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Calculus Homework Assignment 5 1. Find volume of the largest right circular cone that can be inscribed in a sphere of radius 4. [like §4.5 #12] Sol. Let x denote the base radius of the right circular cone inscribed in the sphere with radius 4, and y be the perpendicular distance form the base of the right circular cone to the center of the sphere. As shown in following graph 4 y 4 x p We immediately have that x = 16 −p y 2 . Since the volume of the cone is V = 31 πr2 h, where r = x = 16 − y 2 and h = 4 + y, thus we can express the volume of the cone as a function of y, that is, π V (y) = (16 − y 2 )(4 + y) 3 π 0 So V (y) = 3 [(−2y)(4 + y) + (16 − y 2 )] = − π3 (4 + y)(−4 + 3y), this implies V 0 (y) = 0 ⇔ y = −4 or y = 43 . But by definition of y, y = −4 is impossible, so y = 43 is the only critical point of V (y). Since V 00 (y) = − 2π (4 + 3y) ⇒ V 00 ( 43 ) − 16π < 0, and 3 3 4 −4 ≤ y ≤ 4 ⇒ V (−4) = V (4) = 0, so as y = 3 , we have the . maximal volume of the cone as V ( 43 ) = 2048π 81 2. Find the limits. sin(a + 2h) − sin a h→0 √ h x b. lim+ √ x→0 sin 3x [like §4.6 #20,28] Sol. a. Since sin(a + 2h) − sin a = sin a − sin a = 0 as h = 0, sin(a + 2h) − sin a so is an indeterminate form as h = 0. h Apply L’Hôpital’s Rule, we get a. lim sin(a + 2h) − sin a 2 cos(a + 2h) = lim = 2 cos a h→0 h→0 h 1 lim b. √ x lim+ √ = x→0 sin 3x r lim+ x→0 x 1 1 = lim+ q = lim+ √ q x→0 sin 3x x→0 sin 3x 3 sin3x3x x 1 1 = √ q 3 lim+ x→0 sin 3x 3x 1 1 = √ ·1= √ 3 3 3. Evaluate Z 1 2 a. 2 − cot x dx 2 Z 3 csc θ b. dθ csc θ − sin θ [like §4.8 #52,54] Sol. a. Z Z Z Z Z 1 1 1 2 2 2 − cot x dx = (csc2 x − 1)dx 2dx − cot xdx = 2 dx − 2 2 2 Z Z Z 1 1 = 2 dx − csc2 xdx + dx 2 2 Z Z 5 1 5x + cot x = dx − csc2 xdx = +C 2 2 2 b. Z Z Z 3 3 csc θ 3 sin θ dθ = dθ = dθ 1 csc θ − sin θ − sin θ 1 − sin2 θ sin θ Z Z dθ = 3 = 3 sec2 θdθ = 3 tan θ + C cos2 θ 4. Solve the initial value problem y (4) = − cos x + 8 sin 2x; y 000 (0) = 0, y 00 (0) = y 0 (0) = 1, y(0) = 3 [like §4.8 #86] Sol. Since y (4) (x) = − cos x + 8 sin 2x, so Z 000 y (x) = (− cos x + 8 sin 2x)dx = − sin x − 4 cos 2x + C1 Since y 000 (0) = 0, so C1 = 4 ⇒ y 000 (x) = − sin x − 4 cos 2x + 4. Thus Z 00 y (x) = (− sin x − 4 cos 2x + 4)dx = cos x − 2 sin 2x + 4x + C2 Since y 00 (0) = 1, so C2 = 0 ⇒ y 00 (x) = cos x − 2 sin 2x + 4x. Thus Z 0 y (x) = (cos x − 2 sin 2x + 4x)dx = sin x + cos 2x + 2x2 + C3 Since y 0 (0) = 1, so C3 = 0 ⇒ y 0 (x) = sin x + cos 2x + 2x2 . Thus Z 1 2 y(x) = (sin x+cos 2x+2x2 )dx = − cos x+ sin 2x+ x3 +C4 2 3 Since y(0) = 3, so C4 = 4. Therefore, the solution of the initial value problem is of the form 1 2 y(x) = − cos x + sin 2x + x3 + 4 2 3 5. Find the curve y = f (x) in the xy-plane that passes √ through the point (9, 4) and whose slope at each point is 4 x. [like §4.8 #87] √ Sol. Since the slope of the tangent line at each point is 4 x, so √ dy = f 0 (x) = 4 x. Thus we have that dx Z Z √ 1 8 3 f (x) = 4 xdx = 4 x 2 dx = x 2 + C 3 Since the curve passes through (9, 4), that is, f (9) = 4, so C = −68. Hence the curve is defined by the equation 8 3 y = x 2 − 68 3 6. (a) Inscribe a regular n-sided polygon inside a circle of radius 2 and compute the area of one of the n congruent triangles formed by drawing radii to the vertices of the polygon. (b) Compute the limit of the area of the inscribed polygon as n → ∞. [like §5.1 #22] Sol. (a) Note that for each congruent isosceles triangle has central angle with measure 2π . Thus the area of each isosceles n triangle is 1 2π 2π = 2 sin AT (n) = · 22 sin 2 n n (b) The area of the inscribed regular n-sided polygon is AP (n) = nAT (n) = 2n sin 2π . Hence n sin 2π 2π lim AP (n) = lim 2n sin = 4π lim 2π n = 4π n→∞ n→∞ n→∞ n n 7. Graph f (x) = 2 sin x + 1 over the interval [−π, π]. Partition the interval into four subintervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum 4 P f (ck )∆xk , given that ck is the (a) righthand endpoint, (b) k=1 midpoint of the kth subinterval. (Make a separate sketch for each set of rectangles.) [like §5.2 #32] Sol. (a) 3 2 1 −π –3 –2 −π/ 2 –1 0 0 –1 1 π/2 2 3 π It is clear that ∆xk = π2 for each k = 1, 2, 3, 4. And the righthand endpoint ck for kth subintervals are π π c1 = − , c2 = 0, c3 = , c4 = π 2 2 respectively. Thus the associate Riemann sum is 4 X f (ck )∆xk = k=1 π (−1 + 1 + 3 + 1) = 2π 2 (b) 3 2 1 −π –3 –2 −π/ 2 –1 0 1 π/2 2 3 π –1 It is clear that ∆xk = π2 for each k = 1, 2, 3, 4. And the righthand endpoint ck for kth subintervals are π π 3π 3π , c2 = − , c3 = , c4 = 4 4 4 4 respectively. Thus the associate Riemann sum is c1 = − 4 X k=1 f (ck )∆xk = √ √ √ √ π (1 − 2) + (1 − 2) + (1 + 2) + (1 + 2) 2 = 2π 8. For the function f (x) = 2x + 3x2 , find a formula for the upper sum obtained by dividing the interval [0, 1] into n equal subintervals. Then take a limit of these sums as n → ∞ to calculate the area under the curve over [0, 1]. [like §5.2 #40] Sol. Since f 0 (x) = 2 + 6x which is positive on [0, 1], so f is increasing on [0, 1]. Thus we can use right endpoints to obtain upper sums. Now ∆x = n1 , and xk = nk , k = 1, · · · , n. So upper sum is n n n k1 X X X 2k 3k 2 1 Rn = = + 2 f (xk )∆x = f n n k=1 n n n k=1 k=1 n n 2 X 3 X 2 = k+ 3 k n2 k=1 n k=1 3 n(n + 1)(2n + 1) 2 n(n + 1) + 3· · 2 n 2 n 6 2 4n + 5n + 1 = 2n2 Therefore the area under the curve over [0, 1] is = 4 + n5 + 4n2 + 5n + 1 = lim n→∞ n→∞ 2n2 2 4 = = 2. 2 lim Rn = n→∞ lim 1 n2