MATH 140 Lecture Notes - Lecture 22: If And Only If, Talc, Power Rule

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16 Oct 2018

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If at anygiventimet weknow v t can wefind thedistance travelled. Flt and it have the gamederiv e means that they are thesame up to a constant. Starting distance dad time 0 e d t d t d t t. In general how do we findantiderivs f f x f f. Ahantideriv of f x is a function f x w f f. If fh 0 then f x c f x m then f x mxtc. If 5in x then f x 2 2 0. If 5 x x then f x 3 3to. Antiderivof x f lh 1 1 ells 811 3. 5 x x f x vx x12 n 1 1. Fai 3 0 f x 8 3 ftp x. 3 312 c c f x xe flx is. If f is anantideriv of f then forany. F is an antideriv of f f x 3rx f x 3 5 312 0 2 312 0 flower zalone.

Related Questions

A classical model in mathematical ecology is the LotkaâVolterra predator-prey model. Consider a simple ecosystem consisting of rabbits that have an infinite supply of food and foxes that prey on the rabbits for their food. This is modeled by a pair of nonlinear, first-order differential equations:

where t is time, r(t) is the number of rabbits, f(t) is the number of foxes, and Î± is a positive constant. If Î± = 0, the two populations do not interact, the rabbits do what rabbits do best, and the foxes die off from starvation. If Î± > 0, the foxes encounter the rabbits with a probability that is proportional to the product of their numbers. Such an encounter results in a decrease in the number of rabbits and (for less obvious reasons) an increase in the number of foxes. The solutions to this nonlinear system cannot be expressed in terms of other known functions; the equations must be solved numerically. It turns out that the solutions are always periodic, with a period that depends on the initial conditions. In other words, for any r(0) and f(0), there is a value t = t_p when both populations return to their original values. Consequently, for all t,

(a) Compute the solution with r₀ = 300, f₀ = 150, and Î± = 0.01. You should find that t_p is close to 5. Make two plots, one of r and f as functions of t and one a phase plane plot with r as one axis and f as the other.

(b) Compute and plot the solution with r₀ = 15, f₀ = 22, and Î± = 0.01. You should find that t_p is close to 6.62.

(c) Compute and plot the solution with r₀ = 102, f₀ = 198, and Î± = 0.01. Determine the period t_p either by trial and error or with an event handler.

(d) The point (r₀, f₀) = (1/Î±, 2/Î±) is a stable equilibrium point. If the populations have these initial values, they do not change. If the initial populations are close to these values, they do not change very much. Let u(t) = r(t)â1/Î± and v(t) = f(t) â 2/Î±. The functions u(t) and v(t) satisfy another nonlinear system of differential equations, but if the uv terms are ignored, the system becomes linear. What is this linear system? What is the period of its periodic solutions?

Hi, I need the full Matlab code and answers for all parts of the problem. Thanks!

Page 3 of 5 which gives us y as a function of Theoretically. the integration on the right of t= 1/ - g + alpha f(V)dv can be performed and the result could provide velocity. v,. as a function of t; Then we could perform the integration in y=y0 + V/- g + alpha f(V)dv and substitute v(t) to obtain y as a function of t. Theoretically, that is. Let's look at in example An Example Suppose the air resistance model for a particular object in free fall has where units for v area feet/second Note how f is defined in terms of the sign of v. As usual. downward has been selected as the negative direction so velocities v of an object in free fall will have negative values Thus -v will be positive and the square root makes. sense Similarly if the object is moving upward, v> 0. the value of f(v)) is negative. The object is dropped from a height: of 500 feet with no initial velocity Using g = 32, the integral models from above becomes and where we have used the definition of f(v) for negative values of v since the object is in free fall With a little hard work or the use of a CAS, the first equation can be written as Page 4 of 5 and the second as Now if we could solve the first expression for v in terms of r and substitute the result for v in the expression for y, the result would be very unwieldy to work with. Actually it is impossible to solve the first expression to get a nice closed form for v. Thanks to technology. however, these expressions can still be used to study the object's motion Choose any value of v. say v=-1. Substituting -1 for v in the equations for t and y tells us that t = 0.032 and y = 499.984 so the point (0.032, 499.984) would lie on a graph of distance versus time Similarly, setting v = -50 provides the point (1 839, 452 462) By using a wide range of values for v. we can generate the complete graph of the distance function as a function of time Fortunately, most CAS provides the capability of plotting the points (t(v), y(v)) foe a range of values of v without having to compute each point separately Such a graph appears below. One final note. It is also possible that the integrals representing t and y cannot be integrated into a closed form No problem In this case, simply use the expressions t= 1/ - g + alpha f(V)dV y=y0+ V/ - g + alpha f(V)dV Page 5 of 5 and the numerical integrators provided by a CAS to determine values of t und y Exercises In all of the exercises you will need a CAS or related tool In the example above the graph indicates the object strikes the ground some time between 6 and 7 seconds from release Determine this time to three decimal places In the case where alpha = 1/2 ,f(v) = -v, v0 = 0, y0=100 determine closed form. expressions for v and y as function. Of t. An object is thrown upward at 40 ft/sec from a height of 100 feet The air resistance can be modeled by alpha =0.001 and f(v) = - |v| v Using the trajectory model determine the speed of the object when it strike the ground, determine the time the object strikes the ground after takeoff, construct the graph of the object's height versus time.

MATH 140 Lecture Notes - Lecture 22: If And Only If, Talc, Power Rule

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