The graphs of f and g intersect at (0, 0) and one other point. Find that point: f(y) = g(y) y 2 4y 2y 2 6y = = 2y y 2. 2y(y 3) = 0

Transcription

1 . Compute the area between the curves x y 4y and x y y. Let f(y) y 4y y(y 4). f(y) when y or y 4. Let g(y) y y y( y). g(y) when y or y. x 3 y? The graphs of f and g intersect at (, ) and one other point. Find that point: f(y) g(y) y 4y y 6y y y y(y 3) The graphs intersect at y 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) (y y ) (y 4y) 6y y. The area between graphs is: 3 [ ] 3 6y y dy 3y y 3 3

2 ( ) Find the volume of the solid obtained by revolving the region bounded by the curves y e x, y, and x about the line y. You only need to give a definite integral expressing the volume. Do not solve the integral. y ln x Washer method: ln π(3 ( + e x ) ) dx

3 y ln x Shell method: π(y + ) ln y dy 3. Evaluate each of the following expressions (a) lim n n ( ) i n n i Strategy: interpret this as a Riemann sum and find its value by integrating. Consider the interval [, 3] cut into n parts. Consider the function f(x) ( + x). The right Riemann Sum is: n ( ) i i n n n ( ) So lim + i ( + x) dx n i n n 3 + x + x dx

4 [ ] 3 3 x + x + x 3 ( ). (b) The value f(4) for the continuous function f satisfying x sin πx x f(t) dt Strategy: apply the fundamental theorem of calculus. x d d (x sin πx) f(t) dt dx dx sin πx + πx cos πx f(x ) x f(x ) x sin πx + x πx cos πx π f(4) f( ) sin π + cos π π 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

5 x h r y Note that the hypotenuse of the triangle lies on a line with equation y h hx. r 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 x dx r r [ ] r h h 3 x x 3r ( ) h h 3 r r 3r h r 6 The denominator is just the area of the triangle: of the center of mass is: hr /6 r. hr/ 3 rh. So the x coordinate

6 For the y coordinate, we note that the hypotenuse lies on the line with equation x r r y and so the numerator will be: h h ( y r r ) h h y dy ry r h y dy [ r y r ] h 3h y 3 ( r ) h r 3h h3 r h 6 Dividing by the area of the triangle, we find that the y coordinate of the center of mass is: rh /6 h. hr/ 3 The centroid of ( the ) right triangle with height h and base r shown in the figure 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 about the y axis. The area of the rotated region is rh. The centroid lies distance r/3 from the y axis, so it travels a distance of πr/3 as it is rotated. Hence, by Pappus Theorem, the volume of the cone is: r πr h rh π Given a definite integral b f(x) dx, a let T n be the trapezoid approximation with n intervals, M n the midpoint approximation using n intervals, and S n the Simpson s rule approximation using n intervals. Prove that T n + M n S n. 3 3 We divide the interval [a, b] into n intervals.

7 (b a)i Let x a, x n b, x i a +. n Then: ( ) n b a T n f(x ) + f(x n ) + f(x i ) n i ( ) n b a M n f(xi ) n i b a S n (f(x ) + 4f(x ) + f(x ) + 4f(x 3 ) + f(x 4 ) + + 4f(x n ) + x n ) 6n ( ) n b a n f(x ) + f(x n ) + 4 f(x i ) + f(x i ) 6n i i Therefore: 4 T n + M n T n + M n ( 6 ) n b a n f(x ) + f(x n ) + f(x i ) + 4 f(x i ) 6n S n i i 6. A tank contains L of brine (that is, salt water) with 5 kg of dissolved salt. Pure water enters the top of the tank at a constant rate of 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) dt L/min L kg/min. (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 s(t) dt.

8 Then integrate: ln(s(t)) t + c. To get rid of the logarithm, exponentiate both sides, letting k e c : ln(s(t)) t + c e ln(s(t)) e t+c s(t) e t e c s(t) ke t We know s() 5, so k 5. Hence: s(t) 5e 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.693.) We need: e t ln(e t ) ln( ) t ln t ln 69.3 minutes.

9 MIT OpenCourseWare 8.SC Single Variable Calculus Fall For information about citing these materials or our Terms of Use, visit: