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Directional Derivative ... Tapp, Definition 3.3 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
I am reading Kristopher Tapp's book: Differential Geometry of Curves and Surfaces ... and am currently focused on Chapter 3: Surfaces ... and in particular on Section 1: The Derivative of a Function from \(\displaystyle \mathbb{R}^m\) to \(\displaystyle \mathbb{R}^n\) ... ...

I need help in order to fully understand Definition 3.3 ... ...

Definition 3.3 reads as follows:



Tapp - Defn - Directional Derivative .png



In the above text from Tapp we read the following:


\(\displaystyle df_p(v) = \displaystyle \lim_{ t \to 0 } \frac{ f(p + tv) - f(p) }{t} = (f \circ \gamma) (0)\) ... ... "



Can someone please demonstrate (formally and rigorously) exactly why/how

\(\displaystyle \displaystyle \lim_{ t \to 0 } \frac{ f(p + tv) - f(p) }{t} = (f \circ \gamma) (0)\) ... ...?


Hope someone can help ...

Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,653
Leeds, UK
Can someone please demonstrate (formally and rigorously) exactly why/how

\(\displaystyle \displaystyle \lim_{ t \to 0 } \frac{ f(p + tv) - f(p) }{t} = (f \circ \gamma)' (0)\) ... ...?
By definition, $(f \circ \gamma) (t) = f(\gamma(t)) = f(p+vt)$. Also by definition, $$(f \circ \gamma)' (0) = \lim_{t\to0}\frac{(f \circ \gamma) (t) - (f \circ \gamma) (0)}t = \lim_{t\to0}\frac{f(p+vt) - f(p)}t.$$ I guess the point here is that $f\circ\gamma$ is a function of a real variable $t$. So its derivative is given by the definition of single-variable calculus.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
By definition, $(f \circ \gamma) (t) = f(\gamma(t)) = f(p+vt)$. Also by definition, $$(f \circ \gamma)' (0) = \lim_{t\to0}\frac{(f \circ \gamma) (t) - (f \circ \gamma) (0)}t = \lim_{t\to0}\frac{f(p+vt) - f(p)}t.$$ I guess the point here is that $f\circ\gamma$ is a function of a real variable $t$. So its derivative is given by the definition of single-variable calculus.


Thanks Opalg ...

Appreciate your help ...

Peter