How Do You Calculate the Speed of a Particle Given Its Position Function c(t)?

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Homework Help Overview

The discussion revolves around calculating the speed of a particle based on its position function given by c(t)=(sint+t,cost+t). Participants are exploring the mathematical reasoning involved in deriving the speed function from the position function.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the differentiation of the position function to find the speed, with some presenting their derived expressions for speed and others questioning the accuracy of those expressions.

Discussion Status

There are multiple interpretations of the speed function being discussed, with participants providing different formulations and questioning each other's calculations. Some guidance is offered regarding the correct application of mathematical operations, but no consensus has been reached on the final expression.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of assistance they can provide to one another.

Tubig
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1. Find the speed (As a function of t) of a particle whose position at time t seconds is c(t)=(sint+t,cost+t)
2. speed = [x'(t)^2 + y'(t)^2]^1/2
3.
x'(t) = cost + 1
y'(t)= -sint + 1

speed = [(cos t + 1)^2 +(-sint + 1)^2]^1/2
=[cos^2 t+ 2cost + 1 +sin^2 t - 2sint + 1]^1/2
=[cos^2 t + sin^2 t + 2 cost - 2sint +2]^1/2
=[1 + 2(cost - sint + 1)]^1/2
-Thanks
 
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Or (2*(cos(t)-sin(t))+3)^(1/2). That looks ok to me.
 
No, Dick. It would be (2cos(t)- 2sin(t))+3)^(1/2). The "3" would not be multiplied by 2.
 
HallsofIvy said:
No, Dick. It would be (2cos(t)- 2sin(t))+3)^(1/2). The "3" would not be multiplied by 2.

No, Halls. There's an extra set of parentheses around the cos(t)-sin(t). When you cut and pasted you only deleted one of them.
 

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