# parametric equations - tangents and length of the arc

• Nov 15th 2009, 12:53 PM
calcbeg
parametric equations - tangents and length of the arc
Hi

x = integral sign (S) a= 1 b=t (cos u)/u du

y = S a=1; b=t (sin u)/u du

question is find the length of arc of the curve from the origin to the nearest point where there is a vertical tangent line

So I am thinking I can find dx/du and where it equals 0 is where the vertical tangent is

however when I try and figure out dx/du it gets ugly

dx/du = (1/u)(-sin u) + (cos u)(-1/sqrt u)
= - sin u/u - cos u/sqrt u
= - sin u - cos^2 u

And I don't have a clue what u is going to equal so that dx/du = 0 so I have a vertical tangent. And then the question becomes do I do that # to 1 or to 0 for the length of the arc?

Thanks for any guidance you can provide.

Calculus beginner
• Nov 15th 2009, 01:03 PM
skeeter
Quote:

Originally Posted by calcbeg
Hi

x = integral sign (S) a= 1 b=t (cos u)/u du

y = S a=1; b=t (sin u)/u du

question is find the length of arc of the curve from the origin to the nearest point where there is a vertical tangent line

So I am thinking I can find dx/du and where it equals 0 is where the vertical tangent is

however when I try and figure out dx/du it gets ugly

dx/du = (1/u)(-sin u) + (cos u)(-1/sqrt u)
= - sin u/u - cos u/sqrt u
= - sin u - cos^2 u

And I don't have a clue what u is going to equal so that dx/du = 0 so I have a vertical tangent. And then the question becomes do I do that # to 1 or to 0 for the length of the arc?

Thanks for any guidance you can provide.

Calculus beginner

$x = \int_1^t \frac{\cos{u}}{u} \, du$

using the Fundamental Theorem of Calculus ...

$\frac{dx}{dt} = \frac{\cos{t}}{t}$

$y = \int_1^t \frac{\sin{u}}{u} \, du$

$\frac{dy}{dt} = \frac{\sin{t}}{t}$

vertical tangent line will be where $\frac{dx}{dt} = 0$ ... $t = \frac{\pi}{2}$

arclength ...

$S = \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$