Length of Curve with Parametric Equations x=cost and y=sint in the xy-plane

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SUMMARY

The length of the curve defined by the parametric equations x = cos(t) and y = sin(t) for 0 ≤ t ≤ π can be calculated using the arc length formula S = ∫₀^π √((dx/dt)² + (dy/dt)²) dt. The derivatives dx/dt and dy/dt are -sin(t) and cos(t), respectively. Substituting these into the formula simplifies the integral, leading to a straightforward calculation. The discussion emphasizes that while the integral method is effective, the length can also be derived directly from the properties of the circle represented by the parametric equations.

PREREQUISITES
  • Understanding of parametric equations
  • Knowledge of calculus, specifically integration
  • Familiarity with the arc length formula
  • Basic trigonometric functions (sine and cosine)
NEXT STEPS
  • Study the derivation of the arc length formula for parametric curves
  • Explore the properties of circles in relation to parametric equations
  • Learn about the application of integrals in calculating lengths of curves
  • Investigate other examples of parametric equations and their lengths
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Students studying calculus, mathematics educators, and anyone interested in understanding parametric equations and arc length calculations.

fk378
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Homework Statement


In the xy-plane, the curve with parametric equations x=cost and y=sint, 0<=t=<pi, has what length?

The Attempt at a Solution



I drew the graphs x=cost and y=sint and shaded the area where the graphs intersect between 0 and pi. I don't know where to go from here.
 
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You should have one graph, with (x, y) coordinates, where both x and y depend on t. That's what parametric equations are about. Try plotting about 10 points and see what you get.
 
fk378 said:

Homework Statement


In the xy-plane, the curve with parametric equations x=cost and y=sint, 0<=t=<pi, has what length?

The Attempt at a Solution



I drew the graphs x=cost and y=sint and shaded the area where the graphs intersect between 0 and pi. I don't know where to go from here.

The fact that you shaded in an area indicates that you are thinking in terms of area, not length along the curve that was suggested to you. You should have an equation for arc length to work with. Try this one:

S = \int_0 ^ \pi \sqrt{ dx^2 + dy^2} dt

Now it's up to you to figure out dx and dy. I think you'll find the answer is very simple.
 
AEM said:
The fact that you shaded in an area indicates that you are thinking in terms of area, not length along the curve that was suggested to you. You should have an equation for arc length to work with. Try this one:

S = \int_0 ^ \pi \sqrt{ dx^2 + dy^2} dt

Now it's up to you to figure out dx and dy. I think you'll find the answer is very simple.

I think you mean sqrt((dx/dt)^2+(dy/dt)^2) for the integrand. But otherwise, good advice.
 
Dick said:
I think you mean sqrt((dx/dt)^2+(dy/dt)^2) for the integrand. But otherwise, good advice.

All well and good, but the parametric curve of this problem is simple enough that its length can be obtained without resorting to an integral.
 
Yeah, sure. But the integral is SO EASY. It's not that much easier to remember the formula than to just derive it, if you are doing calculus. It's that easy. Certainly easier than plotting out 10 points.
 
Hint: (Zcos t, Zsin t) where Z is a some arbitrary constant(in the set +R) represents a circle. In your case, substitute Z=1, 0<t<pi.
 

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