Math 181H: Hour Exam 2 (Solutions) Prof. S. Smith: 31 March 1992 Problem 1: Find the length of the curve 9x^{2} = 4y^{3} from (0, 0) to (2\sqrt{3}, 3).

Solve x = \frac{2}{3}y^{\frac{3}{2}} so \frac{dx}{dy} = \sqrt{y} and length is ∫_{0}^{3} \sqrt{1 + \sqrt{y}^{2}} dy = ∫_{0}^{3}(1 + y)^{\frac{1}{2}} dy = \frac{2}{3}[(1 + y)^{\frac{3}{2}} ]_{0}^{3} dy = \frac{2}{3}[4^{\frac{3}{2}} − 1^{\frac{3}{2}} ] = \frac{14}{3}.

Problem 2: Find the length of the parametric curve x = cos t, y = t + sin t between t = 0 and t = π. (Hint: need half-angle to integrate)

We see \frac{dx}{dt} = − sin t, \frac{dy}{dt} = 1 + cos t, so length is ∫_{0}^{π} \sqrt{(− sin t)^{2} + (1 + cos t)^{2}} dt = ∫_{0}^{π} \sqrt{1 + 2 cos t + sin^{2} t + cos^{2} t} dt = ∫_{0}^{π} \sqrt{2 + 2 cos t} dt = ∫_{0}^{π} \sqrt{4 cos^{2} \frac{t}{2}} dt = 2 ∫_{0}^{π} cos \frac{t}{2} dt = 2.2[sin \frac{t}{2}]_{0}^{π} = 4[1 − 0] = 4.

Problem 3: Compute the length of the polar curve r = 2 sin θ from θ = 0 to θ = π. (What is the curve?)

Have \frac{dr}{dθ} = 2 cos θ, so length is ∫_{0}^{π} \sqrt{(2 sin θ)^{2} + (2 cos θ)^{2}} dθ = ∫_{0}^{π} \sqrt{4(sin^{2} θ + cos^{2} θ)} dθ = 2 ∫_{0}^{π} 1dθ = 2[θ]_{0}^{π} = 2π. The curve is the circle with radius 1 centered at (0, 1).

Problem 4: Find the area of the surface obtained by rotating the curve y = \sqrt{x} about the x-axis, between x = 0 and x = 2.

Using y′ = \frac{1}{2\sqrt{x}} area is 2π ∫_{0}^{2} \sqrt{x}\sqrt{1 + (\frac{1}{2\sqrt{x}})^{2}} dx = 2π ∫_{0}^{2} \sqrt{x(1 + \frac{1}{4x})} dx = 2π ∫_{0}^{2} \sqrt{x + \frac{1}{4}} dx = 2π \frac{2}{3}[(x + \frac{1}{4})^{\frac{3}{2}} ]_{0}^{2} = \frac{4π}{3}[(\frac{9}{4})^{\frac{3}{2}} − (\frac{1}{4})^{\frac{3}{2}} ] = \frac{4π}{3}[\frac{27}{8} − \frac{1}{8}] = \frac{13π}{3}.

Problem 5: Find the area of the surface generated by rotating the parametric curve x = cos t, y = 1 + sin t about the x-axis between t = 0 and t = 2π.

Using \frac{dx}{dt} = − sin t, \frac{dy}{dt} = cos t area is 2π ∫_{0}^{2π}(1 + sin t)\sqrt{(− sin t)^{2} + (cos t)^{2}} dt = 2π ∫_{0}^{2π}(1 + sin t)(1) dt = 2π[t − cos t]_{0}^{2π} = 2π[(2π − 1) − (0 − 1)] = 4π^{2}.

Problem 6: The graph of a curve satisfying \frac{dy}{dx} = x\sqrt{y} passes through (0, 1). Find y = f(x). Separate and integrate: ∫ \frac{1}{\sqrt{y}} dy = ∫ x dx, so 2\sqrt{y} = \frac{x^{2}}{2} + C. From 2\sqrt{1} = \frac{0^{2}}{2} + C we get C = 2, so 2\sqrt{y} = \frac{x^{2}}{2} + 2 yields y = (\frac{x^{2}+4}{4})^{2}.

Problem 7: Find the center of mass of the plane lamina bounded by y = x^{3} and the x-axis between x = 0 and x = 2.

First, area is ∫_{0}^{2} x^{3} dx = [\frac{x^{4}}{4} ]_{0}^{2} = \frac{16}{4} = 4. So x_{cm} = \frac{1}{4} ∫_{0}^{2} x.x^{3} dx = \frac{1}{4} ∫_{0}^{2} x^{4} dx = \frac{1}{4}[\frac{x^{5}}{5} ]_{0}^{2} = \frac{8}{5}. And y_{cm} = \frac{1}{4} \frac{1}{2} ∫_{0}^{2}(x^{3})^{2} dx = \frac{1}{8} ∫_{0}^{2} x^{6} dx = \frac{1}{8}[\frac{x^{7}}{7} ]_{0}^{2} = \frac{16}{7}. So cm = (\frac{8}{5}, \frac{16}{7}).

Problem 8: Find the area inside the cardioid r = 2(1 + cos θ). (That is, θ runs from 0 to 2π). Sketch the graph.

\frac{1}{2} ∫_{0}^{2π} 2^{2}(1 + cos θ)^{2} dθ = 2 ∫_{0}^{2π} 1 + 2 cos θ + cos^{2} θ dθ = 2 ∫_{0}^{2π} 1 + 2 cos θ + \frac{1}{2}(1 + cos 2θ) dθ = 2[\frac{3}{2}θ + 2 sin θ + \frac{1}{4} sin 2θ]_{0}^{2π} = 2[\frac{3}{2}2π + 2.0 + \frac{1}{4}0] = 6π.

Problem 9: Find an equation for the parabola with vertex (2, 0) and directrix the y-axis. Sketch, indicating focus.

Directrix is vertical, with vertex to right, so parabola “opens to the right”. Vertex shifted by 2 in x-direction, by 0 in y; so form is (y − 0)^{2} = c(x − 2). Also focus is same distance 2 from vertex but away from directrix, that is at (4, 0). Line x = 4 at focus is distance 4 from directrix, so y-value is 4 at x = 4. See then that c = 8, so y^{2} = 8(x − 2).