1. (13%) Find the orthogonal trajectories of the family of curves y = tan 1 (kx), where k is an arbitrary constant. Solution: For the original curves:

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1 5 微甲 6- 班期末考解答和評分標準. (%) Find the orthogonal trajectories of the family of curves y = tan (kx), where k is an arbitrary constant. For the original curves: dy dx = tan y k = +k x x sin y cos y = +tan y x The curves that we wanted: dy dx = x sin y cos y xdx = sin y cos ydy ( x)dx sin y cos ydy x = cos y + C, where C is an arbitrary constant. Differentiating the original curves to get the slope: points. Cancelling the constant k: points. Writing down the slope of the curves we wanted: points. Solving the differential equation: 6 points. Page of 9

2 (e x y = %). (%) Solve the initial value problem x y y = xe x ln x, x > y() = Multiplying x both sides of the equation, we have y x y = x e x ln x which is a linear equation and the integrating factor I(x) is I(x) = exp ( x dx) = e x. (%) Hence y = I(x) I(x) x e x ln x dx = e x x ln x dx (Let u = ln x, du = x dx) = e x u du = e x (u + c) = e x [(ln x) + c] (%) By initial condition, y() = e c = c = e (%) Therefore the solution is y = e x [(ln x) + e] Page of 9

3 . (%) Find the area of the region that lies inside the curve r = + cos θ but outside the curves r = cos θ and r = cos θ. By symmetry, we only need to compute the area A + A, then the answer will be (A + A ). The intersection points of r = + cos θ and r = cos θ + cos θ = cos θ cos θ = θ =, with r = (%) Note that the dash line is θ =, and A is different from A 5 solution () Compute A first. (method ) A = (A + A ) A = = = = ( θ ( + cos θ) dθ A +A ( + cos θ) ( cos θ) dθ + cos θ cos θdθ ( + cos(θ)) + cos θ dθ + sin θ sin(θ) = ( + ) = 8 (%) (method ) A = because it is half of the area of a circle with radius= (%) ) ( cos θ) dθ A (%) Page of 9

4 A + A = (θ = ( ( + cos θ) dθ + sin θ + sin(θ) + ) (%) ) A = ( + ) A = 8 (%) () Compute A (method ) (method ) A = (A + A ) A = + cos θdθ = (θ + sin θ) = ( 6 + ) = + (%) ( + cos θ) dθ ( cos θ) dθ (%) A = (A + A + A 5 ) A A 5 = ( ( + cos θ) dθ + cos θ + cos(θ) ( cos θ) dθ )dθ ( + cos θ) dθ (%) + cos(θ) dθ ( = (θ sin(θ) + sin θ + ) (θ + sin(θ) ) (θ = ( ) ( + ( 8 )) ( + ( ) + ( 8 )) = ( 8 ) ( 6 ) ( 7 6 ) A +A +A 5 A A 5 = + (%) () Answer = (A + A ) = ( ) = (%) solution A + A = (A + A + A + A ) A A = ( θ ( + cos θ) dθ = ( + + ( ( cos θ) sin(θ) + sin θ + ) (θ + sin(θ) ) 8 6 )) ( ) ( 6 ) ( ) ( 6 ) (6%) A = ( ) A +A +A +A = A Answer = (A + A ) = (%) cos(θ) + cos θ + )dθ + sin θ + sin(θ) ( cos θ) dθ (%) ( θ + sin(θ) 8 ) ) Page of 9

5 . (%) Find the arc length of the curve. x = cos t + ln(tan t), y = sin t, t. x = cos t + ln(tan t) y = sin t dx dt = sin t + sec t tan t = sin t + sin t dy dt = cos t Arc length ( dx dt ) + ( dy dt ) dt csc t dt cot t dt cot tdt cot tdt = ln sin t ln sin t = ln Page 5 of 9

6 5. (%) Let R be the region bounded by the x-axis, x = e and the curve y = ln x. (a) (5%) Find the volume of the solid obtained by revolving R about the x-axis. (b) (5%) Find the volume of the solid obtained by revolving R about the y-axis. (c) (5%) Find the centroid of R. (d) (5%) Find the volume of the solid obtained by revolving R about x + y =. (a) e (ln x) dx(pts) = [x(ln x) e e ln x dx] = (e ) ( e ln x dx = x ln x x e = e e + = ) (b) e x ln x dx(pts) = [ x ln x x ] e = (e + ) (c) A e lnx dx = ( pts) x = A b a xf(x) dx e x ln x dx = e + ȳ y(e ey ) dy ey yey dy = e Thus ( x, ȳ) = ( e +, e ) ( pts) ( pts) (d) By Pappus s centroid theorem V = A d ( pts) Calculate d: ( e + + t, e + t) is on x + y = t = 7 e e d = e + e 7 8 Then V = e + e 7 Page 6 of 9

7 6. (%) Compute the area of the surface generated by rotating the curve y = ln x, x about the y-axis. The surface area S is: S = = e y + e y dy x + dx (6 pts) x + tan θ sec θ dθ ( pt for change variable: x = tan θ) sec θdθ = (sec θ tan θ + ln sec θ + tan θ ) / ( pts) = ( + ln( + )) ( pt) Page 7 of 9

8 7. (%) Evaluate the following integrals. (a) (5%) x x + dx. (b) (7%) x x x (x + ) dx. (a) Let u = x +. Then dx = du. Thus the Substitution Rule gives x x + (u ) u du u u + u du = 5 u 5 u + u u= = (b) The partial fraction decomposition of the rational function is Thus, u= x x x (x + ) = x + x + 5x + x +. x x x (x + ) dx x + x + 5x + x + dx = ln x + x + 5 ln(x + ) + tan x + C. [Grading Criterion] (a) correct change of variable : point, antiderivative : points, answer : point. (b) partial fraction : point, the four terms in the answer :/// points respectively. Page 8 of 9

9 8. (%) Evaluate the following improper integrals. (a) (5%) e x(ln x) dx. (b) (5%) x + x dx. (a) (total 5 points) e t e x(ln(x)) dx t x(ln(x)) dx ( points) Let u = e x, then dx = du, then u ln(t) t u du ( point) [ t u ]ln(t) ( t ln(t) ) = (points) (b) x + x dx t + t x + x dx since x + as x + x Let u = x, dx = udu u t + t u + u du t + t u + du [ln(u + )] t + t = ln() lim ln( t + ) t + = ln() (total 5 points) ( points) ( point) ( points) Page 9 of 9