Video File #01

A bottle rocket begins with a bottle of volume V_bottle containing a volume V₀ of air at a pressure of P₀ (typically a 3-liter bottle with about 2 liters of air at 2 or 3 atmospheres).

• For this model we will assume that V₀ is about 2.5 liters, with total bottle volume V_bottle = 3 liters.

As water is forced through the opening by the pressure inside the bottle, the pressure will decrease and the volume will increase according to the relationship P V^`gamma = constant.

• Since (P₀, V₀) is a point on the pressure vs. volume graph for this situation, the constant must be const = P₀ V₀^{`gamma</sup>.
• The pressure P is therefore P = (P<sub>0</sub> V<sub>0</sub><sup>`gamma}) / V^`gamma.

The pressure in the bottle is related to the velocity with which water exits the bottle by 1/2 `rho v_exit² = P₀ - Patm.

• This relationship is shown in the second-to-last line, and solve for v_exit in the last line.

Video File #02

The pressure P and volume V will be functions of clock time t, and are hence written P(t) and V(t). 

• We will denote the exit velocity by lower-case v; in the figure below the lower-case v has a small 'hook' on the end to ensure the we can distinguish it from the upper-case V.

If at clock time t the velocity of the water leaving the container is v(t), then in a short time interval `dt a column of water of length v(t) `dt will leave the bottle.

• The cross-sectional area of this column is A, so its volume is `dV = A * v(t) `dt.
• We see that v(t) = 1/A `dV / `dt.
• In the limit as `dt -> 0, v(t) = 1/A dV / dt.

We thus have the differential equation    1/A dV / dt = v(t)    for V, in terms of v.

• We have no way of expressing v(t) in terms of V(t), so we cannot solve the equation in this form.

• We do, however, have a way to express V(t) in terms of P(t)
• Since we can express v(t) in terms of P(t) we can rewrite the equation in terms of V(t) and P(t).

• Since v is a function of P, we see that v(t) = `sqrt((2 P(t) - Patm) / `rho ).
• If we let k stand for P₀ V₀^{`gamma</sup>, then P(t) = k / V(t)<sup>`gamma}.

Video File #03

In line below we expressed relationship between v(t) and V(t); in the second line we expressed relationship between v(t) and P(t).

• Replacing P(t) by the equivalent expression k / V(t)^`gamma, we express v(t) in terms of V(t).
• In the third line we set the two expressions for v(t) equal and obtain a differential equation for V.
• We proceed to separate variables, obtain the form in the last line below.

We need now only integrate both sides of the equation to obtain equation for V in terms of t.

• For the specific values of the initial pressure and volume and of `gamma shown below, we obtain the integral in the last line.

Video File #04

It turns out that, though the right hand side is easily integrated, we cannot perform the integration on the left-hand side of this equation.

• We therefore do a Taylor expansion of the expression we wish to integrate.

• We expand this expression about V = 3 liters and simplify our result.

• The second-degree Taylor polynomial is indicated in the integral in the third-to-last line.
• This polynomial is easily integrated and we obtain the equation of the second-to-last line.
• This cubic equation could be solved for V. To the extent that the Taylor expansion is an accurate approximation, the resulting solution would be an accurate approximation of V(t).

We could solve the cubic equation in the preceding problem, but a quadratic would be much more manageable.

• A first-degree Taylor expansion (a linear approximation or a tangent-line approximation, in fact) about V = 2.75 gives us the integral in the figure below.
• The integral is easily evaluated and yields a quadratic equation for V.
• In the second-to-last line, we solve the quadratic for V in terms of t.
• The initial condition V(0) = 2.5 what allows to determine the value of the integration constant c, and we would have a good approximation to volume v. time.

If we use DERIVE to graph the function `sqrt( V^1.4 / (1000 - 100 V^1.4) ), we can compare the graph to that of the second-or first-degree Taylor expansion of this expression.

• Doing so, we see that the second-degree expansion is extremely accurate for V in the neighborhood of 3 liters.
• We see that the first-degree expansion about V = 2.75 is nearly as accurate for the range between V = 2.5 and 3.0, and that this expansion is in fact very close to the actual function.
• We conclude that our approximation to the volume function is reasonably accurate.

Video File #05