cos(x ) migrai+rr(1) But both the leftmost and rightmost limits are 0 and by the Squeeze Theorem, limx_.0+ 3:2 cosh—1) = 0. Taking our attention to the left side of 0 now: lim a$2+bx+c=c=2 => 0:2. (Ii—v0— There are no restrictions on a or b. 8. (a) For a differentiable f, we can write the equation of the tangent line at P(a, b) as y —— b = f ’(a)(-’5 — a)- For the curve y = 1 + 6””, y’ = ex. At the point P(a, b) = P(a,1 + 6“), our tangent line will have equation: y —<1+ea>= ewe — a). If this line passes through the origin, then when we set I = 0 and y = 0 in the equation of the line to get: 0—(1+ea) Zeafo—a) => -1-€a=——aea and we can write that just a little nicer as e“+1—ae“=0. To prove there is only one value of a that satisfies the above equation we will define 9(a) = e“ + 1 — ae“, and show that 9 only has one zero. To show there’s even one solution we’ll use the Intermediate Value Theorem by showing there are a for which 9 is negative, and for which 9 is positive. g(0)=60+1—060=2. Rewriting g as (1 — (1)6“ + 1 it’s clear that as a ——» 00, g —> —00 because that (1 — u)e“ term is dominant. So we know for sure at some point 9 is negative. If you want an exact value of a, just take a : 2 so that 9(2) 2 62 + 1 ~— 262 =1— e2 < O. The point is that g is continuous, so by taking on both a positive and negative value, it must take on every value between, and therefore take on 0. There is at least one solution. To prove uniqueness, we’ll try to Show that g is monotonic (i.e. that g is an increasing or a decreasing function). g’(a) = —e” + (1 — (2)6“ 2 —ae“. This unfortunately changes sign at the origin so 9 isn’t monotonic. However, we do get that g is decreasing on (0, 00), and therefore if we can just show there are no negative roots, we could also arrive at uniqueness (we’d have shown that 9 can only have positive roots, that it has one positive root, but it is monotonic for positive arguments). Observe 9(a) = (1 — (1)6“ + 1 > 16“ + 1 = e“ + l > 0 for a < 0. Thus there are no roots for a < 0. We conclude there is one real a that allows the tangent line to pass through the origin. 12+l’) Figure 1: Graph for problem 4(e): y = f(a:) = 3:2“ ...
View Full Document

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.

Page1 / 9

Math 100 Dec 08 - 2. U! December 2008 Math 100/180 Final...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online