Math 181H: PRACTICE Exam 2 Prof. S. Smith, March 1992 (based on Exam 1 of old Math 132, April 1990) Problem 1: Find the length of the curve y = \frac{1}{3}(x^{2} + 2)^{\frac{3}{2}} from x = 0 to x = 3.

Note y′ = \frac{1}{3} \frac{3}{2}(x^{2} + 2)^{\frac{1}{2}} 2x = x\sqrt{x^{2} + 2}, so length is ∫_{0}^{3} \sqrt{1 + (x\sqrt{x^{2} + 2})^{2}} dx = ∫_{0}^{3} \sqrt{1 + x^{2}(x^{2} + 2)} dx = ∫_{0}^{3} \sqrt{x^{4} + 2x^{2} + 1} dx = ∫_{0}^{3} x^{2} + 1 dx = [\frac{x^{3}}{3} + x]_{0}^{3} = 12.

Problem 2: Find the length of the parametric curve x = t^{3}, y = t^{2}, 0 ≤ t ≤ 4. Have \frac{dx}{dt} = 3t^{2}, \frac{dy}{dt} = 2t, so length is ∫_{0}^{4} \sqrt{(3t^{2})^{2} + (2t)^{2}} dt = ∫_{0}^{4} \sqrt{9t^{4} + 4t^{2}} dt. = ∫_{0}^{4} t\sqrt{9t^{2} + 4} dt (Subst: u = 9t^{2}+4) = \frac{1}{27}[(9t^{2}+4)^{\frac{3}{2}} ]_{0}^{4} = \frac{1}{27}[(9.4^{2}+4)^{\frac{3}{2}} −4^{\frac{3}{2}} ] = \frac{8}{27}[37^{\frac{3}{2}} −1].

Problem 3: Find the length of the polar curve r = e^{−θ}, 0 ≤ θ ≤ 3π.

Have \frac{dr}{dθ} = −e^{−θ}, so length is ∫_{0}^{3π} \sqrt{(e^{−θ})^{2} + (−e^{−θ})^{2}} dθ = ∫_{0}^{3π} \sqrt{2e^{−2θ}} dθ = ∫_{0}^{3π} \sqrt{2}e^{−θ} dθ = \sqrt{2}[−e^{−θ}]_{0}^{3π} = \sqrt{2}[−e^{−3π} − (−1)] = \sqrt{2}[1 − e^{−3π}].

Problem 4: Find the area of the surface generated by rotating about the x-axis the arc of the curve y = x^{3} between x = 0 and x = 1.

Have \frac{dy}{dx} = 3x^{2}, so area is 2π ∫_{0}^{1} x^{3}\sqrt{1 + (3x^{2})^{2}} dx = 2π ∫_{0}^{1} x^{3}\sqrt{1 + 9x^{4}} dx. (Subst: u = 1 + 9x^{4}) = \frac{2π}{54}[(1 + 9x^{4})^{\frac{3}{2}} ]_{0}^{1} = \frac{π}{27}[10^{\frac{3}{2}} − 1].

Problem 5: Find the area of the surface generated by rotating about the x-axis the parametric curve x = \frac{t^{2}}{2} , y = t between t = 1 and t = 3.

Have \frac{dx}{dt} = t, \frac{dy}{dt} = 1, so area is 2π ∫_{1}^{3} t\sqrt{t^{2} + 1^{2}} dt. (Subst: u = t^{2} + 1) = \frac{2π}{3}[(t^{2} + 1)^{\frac{3}{2}} ]_{1}^{3} = \frac{2π}{3}[10^{\frac{3}{2}} − 1].

Problem 6: Find the area of the region enclosed by one loop of the polar curve r = 3 sin 2θ. Draw to see 4 loops, one in [0, \frac{π}{2}]. So area is \frac{1}{2} ∫0^{\frac{π}{2}} (3 sin 2θ)^{2} dθ

= \frac{1}{2} ∫0^{\frac{π}{2}} 9 sin^{2} 2θ dθ = \frac{9}{2} ∫0^{\frac{π}{2}} \frac{1}{2}(1 − cos 4θ) dθ = \frac{9}{4}[θ − \frac{sin 4θ}{4}]0^{\frac{π}{2}} = \frac{9}{4}[(\frac{π}{2} − 0) − 0] = \frac{9π}{8}.

Problem 7: Sketch the polar curve r = 2(1 − sin θ).

Problem 8: Determine the function y of x satisfying the differential equation \frac{dy}{dx} = 2xy^{2}, subject to the condition y(1) = 1.

Separate and integrate: ∫ \frac{1}{y^{2}} dy = ∫ 2x dx, so −\frac{1}{y} = x^{2} + C. Then 1 = y(1) gives −\frac{1}{1} = 1^{2} + C so C = −2. Hence −\frac{1}{y} = x^{2} − 2 yields y = −\frac{1}{x^{2}−2}.

Problem 9: Find the center of mass of the triangle with vertices (0,0) (1,0) (1,2). So right triangle, bounded by y = 0, x = 1, y = 2x; area \frac{1}{2}1.2 = 1. Then x_{cm} = \frac{1}{1} ∫_{0}^{1} x(2x) dx = 2 ∫_{0}^{1} x^{2} dx = 2[\frac{x^{3}}{3} ]_{0}^{1} = \frac{2}{3}. And y_{cm} = \frac{1}{2} ∫_{0}^{1}(2x)^{2} dx = 2 ∫_{0}^{1} x^{2} dx = 2[\frac{x^{3}}{3} ]_{0}^{1} = \frac{2}{3}. So cm at (\frac{2}{3}, \frac{2}{3}).

Problem 10: Find an equation for the ellipse with foci (±1, 0) and vertices (±2, 0). As foci and vertices have y = 0, axis is on x-axis. From vertices with x = ±2 get major axis a = 2. Also 4 is sum of distances from foci to (2,0), so 2 is distance from each focus to (0,b). This gives hypotenuse length 2 for right triangle (0,0) (1,0) (0,b) so b = \sqrt{3}. So equation is \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1.