Integrated Calculus II Exam 2 Solutions 3/28/3

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1 Integrated Calculus II Exam 2 Solutions /28/ Question 1 Solve the following differential equation, with the initial condition y() = 2: dy = (y 1)2 t 2. Plot the solution and discuss its behavior as a function of time. This equation is separable. We separate, writing it as an equation of differentials: dy (y 1) = 2 t2. Now we integrate both sides: dy (y 1) = 2 t 2, (y 1) 1 = t + C. Now we put in the initial conditions that y = 2, when t =, determine C and solve for y: (2 1) 1 = + C, C = 1, (y 1) 1 = t + 1, y 1) = t, y = t = 2 + t 1 + t. We have y = t2 (1+t ) 2 and y = 6t(1 2t ) (1+2t ), which gives a critical point at (, 2), no local maxima or minima and inflection points at (, 2) and at (2 1, 5 ).

2 The solution curve is monotonically decreasing in t. As t, the solution curve approaches the horizontal asymptote y = 1, from above. As t 1 +, the solution curve approaches the vertical asymptote x = 1, with y going to infinity. The curve is concave up in the intervals ( 1, ] and [2 1, ). The curve is concave down in the interval [, 2 1 ]. Question 2 A spring of rest length cm is stretched to a length of 5cm by a force of 5 Newtons. How much work is done in stretching the spring from a length of 4cm to a length of one meter? Let x denote the extension of the spring in meters from its rest length. By Hooke s law, if a force F extends the spring an amount x, then we have F = kx, where k is the spring constant. The spring is stretched by a force of F = 5 Newtons a distance x = 5 =.2 meters. So we have: 5 = k(.2), or k = 5 = 25 in units of 1.2 Newtons per meter. So Hooke s law for the spring now reads: F = 25x. Then the work W Joules done in stretching the spring from a length of 4cm to a length of 1 meter, which is the work done in streching the spring from an extension of x = (4 ) =.1m to an extension of (1 ) =.7m is: 1 1 W =.7.1 F dx = xdx = [125x 2 ].7.1 = 125(.49.1) = =

3 Question A region R is bounded by the curves y = x 2 6x + 5 and y = 2x 2. Sketch the region R. The region R is bounded below by the concave up parabola, which has vertex at (, 4) and crosses the x-axis at (1, ) and (5, and above by the straight line of slope 2 through the point (1, ). The two curves meet where both equations hold, so where x 2 6x + 5 = 2x 2, or x 2 8x + 7 =, or (x 1)(x 7) =, so at the points (1, ) and (7, 12). Find the area of the region R. We use a vertical strip at position x of wih dx and height h = 2x 2 (x 2 6x + 5) = 8x x 2 7. So the strip area is da = hdx = (8x x 2 7)dx. To cover the region R, x ranges over the interval [1, 7]. Then the total area A of R is: A = x=7 x=1 da = 7 1 (8x x 2 7)dx = [4x 2 x 7x]7 1 = 4(49 1) 1 (4 1) 7(7 1) = = 6. If the region R is rotated about the y-axis, what is the volume of the solid it traces out? We rotate the vertical strip about the y-axis, giving a cylindrical shell of wih dx, height h = 8x x 2 7 and radius x, so it has volume: dv = 2πxdA = 2πx(8x x 2 7)dx = 2π(8x 2 x 7x)dx. To cover the region R, x ranges over the interval [1, 7]. Then the total volume V of the solid is: V = 2π 7 1 (8x 2 x 7x)dx = 2π[ 8x x4 4 7x2 2 ]7 1 = 2π(7 )( ) 2π( ) = 2π(4)( 1 12 )(2 21 6) 2π( 1 )(2 42) 12 = π 6 (4(5)+1) = π 6 (1715+1) = π (1728) = 288π =

4 Question 4 A circuit has connected in series: A resistance of 1Ω. An inductance of 2H. A voltage source 5t Volts at time t seconds. Given that initially there is no current flowing, find the current at time t. The circuit will burn up when there are 2 Amperes flowing in it. Approximately when will that be? The differential equation governing the current I Amps at time t seconds is: H di + RI = E. Here H Henry is the inductance, E Volts is the voltage and R Ohms is the resistance. Putting in the numbers, the differential equation is: 2 di + 1I = 5t. Dividing by 2, the differential equation is put in standard linear form: di + 5I = 25t. We have P (t) = 5 and Q(t) = 25t, so the integrating factor is J(t) = e R P (t) = e R 5 = e 5t. The equation is now: d (IJ) = QJ, or: d (IJ) = 25te5t, Integrating both sides, we get: IJ = 25te 5t. 4

