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course
Here's a simple, standard problem, which I think (and certainly hope) is standard in a first course in differential equations:An object is projected upward with initial velocity v_0. It is subject to the forces exerted by gravity and air resistance, the latter being proportional to the square of its speed.
We wish to analyze its motion.
The problem is at this point straightforward enough. Having recently bought my grandson an airsoft gun and watched the paths of BB's fired at various angles, including straight up, I innocently tossed in the following condition:
A .12 gram BB is fired upward at 80 meters / second and rises to maximum height 45 meters. What is the drag force constant?
To set up the problem we first obtain an expression for the net force on the rising BB, which is of the form
F_net = - m g - k v^2,
where m is its mass, g the acceleration of gravity, v its velocity function, and k the drag force constant. The upward direction is implicitly (in light of the - signs) chosen as the positive direction.
Since F_net = m a = m dv/dt = m d^2 y / dt^2, where y is the vertical position of the projectile, we could then write our equation in either of the following forms:
m dv/dt = - m g - k v^2
m d^2 y / dt^2 = - m g - k (dy/dt)^2.
In these forms our solution for v will be in terms of t, the variable with respect to which our derivatives are taken.
Given the conditions of the problem, however, there are advantages to solving the equation for the velocity v in terms of the position y.
Noting that
dv/dt = dv/dy * dy/dt and, since v = dy/dt, that
dv/dt = v dv/dy,
we can write the equation as
m v dv/dy = - m g - k v^2.
This equation leads quickly to the form
m v dv / (m g + k v^2) = -dy, then to
v dv / (g + k / m v^2) = - dy.
The left-hand side is easily integrated using the change of variable u = g + k / m v^2, so that du = 2 k/m * v dv and our v dv becomes 1/2 m / k du, giving use the equation
1/2 m/k du / u = - dy,
which we integrate to obtain
1/2 m/k ln | u | = -y + c, with arbitrary integration constant c.
Returning to our original variable v we have
1/2 m/k ln | g + k/m v^2 | = -y + c.
Rearranging we have
ln | g + k/m v^2 | = -2 k/m y + c
and inverting the natural log function we have
g + k / m v^2 = +-e^(-2 k/m y + c) = e^c e^(-2 k/m y), where c is arbitrary, the to
g + k/m v^2 = +-A e^(-2 k/m y), where A = e^c > 0 and finally to
g + k/m v^2 = A e^(-2 k/m y), A <> 0.
Solving for v we obtain
v = +- sqrt( m/k A e^(-2 k/m y) - m g / k) ).
Since v and y are the only varying quantities, this equation gives v as a function of y. We therefore write this a little more explicitly as
v(y) = +- sqrt( m/k A e^(-2 k/m y) - m g / k) ).
We are given the value of m and know the value of g, and (assuming that we measure y relative to the initial position) we know that v(0) = 80 m/s. We know also that v(45 m) = 0.
Using this information we wish to solve for k.
Let's first notice that k appears within the exponent of our exponential function as well as in a factor of the exponential function (in the denominator of that factor, no less), and in another term of our expression for v(y) (again in the denominator). We anticipate the possibility that it will therefore be impossible to obtain a closed-form solution for k. So we're going to have to be prepared to get creating in order to determine k.
However we don't need to worry about that just yet. We want to first use our initial conditions to find our constant A.
The condition y(0) = y_0 would give us
y_0 = +- sqrt( m/k A e^(-2 k/m * 0) - m g / k) = +- sqrt ( A m/k - m g / k)
so that
A m/k - m g / k = y_0^2
and
A = k/m y_0^2 + g.
Our expression for v becomes
v(y) = +- sqrt( m/k ( k/m y_0^2 + g) * e^(-2 k/m y) - m g / k), or with simplification
v(y) = +- sqrt( (y_0^2 + m g / k) * e^(-2 k/m y) - m g / k).
Now, using the fact that velocity is zero when y = 45 meters, we obtain the equation
+- sqrt( (y_0^2 + m g) / k * e^(-2 k/m * 45 m) - m g / k) = 0, which since a square root is zero if and only if the quantity of which we are finding the square root is zero, will be valid if and only if
(y_0^2 + m g / k) * e^(-2 k/m * 45 m) - m g / k = 0
All quantities except k are known. So all we need to do is solve the equation.
However, as we feared, there is no combination of algebraic manipulations that will allow us to solve this equation for k. So we're left with some options. Some options are:
We first Plug in the values of y_0, m and g. This gives us the left-hand side of the equation as a function of k.
We can use our graphing skills to construct a reasonable graphical approximation to the graphs of the various expressions in our equation.
We can plug the resulting expression into a computer algebra program and let the program solve for k.
We can graph the value of this function vs. k and determine the approximate k-axis intercept(s).
We can graph the expressions y_0^2 + m g / k * e^(-2 k/m * 45 m) and m g / k on the same set of coordinate axes and find the point(s) of intersection.
We can set up a spreadsheet in which the expression v(y) = y_0^2 + m g / k * e^(-2 k/m * x) - m g / k is evaluated for values of x ranging from 0 to 45 meters and perhaps beyond, with the k value set by reference to a fixed cell. The k value can then be manipulated until the v(y) function reaches 0 for the first time at or very close to y = 45 meters.
The first of our five methods can be done quickly and easily, given good graphing skills, and should provide significant insights.
The middle three methods should work, and would be easy to implement.
Some of these methods have the potential to provide us with deeper insight into the behavior of the BB, and into the mathematical behavior of our solution, while some just get the answer, and all will tend to be used for the latter purpose without invoking deeper thinking.
