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Solutions

# Solutions - 1 Solutions to Section 1.1 True-False Review 1...

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1 Solutions to Section 1.1 True-False Review: 1. FALSE. A derivative must involve some derivative of the function y = f ( x ), not necessarily the first derivative. 2. TRUE. The initial conditions accompanying a differential equation consist of the values of y, y , . . . at t = 0. 3. TRUE. If we define positive velocity to be oriented downward, then dv dt = g, where g is the acceleration due to gravity. 4. TRUE. We can justify this mathematically by starting from a ( t ) = g , and integrating twice to get v ( t ) = gt + c , and then s ( t ) = 1 2 gt 2 + ct + d , which is a quadratic equation. 5. FALSE. The restoring force is directed in the direction opposite to the displacement from the equilibrium position. 6. TRUE. According to Newton’s Law of Cooling, the rate of cooling is proportional to the difference between the object’s temperature and the medium’s temperature. Since that difference is greater for the object at 100 F than the object at 90 F , the object whose temperature is 100 F has a greater rate of cooling. 7. FALSE. The temperature of the object is given by T ( t ) = T m + ce - kt , where T m is the temperature of the medium, and c and k are constants. Since e - kt = 0, we see that T ( t ) = T m for all times t . The temperature of the object approaches the temperature of the surrounding medium, but never equals it. 8. TRUE. Since the temperature of the coffee is falling, the temperature difference between the coffee and the room is higher initially, during the first hour, than it is later, when the temperature of the coffee has already decreased. 9. FALSE. The slopes of the two curves are negative reciprocals of each other. 10. TRUE. If the original family of parallel lines have slopes k for k = 0, then the family of orthogonal tra- jectories are parallel lines with slope - 1 k . If the original family of parallel lines are vertical (resp. horizontal), then the family of orthogonal trajectories are horizontal (resp. vertical) parallel lines. 11. FALSE. The family of orthogonal trajectories for a family of circles centered at the origin is the family of lines passing through the origin. Problems: 1. d 2 y dt 2 = g = dy dt = gt + c 1 = y ( t ) = gt 2 2 + c 1 t + c 2 . Now impose the initial conditions. y (0) = 0 = c 2 = 0 . dy dt (0) = c 1 = 0 . Hence, the solution to the initial-value problem is: y ( t ) = gt 2 2 . The object hits the ground at time, t 0 , when y ( t 0 ) = 100. Hence 100 = gt 2 0 2 , so that t 0 = 200 g 4 . 52 s, where we have taken g = 9 . 8 ms - 2 . 2. From d 2 y dt 2 = g , we integrate twice to obtain the general equations for the velocity and the position of the

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2 ball, respectively: dy dt = gt + c and y ( t ) = 1 2 gt 2 + ct + d , where c, d are constants of integration. Setting y = 0 to be at the top of the boy’s head (and positive direction downward), we know that y (0) = 0. Since the object hits the ground 8 seconds later, we have that y (8) = 5 (since the ground lies at the position y = 5).
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Solutions - 1 Solutions to Section 1.1 True-False Review 1...

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