Calculus Problem Sheet Prof Paul Sutcliffe. 2. State the domain and range of each of the following functions

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1 f( f( f( Calculus Problem Sheet Prof Paul Sutcliffe. By applying the vertical line test, or otherwise, determine whether each of the following equations gives a function y( (a + (y = (b y = + (c + y = (d + y = (e y = 4 Applying the vertical line test yields gives the following answers to whether the equation determines a function: (a no, (b yes, (c no, (d yes, (e no.. State the domain and range of each of the following functions (a f( = 7 (b f( = 5 (c f( = 3 (d f( = /( + 4 (e f( = /( 6 (f f( = e /( 4 (a om f = R, Ran f = [ 7, (b om f = [,, Ran f = (, 5] (c om f = (, 3] [ 3,, Ran f = [, (d om f = R, Ran f = [, (e om f = R\{±4}, Ran f = R (f om f = R\{±}, Ran f = (, e /4 ] (, 3. Let f( = + a +, where a R. erive the allowed values of a such that R, f( = f(. f = f iff f so the required condition is equivalent to R, f(. By completing the square f( = ( + a + 4 a hence a 4 giving a [, ]. 4. Graph the following functions (a f( = + 3 (b f( = e (+ (c f( = 4

2 9 h( g( f( For [, 4π], on the same drawing, graph the following three functions h( = sin, g( = + sin, f( = /( + sin. 6. For [ π, π], on the same drawing, graph the following three functions h( = cos, g( = cos(, f( = / cos(. 4 3 h( g( f( For [, ], on the same drawing, graph the following three functions h( = e, g( = e, f( = /(e 5 4 h( g( f( Are the following functions even, odd or neither? Justify your answers. (a f( = ( ( (b f( = n k= k+ (c f( = sin( (d f( = ( + cos (a f( = 3 +, so f( = Since f( f( and f( f( this function is neither even nor odd. (b f( = n k= ( k+ k+ = n k= k+ = f( hence this function is odd. (c f( = sin(( = sin( = f( hence this function is even. (d is odd, but both + and cos are even, hence f( is the product of one odd function and two even functions and is therefore an odd function.

3 9. Are the following functions even, odd or neither? Justify your answers. (a f( = e (b g( = tan (c h( = e log (d k( = log (e p( = ( 3 + /( 3 (f q( = sin (4 (g r( = 4 sin (a f( = e. Since f( f( and f( f( this function is neither even nor odd. (b g( = sin( / cos( = sin / cos = tan = g( hence this function is odd. (c h( = e log( = e log = h( hence this function is odd. (d k( = log = log = k( hence this function is even. (e 3 + is an odd function and so is 3 hence p( is the ratio of two odd functions and is therefore even. (f q( = sin ( 4 = (sin( 4 = ( sin(4 = sin (4 hence this function is even. (g r( = ( 4 sin( = + 4 sin. Since r( r( and r( r( this function is neither even nor odd.. If f : R R is an even function and g : R R is an odd function then determine whether the following functions are even, odd or neither? Justify your answers. { f( if > (a f ( = f( if < (b f ( = f( + f( (c f 3 ( = (g f( (d f 4 ( = (f g( (e f 5 ( = (g g( (a On R\{} { f( if > f ( = f( if < is odd. = { f( if < f( if > = f ( hence this function (b f ( = f( + f( = f( + f( = f ( hence this function is even. (c f 3 ( = (g f( = g(f( = g(f( = (g f( = f 3 ( hence this function is even. (d f 4 ( = (f g( = f(g( = f( g( = f(g( = (f g( = f 4 ( hence this function is even. (e f 5 ( = (g g( = g(g( = g( g( = g(g( = (g g( = f 5 ( hence this function is odd.

4 . Write each of the following functions as the sum of an even function f even ( and an odd function f odd (. (a f( = (b f( = e 3 (c f( = log + (d f( = 3 /( (a By inspection f even ( = 4 and f odd ( = 3 5. (b f even ( = (f( + f( = (e3 + e 3 and f odd ( = (f( f( = (e3 e 3 (c f even ( = (f(+f( = (log + +log and f odd( = (f( f( = (log + log (d Since f( = 3 /( = f( is an odd function f even ( = and f odd ( = f( = 3 /(. In the following cases write a formula for the functions f g and g f and find the domain and range of each of them. (a f( = + and g( = /. (b f( = and g( =. (a (f g( = f(g( = f(/ = Ran (f g = [, \{ }. +. om (f g = (, ] (, (g f( = g(f( = g( + = / + om (g f = (,, Ran (g f = (,. (b (f g( = f(g( = f( = ( = + om (f g = [,, Ran (f g = [,. (g f( = g(f( = g( = = om (g f = R, Ran (g f = (, ]. 3. Given f( = and g( = /( +, find (a (f g( (b (f f( (c (g f( (d (g g( (a (f g( = f(/3 = /3 (b (f f( = f( = (c (g f( = g( = / (d (g g( = g(/3 = 3/4

5 4. Given u( = 3, v( = 4 and f( = /, find (a (u (v f( (b (v (u f( (c (f (v u( (d (v (f u( (a (u (v f( = u(v(/ = u(/ 4 = / 4 3 (b (v (u f( = v(u(/ = v(/ 3 = (/ 3 4 (c (f (v u( = f(v( 3 = f(( 3 4 = /( 3 4 (d (v (f u( = v(f( 3 = v(/( 3 = /( For each f( given below, find the inverse function f ( and identify its domain and range. (a f( = 5 (b f( = 3 + (c f( = /, > (d f( = 4, (e f( = 7 (f f( = / 3, Write y = f ( and use f(y =. (a f(y = y 5 = hence y = 5 = f (. om f = Ran f = R and Ran f = om f = R. (b f(y = y 3 + = hence y = ( 3 = f (. om f = Ran f = R and Ran f = om f = R. (c f(y = /y = hence y = / = f (. om f = Ran f = (, and Ran f = om f = (,. (d f(y = y 4 = hence y = 4 = f (. om f = Ran f = [, and Ran f = om f = [,. (e f(y = y 7 = hence y = + 7 = f (. om f = Ran f = R and Ran f = om f = R. (f f(y = /y 3 = hence y = 3 = f (. om f = Ran f = R\{} and Ran f = om f = R\{}. 6. Which of the following functions are injective? Find the inverses of those which are and specify the domain of the inverse. (a f( = ( on R (b f( = ( on [, ] (c f( = ( on [, ] (d f( = ( /( + on R\{ } (e f( = + on [, ] (f f( = + on [, ]

6 (a. It is injective. Apply horizontal line test or f( = f( iff ( = ( iff ( + 3 = ( + 3 iff =. Write y = f ( and use f(y =. So f(y = (+3y 3 = hence y = 3 ( 3 = f (. om f = Ran f = R. (b. It is injective. Apply horizontal line test. Write y = f ( and use f(y =. So f(y = (y = hence y = + = f (. om f = Ran f = [, ]. (c. It is not injective. Apply horizontal line test or eg. f( = = f(. (d. It is injective. Apply horizontal line test. Write y = f ( and use f(y =. So f(y = (y /(y + = hence y = ( + /( = f (. om f = Ran f = R\{}. (e It is injective. Apply horizontal line test or f( = + = ( + so [, ] is only on one side of the turning point = of the quadratic. Write y = f ( and use f(y =. So f(y = (y + = hence y = + = f (. om f = Ran f = [, ]. (f It is not injective. Apply horizontal line test or eg. f( = = f(. 7. Complete the following tables. g( f( (f g( g( f( (f g( / + g( f( (f g( g( f( (f g( / For, define the following si functions f ( =, f ( =, f 3( =, f 4 ( =, f 5( =, f 6( =. These have the property that the composition of any two of these functions is again one of these functions. Complete the following table f f f 3 f 4 f 5 f 6 f f f 4 f 3 f 4 f 5 f 6

7 f f f 3 f 4 f 5 f 6 f f f f 3 f 4 f 5 f 6 f f f f 4 f 3 f 6 f 5 f 3 f 3 f 5 f f 6 f f 4 f 4 f 4 f 6 f f 5 f f 3 f 5 f 5 f 3 f 6 f f 4 f f 6 f 6 f 4 f 5 f f 3 f 9. State any vertical and horizontal asymptotes of the following functions (a f( = /( 9 (b f( = 3/( + (c f( = 3 /( + (d f( = ( ( (a Vertical asymptotes = ±3, horizontal asymptote y =. (b Horizontal asymptote y =. (c No asymptotes. (d Vertical asymptote =, horizontal asymptote y = 3/7.. Write each of the following rational functions as the sum of a polynomial and a proper rational function (a , (b , (c , (d (a = (b = (c = (d = Evaluate each of the following epressions (without using a calculator (a log e (b log 3 9 (c log 4 6 (d log (e log 5 65 (f log n, (g n Z log(ne m log n+log e m, n, m > (a. (b. (c. (d 3. (e 4. (f n. log(ne log n+ (g m log n+log e = m m log n+m = m.. emonstrate graphically that, R. Prove this inequality (the graphical demonstration should provide a hint.

8 y Figure : The red curve is the graph of. This is obtained by shifting the graph of vertically down by unit and then reflecting in the -ais any portion of the curve below this ais. The green curve is the graph of. This is obtained by shifting the graph of to the right by unit. The inequality corresponds to the fact that the red curve is never above the green curve. efine f( =. We need to show that f(, R. The graph suggests we consider three regions. If then f( = =. If then f( = ( = + ( + =. If then f( = ( = + ( =. Together these three regions cover R hence we have shown that f(, R. 3. Consider the given graph of the function f(. Are the following statements true or false? (a lim f( eists, (b lim f( =, (c lim f( = (d lim f( =, (e lim f( =, (f lim a f( eists a (,. (a true, (b true, (c false, (d false, (e false, (f true.

9 y 4. Consider the given graph of the function f(. Are the following statements true or false? (a lim f( does not eist, (b lim f( =, (c lim f( does not eist, (d lim a f( eists a (, (e lim a f( eists a (, 3. (a false, (b true, (c true, (d true, (e true. 5. If f( > a and lim a f( = L, can we conclude that L >? Justify your answer. No. An eample is provided by f( = with a = so that L = which is not positive. 6. Justify whether the following statement is true or false. If lim a f( eists then so does lim a f(. False. An eample is provided by f( =, with a =. Here lim f( eists (and is equal to but f( is not a real function. 7. Calculate the following limits (a lim (, (b lim 3, (c lim π/ sin, (d lim π cos π. (a, (b, (c π/, (d π. 8. Calculate the following limits (a lim +, (b lim 4 3, (e lim 4 +3, (f lim 4. (c lim , (d lim 9 3 9, (alim + (+( + = lim ( = lim = 3. (b lim 4 3 = lim ( +(+( ( ( ++ (c lim = lim (d lim (e lim ( ( ++4 (+( ( +4 = lim = lim 9 ( 3( +3 ( 9( = lim ( ( +3+ = lim ( +(+ ++ = 4/ (+( +4 = 3/8. 9 = lim 9 ( 9( +3 = lim ( +3 ( +3+ = lim ( ( +3 4 = 4. (f lim 4 4 = lim 4 (4 (+ ( (+ = lim 4(( + = = /6.

10 9. Calculate the limit as of the following (a cos, (b cos, (c tan, (d cos, (e cos 4. (a lim cos = lim ( cos = lim u ( cos u {( cos ( cos (+cos (b lim = lim = lim (+cos u =. sin { tan sin (c lim = lim cos = lim sin cos cos = lim +cos } = /. ( sin cos ( (d lim cos = lim (+cos ( cos (+cos = lim ( (+cos 4 sin = /. (e lim cos 4 = lim (+cos 4 ( cos 4(+cos 4 = lim (4 (+cos 4 6 sin 4 = /8. cos } =. 3. oes lim sin(+ eist? If so, find it. For >, sin(+ = sin. Hence lim sin(+ sin sin + = lim + = lim + For <, sin(+ =. Hence lim sin(+ =. The left-sided and right-sided limits eist but are not equal, hence the limit does not eist. 3. Calculate lim π/ {( π/ tan }. Set u = π/ then lim π/ ( π/ tan = lim u u tan(u+π/ = lim u u sin(u+π/ cos(u+π/ = lim u u cos u sin u =. 3. In each case either evaluate the limit or state that no limit eists (a lim , (b lim 3 + 3, (c lim 3 ( + 3, (d lim 3 ( + ( 3, (e lim h /h +/h, (a no limit eists, (b lim (f lim h +/h +/h. = lim 3 (+4( 3 3 = 7. (c lim 3 ( + 3 = lim 3 (+4 ( 3 3 = lim 3 ( + 4 ( 3 =. (d ( + ( 3 (e lim h /h +/h = (+4( 3 ( 3 = +4 3 = lim h h h + =. (f lim h +/h +/h = lim h h +h h + =. hence no limit eists. 33. Calculate the limit as of the following (a 6+7, (b ( + ( (+ 3 (+ 3, (c +sin. (a lim 6+7 = lim 6+ 7 = 3. ( (b lim + ( 4 = lim (+ 3 ( = lim 4 = 4/ (c First note that sin. sin As lim = then by the pinching theorem lim =. Thus lim +sin = lim + sin =. =.

11 y Calculate the following limits (a lim sin, (b lim cos(/ +(/, (c lim ( /, (d lim (3 + cos(/, (e lim {( 3 cos(/( + sin(/}. Set u = / in each case (a lim sin = lim u sin u u =. cos(/ (b lim +(/ = lim u cos u +u =. (c lim ( / = lim u u u = lim u ep(log u u = lim u ep(u log u = e =. (d lim (3 + cos(/ = lim u (3 + u cos u = 3. (e lim ( 3 cos(/( + sin(/ = lim u (3u cos u( + sin u =. 35. For each of the following statements, either give a proof that it is true or a counter eample to show that it is false: (a If g( > > and lim (f( g( = then lim (f(/g( =. (b If g( > > and lim (f(/g( = then lim (f( g( =. (a False. A counter eample is provided by f( = / and g( = /. (b False. A counter eample is provided by f( = and g( = In each case either evaluate the limit or state that no limit eists (a lim u 5 u 5 u, (b lim y (y 8 /3, (c lim ( ( cos 3, (d lim t 5 t 5 t 5, (e lim , (f lim , (g lim t 5t 3 +8t 3t 6t 4, (h lim 3 tan(( 3 3, (i lim , (j lim + 4, (k lim t t +t t, (l lim t t3 + t +. (a 5/, (b, (c, (d /, (e 3/, (f 3/5, (g 8/3, (h, (i /, (j /, (k 3/, (l no limit eists. 37. Sketch the graph of the function f( and classify any discontinuities, where if < f( = if = / if > Removable discontinuity at =.

12 y y Sketch the graph of the function f( and classify any discontinuities, where { ( 3/( 9 if ±3 f( = /6 if = ±3 Infinite discontinuity at = Sketch the graph of the function f( and classify any discontinuities, where if < f( = 3 if if > No discontinuities. 4. Use the limit definition of the derivative to calculate the derivative of the following functions (a f( = sin, (b f( =, (c f( = cos. (a f sin(+h sin ( = lim h h cos h = sin lim h h (b f ( = lim h (+h +h h (+h = lim 3 3 h ( h (+h +h+ (c f ( = lim h cos (+h cos h = lim h sin cos h+cos sin h sin h sin h + cos lim h h = sin + cos = cos. ( ( = lim h = lim h 3 +3h+h (+h +h h ( (+h +h+ (+h +h+ = 3 = 3 = lim h (cos cos h sin sin h cos h = lim h sin cos h+cos sin h sin h (+h +h+ cos = lim (cos h sin cos sin h cos h+sin sin h h h (sin = lim cos sin h sin cos sin h cos h (sin h h = lim cos ( cos(h sin cos sin(h h h = (sin cos + ( sin cos = sin cos..

13 4. Find the Cartesian equation for the tangent to the graph of f : R R : 5 4 at the point (,. f( = 5 4, hence f ( = 4, with f ( = 6. The Cartesian equation for the tangent at (, is y = f( + f (( = + 6( = Find the slope of the straight line which passes through the point (, and is also tangential to the graph of f( = at some point. Let = a > be the point at which the line is tangential to the graph of f( =. As f ( = /( then the tangent line is given by y = f(a + f (a( a = a + ( a. For (, to be on this point requires that a = a + a ( a, with solution a =. The slope is then f ( = /(. 43. Show that if g( is continuous at = then g( tan is differentiable at =. f(h f( Let f( = g( tan then we need to show that lim h h g(h tan h g( tan g(h tan h h = lim h h = lim h h lim h f(h f( = ( lim h g(h ( lim h cos h ( lim h sin h h = g((( = g( where we have made use of the continuity of g( and / cos at =. Hence f( = g( tan is differentiable at = with f ( = g(. eists. 44. Calculate the following derivatives y + y, (a d dy (b d da (a + b + c, (c d dt ( t, t + (d d d a, where a >, (a d dy (e d da a, where is a rational number. y + y = y+ y +y, (b d da (a + b + c =, (d d d a = d d elog a = d d e log a = log a e log a = log a a, (c d dt ( t t + (e d da a = a. = 4t (t Let f : R R be a differentiable function that satisfies f( + e f( sin f( = and has a continuous derivative. Find f( and f (. Evaluating the given equation at = yields f( + sin f( =, that is f( = sin f(. The only solution of this equation is f( =. ifferentiating the given equation with respect to gives f ( + (f( + f (e f( f ( cos =, and after setting = this becomes f ( f ( =, that is, f ( =. 46. Eplicitly write out the Leibniz rule for d4 d 4 (f(g( and use this to calculate the fourth derivative of 4 cos. d 4 d 4 (f(g( = f (4 (g(+4f (g (+6f (g (+4f (g (+f(g (4 (. f( = 4, f ( = 4 3, f ( =, f ( = 4, f (4 ( = 4. g( = cos, g ( = sin, g ( = cos, g ( = sin, g (4 ( = cos. d 4 d 4 ( 4 cos = ( cos + ( sin.

14 47. Given f( = and g( = /(4 +, find (f g( and (g f(. Calculate (f g ( and (g f (. (f g( = f(/(4 + = 4 (4+ + 3, (g f( = g(4 + 3 = (4+(4+3. f ( = 4 and g ( = g ( =. 4 (4+ 3, hence (f g ( = f (g(g ( = 6 (4+ 3, giving (f (g f ( = g (f(f ( = 6(4+3 (4+(4+3 3, giving (g f ( = 48/3 3 = 48/ Use L Hopital s rule to calculate the limit as of the following (a cos, (b cos, (c tan, (d cos, (e cos 4. (a f( = cos(, g( =, are differentiable and satisfy f( = g( =. f ( = sin(, f ( =, g ( =. lim f(/g( = lim f (/g ( = f (/g ( = / =. (b f( = cos, g( =, are twice differentiable and satisfy f( = g( =. f ( = sin, f ( =, g ( =, g ( =. f ( = cos, f ( =, g ( =. lim f(/g( = lim f (/g ( = lim f (/g ( = f (/g ( = /. (c f( = tan(, g( =, are differentiable and satisfy f( = g( =. f ( = sec (, f ( =, g ( =. lim f(/g( = lim f (/g ( = f (/g ( = / =. (d f( =, g( = cos(, are twice differentiable and satisfy f( = g( =. f ( =, f ( =, g ( = sin(, g ( =. f ( =, g ( = 4 cos( for sufficiently close to =. Also, g ( = 4. lim f(/g( = lim f (/g ( = lim f (/g ( = f (/g ( = 4 =. (e f( =, g( = cos(4, are twice differentiable and satisfy f( = g( =. f ( =, f ( =, g ( = 4 sin(4, g ( =. f ( =, g ( = 6 cos(4 for sufficiently close to =. Also, g ( = 6. lim f(/g( = lim f (/g ( = lim f (/g ( = f (/g ( = Find the radius of a sphere at which the rate of change of its surface area with respect to its radius is cm. Use length units of cm and area units of cm. If the sphere has radius r then its area is A = 4πr. The rate of change is da dr = 8πr. This is equal to when r = /(4π =.8 (to decimal places. Hence the required radius is.8cm. 5. Find an epression for dy in the following cases d (a y 3/ + y =, (b y 4 3/ y =, (c + sin y = y, (d (3y + 7 = 6y, (e + tan(y =, (f cosh + sinh(y =. (a 3 yy + y + y = hence y = ( y/(3 y +. (b y + yy 6 y = hence y = (6 y /(y.

15 (c + y cos y = y + y hence y = (y /(cos y. (d (3y + 7(3y + 3y = 6y hence y = y(3y + 7/( 7 3 y. (e + sec (y(y + y = hence y = (cos (y + y/. (f sinh + cosh(y(y + y = hence y = (y + sinh / cosh(y. 5. Assume that the following equations each define y as a differentiable function of. Calculate dy at the given point d (a 3 y + y =, (,. (b y + y 3 3 =, (,. (c e y + sin(y + y = log, (, log. (a 3 4yy + y + y = at (, y = (, this becomes 3 4y + + y = so y = 4/3. (b y + y + yy 3 = at (, y = (, this becomes y + y 3 = so y =. (c e y +y e y +(y+y cos(y+y = at (, y = (, log this becomes +log +y = so y = log. 5. Find all local maima and minima, and hence the global etreme values, of 4 + in the interval [, ]. Note that f( = 4 + is twice differentiable at all points. f ( = 4(, hence f ( = at =, ±, which are all in [, ]. Also, f ( = 4(3. f ( = 4 <, hence = is a local maimum with f( =. f (± = 8 >, hence = ± are local minima with f(± =. f ( = 4 <, hence = is an endpoint maimum with f( = 9. f ( = 4 >, hence = is an endpoint maimum with f( = 9. The global minimum is therefore and the global maimum is Find the global etreme values of f( = in [, 4]. Note that f( = is differentiable everywhere ecept at = ± 6, although = 6 is outside the interval [, 4]. First restrict to (, 6, then f( = ( with f ( = = 3( + (. Hence f ( = in (, 6 iff =. f( =. Now restrict to ( 6, 4, then f( = ( with f ( = = 3( ( +. Hence f ( in ( 6, 4. The only other possibilities for the global etreme values are = 6 (where f( is not differentiable and the endpoints =, 4. Now f( 6 = 7, f( = 8, f(4 = 8. The global maimum is ma{, 7, 8, 8} = 8. The global minimum is min{, 7, 8, 8} = Find the global etreme values of f( = in [, 4].

16 Note that f( = is differentiable everywhere ecept at = ±. For (,, f( = with f ( = 3 = ( 3( + = iff =. f( = 7 3. For (, 4, then f( = with f ( = 3 + = ( + 3(. Now f( = 3, f( = 3, f(4 = The global maimum is 64 3 and the global minimum is Find the global maimum (if it eists of each of the following (a f( = 4 in [ 3, 4 3 ], (b f( = 4 in [ 3, 4 3 ], (c f( = 4 in [ 3, ], (d f( = 4 in (, ], (e f( = in [, ], (f f( = /( + in, (g f( = cos( /( + in. (a f ( = 4( = in ( 3, 4 3 iff =. f( =, f( 3 = 7 8, f(4 3 = Global maimum is 8. (b f ( = 4( = in ( 3, 4 3 iff =,. f( =, f( =, f( 3 = 7 8, f(4 3 = 3 8. Global maimum is. (c f ( = 4( = in ( 3, iff =,. f( =, f( =, f( 3 = 7 8, f( = 8. Global maimum is 8. (d f ( = 4( < in (, so f( decreases in this interval. As the left-hand point is not contained in the given interval (, ] then there is no global maimum in (, ]. (e f( = is differentiable in (, ecept at =. For (,, f( = ( =, so f ( =. For (,, f( = + ( =, so f ( =. f( =, f( =, f( =, hence the global maimum is. (f For >, f ( = ( /( + = iff =. f( =, f( =, lim f( =, hence the global maimum is. (g For, f ( = cos( + sin( (+ (+ >, thus f( is increasing for. In fact lim f( = lim { cos( /( + } = lim u cos u +u = cos =. There is no global maimum in. 56. A group of Chilean miners are trapped underground at a depth of 3 metres. A rescue team starts at the bottom of an abandoned mine shaft that is 6 metres West of the trapped miners and has a depth of metres. The rescue team must dig a tunnel to the trapped miners that has an initial horizontal segment followed by a segment directly towards the trapped miners. At a depth of metres the rock is soft and it takes only 5 minutes to dig one horizontal metre. However, at any depth below this, the rock is hard and it takes 3 minutes to dig a distance of one metre. Calculate the minimal number of hours it takes to tunnel to the trapped miners. Use length units of metres and time units of minutes. Let the horizontal tunnel have a length 6, where [, 6]. Then the distance from the end of the horizontal tunnel to the trapped miners is + (. The time taken is T ( = 5( (.

17 dt d = dt, therefore +( d = iff 5( + 4 = 69, ie. 6 = 44 giving = / = 5/3. At this value T (5/3 = ( (5 5 + ( = 5 3 (55 + (33 = 5 3 ( = 5 3 (34 = 54. Check the endpoints: T ( = 56 and T (6 = 6 > 54. Thus the minimal time is T = 54 minutes ie. 54/6 = 9 hours. 57. Let F ( = π t sin t dt. Calculate F (π, F ( and F (π/. 58. Let F (π =, F ( = sin and F (π/ = π/. F ( = 3 + e t dt. Find all critical points of F ( and determine whether they are local minima, maima or points of inflection. Prove that F (3 > F (3. F ( = 4/(3 + e hence F ( = iff =. For <, F ( > whereas for >, F ( <, hence = is a local maimum. As F ( < for > then F ( is strictly monotonic decreasing in (, hence F (3 < F ( Calculate the derivative of F ( where (a F ( = (c F ( = 3 cos t cos t dt, (b F ( = t dt, (d F ( = (t sin t dt, t + t 4 dt (a F ( = 3 5 cos( 3, (b F ( = ( sin (, (c F ( = sin sin, (d F ( = 6. Calculate the derivative of the following functions (a G(t = t 4 t u du, (b H( = sin( (a G (t = t (4t 3 t t = 4t 5 t t, = + (+. 3t dt, (c y( = (b H ( = 3 sin ( cos( 3( = 6 sin ( cos( 4, (c y ( = ( ( = The function f(t is defined in terms of a constant k by k if t f(t = k if < t if t > Calculate F ( = f(t dt at =, =, =, and for all >. t dt

18 y F ( = f(t dt =. F ( = f(t dt = k dt = k. F ( = f(t dt = k dt + k dt = k + k = k. If > then F ( = f(t dt = k dt + k dt + 6. Sketch the graph of the function f( = { + if if < 3. dt = k + k + = 3k. Calculate the function F ( = f(t dt, with om F = [, 3], and sketch its graph. Eamine the continuity and differentiability properties of both f( and F ( at =. For, F ( = [ ] (t t + t dt = t = f( F( For < 3, F ( = (t + t dt + t dt = [ t ] = = 6. lim f( = lim ( + = and lim + f( = lim + =. Hence f( is continuous at = since lim f( = lim + f(. lim h +(f( + h f(/h = lim h +(( + h /h = and lim h (f( + h f(/h = lim h (( + h + + h /h = 3. As these two limits are not equal then lim h (f( + h f(/h does not eist. Therefore, f( is not differentiable at =. F ( is differentiable at = since F ( = f( is continuous at =. F ( is continuous at = since it is differentiable at =. 63. Evaluate the following definite integrals (a 4 ( d, (b 6 t t 3 dt, (c (a Put u = + 3 then 4 ( + 34 d = u4 du = (b 6 t t 3 dt = (6t 3 t dt = (c Put q = tan u then dq = sec u du and [ 8 + 4q dq. ] u 5 5 = 5. [ 3t + t ] = = dq = [ ] π/4 π/4 +4q 4 du = 4u = π.

19 64. Calcuate the following indefinite integrals (a d, (b cot d, + 3 (c + d. (a Put u = + 3 then du = 6 d d = +3 6 u / du = 3 u/ + c = c. (b Put u = sin then du = cos d cot d = cos sin d = du u = log u + c = log sin + c. (c Put u = then du = d + d = u +u du = ( +u du = u log + u + c = log + + c. 65. Calculate the following integrals (a + d, (b cos d, (c n log d, where n is a positive integer, (d e sin(3 d, (e e sinh d. (a + d = 3 ( + 3/ 3 ( + 3/ d = 3 ( + 3/ 4 5 ( + 5/ + c (b cos d = sin sin d = sin + cos cos d = sin + cos sin + c. (c n log d = log n+ n+ n+ n+ d = log n+ n+ n+ (n+ + c. (d e sin(3 d = 3 e cos(3 + 3 e cos(3 d = 3 e cos(3 + 9 e sin(3 9 e sin(3 d and hence e sin(3 d = e ( 3 cos(3 + sin(3 + c. (e e sinh d = e (e e = ( e d = + 4 e + c. 66. For integer n define F n ( = n+ 3 + d. Find a recurrence relation between F n ( and F n 3 ( and hence calculate F 3 ( and F 6 (. F n ( = n+ 3 + d = 3 n 3 + n 3 = 3 n 3 + n 3 (F n( + F n 3 (. Hence (3 + nf n ( = n 3 + nf n 3 (. n 3 + d = 3 n 3 + n 3 From the first line of this solution we have that F ( = c. Setting n = 3 in the recurrence relation gives 9F 3 ( = F ( hence F 3 ( = 9 ( c 3. Setting n = 6 in the recurrence relation gives 5F 6 ( = F 3 ( hence F 6 ( = 45 ( c 6. n d

20 67. For integer n define I n = π/4 cos n+ d. Find a recurrence relation between I n and I n and hence evaluate I and I 4. I n = π/4 cos n+ d = π/4 cos n cos d = [cos n sin ] π/4 + π/4 n cos n sin d = n+ + n π/4 (cos n ( cos d = n+ + n(i n I n hence I n = ( n+ + ni n /(n +. From above I = / therefore I = ( 3 + I /3 = 5/(6 and I 4 = ( 5 + 4I /5 = 43/( Write the following in partial fraction form (a 7 ( (, (b 7 ( + (, (c 8 ( ( + ( +. 7 (a ( ( = A + B + + C A = 7 ( + ( =, B = 7 = ( ( =, C = 7 = ( ( + =. =/ 7 ( + ( = A (b + + A = 7 ( =, = B + C ( C = 7 + =, = Evaluating at = gives 7 4 = B + 4 hence B =. 8 (c ( ( + ( + = A + B ( + C E + 8 B = 8 ( + ( + =, C = = ( ( + =, = Evaluating at = gives E = A + 5. Evaluating at = gives 3A + = 7. Evaluating at = gives A + 3 = 3. Hence =, A = 3, E =. 8 ( ( + ( + = 3 + ( Calculate the following integrals + 4 (a d, (b + ( d, (c d.

21 (a + 4 = = d = d = log + log + c. + ( d = (b + ( = d = log + log + log + + c. + (c = A + B + C E +. = C+B+(A+C +(B+E 3 +(A+ 4 hence C =, B =, A =, E =, = d = d = log + + log( + + c. 7. Evaluate the following integrals (a π/4 π/4 e sec cosh d, (b π/3 π/3 ( + 4 cos d, (c d, (d + esin d. + etan (a (b (c (d π/4 In each case replace the integrand by its even part π/4 π/3 π/3 e sec cosh d = π/4 (e + e sec π/4 [ ] π/4 d = sec d = tan =. π/4 cosh π/4 π/4 ( + 4 cos π/ d = ( cos π/3 + e sin d = + e tan d = d = + e sin + + e sin d = + e tan + + e tan d = [ ] = d = 6 3 = Solve the following differential equations for y( π/3 π/3 cos d = [ ] π/3 sin = 3. π/3 + e sin + e sin + e sin + e sin d = ( + e tan + e tan + e tan + e tan (a yy = + y, (b y + y = e, (c ( y + 6d + (log dy =, (d (y + y 3 dy = log d, (e ( + yy = ( y, (f y = ( 3 y/, (g 4yy = + 3y, (h y y =, (i y + y(ye / + =. y (a + y dy = d, log( + y = log + log c, y = ± c. (b I = ep d = e, y = e e e d = e ( e + c = e + ce. (c M = y + 6, N = log, M y = = N, hence eact. d d =.