The last method provides us directly with a set of velocity values, which can be further analyzed and compared with our physical intuition.
80 is a respectable score, but I don't think it reflects what you know.
I'll give you specific feedback on individual problems, but first let me give you a general overview of the nature of your errors on this test:
1. You graphed a number of functions by evaluating them at several points, plotting points and connecting them with a curve. This is not an appropriate method for this course, where we use the basic points scheme and other graphing techniques to build a graph by understanding its properties. I just got done working with a differential equations student, a good student with a good background, who 'hit the wall' on a series of problems because he doesn't know the graphing techniques he should have learned in precalculus.
2. When solving an equation, the steps in finding the solution should be spelled out, and an exact solution is required when an exact solution can be found. An approximate solution should also be calculated.
3. Always, always, be really careful about the laws of exponents. They even trip me up sometimes, in complicated expressions, if I'm not careful and methodical.
Now for some specifics:
To graph 6 * 2^x using basic points: The basic points of an exponential function are the x = -1, 0 and 1 points, along with the horizontal asymptote. You know that this function approaches the negative x axis as a horizontal asyptote because 2^(-x) = 1 / 2^x, and 2^x rapidly approaches infinity for large positive x. The basic points are (-1, 3), (0, 6) and (1, 12). Knowing by experience what the graph of an exponential looks like, you can easily construct a graph depicting the general behavior of the function by plotting these three points. There are certainly cases where you would want to use more points to get more specific information, and for this function it is easy to find the coordinates of more points, but the question specifically asked you to identify the basic points and use them to graph the function.
On the second problem you used an interesting and not quite correct test to identify the series which was exponential in its behavior. You showed that each number in the first series was the 1.19 power of the preceding. However that test doesn't reveal exponential behavior (any exponential sequence consists of numbers A b^n for some A and b, and some integer n; if A b^(n+1) = (A b^n)^p then A b^(n+1) = A^p b^(n p), an equation that can be solved for n in terms of A, b and p; it follows that there is only one value of n for which this is so. This is a fairly challenging argument which you are capable of understanding but that would probably take time you don't have at the moment). In any case it was an interesting idea but it doesn't do the job. In fact, in light of the preceding, your result shows that the first sequence is not exponential in nature. The test we use in this course is simply that the ratio of each number to the preceding be the same (since A b^(n+1) / A b^n = b for all values of n, a statement you should understand). It turns out that this is the case for the second sequence.
You did well on the third problem.
On the fourth problem you did pretty well didn't always provide an exact solution, and you did have one outright error. On the fourth equation you skipped from 3^(5x+1) = 12 to x = .252371904. The intermediate steps were missing. Taking the log of both sides you would get (5x+1) log(3) = log(12) so that 5x+1 = log(12) / log(3) and x = (log(12) / log(3) - 1) / 5. On the fifth problem you went from x * 2^x = 9/5 to x = .9389210889 without any explanation. It turns out that this equation can't be solved exactly, but you didn't indicate what method you used to get your result. On the second equation you got 9^x = 7^(x-4) and asserted that the equation can't be solved; however taking logs of both sides yields x log(9) = (x - 4) log(7), which can easily be solved for x.
To sketch the graph of (x-2)^4 (x-2.5) (x^2 + 2 x - 5) you would find the zeros of the quadratic, which are x = -1 +- sqrt(6), either by the quadratic formula or completing the square. Your solution should be documented. You did indicate correct approximations 1.449 and -3.449 to these solutions. You would then plot the zeros of the function. You created a more extensive table of values, which did include the zeros, so you did plot the zeros. Your graph indicated a negative concavity at both ends, which would not be characteristic of a graph whose far-left and far-right behavior approaches that of the function y = x^8. The zero at x = 2.5 does not pass through the x axis, nor does the zero at x = 2, and this behavior should have been indicated. In all you didn't do too badly here, but the methods of the course would dicate that rather than being based on about 12 points (which were well chosen) the graph be based on just the zeros and y intercept, the nature of the zeros, and the behavior for large | x |.
You did well with linearizing the data on problem 6, finding the correct linearization function and solving it correctly up to the point where you simplified 10^(-.32 x + .85) as 10^(-.32 x) + 10^.85. I'm sure you'll see the nature of your error there, but let me know if not. You lost 2 of 10 points for your error, so on the problem as a whole you didn't do badly (especially seeing that most students get about 2 points of the 10 on this sort of problem, if that).
You made a similar error on the next problem, simplifying .89^(x+1) as .89^x + .89^1.
The statement of Problem 8 was garbled so I'll omit it, but note that if A(t) = A_0 e^(k t) the time to fall to half the value is the solution to the equation 1/2 A_0 = A_0 e^(k * t_half), which yields e^(k * t_half) = 1/2; talking the natural log of both sides we get k * t_half = ln(1/2) so t_half = ln(1/2) / k.
The last problem asked for graphs of (x-5)(x+3)(x-5) and 1/75 (x-5)(x+3)(x-5). You constructed your graphs by plotting a lot of points (see previous notes) and missed two important behaviors, the behavior of the graph near x = 5 and the far-right behavior. x = 5 is a quadratic zero so the graph does not pass through the x axis at this point but rather has near the zero a nearly parabolic shape with vertex (5, 0), and the y value approaches infinity with increasing slope at far right. You indicated a graph passing through the x axis at the zero with no indication of the correct far-right behavior.
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Self-critique (if necessary):
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Self-critique rating:
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