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Any curve with the property that whenever it intersects a curve of a given family it does so at an angle $a \neq \pi / 2$ is called an oblique trajectory of the given family. (See Figure 1.1.7.) Let $m_{1}$ (equal to tan $a_{1}$ ) denote the slope of the required family at the point $(x, y),$ and let $m_{2}$ (equal to tan $a_{2}$ ) denote the slope of the given family. Show that$$m_{1}=\frac{m_{2}-\tan a}{1+m_{2} \tan a}$$[Hint: From Figure $\left.1.1 .7, \tan a_{1}=\tan \left(a_{2}-a\right)\right] .$ Thus,the equation of the family of oblique trajectories is obtained by solving$$\frac{d y}{d x}=\frac{m_{2}-\tan a}{1+m_{2} \tan a}$$(FIGURE CAN'T COPY)

$m_{1}=\frac{m_{2}-\tan (a)}{1+m_{2} \tan (a)}$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

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instead of dealing with orthogonal trajectories in this problem, we're gonna deal with it trajectories that intersect at angle A which is not equal to pi or two. We're going to let the blue curve represent the given family of curves, and we'll let the green curve represent the family of curves that we seek, which will always intersect in that angle. A. Furthermore, will let em one denote tangent of a one where a one is indicated here and m to denote tension of a to were aces up to Is this single indicated here to start this process of showing that M one is equal to end to my extension of a divide by one plus two times tension of A. We'll begin by writing tangent of the quantity a one and working with this quantity, a one knows that tangent of a one can also be expressed as taking the larger angle a to so we'll have tangent of a to subtracting off the smaller angle. A so tender day one is equal to this tangent of the difference of angles, but that allows us to use the falling formula from trigonometry that whenever we have a difference of Ingles. It can be broke broken down into this format. So now we can express that attention to of a one is equal to tangent of a To with a minus will use a minus in the numerator. So my extension of a divide by one plus the product of these same tangents. So tangent day to times tangent A for the next step of this process will make some substitution. Recall that tangent of a one was meant to be m one. We obtained m one on the left hand side and tension of a two was meant to be m two. So this tangent of M two or a two can be expressed as m two mice The tangent of a divide by one plus tangent of a two which is m to itself times, tangent A And this is expression that we were looking to find

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