What if it's not a unit vector in directional derivatives

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The discussion explores the implications of using a non-unit vector for directional derivatives in the context of the Ideal Gas Law. It highlights that if the vector U is not a unit vector, it alters the rate of change of the function f in any direction, specifically affecting pressure changes in relation to temperature and volume. The calculations show that with specific increments in temperature and volume, the pressure is decreasing at a rate of -4/5 units. Questions raised include the nature of pressure change in the direction of the gradient, optimal rates for temperature and volume changes to maximize pressure change, and the physical significance of scaling the gradient vector. Understanding these concepts can enhance experimental approaches in thermodynamics.
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i came to this topic and they said

Duf(x) = ||gradient vector|| * ||U|| * cos 0

if ||U|| were not a unit vector it would give different rate of change of f in any direction

what would happens if used ||U|| = 10 ?
 
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Are you surprised? \nabla f(x)*cos(\theta), where \theta is the direction of vector u, is defined as the rate of change of vector function f(x) in the direction of vector v. D_uf(x) is the rate of change of vector function f "with respect to vector u" which is essentially using u as the unit of measurement in the same way we define "df/du", with f and u scalar functions, using the "chain rule" in Calculus I. To go from "in the direction" to "with respect to the vector" we have to include the length of the vector, ||u||, as well.
 
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could you help me discuss with this problem , considering a real physical problem would get me a better understanding of what does this mean

from Ideal gas law

PV=nRT

where P = pressure , V= volume , T = temperature, n= constant=1 , R = constant =1
let n = R = 1 , we have P=T/V

what rate should the temperature and volume be changing to make the rate of change of pressure fastest at point T = 25 , V = 10 ?

differentiate with respect to t

dP/dt = ∂P/∂T * dT/dt + ∂P/∂V * dV/dt......∂P/∂T = 1/V , ∂P/∂V = -T/V2


if i set T = 25 +t , V = 10+2t , i.e tempertature starting at 25 degree Celcius with increment of 1 degree per minute and volume of 10 cm3 with increment of 2 cm3 per minute

from this point ∇f(x) would be < 1/V , -T/V2> = < 1/10 ,-25/100>

and U = < dT/dt, dV/dt > = < 1,2 >

Ok now, if i substitute all above and calculate dP/dt = D(1,2)(25,10) = 1/10-1/2 = -4/5 so the pressure is decreasing at rate of -4/5 unit when volume is raised 2 times faster than temperature at the point V= 10 cm3 , T= 25 degree

∇f(x)= < 1/10 ,-25/100> is the direction where rate of change of pressure is fastest which is

√(1/100+1/16) = 0.27 (i don't even know if it's decreasing or increasing at this rate) and it's less than 4/5 if i scale down <1 , 2> to unit vector it will be -0.18 but V(t) = 10+2t/√5 same apply to T(t) also

you see my point?

here are list of questions

1. Is pressure increasing or decreasing at the direction of ∇f(x)
2. How fast should i change my temperature and volume(suppose I am doing an experiment where i can change the rate of these 2 factors)
to give the maximum dP/dt at the given initial point T=T0 and V=V0
3. is there always a minimum dP/dt in the direction of -∇f(x)?

4.what does scaling the magnitude of ∇f(x) and U up and down mean(which of course, affect the value of dP/dt) to my experiment does it make any sense in physical point of view?
 
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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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