Find a formula for the velocity at any time ( in terms of k ): v(t) = (Ce-((k/4)(t)) -39.2)/k Find the limit of this velocity for a fixed time to as the air resistance coefficient k goes to 0. (Write to as to in your answer.) v(to) = -9.8t+9

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Chapter1: Units, Trigonometry. And Vectors
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A body of mass 4 kg is projected vertically upward with an initial velocity 9
meters per second.
The gravitational constant is g = 9.8m/s². The air resistance is equal to k|v| where
k is a constant.
Find a formula for the velocity at any time (in terms of k ):
v(t) = (Ce-((k/4)(t)) -39.2)/k
Find the limit of this velocity for a fixed time to as the air resistance coefficient k
goes to 0. (Write to as to in your answer.)
v(to) = -9.8t+9
How does this compare with the solution to the equation for velocity when there is
no air resistance?
This illustrates an important fact, related to the fundamental theorem of ODE and
called 'continuous dependence' on parameters and initial conditions. What this
means is that, for a fixed time, changing the initial conditions slightly, or changing
the parameters slightly, only slightly changes the value at time t.
The fact that the terminal time t under consideration is a fixed, finite number is
important. If you consider 'infinite' t, or the 'final' result you may get very different
answers. Consider for example a solution to y'= y, whose initial condition is
essentially zero, but which might vary a bit positive or negative. If the initial
condition is positive the "final" result is plus infinity, but if the initial condition is
negative the final condition is negative infinity.
Transcribed Image Text:A body of mass 4 kg is projected vertically upward with an initial velocity 9 meters per second. The gravitational constant is g = 9.8m/s². The air resistance is equal to k|v| where k is a constant. Find a formula for the velocity at any time (in terms of k ): v(t) = (Ce-((k/4)(t)) -39.2)/k Find the limit of this velocity for a fixed time to as the air resistance coefficient k goes to 0. (Write to as to in your answer.) v(to) = -9.8t+9 How does this compare with the solution to the equation for velocity when there is no air resistance? This illustrates an important fact, related to the fundamental theorem of ODE and called 'continuous dependence' on parameters and initial conditions. What this means is that, for a fixed time, changing the initial conditions slightly, or changing the parameters slightly, only slightly changes the value at time t. The fact that the terminal time t under consideration is a fixed, finite number is important. If you consider 'infinite' t, or the 'final' result you may get very different answers. Consider for example a solution to y'= y, whose initial condition is essentially zero, but which might vary a bit positive or negative. If the initial condition is positive the "final" result is plus infinity, but if the initial condition is negative the final condition is negative infinity.
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