Final Exam Solved - Calculus for Life Sciences Students | MATH 3B, Exams of Mathematics

Download Final Exam Solved - Calculus for Life Sciences Students | MATH 3B and more Exams Mathematics in PDF only on Docsity! MATH 3B (Butler) Practice for Final (II, Solutions) 1. Let G(x) = ∫ x2 sinx et 2 dt. Find the degree 2 Taylor polynomial about x = 0 for G(x). The degree 2 Taylor polynomial is given by P2(x) = G(0) +G ′(0)x+ G′′(0) 2 x2. We now need to calculate G(0), G′(0) and G′′(0). We have G(0) = ∫ 0 0 et 2 dt = 0, since the upper and lower bounds match. Next, by Leibniz’s rule we have G′(x) = d dx ( ∫ x2 sinx et 2 dt ) = e(x 2)2(2x)− e(sinx)2(cosx) = 2xex4 − cosxesin2 x and so G′(0) = −1. Finally we have G′′(x) = 2ex 4 + 8x4ex 4 + (sinx)esin 2 x − 2 sinx cos2 xesin2 x and so G′′(0) = 2. Putting this altogether we have P2(x) = −x+ x2. 2. For y between 0 and 4 find the area between the curves g(y) = 4y2 + 8y + 23 and h(y) = −2y2 + 26y + 23. First we check for intersections. We have 4y2 + 8y + 23 = −2y2 + 26y + 23 or 6y2 − 18y = 0 or 6y(y − 3) = 0. So they intersect at 0 and 3, so to find the area we need to break it up to finding the area between 0 and 3 and also the area between 3 and 4. By looking at the functions it is easy to check that h(1) ≥ g(1) so that h(y) ≥ g(y) for 0 ≤ y ≤ 3 and g(y) ≥ h(y) for y ≤ 0 and y ≥ 3. So we now have Area = ∫ 3 0 (h(y)− g(y)) dy + ∫ 4 3 (g(y)− h(y)) dy = ∫ 3 0 (18y − 6y2) dy + ∫ 4 3 (6y2 − 18y) dy = (9y2 − 2y3) ∣∣∣∣3 0 + (2y3 − 9y2) ∣∣∣∣4 3 = ((81− 54)− (0− 0)) + ((128− 144)− (54− 81)) = 38. 5. (a) Use substition to show that for a function f(x) that ∫ a 0 f(x) f(x) + f(a− x) dx = ∫ a 0 f(a− x) f(x) + f(a− x) dx. If we make the substitution u = a− x, or x = a− u then we have du = −dx and so∫ a 0 f(x) f(x) + f(a− x) dx = − ∫ 0 a f(a− u) f(a− u) + f(u) du = ∫ a 0 f(a− u) f(a− u) + f(u) du Since the u is a “dummy variable” we can replace all the u’s in the last integral by x’s and we get the result. (b) Show that ∫ a 0 f(x) f(x) + f(a− x) dx = 1 2 a. (Hint: ∫ a 0 g(x) dx = 1 2 ( ∫ a 0 g(x) dx+ ∫ a 0 g(x) dx), and use part (a).) Using the hint we have (along with the previous part) ∫ a 0 f(x) f(x)+f(a−x) dx = 1 2 ( ∫ a 0 f(x) f(x)+f(a−x) dx+ ∫ a 0 f(x) f(x)+f(a−x) dx ) = 1 2 ( ∫ a 0 f(x) f(x)+f(a−x) dx+ ∫ a 0 f(a−x) f(x)+f(a−x) dx ) = 1 2 ∫ a 0 f(x) + f(a− x) f(x) + f(a− x) dx = 1 2 ∫ a 0 1 dx = 1 2 a. (c) Find ∫ π/2 0 1 1 + tan x dx. (Hint: sinx = cos (π 2 − x).) We first rewrite this and using the hint we have∫ π/2 0 1 1 + tan x dx = ∫ π/2 0 cosx cosx+ sinx dx = ∫ π/2 0 cosx cosx+ cos (π 2 − x) dx. This is the integral done in part (b) where we have a = π/2 and f(x) = cos x; so we can conclude that the integral is π/4. 6. Find ∫ ex − 1 ex + 1 dx. Let us first rewrite it by making the substitution u = ex so that du = ex dx = u dx or dx = du/u. In particular, we have∫ ex − 1 ex + 1 dx = ∫ u− 1 u(u+ 1) du. In this form we see that this is a straightforward integration using the method of partial fractions. So we first check and see that no long division is required. So the next step is for us to break it into smaller parts, namely u− 1 u(u+ 1) = A u + B u+ 1 . Clearing the denominators this is the same as u−1 = A(u+ 1) +Bu, we can now either group coefficients of by choosing “nice” values of u we see that A = −1 (u = 0) and B = 2 (u = −1). So we now have∫ u− 1 u(u+ 1) du = ∫ ( − 1 u + 2 1 u+ 1 ) du = − lnu+ 2 ln(u+ 1) + C. So we have∫ ex − 1 ex + 1 dx = − ln ex + 2 ln(ex + 1) + C = −x+ 2 ln(ex + 1) + C 7. Consider the autonomous differential equation dy dt = 1− y2 1 + y2 . (a) Find the equilibrium solutions to the differential equation, and determine the stability of each equilibrium point. The equilibrium solutions. or constant solutions are found by solving 1− y2 1 + y2 = 0 or y2 = 