22 g = y +6, g = y log +3 +φ(y, g y = log +φ = log, φ =, φ = y. g = y log + 3 y = c, y = c 3 log. (d (y + y 3 dy = log d. To integrate the right-hand-side put u = log with du = d. y + 4 y4 = u du = u + c = (log + c. (e y = f(, y = ( y/( + y, f(t, yt = f(, y hence homogeneous. Put y = v, y = v+v = ( v/(+v = ( v/(+v, v = ( v v /(+v, + v v + v dv = d, log(v + v = log + log c, v + v = c/, y / + y/ = c/, y + y = c. (f y + y/ =, I = ep d = ep( log =, y = 4 d = ( c = c (g y = f(, y = ( + 3y /(4y,. f(t, yt = f(, y hence homogeneous. Put y = v, y = v + v = ( + 3y /(4y = ( + 3v /(4v, v = ( v /(4v, 4v v dv = d, log(v = log + log c, (v = c /, y = c 3/, y = ± + c 3/. (h y / dy = / d, 3 y3/ = + 3 c, y = (3 / + c /3. (i Bernoulli with n = so put v = y n = /y with y = v /v ( + v v e/ + =, v v = e/, I = ep d = e /. v v v = e / e / e / d = e/ ( + c, y = /v = e / /( + c. 7. Solve the following initial value problems for y( (a ( + ye y/ d e y/ dy =, with y( =, (b y + y =, with y( =, (c (e y y cos(yd + (e y cos(y + ydy =, with y( =, (d y log + y = log, with y(e =. (a y = f(, y = +yey/, f(t, yt = f(, y hence homogeneous. e y/ Put y = v, y = v + v = ( + ve v /(e v = e v + v, v = e v, e v dv = d, ev = log + c, v = log(log + c, y = v = log(log + c, y( = = log c, hence c = giving y = log(log +. (b I = ep d = e, y = e e d = e ( e + c, y( = = + c, hence c = to give y = ( + e. (c M = e y y cos(y, N = e y cos(y+y, M y = ey cos(y+y sin(y = N g hence eact. = ey y cos(y, g = e y sin(y + φ(y, g y = ey cos(y + φ = e y cos(y + y, φ = y, φ = y.

23 g = e y sin(y + y = c, but y( = so c = and e y sin(y + y =. (d y + y log =, y = log d = log log I = ep d = ep(log(log = log, log, where u = log, du = d, u du = ( log (log +c y(e = = ( + c hence c = so y = log + log. 73. Solve the following initial value problems for (t (a ẋ = 4, with ( =, (b 4ẋ =, with ( =, (c tẋ = t + 3, with ( =, (a d = 4 dt, log = 4t + log c, = ce 4t, ( = = c, = e 4t. (b 4 d = dt, = t + c, but ( = hence c =, and = (c ẋ 3 t = t, I = ep 3 t dt = ep( 3 log t = t 3, + t. = t 3 t dt = t 3 ( t + c = t + ct 3, = t + t 3. ( = = 4 + 8c, hence c = and 74. Solve the following initial value problems for y( (a y + 4 y =, y(π =, y (π =. (b y y + 5y =, y(π/ =, y (π/ =. (a (Char λ + 4 =, λ = ± i, y = A cos(/ + B sin(/, y(π = = B, y = A sin(/ + cos(/, y (π = = A/, A =, y = cos(/ + sin(/. (b (Char λ λ + 5 =, λ = ± i, y = e (A cos( + B sin(, y(π/ = = e π/ A, A =, y = Be (sin( + cos(, y (π/ = = Be π/, B = e π/, y = e π/ sin(. 75. Solve the following boundary value problems for y( (a y 5y + 6y =, y( =, y(/ = e. (b y y + y =, y( = 3, y(π/ =. (a (Char λ 5λ + 6 = = (λ (λ 3, λ =, 3, y = Ae + Be 3, y( = = A + B, y(/ = e = Ae + Be 3/, so = A + B e, giving B =, A =, y = e. (b (Char λ λ + =, λ = ± i, y = e (A cos + B sin, y( = 3 = A, y(π/ = = e π/ B, so B =, giving y = 3e cos.

24 76. Solve the following initial value problems for y( (a y 3y + y = cos, y( =, y ( =, (b y + y 3y = e + e cos, y( =, y ( =, (c y 5y + 4y = 3e 4, y( =, y ( =, (d y + 4y + 4y = e + 3e, y( =, y ( =. (a CF: (Char λ 3λ + = = (λ (λ, λ =,, y CF = Ae + Be. PI: y = c cos +d sin, c cos d sin 3( c sin +d cos +(c cos +d sin = cos, c 3d =, d = 3c, c =, d = 3, y P I = cos 3 sin. y = Ae + Be + cos 3 sin, y( = = A + B +, y = Ae + Be sin 3 cos, y ( = = A + B 3 y = e 3 5 e + cos 3 sin., A =, B = 3 5, (b CF: (Char λ + λ 3 = = (λ (λ + 3, λ =, 3, y CF = Ae + Be 3. PI: y = ae + e (b cos + c sin, gives ae + e ((3b + 4c cos + (3c 4b sin ae + e ((4b + c cos + (4c b sin 3ae + e ( 3b cos 3c sin = e + e cos, a = 4, 6c + 4b =, c = 3b, b = 3, c = 3 6, y P I = 4 e + e ( 3 cos sin, y = Ae + Be 3 4 e + e ( 3 cos sin, y( = = A + B 9 5, y ( = = A 3B + 7 5, A =, B = 9 y = e e 3 4 e + e ( 3 cos sin. (c CF: (Char λ 5λ + 4 = = (λ (λ 4, λ =, 4, y CF = Ae + Be 4. PI: y = ae 4, ae 4 ( ( = 3e 4, a =, y P I = e 4. y = Ae +Be 4 +e 4, y( = = A+B, y ( = = A+4B+, A = 3, B = 3, y = 3 e 3 e4 + e 4. (d CF: (Char λ + 4λ + 4 = = (λ +, λ =, y CF = e (A + B. PI: y = a e + be, ae ( be = e + 3e, a =, b =, y P I = e + e. y = e (A + B e + e, y( = = B +, y ( = = A B + 4, A = 8, B = 3. y = e ( 8 3 e + e. 77. Solve the following pairs of first order differential equations for y(, z( by first finding a second order differential equation for y( (a y = y + 3z, z = 3y + z, (b y + 6y z =, z + 8y 4z =. (a z = 3 (y y, 3 (y y = 3y + 3 (y y, y y + y =, (Char, λ λ + =, λ = ± 3i, y = e (A cos(3 + B sin(3. z = 3 (y y = e (B cos(3 A sin(3. 5,

