Per-, Perpendicular... Derivative?

[Amaton]

Just came across an idea out of curiosity, so albeit this discussion will be quite "soft". When we differentiate a single-variable function, we are intuitively finding the slope of the tangent line given an arbitrary point.

 

What if we instead observe the slope of the line perpendicular to the tangent? For example, the linear function [math]f(t)=2t[/math]. Differentiating yields [math]\frac{d}{dt} 2t=2[/math]. At any given point in the function, the slope of the tangent line is 2.

 

However, if we looked at the slope of the line perpendicular to the tangent, we'd come across [math]-\frac{1}{2}[/math]... Because at any given point on the function, the slope perpendicular to that infinitesimal line is -1/2.

 

Of course, this is stupidly nontrivial, and I was wondering if I could apply the same concept to non-linear functions. Would I just have to find the negative reciprocal of the function's derivative? Does this idea have any real use in mathematics?

Edited November 17, 2012 by Amaton

[mathematic]

You are looking at the direction of the normal to the curve, which varies for curves which are not straight lines.

[ajb]

Of course, this is stupidly nontrivial, and I was wondering if I could apply the same concept to non-linear functions. Would I just have to find the negative reciprocal of the function's derivative? Does this idea have any real use in mathematics?

 

 

The same thing applies to smooth curves. You can look at the tangent line at a given point and consider the line perpendicular to the tangent line at that point. This is the normal to a curve (at a specified point).

[Amaton]

You are looking at the direction of the normal to the curve' date=' which varies for curves which are not straight lines. [/quote']

The same thing applies to smooth curves. You can look at the tangent line at a given point and consider the line perpendicular to the tangent line at that point. This is the normal to a curve (at a specified point).

 

Thanks guys. I thought this concept had to have some application.

 

Can one derive a function modelling the slope of normals for all points on some (smooth) function?

Edited November 21, 2012 by Amaton

[QuantumBullet]

For the function f(x), finding d/dx f(x) = f'(x) will yield the gradient of the curve at the point x.

Simply take the negative reciprocal to this value to find the gradient of the normal to the curve at this point. (i.e. m = - 1/f'(x) )

Edited December 12, 2012 by QuantumBullet