Energy expenditure in a climb to the Moon:  A reminder of proportionality and the Fundamental Theorem of Calculus.

At class time John Glenn was near the end of his first day in space since his original orbital mission. In relation to a question that arose in physics class, where we analyzed the energy acquired by Sen. Glenn's mass in the process of achieving orbit, we wish to determine how long it would take an individual of mass 75 kg to climb a hypothetical tower to the Moon.

We begin by assuming that the individual could produce 1 kWHr of useful climbing energy daily. This is a fairly high estimate, but a well-conditioned individual of this mass could manage it, provided that the individual's joints and other supporting physical structures stayed healthy. You don't have to know what a kWHr is, but you should know that it contains 3.6 megaJoules of energy and that a Joule is simply a unit of energy. We will later do an integral that gives us the total number of Joules required for this individual to climb to the Moon.

We can use the proportionality for g(r) = k / r^2, given above, to determine the formula for g(r), provided we have a value of g(r) at a known r. We will use the fact that when r = rE = 6.4 * 10^6 meters, g(r) = 9.8 m/s^2. Note: the distance is 6.4 * 10^6 meters, not 6.4 * 10^9 meters as indicated on the figure below. We would all be in big trouble if the Earth was that size.

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We can solve the equation k / (6.4 * 10^6 m) ^ 2 = 9.8 m/s^2 for k to obtain the value of k given below. Then, since F(r) = 75 kg * g(r) = 75 kg * k / r^2, we have F(r) = 75 kg * 9.8 m/s^2 (6.4 * 10^6 m) ^ 2 / r^2. We can multiply 75 kg * 9.8 m/s^2 to get 750 Newtons, approximately, and we can consolidate the squared terms to obtain the final expression for F(r) given below.

Error: Note that in the figure below, 6.4 * 10^9 m, which is supposed to represent the radius of the Earth, should be 6.4 * 10^ 6 meters.

We now use the fact that `dW = F(r) `dr to geometrically model `dW as the area of a rectangle whose dimensions are F(r) by `dr. This indicates that on a graph of F(r) vs. r, as shown below, the total work W required to move between rE and rM (the radii of the Earth and of the Moon's orbit) is just the area under the graph of F(r), between r = rE and r = rM. As we know, this area can be found by integrating F(r) from r=rE to r = rM.

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The figure below indicates the preceding observations. The area is labeled as an integral, and the explicit form of F(r) is given.

We see them that the work is given by the integral at the bottom of the figure below.

This integral is actually fairly easy to compute. We first note that everything inside the integral is a constant except for r. We therefore factor all the constants outside of the integral, then multiply the constants to see that the integral is 3 * 10^16 N m^2 times the integral of 1/r^2 between rE and rM. Thus all we have to do is integrate 1 / r^2 between these limits.

Recall that the integral of a function f(r) between limits r = a and r = b is the difference of any antiderivative function evaluated between a and b. To remind you of this idea, this relationship between the integral and the antiderivative is noted in the figure below. Here the antiderivative function, which is usually denoted by F, is denoted by F with a little pointy hat to distinguish it from the force function F.

The antiderivative function for 1 / r^2 is - 1 / r, as you can easily verify by taking the derivative of this antiderivative. Evaluating this antiderivative function at rM and rE, then taking the difference, we obtain - 1 / rM - (-1 / rE) = -1 / (4 * 10^8 m) - (-1 / 6.4 * 10^6 m); multiplying this by the 3 * 10^16 N m^2 we obtained for the constant part of the integral, refine the end up with the total work equal to 4.7 * 10^9 J.

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At 3.6 * 10^6 J / day, it will require 4.7 * 10^8 J / (3.6 * 10^6 J/day) = 1200 days (approximately) for a 75 kg individual to climb to the Moon. It is not clear whether this is feasible for anyone 77 years old, but Sen. Glenn use in superb condition for his age and could probably give it a shot if he was so inclined, and if such a tower was available.

Note that there are factors we have not considered here. For instance, the Earth is rotating. If we had a tower like the one described here, we would reach a point where the centrifugal force of the rotating Earth and tower would take over and propel an individual further and further 'up' the tower. The gravitational attraction of the Moon would also have to be considered. And after a certain point no spiraling ramp or stairway would be steep enough to tax of the ability of the individual to do work and some means of ascending other than walking or climbing would have to be devised. None of these considerations are really important for calculus class, but they might be of interest to some readers.