5 We calculate the integral on the right-hand side by integration by parts: 25te 5t = udv, u = t, du =, dv = 25e 5t, v = Ie 5t = 25te 5t = = 5te 5t dv = udv = uv 5e 5t = 5te 5t e 5t + C = e 5t (5t 1) + C. 25e 5t = 5e 5t, vdu Putting in the initial conditions, I =, when t =, we get: = 1 + C, C = 1. Ie 5t = e 5t (5t 1) + 1. Multiplying both sides of the equation by e 5t gives the solution for the current as: I = 5t 1 + e 5t. As t increases, the exponential term falls rapidly until it is very close to zero and the term 5t 1 dominates. In particular, we see that I is approximately 2, when 5t 1 = 2, so t = 21 = 4.2 seconds. 5 In fact if we use Maple to solve the equation 5t 1+e 5t = 2, numerically, we get: t =

6 Question 5 Consider the integral J = 5 ln(1 + x)dx. Determine the trapezoidal rule approximation, T 1 to J with ten intervals. The data for T 1, with f(x) = ln(1 + x) and the weights w are as follows: x f ln( ) ln(2) ln( 5) ln() ln( 7) ln(4) ln( 9) ln(5) ln( 11) ln(6) w wf 2 ln( ) 2 ln(2) 2 ln( 5) 2 ln() 2 ln( 7) 2 ln(4) 2 ln( 9) 2 ln(5) 2 ln( 11) ln(6) We add the weighted data and multiply by one half of the interval wih 1 2 to give T 1 : T 1 = 1 ( 4 ln ( 9 ) 4 )(4)(25 4 )(9)(49 4 )(16)(81 4 )(25)(121 4 )(6) = 1 ( ) 9 4 ln = 1 4 ln( ) 8 = Is the estimate T 1 too high or too low? Explain your answer. For f(x) = ln(1 + x), we have f (x) = 1 = (1 + 1+x x) 1 and f (x) = (1 + x) 2 <, so the graph of the function f is everywhere concave down. So the trapezoids of the trapezoidal rule lie below the curve, so T 1 is an underestimate of the true integral. Determine the Simpson s rule approximation, S 1 to J with ten intervals. The data for S 1, with f(x) = ln(1 + x) and the weights w are as follows: x f ln( ) ln(2) ln( 5) ln() ln( 7) ln(4) ln( 9) ln(5) ln( 11) ln(6) w wf 4 ln( ) 2 ln(2) 4 ln( 5) 2 ln() 4 ln( 7) 2 ln(4) 4 ln( 9) 2 ln(5) 4 ln( 11) ln(6) 6

7 We add the weighted data and multiply by one third of the interval wih 1 2 to give S 1 : S 1 = 1 ( 6 ln ( 81 ) 16 )(4)( )(9)( )(16)( )(25)( )(6) = 1 6 ln ( ) = 1 ( ln 8192 = Use appropriate error formulas to estimate the errors in T 1 and S 1. The error formula E T for the trapezoidal rule is: E T = K 2(b a) 12n 2. The error formula E S for Simpson s rule is: E S = K 4(b a) 5 18n 4. Here n is the number of intervals, [a, b] is the interval I of integration, K 2 is the maximum of f (x) on the interval I and K 4 is the maximum of f (x) on the interval I. For the given integral approximations, we have n = 1, a = and b = 5. We have f (x) = (1 + x) 2, so f (x) = (1 + x) 2, which decreases on the interval [, 5], with its maximum value of 1 at x =. So K 2 = 1. Also, we have f (x) = 2(1 + x), so f (x) = 6(1 + x) 4 and f (x) = 6(1 + x) 4, which decreases on the interval [, 5], with its maximum value of 6 at x =. So K 4 = 6. So we get: 1(5 ) E T = 12(1 2 ) = = Also we get: E S = 6(5 )5 18(1 4 ) = 1 (2 5 ) = )

