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Part (a)

u. v. Part (a). We’ll need the product rule. They want us to find acceleration, which is the derivative of velocity. u’ = -1. v’ = t cos (t 2 /2). a(t) = v’(t) = (-1) (sin (t 2 /2)) + (-t-1) (t cos (t 2 /2)). a(2) = =. -sin 2 - 6 cos 2. 1.587 or 1.588.

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Part (a)

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  1. u v Part (a) We’ll need the product rule. They want us to find acceleration, which is the derivative of velocity. u’ = -1 v’ = t cos (t2/2) a(t) = v’(t) = (-1)(sin (t2/2)) +(-t-1)(t cos (t2/2)) a(2) = = -sin 2 - 6 cos 2 1.587 or 1.588

  2. “Is the speed of the particle increasing at t=2 ? Why or why not ?” v(2) = -(2+1) sin(4/2) v(2) = -3 sin 2 Since a(2) was positive, but v(2) is negative, the SPEED of the particle is DECREASING at t=2.

  3. Part (b) The velocity changes direction when v(t) = 0. -(t+1) sin (t2/2) = 0 t could be -1, since that would make the first parenthesis equal to zero. Unfortunately, -1 is not in the interval 0 < t < 3.

  4. The velocity changes direction when v(t) = 0. -(t+1) sin (t2/2) = 0 t could also be 0, since the sine of 0 is equal to zero. Unfortunately, 0 is outside the interval 0 < t < 3.

  5. NOTE: t= - 2pis not in the interval. t= 2p The velocity changes direction when v(t) = 0. -(t+1) sin (t2/2) = 0 Since sin (t2/2) must be equal to 0, t2/2 has to be equal to p. This works because sin p = 0. t2/2 = p t2 = 2p

  6. 2p 3 2p 0 0 TD = -(t+1) sin (t2/2) dt 3 TD = - -(t+1) sin (t2/2) dt + -(t+1) sin (t2/2) dt Using the graphing calculator... + -(-3.265) 1.069 Part (c) TD = 4.333 or 4.334

  7. s(t) will be a maximum either at the endpoints(t = 0 & t =3), or when its derivative [v(t)] is zero. From Part (c), we know that v(t) = 0 at t = 2p. 2p -(t+1) sin (t2/2) dt = -3.265 0 Part (d) The greatest distance from the particle to the origin occurs when s(t) is a MAXIMUM. From Part (c)... 3 -(t+1) sin (t2/2) dt = 1.069 2p

  8. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: From Part (c)... 3 -(t+1) sin (t2/2) dt = 1.069 2p

  9. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 t = 0x = 1 3 2p

  10. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 t = 0x = 1 3 2p

  11. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  12. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  13. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  14. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  15. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  16. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  17. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  18. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  19. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  20. 2p -(t+1) sin (t2/2) dt = -3.265 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  21. 2p 3 -(t+1) sin (t2/2) dt = -3.265 2p 0 Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 t = 2px = -2.265

  22. Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  23. Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  24. Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  25. Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  26. Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 3 2p

  27. Here’s what the actual movement of the particle would look like on the x-axis: -4 -2 2 0 4 t = 3x = -1.196 3 2p

  28. Therefore, the greatest distance from the origin would be 2.265 units. As you could see, the greatest distance from the origin happened right here. -4 -2 2 0 4 Since the entire distance traveled by the particle was only 4.334, it will not make it all the way back to the origin.. t = 2px = -2.265 3 2p

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