The figure below depicts a notationally simplified version of the integral we have used in this example. Since, as our proportionality assumption implies, F(r) is equal to a constant number times 1 / r^2. If we let a and b be the two distances between which we integrate, and let c be the constant number in F(r), we obtain the integral in the second line below. (The note in purple tells us that c is equal to the weight of the individual multiply by the Square of the radius of the Earth).

In the third line below we have factored the constant c out of the integral. In the fourth line we have evaluated the antiderivative between a and b. The final integral is therefore c (-1/b - (-1.a)) = weight * rE^2 ( 1 / rE - 1 / rM).

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Derivatives of Trigonometric Functions

We began by simply stating that the derivative of the function y = sin(x) is the function y = cos(x), while the derivative of y = cos(x) is y = -sin(x). We will prove these formulas later.

To see the plausibility of these assertions, we begin with a graph of y = cos(x), shown in green on the top graph below. As indicated by the slopes of this graph, the derivative function starts out with velocity 0, then becomes increasingly negative up to the point for the original function crosses the x axis, then approaches zero again through negative values up to the point where the original function again levels off. After this the values of derivative function become increasingly positive up to the second point where the graph of the original function crosses the x axis. After this the slopes gradually return again to 0. The graph of the slopes, or of the derivatives, is shown in red in the figure below. This graph looks much like that of the function y = - sin(x); this should at least make a plausible that the derivative of the cosine function is y = -sin(x).

A similar analysis shows that the derivative graph of the function y = sin x, depicted below in purple, looks like the graph depicted in black below. This graph is easily seen to have the same basic shape is that of y = cosine (x).

The derivative of the tangent function is worked out below. We note that tan(x) = sin(x) / cos(x). We can easily use the quotient rule to find the derivative of this function. The details are straightforward and are covered in the text as well as being shown in the figure below. In the end we end up with (tan(x)) ' = ` / cos^2(x).

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Using the chain rule with trigonometric functions

For practice in applying the chain rule, we compute the derivative of sin(x^2). We see that this function is the composite f(g(x)) of g(x) = x^2 and f(x) = sin(x); the process is shown in the usual way in the figure below.

We can also use the dy / dx = dy / dz * dz/ dx 'chain' form of the rule. If y(z) = sin(z) and z(x) = x^2, then y(z) = sin(z) = sin(x^2). Then, since y'(z) = cos(z) and z'(x) = 2x, we have dy / dx = dy/dz * dz/dx = cos(z) * 2x = 2x * cos(x^2). This value of dy/dx agrees with that found by the previous method.

The 'chain' notation is usually preferable to the f(g(x)) version of the chain rule when we have a composite of more than two functions. For example, the function y = sin(e^(x^2)) is a composite of three functions. If we let y(z) = sin(z) and z(w) = e^w, we see that y(z) = sin(z) = sin(e^w). If we then let w = x^2, we see that sin(z) = sin(e^(x^2)). We calculate the derivative of this composite using the chain dy / dx = dy / dz * dz / dw * dw / dx. The chain starts with dy and ends with / dx; the intermediate terms do not really cancel out but can be loosely thought of as if they did.

Since dy / dz = cos(z), dz / dw = e^w and dw / dx = 2x, we obtain dy / dz = cos(z) * e^w * 2x = cos( e^(x^2)) * e^(x^2) * 2x. When simplified this expression would be a rear range to read 2x e^(x^2) cos(e^(x^2)).

In preparation for the proof that the derivative of the sine function is the cosine function, we begin with the picture below, depicting a small angle `theta on a unit circle. Since the arc distance corresponding to an angle of one radian on a unit circle is equal to the radius, the arc distance corresponding to an angle of `theta radians will be `theta * r, where r is the radius of the circle. It follows that on a circle of radius 1, the arc distance corresponding to an angle `theta must be equal to `theta. This in no way depends on how small or large the angle `theta is.

The figure below depicts the situation for a small angle `theta. The angle is `theta, and the arc distance is `theta. The sine of the angle is y / 1 = y, where y is the y coordinate on the circle corresponding to angle `theta.

As the angle `theta gets smaller and smaller, the arc segment approaches a vertical straight line. Since the arc segment ends at the same point (x, y) as the vertical segment of the right triangle, it follows that the values of `theta and y approach each other in such a way that their ratio approaches 1. So the smaller `theta is, the more precisely it can be approximated by y = sin(`theta). It follows that for small `theta, sin(`theta) is very close to `theta. This will be useful in our derivation of the formula for the derivative of the sine function.

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