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

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The discussion focuses on calculating 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 between t=0 and t=9. The correct formula for the distance traveled is the integral of the square root of the sum of the squares of the derivatives, specifically ∫ from 0 to 9 of √((dx/dt)² + (dy/dt)²) dt. A participant initially set up an incorrect integral that calculated area instead of distance. The conversation clarifies the distinction between calculating area under the curve and the actual length of the curve. The correct approach was confirmed, leading to a successful calculation of the distance.
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Consider the parametric curve given by the equations
x(t) = t^2+30t-11
y(t)=t^2+30t+38
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:

dx/dt = 2t+30

my integral:

\int_{0}^{9}(t^2+30t+38)*(2t+30)

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
x(t) = t^2+30t-11
y(t)=t^2+30t+38
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:

dx/dt = 2t+30

my integral:

\int_{0}^{9}(t^2+30t+38)*(2t+30)

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

The distance traveled is

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

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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