Slope fields, solution curves, isoclines. |
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Did you understand just how the isoclines of a DE act like venetian blinds. The flow of "parallel" solution curves (think of a fluid flowing) must cross the isoclines tangent to the tangent field elements (the shutters). Of course, if the venetian blind is closed then the isocline is also a solution curve (or flow line). Each isocline is labeled by the slope m of the slope field along that curve.
Question 1: For the m=0 isocline to be a flow line it must be ......?
To solve for the isoclines of a DE F(x,y,y')=0 the trick it to substitute a constant m in place of y' (because, by definition, an isocline is where the slope of the field is constant).
Question 2: What if F(x,y,m)=0 is never true? [Hint: recall the example xy'=1.]
Here is Torricelli's answer: The (v)elocity of the liquid exiting the hole is proportional to the sqare-root of the vertical distance (y) from the hole to the surface.
Consider a drop falling through a distance y from the surface to the hole. From Newton's Law, F=ma, and the F = mg, we get a = g for the constant acceleration. From dv/dt =a we get v =gt. From v = ds/dt, where s is the vertical di(s)placement from the top, we get s = (g/2)t*2. Solving y = s_final = (g/2)(t_final)^2 = (g/2)(v_final/g)^2 = (v_final)^2/2g , whence v_final = \(2gy).
Note: You should copy this to a piece of paper, using standard rather than "ascii" mathematical notation. In ascii mathematical notation ^ means superscript and _ means subscript. Other notation we make up as we go along. It is useful for text editors that can use only the standard keyboard (ascii). You should use it when you email.
If the question is how long it takes for the tank to empty, we need some functional relation between the (V)olume to be emptied and the he(y)ght. There are examples in the text of this kind of question.
An equally interesting question is how one might measure the velocity of the escaping liquid. Imagine a spigot attached to the barrel so that the water must exit horizontally. So now our unknown v is its horizontal velocity. Neglecting air resistance, after a (T)ime, a droplet has horizontal displacement X = vT. So if the vertical displacement above a flat floor of the spigot is Y, then (from our free-fall solution above, t_final = \(2y/g) ) we have T = \(2Y/g). Solving this for v we find, with Torricelli, that the exit velocity is proportional to the horizontal stream displacement X. So if the height y above the hole varies directly with the squared distance, X^2, we have verified Torricelli's law. Anyone have a water-cooler handy? And a mop?
Question 3: Verify the above to make sure I didn't make any more mistakes than I found already.