Practice Problems: Exam 2 MATH 230, Spring 2011 Instructor: Dr. Zachary Kilpatrick Show all your work. Simplify as much as possible.

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1 Practice Problems: Exam MATH, Spring Instructor: Dr. Zachary Kilpatrick Show all your work. Simplify as much as possible.. Write down a table of x and y values associated with a few t values. Then, graph the following parametric equations for the given t values, drawing an arrow in the direction of increasing t. Then, find a Cartesian equation of the curve. (a) x = (t ), y = t t, t 4 t x y 7/ 7/ 4 4 /4 y solving x equation for t: t = ± x+ y = ± x+4 ± x+ 4 x (b) x = e t +, y = e t, t t x y / e+ e / e + e / e + e / e 4 + e y x solving x equation for t: t = ln(x ) e t = e (ln(x ))/ = (x ) / = x y = x

2 (c) x = cost, y = sint, t π t x y - π/4 / π/ π/4 / π y x rewriting the x and y equations: x/ = cost, y + = sint (x/) +(y +) = cos t+sin t = ( x ) +(y +) = (d) x = sint, y = 4cost+, t π t x y 5 π/ π π/ 5 π 5 y x rewriting the x and y equations: (x+)/ = sint, ( ) ( ) x+ y + = (y )/4 = cost 4

3 . Find the general equations for dy/dx and d y/dx. Then, find where the curve s tangent is horizontal and where it is vertical in the given domain. Use this information to graph the curve. (a) x = cost, y = sintcost, t π dy dx = t sin t cos sint = sin t cos t sint = cos(t) sint d y dx = ( d cos(t) dt sint sint ) = sin(t)sint cos(t)cost sin t curve s tangent is horizontal where cos(t) = so where t = π/,π/,5π/,7π/ or t = π/4,π/4,5π/4,7π/4, the points (/, /); ( /, /); ( /,/); (/, /) curve s tangent is vertical where sint = so where t =,π,π, the points (,); (-,); (,).5 y x

4 (b) x = cost+cost, y = sint sint, t π dy dx = cost cos(t) sint sin(t) = cos(t) cost sin(t)+sint = (cost )(cost+) (cost+)sint = cost sint d y dx = sin t cos t+cost sin t sint sin(t) = ( cost)/sin t (sin(t)+sint) = /(+cost) (sin(t)+sint) = (+cost)(sin(t)+sint) curve s tangent may be horizontal where cost = so where t =,π, but notice the denominator of the slope is here too. thus, we must use L Hospital s rule for t = : cost lim t sint sint = lim = t cost and the same is true for t = π. thus, the point (,) has a horizontal tangent curve s tangent may be vertical where sint = but recall it is horizontal at t = and t = π. thus, we have it is vertical only at the point (,) y x 4

5 (c) x = 4t t, y = t 4 t, t 4 dy dx = 4t 4t = t t 4 t t d y = ((t )( t)+(t t))/( t) dx 4 t = t +6t +t +t t ( t) = t +6t ( t) curve s tangent may be horizontal where t = t so where t =,. the denominator is not zero at any of these places. thus, the points (,) and (, ) have horizontal tangent curve s tangent may be vertical where t = so at the point (4,8) 5 y 5 4 x 5

6 (d) x = e t t, y = t, t dy dx = e t d y dx = e t (e t ) = curve s tangent is not horizontal anywhere since curve s tangent is vertical where e t = so where t = ln. so the points ( ln, ln ) 6 4 y 4 5 x 6

7 . Find the area bounded by the given parametric curves. (a) x = cost+cost, y = sint sint curve is periodic, when starting at t =, repeats itself starting at t = π, so these are limits using f(t) = cost+cost, f (t) = sint sint, and g(t) = sint sint so integral is π (sint sint)( sint sint)dt = = = π π sin t+sintsint sin (t)dt ( cost)+ (cost cost) ( cos4t)dt [ t sint+ sint 6 sint+ 8 sin4t ] π = π (b) x = t, y = t +t, x axis the curve intersects the x-axis when y =, so when t +t = or (t )(t ) = so t =,. thus we integrate between these limits using f(t) = t, f (t) = t, and g(t) = t +t so integral is [ t ( t +t )(t )dt = t 4 t +t 5 dt = 5 t4 4 + t ( = ) ] = 7

8 4. Find the length of the curve. (a) x = e t t, y = 4e t, 4 t 4 x (t) = e t y (t) = 4e t (x (t)) +(y (t)) = (e t ) +(4e t ) = 4e 4t 8e t +4+6e t = 4e 4t +8e t +4 = 4(e t +) L = = (e t +) dt 4 e t +dt [ e t = +t ] 4 4 = e 8 e 8 +6 (b) x = cost cost, y = sint sint, t π x (t) = sint+sin(t), y (t) = cost cost (x (t)) +(y (t)) = sin (t) sintsin(t)+sin t+cos (t) costcos(t)+cos t = (sint(sin(t)cos(t))+cost( sin t)) = cost = 4sin (t/) L = π π 4sin (t/)dt = sin(t/)dt = [ cos(t/)] π = 8 8

9 5. For the following polar equations: (i) find an associated Cartesian equation; (ii) calculate the slope of the tangent line to the curve for general θ; (iii) find the points where horizontal and vertical tangents exist; and (iv) sketch the curve. (a) r = sinθ multiplying through by r, r = rsinθ, which in Cartesian coordinates is x +y = y we complete the square to find x +(y +) = dy dx = cosθsinθ sinθcosθ cos θ+sin θ cosθsinθ = cos θ sin θ = (sinθ)/ = tanθ cosθ thus, horizontal tangents occur where sinθ = so θ =,π/,π, at the points (,); (,π/); (,π). vertical tangents occur where cosθ = or θ = π/4,π/4, at the points (,π/4) and (,π/4)

10 (b) r = tanθ replacing r = x +y and tanθ = y/x: x +y = y x x +y = y y = ± x x x (x )y = x 4 y = x4 x dy dx = sec θsinθ +tanθcosθ sec θcosθ tanθsinθ = sinθ(+sec θ) secθ( sin θ) = sinθ sin θ cos θ thus horizontal tangents occur where sinθ = sin θ so where θ =,π,π, at the points (,) and (,π). vertical tangents occur where cos θ =, so where θ = π/,π/, at the points (,π/) and (,π/)

11 (c) r = cosθ replacing r = x +y and cosθ = x/ x +y, we have x x +y = x +y x +y +x x +y = (x +y +x) x +y = (x +y x) = x +y dy dx = sinθ(sinθ)+( cosθ)cosθ sinθcosθ +(cosθ )sinθ = cosθ cosθ sinθ sinθ thus horizontal tangents occur where cosθ = cosθ so where θ =,π/,4π/, at the points (,), (/, π/), and (/, 4π/). vertical tangents occur where sinθ = sinθ so where θ = π/,π,5π/, at the points (/, π/), (, π), and (/,5π/)

12 6. Find the area of the region bounded by the given curve in the specified sector. (a) r = sin(θ), π θ π A = π/ π (sinθ) dθ = 4 π/ π ( cos4θ)dθ = 4 [ θ sin4θ ] π/ = π 4 π 8 (b) r = +cos(θ), π θ π A = = π/ π/(+cosθ) dθ = π/ π/ π/ π/+cosθ+ (+cos4θ)dθ = +cosθ+cos θdθ [ θ +cosθ+ ] π/ 8 cos4θ = π π/ 4 (c) r = θ+e θ, θ π π π A = (θ+e θ ) dθ = θ +θe θ +e θ dθ = = [ π +eπ (π )++ eπ [ θ +eθ (θ )+ eθ ] = π 6 +(π )eπ + eπ ] π (d) r = θ sinθ, π θ π π/ π/ A = π/(θ sinθ) dθ = θ θsinθ+sin θdθ π/ = [ θ sinθ+θcosθ+ θ cosθ ] π/ = [ π 4 π/ 4+ π ] = π 4 + π 4

13 7. Find the area of the specified region. (a) inside top lobe of r = sin θ and inside r = 4 firstfindintersectionpoints: /4 = sin θiswheresinθ = ± /orwhereθ = π/,π/,4π/,5π/. thoseinthetoplobeareθ = π/,π/since thisisabovethex-axis. noticesin θ > /4when θ (π/,π/) and sin θ < /4 when θ (,π/) and θ (π/,π). thus, we integrate from to π, making r = /4 the radius when θ (π/,π/) and r = sin θ otherwise: A = = 8 π/ π/ π/ (sin θ) dθ+ π/ π/ ( cosθ) dθ+ (/4) dθ+ [ ] π/ 9θ + 6 π/ 8 π π/ π π/ (sin θ) dθ ( cosθ) dθ π = ( cosθ+(+cos4θ)/)dθ+ π = [ ] π/ θ sin4θ sinθ+ + [ θ sin4θ sinθ [ = π 8 9 ] [ + π ] + π 6 = 7π 9 64 π/ ] π ( cosθ+(+cos4θ)/)dθ π/ + π (b) outside right lobe of r = cos(θ) and inside r = cosθ notice, the right lobe of the curve r = cosθ begins at θ = π/4 and ends at θ = π/4. the circle r = cosθ totally encompasses this lobe, so subtract the area of the lobe from the area of the circle. note circle goes from θ = π/ to θ = π/ π/ π/4 π/ A = π/cos θdθ π/4cos θdθ = 4 π/+cosθdθ 4 = [ θ + sinθ ] π/ [ θ+ sin4θ ] π/4 = π 4 π/ 4 4 π/4 8 π/4 π/4 +cos4θdθ

