Solving a Math Problem: ΣKQ^n & ΔQΣKnQ^(n-1)

  • Thread starter Thread starter foo9008
  • Start date Start date
AI Thread Summary
The discussion focuses on understanding the mathematical expression ΔQΣKnQ^(n-1) in relation to ΣKQ^n. The key point is that the transformation relies on the approximation that ΔQ is much smaller than Q, allowing the expansion of (Q + ΔQ)^2 to simplify to Q^2 + 2QΔQ. This approximation is valid in later iterations, as ΔQ decreases. The conversation also touches on differentiation, relating it to the Newton method for solving equations. Overall, the participants are working through the nuances of these mathematical concepts to clarify the problem.
foo9008
Messages
676
Reaction score
4

Homework Statement


in the second picture (refer to the circled part) , i can understand the ΣKQ^n , but i don't understand the second one , why it will become ΔQΣKnQ^(n-1) ?

Homework Equations

The Attempt at a Solution

 

Attachments

  • 01.jpg
    01.jpg
    45.8 KB · Views: 344
  • 02.jpg
    02.jpg
    45.1 KB · Views: 372
  • 03.jpg
    03.jpg
    32.6 KB · Views: 329
Physics news on Phys.org
Can't read the exponents. Did you understand 5.13 ?
 
BvU said:
Can't read the exponents. Did you understand 5.13 ?
not really
 
It all hinges on $$\Delta Q << Q \ \ \Rightarrow \ \ (Q + \Delta Q)^2 = Q^2 + 2 Q\, \Delta Q + (\Delta Q)^2\approx Q^2 + 2 Q \Delta Q$$ which is satisfied (maybe not in the first iteration, but later on it is) -- ##\Delta Q## gets smaller and smaller if you do things right.
(I filled in n = 2 for simplicity)

You could also see this as a differentiation ( ##{dh\over dQ} = 2KQ {\rm \ \ or \ \ } \Delta h = 2 Q \Delta Q ## ) and then the method is basically the Newton method
 
BvU said:
It all hinges on $$\Delta Q << Q \ \ \Rightarrow \ \ (Q + \Delta Q)^2 = Q^2 + 2 Q\, \Delta Q + (\Delta Q)^2\approx Q^2 + 2 Q \Delta Q$$ which is satisfied (maybe not in the first iteration, but later on it is) -- ##\Delta Q## gets smaller and smaller if you do things right.
(I filled in n = 2 for simplicity)

You could also see this as a differentiation ( ##{dh\over dQ} = 2KQ {\rm \ \ or \ \ } \Delta h = 2 Q \Delta Q ## ) and then the method is basically the Newton method
removed
 
??
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top