25 (b z = (y + 6y, (y + 6y + 8y (y + 6y =, y + y 8y =, (Char, λ + λ 8 = = (λ (λ + 4, λ =, 4, y = Ae + Be 4. z = (y + 6y = 4Ae + Be Calculate the second order Taylor polynomial of f( = e sin about = π/. f(π/ = πe, f ( = e sin ( + cos, f (π/ = e, f ( = e sin ( cos sin + cos, P ( = πe + e( π πe 4 ( π. f (π/ = πe, 79. Calculate the first order Taylor polynomial of f( = e sin about = and use this result to show that e sin + 3 e. f( = e sin, f( =, f ( = e sin sin(, f ( =, f ( = e sin ( cos( + sin (, therefore f ( 3e. Hence P ( = and f( = + R ( where R ( 3 e. Using the fact that f(, the above yields the required result e sin + 3 e. 8. Use Taylor polynomials to estimate the following to within. (a e (which is less than, (b sin(.3, (c log(.. (a e = n! n +R n (, where R n ( = (n+! ec n+, for some c (,. e = e / = n! + R n n (, R n( (n+! e n+ (n+! = Q n n Q = 4 >., Q 3 = 9 <. so we can use the third order Taylor polynomial ( e = Note: = , e = (b At =.3 the sine series gives sin(.3 =.3 3! ( ! (.35 7! ( This is a (convergent alternating series and the magnitude of each term is decreasing. ( As 3! (.33 =.45 <. then sin(.3.3. Note: sin(.3 = (c At =. the log series gives log(. = log( +. =. (. + 3 (.3 4 ( This is a (convergent alternating series and the magnitude of each term is decreasing. As (. =. >., 3 (.3 =.6.. < (., then we need the second order Taylor polynomial log(.. (. =.8. Note: log(. = Let P n ( be the n th order Taylor polynomial of the function f( = sin. Find the least integer n for which (a P n ( approimates sin( to within. (b P n ( approimates sin( to within.

26 The Taylor polynomial of sin of order m + is P m+ ( = 3! 3 + 5! ( m (m+! m+. (a 5! = >. and 7! = 54 <. hence n = 5. (b 9 9! =.4.. >. and! =.5.. <. hence n = Compute the Taylor series of f( = 3 ++ about = and comment on the result. f( = 3, f ( = 3 +, f ( = 4, f ( = 6, f ( =, f ( = 6. f (n ( = for n 4. f( = 3 + 4( +! ( + 6 3! ( 3 = 3 + 4( + 6( + ( 3. The Taylor series has a finite number of terms and is just a rearrangement of the original polynomial. 83. Compute the third order Taylor polynomial of f( = log( + about = and give the remainder term in Lagrange form. f( = log(3, f ( = /(+, f ( = /3, f ( = /(+, f ( = /9, f ( = /( + 3, f ( = /7, f (4 ( = 6/( + 4 log( + = log(3 + 3 ( 8 ( + 8 ( 3 + R 3 ( where R 3 ( = ( 4 for some c (,. 4(+c Show that e sin = R ( where R ( 5 6 e 3. f( = e sin, f( =, f ( = e sin cos, f ( =, f ( = e sin ( sin + cos, f ( =, f ( = e sin ( cos 3 cos sin + cos 3. e sin = R ( R ( = 3! esin c ( cos c 3 cos c sin c + cos 3 c e 3. (+ 85. Use Taylor series to calculate lim k k, where k is a positive real number. From the Taylor series ( + k = k + k k + o( hence ( + k k lim k k + o( = lim = k k. 86. Calculate the third order Taylor polynomial of f( = log(+ result to calculate lim f( about = and use this log( + = ( + 3 (3 + o( 3 = o( 3, and ( = o( 3 hence f( = log(+ = o( 3 = o( 3. lim f( = lim o( = lim ( o( 3 ( ( + 3 ( + 3 ( 8 3 = lim 3 + o( 3 ( ( o(( = 6/3. = lim

27 87. Simplify the following epressions involving the positive integer n (a sin 55π + sin ( nπ 43π + cos + cos 3π, (b cos(nπ + cos( nπ, (c n+ cos((n + π + n cos(nπ. (a sin 55π + sin ( nπ 43π 3π + cos + cos = sin 3π + ( ( n + cos 3π + cos π = + ( ( n + = (3 + ( n. (b cos(nπ + cos( nπ = cos(nπ = ( n. (c n+ cos((n + π + n+n+ (n+ n cos(nπ = (n+n cos(nπ = n(n+ ( n. 88. Calculate the Fourier series of f( = { if < < if < and state the value that the series converges to when =. The function is defined on ( L, L with period L where L =. a n = b n = cos(nπ d+ sin(nπ d+ a = = (nπ f( d = cos(nπ d = nπ [ ] cos(nπ sin(nπ d = nπ d + = nπ ( ( n nπ ( n + (nπ Hence the Fourier series of f( is d = 3. [ ] sin(nπ + [ ] sin(nπ nπ nπ = (nπ (( n [ ] cos(nπ nπ [ sin(nπ ] [ ] cos(nπ + nπ = nπ { (nπ (( n cos(nπ } nπ sin(nπ. n= f( has a jump discontinuity at = with lim f( = and lim + f( = hence at = the Fourier series converges to ( + =. 89. Calculate the Fourier series of f( = { if π < < if < π sin(nπ d cos(nπ d and use this to evaluate the series ( n+ n=, n n=, n and n=. (n

28 The function is defined on ( L, L with period L where L = π. a n = π b n = π π π a = f( d = d = π π π π 3. f( cos(n d = π [ ] π cos(n d = π π nπ sin(n nπ = ( [ ] π nπ n cos(n + π cos(n d = n n ( n π f( sin(n d = π π [ sin(n = π n ( n + n π π ] π π sin(n d = nπ n π π [ ] π cos(n + nπ sin(n d = π n ( n + n 3 π = π n ( n+ + n 3 π (( n Hence the Fourier series of f( is π 6 + n= { n ( n cos(n + π π sin(n d [ cos(n ( π n ( n+ + } n 3 π (( n sin(n. f( is continuous at = hence at = the Fourier series converges to f( =, ie. = π 6 + n ( n, n= which gives ( n+ n= n = π. cos(n d f( has a jump discontinuity at = π since lim π f( = π and lim π + f( = hence at = π the Fourier series converges to π. Therefore n= π = π 6 + n (n = n= n= n + which gives ( n+ n= n= n = 9. Calculate the Fourier series of if < < f( = if < if < n = π 6. π 6 + π = π 8 The function is defined on ( L, L with period L where L =. a n = a = f( cos( nπ d = f( d = d =. cos( nπ d = [ sin( nπ ] nπ = nπ sin(nπ Hence a k = and a k+ = (k+π ( k+. As f( is an even function then it has a cosine series (ie. b n = + (n + π ( n+ cos n= ( (n + π. ] π

29 9. Calculate the Fourier series of f( = 3 π over ( π, π and apply Parseval s theorem to calculate n=. n 6 As f( is an odd function we get a sine series with b n = π ( 3 π sin(n d = [ ] π ( 3 π cos(n + π (3 π cos(n d π π πn π πn π = [ ] π πn (3 π sin(n 6 π π πn sin(n d = 6 [ ] π π πn 3 cos(n 6 π π πn 3 cos(n d π By Parseval s theorem π π π( 3 π d = Hence 3 π = n= hence = ( n n 3. ( n sin(n n= n 3 44 n 6 = [ π π 5 + ] π 3 π4 3 = 8π6 π 5 n= n 6 = π Calculate the half range cosine series of the function f( = sin defined on (, π. For the half range cosine series on (, L we work with the even etension of the function on the interval ( L, L, where L = π. a = π π a = π π sin cos( d = π π sin d = π π [ ] π cos = 4 π. sin( d = π π [ ] π cos( =. For n >, a n = sin cos(n d = (sin((n + sin((n d π π = [ π n + cos((n++ ] π n cos((n = ( ( n + ( n + = (( n + π n + n (n. π Hence the half range cosine series of f( is π n= (( n + (n π cos(n = π 4 π 93. Calculate the comple Fourier series of { if < < f( = if < < n= 4n cos(n.

