Find the length of curve r=cos(theta)-sin(theta)

[mill]

Homework Statement

Find the length of the curve r=cosΘ - sinΘ, 0≤Θ≤∏/4

Homework Equations

Arc length = ∫|v| dt

The Attempt at a Solution

I found r'(θ), then used the arc length formula.

Arc length = ∫ sqrt (sin^2 Θ + cos^2 Θ) = ∫ dΘ

and integrated it to find ∫dΘ = ∏/4

The correct answer is however ∏/(2sqrt(2)). Where did I go wrong?

 

[SteamKing]

Homework Statement

Find the length of the curve r=cosΘ - sinΘ, 0≤Θ≤∏/4

Homework Equations

Arc length = ∫|v| dt

The Attempt at a Solution

I found r'(θ), used the arc length formula, and integrated it to find ∫dΘ = ∏/4. The correct answer is however ∏/(2sqrt(2)). Where did I go wrong?

We can't guess unless you post your work.

 

[LCKurtz]

I will guess that he is confusing a parametric (vector) equation of a curve with a polar coordinate equation of a curve.

 

[mill]

I will guess that he is confusing a parametric (vector) equation of a curve with a polar coordinate equation of a curve.

Could you possibly expand on this? I am not sure what you mean.

 

[Curious3141]

Could you possibly expand on this? I am not sure what you mean.

There is quite a different formula for the length of a curve in polar coordinates. It is $\displaystyle s = \int_{\theta_1}^{\theta_2}\sqrt{r^2 + {(\frac{dr}{d\theta})}^2}d\theta.$ When you apply some trig identities, the problem becomes very simple.

 

[LCKurtz]

I will guess that he is confusing a parametric (vector) equation of a curve with a polar coordinate equation of a curve.

Could you possibly expand on this? I am not sure what you mean.

Your given equation is not a parametric equation like $\vec r(t)=\langle x(t),y(t)\rangle$ so your arc length formula doesn't apply. It is a polar coordinate equation like $r = f(\theta)$. Look up the formula for arc length for a polar coordinate equation.

 

[mill]

Thanks so much. I didn't realize there was another formula.