Find Distance Covered by Point P on Parametric Curve: 0 to 9

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Homework Help Overview

The discussion revolves around finding the distance covered by a point on a parametric curve defined by the equations x(t) = t^2 + 30t - 11 and y(t) = t^2 + 30t + 38, specifically between the bounds of t = 0 and t = 9.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up an integral for calculating the distance but expresses confusion over the setup and results. They initially calculate dx/dt and propose an integral involving the parametric equations. Another participant points out that the original setup may have calculated area instead of length, prompting further clarification on the correct formula for curve length.

Discussion Status

The discussion includes attempts to clarify the correct approach for calculating the distance along the curve. Some guidance has been offered regarding the formula for curve length, but there is no explicit consensus on the correct setup as the original poster initially struggles with their integral.

Contextual Notes

Participants are working within the constraints of a homework problem, focusing on the correct application of calculus concepts to find the distance along a parametric curve.

ILoveBaseball
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Consider the parametric curve given by the equations
[tex]x(t) = t^2+30t-11[/tex]
[tex]y(t)=t^2+30t+38[/tex]
How many units of distance are covered by the point P(t) = (x(t),y(t)) between t=0, and t=9 ?

well since the bounds are already given (0->9), i just need help on setting up the integral. here's what i done:

[tex]dx/dt = 2t+30[/tex]

my integral:

[tex]\int_{0}^{9}(t^2+30t+38)*(2t+30)[/tex]

but i get the incorrect answer when i integral it, can someone help me set it up?
 
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ILoveBaseball said:
Consider the parametric curve given by the equations
[tex]x(t) = t^2+30t-11[/tex]
[tex]y(t)=t^2+30t+38[/tex]
How many units of distance are covered by the point P(t) = (x(t),y(t)) between t=0, and t=9 ?

well since the bounds are already given (0->9), i just need help on setting up the integral. here's what i done:

[tex]dx/dt = 2t+30[/tex]

my integral:

[tex]\int_{0}^{9}(t^2+30t+38)*(2t+30)[/tex]

but i get the incorrect answer when i integral it, can someone help me set it up?

The distance traveled is

[tex]\int_{0}^{9}\sqrt{(dx/dt)^2+(dy/dt)^2}dt[/tex]

You either do this integral, or notice that the curve is a straight line.

ehild
 
What is the formula for calculating curve's length? You have calculated the area under the curve, not the length.
 
thank you, i got it correct
 

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