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

In summary, the conversation discusses the concept of the rate of change of a vector function in a specific direction, defined as D_uf(x). The formula for D_uf(x) is given as ||gradient vector|| * ||U|| * cos 0, where ||U|| is the length of the vector and cos 0 is the direction of the vector. The conversation explores the relationship between D_uf(x) and the Ideal Gas Law, specifically in relation to changes in pressure, temperature, and volume. The questions raised include the direction of change in pressure, the optimal rate of change for temperature and volume to achieve maximum dP/dt, and the significance of scaling the magnitude of ∇f(x) and U in a
  • #1
kochibacha
14
0
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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  • #2
Are you surprised? [itex]\nabla f(x)*cos(\theta)[/itex], where [tex]\theta[/tex] is the direction of vector u, is defined as the rate of change of vector function f(x) in the direction of vector v. [itex]D_uf(x)[/itex] 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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  • #3
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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Related to What if it's not a unit vector in directional derivatives

1. What is a unit vector in directional derivatives?

A unit vector in directional derivatives is a vector that has a magnitude of 1 and is used to represent the direction of change in a function.

2. Why is it important for a directional derivative to be a unit vector?

When calculating a directional derivative, using a unit vector ensures that the result is independent of the scale of the vector and only represents the direction of change in the function.

3. What happens if the vector used in a directional derivative is not a unit vector?

If the vector used is not a unit vector, the resulting directional derivative will not accurately represent the direction of change in the function and will be dependent on the scale of the vector.

4. How can I convert a vector into a unit vector for use in directional derivatives?

To convert a vector into a unit vector, divide each component of the vector by its magnitude. This will result in a vector with a magnitude of 1, which can then be used as a unit vector in directional derivatives.

5. Can a directional derivative be calculated using a non-unit vector?

Yes, a directional derivative can be calculated using a non-unit vector, but the result will not accurately represent the direction of change in the function and will be dependent on the scale of the vector.

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