Directions: Work with your group, to complete the following three problems. All

Question

Directions: Work with your group, to complete the following three problems. All work must be completed before leaving the classroom. Show all supporting work on separate paper One final version will be graded from the group. Turn in your scrap paper as well. A curve is defined parametrically by the parametric equations Find the length of the arc of the curve from the origin to the nearest point where there is a vertical tangent. (You will need to find the values of t that will be the bounds) Recall: The Arc Length formula for a parametric curve is: L = integral^b_a Squareroot x'(t) + y'(t)^2 dt

Explanation / Answer

given x(t)=[1 to t](cosu)/u du ,y(t)=[1 to t](sinu)/u du

at origin x(t)=0,y(t)=0 => t =1

x(t)=[1 to t](cosu)/u du ,y(t)=[1 to t](sinu)/u du

differentiate with respect to t

x'(t)=(cost)/t ,y'(t)=(sint)/t

vertical tangent =>x'(t)=0

(cost)/t =0

=>t =/2

arclength L=[1 to /2][(x'(t))2+(y'(t))2] dt

arclength L=[1 to /2][((cost)/t)2+((sint)/t)2] dt

arclength L=[1 to /2][(cos2t+sin2t)/t2] dt

arclength L=[1 to /2](1/t) dt

arclength L=[1 to /2]ln(t)

arclength L=ln(/2) -ln(1)

arclength L=ln(/2) -0

arclength L=ln(/2)

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