1. So the equilibrium solutions are at y = ±1. To determine their stability we can draw a graph or test the “eigenvalues”. In the latter case if we let g(y) = 1− y2 1 + y2 then g′(y) = (1 + y2)(−2y)− (1− y2)(2y) (1 + y2)2 = − 4y (1 + y2)2 . In particular we see that g(−1) = 1 > 0 and so −1 is an unstable equilibrium solution, while g(1) = −1 < 0 and so 1 is a stable equilibrium solution. (b) If the initial condition y(0) = 2 is given, what happens to y as t → ∞? (Hint: you do not need to solve the differential equation to answer this part.) We are starting at y = 2, by looking at g(y) we see that for y > 1 that g(y) < 0. This shows that for y > 1 that dy/dt < 0 and in particular the value of y will decrease. And so the solution will be decreasing and continue to decrease towards the equilibrium solution y = 1, i.e., as t → ∞ we have that y → 1. 10. One way to describe a surface is as z = f(x, y), but we can also describe a surface by F (x, y, z) = c, where c is a fixed constant. The tangent plane of F (x, y, z) = c at the point (x0, y0, z0) is given by ∂F (x0, y0, z0) ∂x (x− x0) + ∂F (x0, y0, z0) ∂y (y − y0) + ∂F (x0, y0, z0) ∂z (z − z0) = 0 Find the tangent plane to the ellipsoid (the three dimensional analog of an ellipse) x2 + 4y2 + 9z2 = 34 at the point (3,−2,−1). This question is really not about tangent planes, but is really about partial derivatives and seeing if we know how to follow directions. First we compute the partial derivatives, in our case we have F (x, y, z) = x2 + 4y2 + 9z2 and the point (x0, y0, z0) = (3,−2,−1). So we have ∂F (x, y, z) ∂x = 2x so ∂F (3,−2,−1) ∂x = 6 ∂F (x, y, z) ∂y = 8y so ∂F (3,−2,−1) ∂y =−16 ∂F (x, y, z) ∂z = 18z so ∂F (3,−2,−1) ∂z =−18 We now have all the terms in the tangent plane formula given us and so we have that the tangent plane is 6(x− 3)− 16(y + 2)− 18(z + 1) = 0 or 3x− 8y − 9z = 34. 11. Find the gradient for g(x, y) = 3x2y − sin(xy) + ey − ln(x2 + 1). We have ∇g(x, y) =  ∂g(x, y) ∂x ∂g(x, y) ∂y  =  6xy − y cos(xy)− 2xx2 + 1 3x2 − x cos(xy) + ey  . (This almost seems too easy!) 12. You have been hired to design a new shipping container. The design is simple enough, a rectangular box, so your main job is to decide what dimension to make the box. The materials to make the top/bottom sides cost $2/m2 the materials to make the front/back sides cost $3/m2 and the material to make the left/right sides cost $5/m2. If the container must hold 240m3, find the dimensions of the box that minimize the cost. First we translate this into a more mathematical form. There are three dimensions for the box which we will denote by length by x, width by y and height by z. The area of a top/bottom side is xy, the area of a front/back side is xz and the area of a left/right side is xz. So then using the cost of material for the sides (and remembering we have two of each type of side) the total cost of the shipping container is f(x, y, z) = 4xy + 6xz + 10yz. We also have that the container must hold 240m3 so that tells us that xyz = 240, this acts as our constraint. So we now see that we are maximizing a function f(x, y, z) given g(x, y, z) = c, a Lagrange multiplier problem. So we need to find the critical points for F (x, y, z, λ) = 4xy + 6xz + 10yz − λ(xyz − 240). We have ∂F (x, y, z, λ) ∂x = 4y + 6z − λyz = 0 ∂F (x, y, z, λ) ∂y = 4x+ 10z − λxz = 0 ∂F (x, y, z, λ) ∂z = 6x+ 10y − λxy = 0 ∂F (x, y, z, λ) ∂λ = −xyz + 240 = 0 We now can solve the first three equations for λ and doing so we can conclude that 4 z + 6 y = 4 z + 10 x = 6 y + 10 x . From this it is easy to see that 6/y = 10/x or y = (3/5)x and 4/z = 10/x or z = (2/5)x. Finally using the last equation we have 240 = xyz = x( 3 5 x)( 2 5 x) = 6 25 x3 or x3 = 1000 or x = 10. So the cost minimizing dimensions are x = 10m, y = 6m and z = 4m.