- #1
Help for Dynamics Continuous Motion
- Thread starter JamieShumJr
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- Continuous Dynamics Motion
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Discussion Overview
The discussion revolves around obtaining specific figures and results from integration in the context of dynamics and continuous motion. Participants seek clarification on integration techniques and evaluation methods related to their calculations.
Discussion Character
- Homework-related
- Mathematical reasoning
Main Points Raised
- One participant asks how to obtain a specific figure indicated in a red circle, indicating a need for assistance with integration.
- Another participant suggests applying a specific integration technique and notes that differentiating the right-hand side yields additional factors that must be considered.
- A participant acknowledges a mistake in neglecting a factor during integration, indicating a common challenge in applying integration rules.
- Another participant inquires about obtaining a specific value (-1/60) at the end of their calculations, seeking clarification on the evaluation process.
- A response explains that the notation requires evaluating the expression at two points and subtracting the results, which is a common step in definite integrals.
- A participant admits to making an error in their evaluation, leading to confusion about the result.
Areas of Agreement / Disagreement
Participants appear to agree on the need for clarification regarding integration techniques and evaluation methods, but there is no consensus on the specific approaches or results discussed.
Contextual Notes
Participants express uncertainty regarding the application of integration rules and the evaluation of expressions, indicating potential gaps in understanding or missing assumptions about the integration process.
- #2
tiny-tim
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Welcome to PF!
Hi JamieShumJr! Welcome to PF!
You're asking why is ∫ (0.8t + A)-1/2 dt = (2/0.8)(0.8t + A)1/2 …
apply the https://www.physicsforums.com/library.php?do=view_item&itemid=353" twice …
if you differentiate the RHS, you get an extra 0.8 from the t, and an extra 1/2 from the power, = (0.8/2)
Hi JamieShumJr! Welcome to PF!
JamieShumJr said:
You're asking why is ∫ (0.8t + A)-1/2 dt = (2/0.8)(0.8t + A)1/2 …
apply the https://www.physicsforums.com/library.php?do=view_item&itemid=353" twice …
if you differentiate the RHS, you get an extra 0.8 from the t, and an extra 1/2 from the power, = (0.8/2)
Last edited by a moderator:
- #3
JamieShumJr
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tiny-tim said:Hi JamieShumJr! Welcome to PF!
You're asking why is ∫ (0.8t + A)-1/2 dt = (2/0.8)(0.8t + A)1/2 …
apply the https://www.physicsforums.com/library.php?do=view_item&itemid=353" twice …
if you differentiate the RHS, you get an extra 0.8 from the t, and an extra 1/2 from the power, = (0.8/2)
Thanks a lot!
I neglected the 0.8 during the integration. Got it already. Forgotten the basic rule of integration. 
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- #4
mintz
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may i know from same question as above shown, how to get the -1/60 at the end there?
- #5
tiny-tim
Science Advisor
Homework Helper
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welcome to pf!
hi mintz! welcome to pf!
the ##|^t_0## at the end of the previous line means you have to evaluate it at t and at 0, and subtract
hi mintz! welcome to pf!
mintz said:
the ##|^t_0## at the end of the previous line means you have to evaluate it at t and at 0, and subtract
- #6
mintz
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tiny-tim said:hi mintz! welcome to pf!
the ##|^t_0## at the end of the previous line means you have to evaluate it at t and at 0, and subtract
oh~okie~thanks! i made mistake that evaluate at 0 whole things became 0~
thanks anyway!=D
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