30 The function is defined on ( L, L with period L where L =. c = 4 f( d = 4 For n, c n = 4 = i [ ] e inπ + i nπ nπ Hence c n = and c n+ = d + 4 f(e inπ d = 4 [ e inπ i (n+π i ] n= d =. e inπ d + 4 = i nπ (einπ + e inπ = and the comple Fourier series of f( is (n + π ei(n+π/. e inπ d i nπ (( n. 94. Taylor epand f(, y = quadratic order. y about (, y = (, up to and including terms of f(, =, f = ( y, f (, =, f y = ( y, f y (, =, f = ( y 3, f (, =, f yy = ( y 3, f yy (, =, f y = ( y 3, f y (, =, hence f(, y = + + y + + y + y Taylor epand f(, y = y / 3 about (, y = (, up to and including terms of quadratic order. f(, =, f = 3 4 y, f (, = 3, f y = 3 y, f y (, =, f = 5 y, f (, =, f yy = 3, f yy (, =, f y = 6 4 y, f y (, = 6, hence f(, y = 3( (y + + 6( + 6( (y + + (y Given f(, y = sin y with (t = cos t and y(t = t obtain df as a function of t using dt the chain rule and check the result by epressing f in terms of t and differentiating directly. Using the chain rule or directly df dt = f d dt + f dy y dt = sin y d dt f = cos t sin(t, + cos y dy dt = sin(t sin t + cos t cos(t t. df dt = sin t sin(t + t cos t cos(t. 97. Given f(, y = 4e log y with (u, v = log(u cos v and y(u, v = u sin v, obtain f and u f as functions of u, v using the chain rule and check the result by epressing f in terms v of u, v and performing the partial differentiation directly. Using the chain rule f u = f u + f y y u = 4e 4e log y + sin v = 4 cos v log(u sin v + 4 cos v u y f v = f v + f y y v = 4e log y sin v cos v + 4e y u cos v = 4u sin v log(u sin v + 4u cos v. sin v

31 or directly f = 4u cos v log(u sin v, f f = 4 cos v log(u sin v+4 cos v, u v = 4u sin v log(u sin v+4u cos v. sin v 98. Given z(, y satisfies z + y log z z + 4 =, calculate z at (, y, z = ( 3,,. 99. Calculate ifferentiating the given equation yields z + z + y z z z z =, at (, y, = ( 3,, this gives z = 6. e y ddy, where = [, a] [, b].. Calculate e y ddy = = a cos( + y ddy, a ( b [ e (e b b d = b e y dy d = a [e y ] y=b ] a = eab a. b where = [, π] [, π]. d y=. Calculate cos( + y ddy = π ( π π [ ] y=π cos( + y dy d = sin( + y d y= π [ ] π = (sin( + π sin d = cos( + π + cos = 4. 3 y ddy, where is the triangle with vertices (,, (,, (,.. Calculate y ddy, 3 y ddy = = ( 5 d = 3 y dy d = [ 6 ] =. [ 3 y where is the finite region between the curves y = and y =. ] y= y= d 3. Calculate ( y ddy = y dy d = = 3 e y cosh ddy, [ ( 7 3 d = [ ] y= 3 3 y d y= ] = 7. where = {(, y :, y }.

32 e y cosh ddy = ( (e y+ + e y dy d = [e y+ + e y ] y= d y= = (e e d = 4. Calculate y ddy, y ddy = sinh( d = [ ] cosh( = (cosh 4 cosh. where is the unit disc centred at the origin. ( y dy d = [ y 3 y= ] 3 y= d = 3 ( d = 3 ( + 4 d = [ ] 5 5 = Calculate e y ddy, where is the triangle with vertices (,, (,, (,. = [e y ] =y e y ddy = dy + =y/ = [ e y + e y/ ] + ( y y/ ] =3 y [e y dy = =y/ e y d dy + ( 3 y y/ (e y e y/ dy + [ ] e3 y + e y/ = 3 ( e + e e y d dy (e 3 y e y/ dy Calculate π/ ( π/ sin y y dy d. The given iterated integral can be written as a double integral over the region between the curves y = and y = π/ for π/, hence π/ ( π/ = π/ sin y y [ sin y y dy d = ] =y = dy = sin y y π/ ddy = sin y dy = π/ ( y sin y y [ ] π/ cos y =. d dy 7. Calculate ( 4 + y dy d. 3

33 The given iterated integral can be written as a double integral over the region between the curves y = and y = for, hence ( 4 + y dy d = y ddy = ( y y d dy = [ ] = y 4 + y dy = ( [ ] y dy = y =. = Use polar coordinates to calculate ( + y ddy, where is the intersection of the half plane with the disc of radius 3 centred at the origin. In polar coordinates is the region r 3 and π/ θ 3π/. = 3 ( + y ddy = 9. Use polar coordinates to calculate at the origin. e ( +y ddy = 3 ( 3π/ π/ [ ] θ=3π/ sin θ cos θ r dr = θ=π/ ( π r(cos θ + sin θ dθ rdr 3 e ( +y ddy, e r dθ rdr = π r dr = 3 [ r 3 ] 3 = 8. where is the unit disc centred [ ] ( re r dr = π e r = π. e. Use the change of variables = u sin v and y = u cos v to calculate ddy, +4y where is the region between the two ellipses + 4y = and + 4y = y = u, hence is the region u and v π. (, y J = (u, v = u v = sin v u cos v cos v u sin v = π + 4y ddy = y u [ ] u= 4 u sin v dv = 3 u= 8 y v u sin v u u dudv = π π ( = u. u du sin v dv ( cos v dv = 3 [v ] π 8 sin v = 3π 4.. Use the change of variables = (u+v and y = (v u to calculate cos where is the triangle with (, y vertices (,, (,, (,. ( y +y ddy,

34 In terms of [u, v] coordinates the vertices of the triangle are [, ], [, ], [, ]. The edges of the triangle lie along the lines =, y =, y =, which become the lines v = u, v = u, v =, hence is also a triangle in the uv-plane. cos = J = (, y (u, v = ( y ddy = + y [ ] u=v v sin(u/v u= v u y u v y v = cos(u/v dudv = dv = =. ( v v cos(u/v du dv [ ] v sin( dv = sin( v = sin(