8 Question 6 A storage tank filled with water has a V -shaped cross-section. The tank is 8 meters long. The cross-section is an isosceles triangle with one vertex at a depth of 2 meters and the other two level with the surface of the water and meters apart. How much work is done in removing the water from the tank, through a pipe level with the top of the tank? We consider the work dw done in removing a horizontal layer of water of thickness dx, at a height of x above the bottom of the tank. The length of the layer is 8 meters. If its wih is y, then by similar triangles, we have: Then we have: y x = 2 = 2, y = x 2. dw = (2 x)dw, where dw is the weight of the layer, since 2 x is the distance moved by the layer against gravity. dw = ρgdv, where ρ is the density of water, g is the acceleration due to gravity and dv is the volume of the layer. dv = x (8)(dx) = 12xdx. 2 So dw = (2 x)ρg(12)xdx = k(2x x 2 )dx, k = 12ρg. Then the total work W Joules done in removing the water from the tank is: W = x=2 x= dw = 2 k(2x x 2 )dx = k(1x 2 x ]2 = k(4 8 ) = k 4 = 12ρg 4 = 16ρg = 16(9.81) = = (1 9 ). 8

9 Question 7 A lamina of density kilos/square meter is bounded by the curves y = 6 x 2 and y = x (units are SI). Sketch the lamina. The two curves bounding the lamina are parabolas, one concave up and the other concave down. They meet where both equations hold at once, so where 6 x 2 = x 2 + 1, or 2x 2 = 5, so where x 2 = 25 and x = ±5. Then the meeting points are (±5, 5). Find the centroid of the lamina. We see that the lamina is symmetrical about the y-axis and about the line y = 5. So the centroid is at (, 5). Find the moments of the lamina about the x and y-axes. The moment about the y-axis is zero, since the center of mass lies on the axis. The moment about the x-axis is 5m = 5ρA = 5()A = 15A, where m is the mass of the lamina, ρ is its density and A is its area. We have: A = 2 5 = 2 (6 x 2 (x 2 + 1))dx 5 (5 2x 2 )dx = 2[5x 2x ]5 = 2(25 25 ) = 1. Then the mass is m = ( 1 ) = 1 kilos and the moment about the x-axis is 5(1) = 5 kilo meters. Find the volume of the surface of revolution generated by rotating the lamina around the x-axis. By Pappus Theorem, we have for the volume V : V = 2πxA = 2π5 1 = 7π 9 =

10 Question 8 Consider the following parametrized curve: x = t t, y = t 2. Sketch the part AB of the curve, between the points A = (, ) and B = (2, ) on the curve. The point A has y = t 2 =, so t =. The point B has y = t 2 =, so t 2 = 1, so t = ±1. When t = 1, we get x = 2, which is incorrect for B. When t = 1, we get x = 2, which gives B correctly. So the t interval is [, 1]. If we plot the curve for this interval we find a simple curve, concave up, stretching from A to B, starting with slope zero at A and turning to vertical as it reaches B. Find the length along the curve from A to B. We have: X = [t t, t 2 ], V = dx = [ t2, 6t] = [1 t 2, 2t], ds = V = (1 t 2 ) 2 + (2t) 2 = 1 2t 2 + t 4 + 4t 2 = 1 + 2t 2 + t 4 = (1 + t 2 ). So the length s along the curve from A to B is: s = t=1 t= ds = 1 ds = 1 = [t + t ] 1 = + 1 = 4. (1 + t 2 ) 1

11 Write an integral for the surface area of the curved surface, obtained by rotating the curve AB about the x-axis. At position t we draw a vertical frustum of a cone with slant wih ds and radius y. The frustum has surface area ds = 2πyds. So the total required area S is: B S = 2π yds = 2π A 1 = 18π y ds = 2π 1 1 (t 2 + t 4 ) (t 2 )((1 + t 2 )) = 18π[ t + t5 5 ]1 = 18π( ) = 144π 15 = 48π 5 = Write an integral for the surface area of the curved surface, obtained by rotating the curve AB about the y-axis. At position t we draw a horizontal frustum of a cone with slant wih ds and radius x. The frustum has surface area ds = 2πxds. So the total required area S is: B S = 2π xds = 2π A = 6π 1 1 = 6π 1 x ds = 2π 1 (t + t t t 5 ) (t + 2t t 5 ) (t t )((1 + t 2 )) = 6π[ t2 2 + t4 2 t6 6 ]1 = π(9 + 1) = 11π =