(1 point) Find the arclength s(t) of the curve r(t) Don't forget to submit your answer as a function of t. 9t21+ 8t3j + 4t4k from r(0) to r(t). You can assume that t is positive. s(t) (1 poi...
1 point) Find the arclength of the curve r(t)=(2t2,2y2t,int), for 1-t-6. 1 point) Find the arclength of the curve r(t)=(2t2,2y2t,int), for 1-t-6.
The answer above is NOT correct. (1 point) Find the length of the curve r(t) = i +3t'j + t'k, 0 <t</45 L Preview My Answers Submit Answers Your score was recorded. You have attempted this problem 9 times. You received a score of 0% for this attempt. Your overall recorded score is 0%. You have unlimited attempts remaining. Email WebWork TA WebWork 1996-20
(1 point) Starting from the point (-4,-1,0) reparametrize the curve r(t) = (-4+ 3t)i + (-1+2t)j + (0+2t)k in terms of arclength. r(t(s)) it j+ k
(1 point) Starting from the point (1,1, -3) reparametrize the curve r(t) = (1 – 1t) i + (1 – 1t)j + (-3 – 3t) k in terms of arclength. r(t(s)) = it j+ k
1. Find t(s), n(s), b(s), k(s), T(s) for the following curves (don't forget to reparametrize by arc-length if necessary). (i) a(t) = (e', e' sin(t), e' cos(t)) for te R. (ii) a(t) = (13/2, t, t³/2), on the interval I = (0, o0).
(1 point) Starting from the point (4,3,2)(4,3,2) reparametrize the curve r(t)=(4+3t)i+(3−3t)j+(2−2t)kr(t)=(4+3t)i+(3−3t)j+(2−2t)k in terms of arclength.
Find the arc length parameter along the curve from the point where t=0 by evaluating the integral s | |vIdT. Then find the length of 0 the indicated portion of the curve. The arc length parameter is s(t) (Type an exact answer, using radicals as needed.) Find T, N, and k for the plane curve r(t) (2t+9) i+ (5-t2) j T(t)= (Type exact answers, using radicals as needed.) (Type exact answers, using radicals as needed.) Find the arc length parameter...
Compute the arclength of the following curve. You must SHOW your work to receive full credit. 3 x (t) = 2tz +2, y(t) = 2t -3, 0 <t<1 Your answer may be in the form of a calculator input OR a decimal rounded to 2 decimal places.
(1 point) A parametric curve r(t) crosses itself if there exist t s such that r(t)-r(s). The angle of intersection is the (acute) angle between the tangent vectors r() and r'(s). The parametric curver (2 -2t 3,3 cos(at), t3 - 121) crosses itself at one and only one point. The point is (r, y, z)-5 3 16 Let 0 be the acute angle between the two tangent lines to the curve at the crossing point. Then cos(0.997 (1 point) A...
need help Find the length of the curve defined by the parametric equations y3In(t/4)2-1) from t 5 tot- 7 Find the length of parametized curve given by a(t) -0t3 -3t2 + 6t, y(t)1t3 +3t2+ 0t, where t goes from zero to one. Hint: The speed is a quadratic polynomial with integer coefficients. A curve with polar equation 14 7sin θ + 50 cos θ represents a line. Write this line in the given Cartesian form Note: Your answer should be...