Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • one year ago

Write an equation of a curve that intersects at right angles every curve of the family y = ln x+C for every real value of k.

  • This Question is Open
  1. SithsAndGiggles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    In other words: you're asked to find a general equation of a line that is normal (perpendicular) to any given point of \(\ln x+C\). The normal line is perpendicular to the tangent line, which means the slope of the normal line is the negative reciprocal of the slope of the tangent. The slope of the tangent is given by the derivative. \[y=\ln x+C~~\Rightarrow~~\frac{dy}{dx}=\frac{1}{x}~~\Rightarrow~~\text{slope}_{\text{normal}}=-x\] So for some point \((x_0,y_0)\) on \(y=\ln x+C\), that is, \((x_0,\ln x_0+C)\), the normal line has slope \(-x_0\). Also, since it passes through \(x_0,\ln x_0+C\), it will have the equation \[y-\ln x_0-C=-x_0(x-x_0)~~\iff~~y=\ln x_0-x_0(x-x_0)+C\] I'm not sure where the \(k\) comes in; I don't really have any context for the problem. This is just how I would approach it.

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy