Geometric Applications

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. Take any curve, and draw the tangent at some point on it. Follow that tangent down until it hits the x-axis. Now measure along the x-axis, from where the tangent lands, to the spot straight below our point. That stretch is called the subtangent. Suppose this length stays the same everywhere on the curve. Which curve could do that? Is it a straight line? Maybe a parabola, or a circle? Hard to guess, right? By the end of this video, a differential equation will hand you the answer in a few lines. Questions like this do appear in JEE, so let's begin.

0:43The tangent at (x,y): where it meets the axes

So, let us set up some names first. Take a point x, y on a curve. The slope of the tangent there is d y by d x. We also write it as y prime, where the little dash means the derivative, and here we call it p, for short. Okay? Now, from our point, slide along the tangent all the way to the y-axis. That is a horizontal distance of x. The slope is p, so by how much does the height change on the way? Yes, exactly, it changes by x times p! Going back to the y-axis, the height drops by x times p, so the tangent meets the y-axis at height y minus x times p. Right? Now the x-axis. Here the height must fall all the way from y to zero. With slope p, that needs a horizontal distance of y over p. So the tangent meets the x-axis at x, minus y over p. Keep these two intercepts handy. Clear?

1:42Subtangent and subnormal

Now, remember the subtangent from our puzzle? It runs along the x-axis, from where the tangent lands, at x, minus y over p, to the point straight below ours, at x. So how long is it? Yes, exactly, it is just y over p! The x terms cancel, and only y over p is left. This can come out negative when the tangent lands on the other side, so the actual length is its modulus, as shown in the brackets. With me? Next, the normal. It is the line through our point, perpendicular to the tangent. Perpendicular slopes multiply to minus one, so the slope of the normal is minus one over p. Do the same walk down to the x-axis, and the normal lands at x, plus y times p. Okay? Now you tell me. The subnormal runs along the x-axis, from the point below ours, at x, to where the normal lands. How long is it? That is correct, it is y times p, and again the length is its modulus. Now here is a classic trap, because students often swap these two. So remember it like this: subtangent, y over p, and subnormal, y times p. The tangent divides, the normal multiplies. Got it? Now one more result: when does the normal pass through the origin? The line from the origin to our point has slope y over x, and that must be the normal's slope, minus one over p. Cross multiply, and you get x plus y times p equals zero. The double-headed arrow means if and only if, so this condition is the exact test.

3:26Orthogonal trajectories

Right? Alright, the last idea: orthogonal trajectories. Orthogonal means at right angles. Orthogonal trajectories are curves that cut every curve of a given family at right angles. Okay? Step one: write the family as capital F of x, y and c, equals zero, where c is the constant that picks out one curve. Then form its differential equation, y prime equals small f of x, y. Forming the D E means eliminating c, so this f has only x and y in it. Clear? So at any point, the curve of the family has slope f of x, y. The trajectory through that same point must cross it at right angles. So what must its slope be? Yes, exactly, minus one over f of x, y, because the two slopes must multiply to minus one! So we solve y prime equals minus one over f of x, y, and that gives the whole new family. With me?

4:29Example 1: constant subtangent gives exponential curves

Now, let's crack our puzzle from the start. Find the curves whose subtangent has the same length, small a, at every point. Pause the video and try it first. Done? Okay, let's check. The subtangent is y over p, and p is y prime, so y over y prime equals our constant, small a. Separate the variables, and the double arrow means this gives d y over y, equals d x over a, right? Now integrate both sides. The letters l n stand for the natural log, and the vertical bars mean the modulus, so this is the natural log of mod y. It equals x over a, plus small k, the constant of integration. Okay? Now remove the log. So which curve is it, a line, a parabola, a circle, or something else? It is an exponential curve! y equals capital C times e to the power x over a, where capital C is plus or minus e to the power k, as shown in the brackets. And since a length is a modulus, y over y prime could also be minus a, which just gives e to the power minus x over a. So the answer to our puzzle was none of our guesses. Clear?

5:48Example 2: orthogonal trajectories of y = Cx2 are ellipses

Now a harder one. Find the orthogonal trajectories of the parabolas y equals capital C times x squared. Here is a hint: form the D E first. Pause the video and try it. Done? Alright, let's go. Differentiate, and y prime equals two times capital C times x. So the trajectories have slope minus one over two times C times x, right? Careful! That slope is right at one point, but capital C is still sitting in there! Each parabola has its own C, and the D E must not contain it. From the family, C equals y over x squared. Put it back, and y prime equals two y over x. That is the D E of the family. Okay? Now take minus one over this slope, so the trajectories satisfy y prime equals minus x over two y. Got it? Separate the variables: two y d y equals minus x d x. Integrate both sides, and y squared equals minus x squared over two, plus small k, a constant. Right? Multiply by two, and bring x squared across: x squared plus two y squared equals two k, which we call capital K, a new constant. These are ellipses! And check it: the slopes two y over x and minus x over two y multiply to minus one, which is perfect. So every ellipse cuts every parabola at right angles, wherever they meet. Clear?

7:25Quick revision: subtangent, subnormal, trajectories

So, quick revision. The tangent meets the y-axis at y minus x times p, and the x-axis at x, minus y over p. The subtangent is y over p, the subnormal is y times p, and the normal passes through the origin when x plus y times p equals zero. For orthogonal trajectories, eliminate c to get y prime equals f, then solve y prime equals minus one over f. Now go and try the practice problems on this page. You can do them, all the best!