14 (c) inside r = sinθ and inside r = sinθ intersection points at sinθ = sinθ or sinθ = or sinθ = / so θ = π/6,5π/6 or the points (/, π/6) and (/,5π/6). circle r = sinθ starts at θ = and ends at θ = π and is the innermost curve when θ (,π/6) and θ (5π/6,π). otherwise, when θ (π/6,5π/6), r = sinθ is innermost. so we integrate A = π/6 π/6 sin θdθ+ 5π/6 π/6 5π/6 ( sinθ) dθ+ π 5π/6 sin θdθ = cosθdθ+ sinθ+ 4 π/6 ( cosθ)dθ+ 4 [ = π π ] [ 6 + π ] + = π 4 4 π 5π/6 cosθdθ 4

15 8. For each of the given sequences a n : (i) list the first six terms; (ii) determine whether it converges or diverges; and (iii) find the limit if it converges. Then for the associated series an : (i) list the five terms of the partial sum; (ii) determine whether the series converges or diverges; and (iii) find the sum if it is convergent. (a) a n = n+ a = /, a = /, a = /4, a 4 = /5, a 5 = /6, a 6 = /7 lim n = sequence converges n+ for the series a n, partial sums are: s = a = s = a +a = 5 6 s = a +a +a = s 4 = a +a +a +a 4 = 77 6 s 5 = a +a +a +a 4 +a 5 = 9 note we can bound the series from below with n+ > x+ dx = ( ) lim ln x+ ln = x n= so the series diverges (b) a n = n +( ) n n a = /,a = 5/9,a = 7/7,a 4 = 7/8,a 5 = /4,a 6 = a 6 = 65/79 ( ) n +( ) n n ( ) n lim = lim + = + = sequence converges n n n partial sums of the series are: s = /,s = 8/9,s = /7,s 4 = /8,s 5 = 6/4 since the series is the sum of two geometric series, we can compute its sum as n +( ) n ( ) n ( ) n = + = / n / + / +/ = /4 = 7 4 n= n= n= 5

16 (c) a n = ncos(nπ)+n n+ a =,a = 4/,a =,a 4 = 8/5,a 5 =,a 6 = /7 notice that cosnπ = ( ) n, which does not converge as n, we can rewrite a n = (n+)( )n ( ) n +n n+ lim a n = lim( ) n + n n which does not converge so the whole sequence diverges. partial sums of the associated series are: s =,s = 4/,s = 4/,s 4 = 44/5,s 5 = 44/5 since the sequence diverges, the series also diverges. (d) a n = sin(nπ/) n + n n+ n / a = +, a = /( ), a = /+6/( ), a 4 = 4 5/7, a 5 = /5+5 6/( 5 5), a 6 = 6 7/( 6 6) using the squeeze theorem, notice sinnπ/, so so lim n n = lim n n sin(nπ/) = so lim n n +/n lim a n = + lim = = n n /n / so the sequence converges to. partial sums are: s = + s = + + s = s 4 = s 5 = = note, since the original sequence converge to (not zero), the series must diverge 6

17 (e) a n = n(n 5) (n+) a = 4/,a = /4,a = 6/69,a 4 = /49,a 5 =,a 6 = /8 taking the limit of the sequence lim n n 5n n +n+ = lim n 5/n +/n+/n = = sequence converges partialsumsare: s = 4/,s = 7/94,s = 5497/49776,s 4 = 459/4484,s 5 = 459/4484 note, since the original sequence converge to (not zero), the series must diverge (f) a n = ( ) n a =,a =,a =,a 4 =,a 5 = the sequence has no limit, since it alternates from to endlessly, so it diverges partial sums are: s =,s =,s =,s 4 =,s 5 = the original sequence diverges, so the series diverges (g) a n = n +n n = to a = /4,a = /,a 4 = /6,a 5 = /8,a 6 = /4,a 7 = /8 the limit of the sequence is given lim n n +n = series converges partial sums: s = /4,s = /,s 4 = 7/6,s 5 = 9/5,s 6 = 5/68 we can compute the series using a telescoping sum s n = n i= n i +i = i i+ i= = n n+ + n n+ = + + n n+ n+ = 6 n +6n+ n(n+)(n+) lim s n = lim n n 6 n +6n+ n(n+)(n+) = 6 so the series converges 7

18 n (h) a n = ln n + a = ln/,a = ln/5,a = ln/,a 4 = ln4/7,a 5 = ln5/6,a 6 = ln6/7 the limit of the argument of the ln function is given so lim n n n + = lim ln n n n + = partial sums are given sequence diverges s = ln s = ln 5 s = ln 5 s 4 = ln 85 s 5 = ln 44 and since the original sequence diverges, so does the series 8

19 9. Use either the Comparison Test or the Integral Test to determine whether or not the series is convergent or divergent. (a) this is a series, which can be bounded with the Integral Test as [ + n/ x/dx = + ] = + x = n= so the series converges (b) n= n n + since n + n, we can bound the series from below by Comparison Test n= n n + n n = n= since the series diverges for p n p n= n= n = so series diverges (c) n= n n n n > n so using Comparison Test n= n n n < n= n n = n n= = / = series diverges n / (d) n= n+e n note e n > n for n thus e n +e n > n+e n so using Comparison Test n= n+e < n e n +e = n e so series converges n= /(e) = en /e = e (e ) n= 9