Any girls here like abs?

Find the arc length parameter along the given curve from the point where tequals=0 by evaluating the integral s(t)equals=integral from 0 to t startabsolutevalue bold v left parenthesis tau right parenthesis endabsolutevalue d tau∫0tv(τ) dτ. then find the length of the indicated portion of the curve r(t)equals=1010cosine tcost iplus+1010sine tsint jplus+88t k, where 0less than or equals≤tless than or equals≤startfraction pi over 3 endfraction π 3.

Pick the expression that matches this description: a polynomial of the 5th degree with a leading coefficient of 7 and a constant term of 6 (choice a) 7x^5+2x^2+6 (choice b) 7x^6−6x^4+5 (choice c) 6x^7−x^5+5 (choice d) 6x^5+x^4+7

Fast! find the length of cu. the triangles are similar. show your work.