Orthogonal Trajectories
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiPTnOHfgy0zbEX8M83tYFJj7Wmi5xAeBhBKxPMcLPwSe_q9twUuiwowRlwTzVsN9G_SY66vr_OYRxhspsZK9xc9clVV_m5tEsQXNIIqsNTs_MJdzEwAt0DoklGSurbxfWm0lbb-I1tx18/s320/mypic.jpg)
Say this prayer: it helps!
Orthogonal Curves ....
Properties:
Remember that a family of curves can represent solutions in differential equations. Presented here below are curves that are tangent to the y-axis at the origin. x^2 + y^2 = 2cx the goal is to take derivatives with respect to x on both sides solving the result for 2c (yielding 2x + 2y(dy/dx) = 2c). The resulting equation will yield a term (at least one) with dy/dx. Now take the original equation and solve it in for 2c (yielding x + (y^2)/x = 2c)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhBB44tKSEI66QitAWQKQzTdVBn0NfKUMQdWtSxHREmNRY1hexo82_Us37WKhTm-mS0YoDD6_YkSfc6m9diEHaCn6SAyDWfgen0lG9oMThERcFnUvBxb1ilCPWqIrLZynjBbh7zkszQ308/s320/pic9.jpg)
Now, as shown above equate the two terms for 2c to each other, or 2x + 2y(dy/dx) = x + (y^2)/x, and solve this expression for dy/dx = (y^2 - x^2)/2yx.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiIaWXF18M-sFLjOFIQ52Xu5OUSk6CgKt_43by3QcwmuKisUQOIQacER866-n6rufMDYLiFAfTpz_PiECwMtSlqXVHzAslVIzuQjpSj0Zvyj2y3raecej-AznvJcJrrqCAoTGXUS7Hz5Qo/s320/pic10.jpg)
"In summary we see that we can pass back and forth between a differential equation and its family of solution curves."(Krantz)
You can go to DE Ex2 here...D.E. Ex2.
You can go to DE Ex1 here ...D.E. Ex1.
Comments
Post a Comment