dt = 32t + C1 ( 32t + C1 ) dt = ( 32) t2 + C1 t + C2 = 16t 2 + C1 t + C2 2 in .NET framework

Creator QR Code in .NET framework dt = 32t + C1 ( 32t + C1 ) dt = ( 32) t2 + C1 t + C2 = 16t 2 + C1 t + C2 2

32 dt = 32t + C1 ( 32t + C1 ) dt = ( 32) t2 + C1 t + C2 = 16t 2 + C1 t + C2 2
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in which the values of C1 and C2 are determined by the initial conditions of the problem. In the present case, it is given that v(0) = 256 and s(0) = 0. Hence, 256 = 0 + C1 and 0 = 0 + 0 + C2 , so that v = 32t + 256 s = 16t 2 + 256t (1) (2)
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(a) For maximum height, ds/dt = v = 32t + 256 = 0. So, t= when the maximum height is reached. (b) Substituting t = 8 in (2), s(8) = 16(8)2 + 256(8) = 1024 + 2048 = 1024 feet (c) Setting s = 0 in (2), 16t 2 + 256t = 0 16t(t 16) = 0 t=0 The rocket leaves the ground at t = 0 and returns at t = 16. (d) Substituting t = 16 in (1), v(16) = 32(16) + 256 = 256 feet per second. The speed is the magnitude of the velocity, 256 feet per second. or t = 16 256 = 8 seconds 32
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29.6 Find an equation of the curve passing through the point (2, 3) and having slope 3x 3 2x + 5 at each point (x, y).
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The slope is given by the derivative. So, dy = 3x 3 2x + 5 dx 3 (3x 3 2x + 5) dx = x 4 x 2 + 5x + C 4
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Hence,
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Since (2, 3) is on the curve, 3= Thus, C = 15, and y= 3 4 x x 2 + 5x 15 4 3 4 (2) (2)2 + 5(2) + C = 12 4 + 10 + C = 18 + C 4
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29.7 Find the following antiderivatives: (a) (d) (g) (j) (m) (2x 3 5x 2 + 3x + 1) dx 5 x 2 dx 1 1 5 x x3 dx
(b) (e) (h) (k) (n)
1 5 dx x 3 dx x4 3x 2 2x + 1 dx x (csc2 x + 3x 2 ) dx tan2 x dx
(c) (f ) (i) (l) (o)
24 x dx (x 2 1) x dx (3 sin x + 5 cos x) dx x 3x dx x(x 4 + 2)2 dx
(7 sec2 x sec x tan x) dx 1 dx sec x
[Hint: Use Theorem 28.3 in (n).] 29.8 Evaluate the following antiderivatives by using Rule 5 or Rule 6. [In (m), a = 0.] 1 (a) 7x + 4 dx (b) dx (c) (3x 5)12 dx x 1 cos x x (d) sin (3x 1) dx (e) sec2 dx (f ) dx 2 x x 3 (h) x 2 x 3 + 5 dx (i) (g) (4) 2t 2 )7 t dt dx x+1 x 3 2 (j) x 2x + 1 dx (k) (x 4 + 1)1/3 x 7 dx (l) dx 1 + 5x 2 cos 3x (n) 1 x x 2 dx (m) x ax + b dx dx (o) 2 3x sin sin(1/x) cos (1/x) (q) (4 7t 2 )7 t dt (r) dx (p) (3x 5)12 x dx x2 1 3 (s) sec2 4 dx x5 x 29.9 A rocket is shot vertically upward from a tower 240 feet above the ground, with an initial velocity of 224 feet per second. (a) When will it attain its maximum height (b) What will be its maximum height (c) When will it strike the ground (d) With what speed will it hit the ground 29.10 (Rectilinear Motion, 18) A particle moves along the x-axis with acceleration a = 2t 3 feet per second per second. At time t = 0, it is at the origin and moving with a speed of 4 feet per second in the positive direction. (a) Find a formula for its velocity v in terms of t. (b) Find a formula for its position x in terms of t. (c) When and where does the particle change direction (d) At what times is the particle moving toward the left 29.11 Rework Problem 29.10 if a = t 2 13 feet per second per second. 3 29.12 A rocket shot straight up from ground level hits the ground 10 seconds later. (a) What was its initial velocity (b) How high did it go
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