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# Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Show that the given families of curves are orthogonal trajectories of each other; that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes.$x^2 + y^2 = r^2, ax + by = 0$

## $$a x+b y=0$$

Derivatives

Differentiation

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we're told two curves are orthogonal, their tangent lines are perpendicular at each point of intersection. Were asked to show that the given families of curves are orthogonal trajectories of each other. That is every curve in one family is orthogonal to every curve in the other family. Then we're asked to sketch both families of curves in the same axes. The first family of curves is x squared plus y squared equals r squared. These are circles centered at the origin and the second family of curves is A. X plus B. Y equals zero. These are lines through the origin. Intuitively this seems like orthogonal trajectories. Check this. However, when do three things first will differentiate the equation of the family of curves. Then we'll paint a differential equation and then we'll solve this differential equation. So first well differentiate both sides of our first family with respect to X. So we get two X Plus two y times. Dy DX equals zero. Very common. And therefore solving for dy dx we get dy dx equals negative X over Y. Now, in our second step I'm gonna replace dy dx Oh yeah with negative dx Dy in this way will generate curves which are orthogonal to our original family, X squared plus y squared equals r squared. So we get negative dx dy equals negative X over Y. Therefore dx dy equals X. Y. This is a separable differential equation which is easy to solve. So we get D X over X equals dy over why. We split up the derivative into differentials, integrating both sides. We have the integral of D X over X equals the integral of Dy over why. And therefore the natural log of the absolute value of X equals the natural log of the absolute value of Y plus some constant C. Hi guys now combining terms and using log properties, we get the natural log of absolute value of X divided by Y equals C. Using inverse functions we have absolute value of X over Y equals E. To the constant C. Right. And therefore we obtained the X over Y equals plus or minus E. To the sea. Yeah. Now we'll let plus or minus E to the C. Which is a constant. Be B over a. Sorry, negative B over A. Then it follows that X over Y equals negative B over A. And therefore A X plus B. Y equals zero. Or thanks. We have shown that these two families of curves right, X squared plus Y squared equals R squared and A X plus B. Y equals zero are orthogonal trajectories. I love beef jerky. That was, you know, okay. So I was vegan but I did eat beef jerky.

Ohio State University

Derivatives

Differentiation

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