University of Bergen. Solutions to Exam in MAT111 - Calculus 1

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1 Universit of Bergen The Facult of Mathematics and Natural Sciences English Solutions to Eam in MAT - Calculus Wednesda Ma 0, 07, Eercise. a) Find the real numbers and such that the comple variable z = +i solves the sstem z +i = z, z = 4. Draw on the comple plane the set of the points that satisfies to each of the equation in the sstem. Show on the comple plane the solution of the sstem. b)writethecomplenumber( i) 6 intheformre iθ = rcos ( cosθ+ isinθ ) (polar form) and in the form a+ib (cartesian form). c) Find all the comple numbers w satisfing the equation w 4 = 6 in both polar and cartesian forms and show the answer on the comple plain. Solution to a). We substitute z = + i into the first equation and obtain z +i = z +i( +) = ( )+i = ++ =. Substituting z = +i into the second equation, we obtain z = 6 = + = 6. Thus we need to solve the sstem =, + = 6.

2 We deduce =, + = 6, = So we obtain two solutions =, = 6, = =, = ±. =, =, and =, =. The set of points z+i = z is the straight line = and z = 4 is the circle of radius centered at the origin. (, ) 4 + = (, ) = Solution to b). Wewritez = iinthepolarformasz = e i5π 4 and obtain ( i) 6 = ( ( ) 6 e i5π 4 6 = 8e i3π = 8 cos( 3π ) )+sin(3π ) = 8i. Solution to c). We appl the formula of n-th root of z = r(cosθ+ isinθ) and obtain: 4 ( z = 4 r cos ( θ+πk) (θ+πk) ) +isin. 4 4 Thus w 0 = (cos( π ) 4 )+isin(π 4 ) = +i, w = (cos( 3π ) 4 )+isin(3π 4 ) = +i, w = (cos( 5π ) 4 )+isin(5π 4 ) = i, w 3 = (cos( 7π ) 4 )+isin(7π 4 ) = i.

3 3 (, ) (, ) (, ) 0 (, ) Eercise. a) Show that =. + π( )(ln( )) b) Find the it +i+3 or show that the it does not eists. i + Solution to a). Since the eponential function is continuos, we need to show that π + ( )(ln( )) = or + ( )(ln( )) = 0. It is equivalent to show 0+ (ln) = 0. We rewrite the latter epression and use the L Hopital rule (ln) 0+ (ln) = 0+ = 0+ We use the L Hopital rule again and obtain ln 0+ ln = 0+ = 0+ = = ln. Solution to b). We calculate +i+3 i + Eercise 3. ( i)(+3i) = i ( i)(+i) = +3i i +i =.

4 4 a) Find the values of α and β such that the function αcos+, for < 0, f() = βe 3 +α, for 0 is continuos on the interval (, + ). b) For which values of α and β is the function differentiable? Solution to a). We have f(0) = β. We calculate the it of f() as 0 from the left and from the right: αcos+ = α+ and 0 0+ βe3 +α = β. We conclude that the are equal and coincide with f(0) = β for all α+ = β Solution to b). The right derivative of f() is βe 3 +α β 0+ The left derivative is e 3 = β 0+ 3e 3 = β 0+ = 3β. αcos+ (α+) cos sin = α = α = The right and the left derivatives coincide onl for β = 0. Since the differentiable function have to be continuos we conclude that α =. Eercise 4. Find the indefinite integrals a) tan() ln(cos ) d, b) ( +5)e d. Solution to a). Weuse thesubstitution u = cosanddu = sin. Then ln(cos) lnu tan()ln(cos)d = cos ( sin)d = u du. Now we use the substitution v = lnu and dv = du and obtain u lnu u du = vdv = v = (lnu) = ( ) ln(cos).

5 5 The indefinite integral is equal to ( ln(cos) ) +C. Solution to b). We calculate e d = e. To calculate the integral e d we use the method integration-bparts. We write u =, du = d and dv = e, v = e and obtain e d = e e d = e e 4. To calculate the integral e d we use the method integrationb-parts and reduce to the previous integral. First we write u =, du = d and dv = e, v = e. Then e d = e e d = e Summing all the integrals, we obtain ( +5)e d = e ( e 3 Eercise 5. e + e 4. ) e e +C. Consider the function f() = e ( +5+) defined on (,+ ). a) Find maimum and minimum of the function. b) Find the asmptotes for the function. c) Find the inflection points and indicate the intervals, where the function is concave and where the function is conve. d) Draw the graph of the function. Solution to a). Wecalculatef () = e ( +3 4). Thecritical points are given b the zeros of the derivative and equal to = 4 and =. Since f () = e (+4)( ) we conclude that f decreases on (, 4) (,+ ), f increases on ( 4,), and = 4 is the minimal point and = is the maimal point of f(). Solution to b). There is no vertical asmptotes, since the domain ofthedefinitionofthefunctionis(,+ ). Tofindotherasmptotes

6 6 we calculate e ( +5+) e ( +5+) = 0 and =. + We conclude that it could be a horizontal asmptote for +. We calculate + e ( +5+) = 0, and deduce that there is the horizontal asmptote = 0. Solution to c). We calculate f () = e ( + 7) = e ( 9 )( +9 ). The inflection points are = ± 9 and f is concave below (conve) on (, 9 ) ( + 9,+ ), f is concave on ( 9, + 9). Solution to d). Before we draw the graph of the function we note that f() = 0 at = 5 and intersect the vertical ais at the point = Eercise 6. a) Find as a function of b solving the following initial value problem d = 9a, d (a+) (0) = 3 a,

7 7 where a is a positive constant. b) Find (0) and (0) without solving the initial value problem and write the Talor polnomial P () around = 0. c) Find the quadratic approimate value of the solution of the initial value problem () at = a. d)what is error E () at = a? Solution to a). We have d = 9a d (a+). Integrating, we obtain = 8a +a +C = (0) = 8a+C = 9a = C = 9a. The solution to the initial value problem is given b = 8a +a 9a. Solution to b). We have (0) (0) = 9a a = (0) = 9 (0) = 3 a. We differentiate the equation d = 9a and find that ( ) + = d (a+) 8a. It implies (+a) 3 (0) = (0) ( 8 a ( (0) ) ) = (0) = 3 a a. The second order Talor polnomial around = 0 is given b P () = 3 a 3 a + 3 a a. Solution to c). The quadratic approimate value of the function () at = a is given b P (a) = 3 a. Solution to d). The error value is given b E (a) = P (a) (a) = 3 a 0 = 3 a. Eercise 7.

8 8 LetABC bearighttrianglesuchthat AB = AC and BC =. Let the rectangle be inscribed to the triangle ABC in two was as it is shown on the pictures. C C A B A B In both cases find the length of the sides of the rectangles such that the areas are maimal. What is common in both cases and what is the difference? Solution. B using the Pthagorus theorem we find that AB =. The first case. Let us denote the side of the rectangle b, then the other side will be and the area is the function of given b f() = ( ). The maimum is given b the zero of the derivative: f () = = 0 if =. So the rectangle have to have equal sides and it will be the square. The second case. Wedenotebthesideoftherectanglethatlieson the side BC of the triangle. Then the other side is equal to ( ) and the area is given b the function f() = ( ). We find f () = ( ) = 0 if =. Thus the rectangle has the sides and. We see that the maimal areas are equal to, but in the first case the rectangle has equal sides, and in the second case the are different. Eercise 8.

9 Calculate the volume of the bod obtained b the rotation around the -ais of the region inside of the circle + = 5 between the straight line = and the vertical line. On the picture it is the dashed domain. 9 = = 4 Solution. First we need to fund the point of intersection of the circle and the straight line = 5 = = ±. Thus the intersection point that we are interested in is =, =. The volume of the bod is given b the integral ( 5 ) π 4 d + (5 )d = 0π 3 ( 5 ). Thanks for the eam! Irina Markina Eirik Berge