$Explain proof of directional derivative$

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$Explain proof of directional derivative$

I will try to mark the exact thing I don't understand, it comes from this thread https://matchmaticians.com/questions/u6me4r#answers :
What I don't understand is when we have $$g(h)=f(x_0+ha,y_0+hb)$$
That means that when 'h'(independent variable) changes in g(h), it will produce the same change in 'h' of f(x0+ha,y0+hb).
Derivative formula says that:
$$g'(x) = \lim_{{h \to 0}}\frac{g(x+h)-g(x)}{h}$$
so just changing variable name I receive
$$g'(h) = \lim_{{\Delta{h} \to 0}}\frac{g(h+\Delta{h})-g(h)}{\Delta{h}}$$
And when I want to find it's derivative at 0 I receive:
$$g'(0) = \lim_{{\Delta{h} \to 0}}\frac{g(0+\Delta{h})-g(0)}{\Delta{h}} = \lim_{{\Delta{h} \to 0}}\frac{g(\Delta{h})-g(0)}{\Delta{h}}$$
So the question is since $$g(h)=f(x_0+ha,y_0+hb)$$
Why can I replace $$g(\Delta{h}) by f(x_0+ha,y_0+hb)$$
Since by replacing it I will receive not $$f(x_0+ha,y_0+hb)$$ but $$f(x_0+\Delta{h}a,y_0+\Delta{h}b)$$
P.S. I understand how directional derivatives work, I don't want you to explain it, I just want the explanation of this particular case please, feel free to ask for more money,.

Calculus

Babaduras

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I think you are confsusing $h$ and $\Delta h$. Both these variable are used as variables which will go to zero in the limit, and you should only use one. Otherwise it will be confusing. You can NOT replace

$$g(\Delta{h}) by f(x_0+ha,y_0+hb).$$

Since $$g(h)=f(x_0+ ha,y_0+ h b),$$
by replacing $h$ by $\Delta h$ in the above equation we get

$$g(\Delta{h}) =f(x_0+\Delta ha,y_0+\Delta h b).$$

Note also that there should not be any $h$ in the following computations

$$g'(0) = \lim_{{\Delta{h} \to 0}}\frac{g(0+\Delta{h})-g(0)}{\Delta{h}} = \lim_{{\Delta{h} \to 0}}\frac{g(\Delta{h})-g(0)}{\Delta{h}}.$$

Indeed the proof will not make much sense if you replace $$g(\Delta{h}) by f(x_0+ha,y_0+hb).$$

Erdos

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Erdos

0

Let me know if you need any clarifications.
Babaduras

0

yeah, I think I was confusing delta(h) and (h) all the time. so correct case for derivative of g at 0 is g'(0) = limit of (g(delta(h) - g(0))/delta(h) with delta(h) approaching zero, replaceing g(h) with f(x0+ha,y0+hb) will give me limit of (f(x0+delta(h)a,y0+(delta(h)b) - f(x0,y0))/delta(h) with delta(h) approaching zero ?
- Erdos
  
  0
  
  Replacing g(h) with f(x0+ha,y0+hb) will give me limit of (f(x0+ha,y0+hb) - f(x0,y0))/h with h approaching zero. You statement "replaceing g(h) with f(x0+ha,y0+hb) will give me limit of (f(x0+delta(h)a,y0+(delta(h)b) - f(x0,y0))/delta(h) with delta(h) approaching zero" has the same problem. Remember, you either use h or \Delta h. It does not make sense to use both in one computation.
- Babaduras
  
  0
  
  yeah, I got it, I mean that since g(h) = f(x0+ha,y0+hb), then replacing h with delta(h) will give me imit of (f(x0+delta(h)a,y0+(delta(h)b) - f(x0,y0))/delta(h) with delta(h) approaching zero
- Erdos
  
  0
  
  That's correct.

$Explain proof of directional derivative$

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