How to Calculate the Gradient of a Separation Vector?

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The discussion focuses on calculating the gradient of the separation vector from a fixed point to a variable point in Cartesian coordinates. The separation vector, denoted as c', is defined, and its length c is expressed as the square root of the sum of squared differences in coordinates. The key formula derived is Gradient(1/c) = -c'(hat)/c^2, where "hat" indicates the unit vector in the direction of c'. Participants clarify the differentiation process using the chain rule, emphasizing that primed terms remain constant during differentiation. The conversation concludes with a brief explanation of the term "hat" in this context.
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Let c' be the separation vector from a fixed point(x'',y'',z'') to the point (x,y,z) and let c be its length. show that

Gradient(1/c) = -c'(hat)/c^2

Thnaks for the help
 
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your work?
 
begin by writing 1/c in terms of cartesian coordinates.

c = sqrt[(x - x`)^2 + (y - y`)^2 + (z - z`)^2]
1/c = ?

then differentiate using multiple applications of the chain rule. Remember that the primed terms are constant when differentiating respect to x, y or z. This was the part that confused me at the beginning as I didn't know how to differentiate those.
 
starbaj12 said:
Let c' be the separation vector from a fixed point(x'',y'',z'') to the point (x,y,z) and let c be its length. show that

Gradient(1/c) = -c'(hat)/c^2

Thnaks for the help

What is "hat"?
 
^

mathwizarddud said:
What is "hat"?

"hat" is ^

it means the unit vector in the direction of c' :smile:
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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