18.01SC Exam 3 Solutions

1. Compute the area between the curves x = y 2 − 4y and x = 2y − y 2 . Let f (y) = y 2 − 4y = y(y − 4). f (y) = 0 when y = 0 or y = 4. Let g(y) = 2y − y 2 = y(2 − y). g(y) = 0 when y = 0 or y = 2. x 2 3 y ? The graphs of f and g intersect at (0, 0) and one other point. Find that point: f (y) y − 4y 2y 2 − 6y 2y(y − 3) 2 = = = = g(y) 2y − y 2 0 0 The graphs intersect at y = 0 and at y = 3. When y = 3, f (y) = −3 so the second point of intersection is (3, −3). (Check this by finding g(3).) Over the interval between intersections of the graphs, g(y) > f (y). The distance between graphs is: g(y) − f (y) = (2y − y 2 ) − (y 2 − 4y) = 6y − 2y 2 . The area between graphs is: � �3 � 3 2 3 2 2 6y − 2y dy = 3y − y 3 0 0 � � 2 3 2 = 3·3 − ·3 −0 3 = 27 − 18 = 9. 2. Find the volume of the solid obtained by revolving the region bounded by the curves y = ex , y = 2, and x = 0 about the line y = −1. You only need to give a definite integral expressing the volume. Do not solve the integral. y 2 ln 2 x � Washer method: 0 ln 2 π(32 − (1 + ex )2 ) dx y 2 1 ln 2 x � Shell method: 2 2π(y + 1) ln y dy 1 3. Evaluate each of the following expressions (a) lim n→∞ n � � i=1 3 1+i· n �2 3 n Strategy: interpret this as a Riemann sum and find its value by integrating. Consider the interval [0, 3] cut into n parts. Consider the function f (x) = (1 + x)2 . The right Riemann Sum is: �2 n � � 3 3 1+i n n i=1 �2 � 3 n � � 3 3 So lim 1+i = (1 + x)2 dx n→∞ n n 0 i=1 � 3 = 1 + 2x + x2 dx 0 �3 1 3 = x+x + x 3 0 = (3 + 9 + 9) − 0 = 21. � 2 (b) The value f (4) for the continuous function f satisfying � x sin πx = x2 f (t) dt 0 Strategy: apply the fundamental theorem of calculus. � x2 d d (x sin πx) = f (t) dt dx dx 0 ⇒ sin πx + πx cos πx = f (x2 ) · 2x 1 1 ⇒ f (x2 ) = sin πx + πx cos πx 2x 2x 1 π ⇒ f (4) = f (22 ) = sin 2π + cos 2π 2·2 2 π = 2 4. (a) Find the centroid (i.e. center of mass) of a right triangle with height h and base r (assuming the triangle has uniform density). For a plane figure with uniform density, the coordinates of the center of mass are given by weighted averages, where the weighting function is the moment of inertia: � �� � xf (x) dx yg(y) dy � , � . f (x) dx g(y) dy x h r y Note that the hypotenuse of the triangle lies on a line with equation y = h − hr x. You may know from the homework that the center of mass lies at the centroid (h/3, r/3). If not, you will need to calculate the x and y coordinates of the center of mass separately. The formula for the x coordinate of the center of mass looks something like: � xf (x) dx � . f (x) dx In this case, the numerator of this expression is: � � r � � r h h x h − x dx = hx − x2 dx r r 0 �0 �r h 2 h 3 = x − x 2 3r � �0 h 2 h 3 = r − r −0 2 3r h 2 = r 6 1 The denominator is just the area of the triangle: rh. So the x coordinate 2 of the center of mass is: 2 r hr /6 = . hr/2 3 For the y coordinate, we note that the hypotenuse lies on the line with equa­ tion x = r − hr y and so the numerator will be: � 0 h � h � r � r y r − y dy = ry − y 2 dy h h 0 �r r �h = y2 − y3 3h �0 �2r r 2 h − h3 − 0 = 2 3h r 2 h = 6 Dividing by the area of the triangle, we find that the y coordinate of the center of mass is: h rh2 /6 = . hr/2 3 The centroid of right triangle with height h and base r shown in the figure � the � r h above lies at , . 3 3 (b) Pappus’ Theorem says that the volume of the solid formed by rotating a region is the area of the region times the distance traveled by the rotating centroid. Use Pappus’ Theorem and your answer in the previous part to find the volume of a cone with height h and base radius r. We can form a cone with height h and base radius r by rotating the triangle above 1 about the y axis. The area of the rotated region is rh. The centroid lies distance 2 r/3 from the y axis, so it travels a distance of 2πr/3 as it is rotated. Hence, by Pappus’ Theorem, the volume of the cone is: 1 r πr2 h rh · 2π = . 2 3 3 5. Given a definite integral � b f (x) dx, a let Tn be the trapezoid approximation with n intervals, Mn the midpoint approx­ imation using n intervals, and S2n the Simpson’s rule approximation using 2n intervals. Prove that 1 2 Tn + Mn = S2n . 3 3 We divide the interval [a, b] into 2n intervals. Let x0 = a, x2n = b, xi = a + (b − a)i . 2n Then: Tn b−a = 2n Mn = S2n � f (x0 ) + f (x2n ) + 2 � f (x2i ) i=1 � n b−a � n n−1 � � f (x2i−1 ) i=1 b−a = (f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + 2f (x4 ) + · · · + 4f (x2n−1 ) + x2n ) 6n � � n n−1 � � b−a = f (x0 ) + f (x2n ) + 4 f (x2i−1 ) + 2 f (x2i ) 6n i=1 i=1 Therefore: 1 2 1 4 Tn + M n = Tn + Mn 3 3 3 �6 � n−1 n � � b−a = f (x0 ) + f (x2n ) + 2 f (x2i ) + 4 f (x2i−1 ) 6n i=1 i=1 = S2n 6. A tank contains 1000 L of brine (that is, salt water) with 15 kg of dissolved salt. Pure water enters the top of the tank at a constant rate of 10 L / min. The solution is thoroughly mixed and drains from the bottom of the tank at the same rate so that the volume of liquid in the tank is constant. (a) Find a differential equation expressing the rate at which salt leaves the tank. Let s(t) = amount of salt in kg. at time t. Then ds s(t) kg s(t) = −10 L/min · =− kg/min. dt 1000 L 100 (b) Solve this differential equation to find an expression for the amount of salt (in kg) in the mixture at time t. Use separation of variables: ds 1 =− dt. s(t) 100 Then integrate: ln(s(t)) = − 1 t + c. 100 To get rid of the logarithm, exponentiate both sides, letting k = ec : ln(s(t)) = − 1 t+c 100 1 eln(s(t)) = e− 100 t+c 1 s(t) = e− 100 t ec 1 s(t) = ke− 100 t We know s(0) = 15, so k = 15. Hence: 1 s(t) = 15e− 100 t . (c) How long does it take for the total amount of salt in the brine to be reduced by half its original amount? (Recall ln 2 ≈ .693.) We need: 1 e− 100 t = 1 2 1 1 ln(e− 100 t ) = ln( ) 2 1 − t = − ln 2 100 t = 100 · ln 2 ≈ 69.3 minutes.