Question

QUESTION 16 Find the arc length of the helix traced by r(t) = <p cost, p sint,pt> for Osts2.when the value of p-3. QUESTION 17 The equation passes through the points P (6,1,-1). Q(0, -2.0) and R(1.-1.7) is given by ax+by+czek Follow the steps below and find value of k. Lick Save and Submit to save and submit. Click Save All Answers to save all answers. A ) 17 F5 F4 F3 F2 % 7

Answer 1

From equation ③ and P=1 20 a-(-K/2) +76K at k/2 +7C=K atrc =k-K/2 atte = k/2 hoy ☆a= k 7e Plug in the value of a in egration © , '6 ( 4 2 - 7 C) -e = 3k » 3k - 42c-c = 3 43 C = K - 3k 7 43 c = 3k ; Plug in the value of & in equation & weget, Ķ -7035 3k AN dt given: r(t) = <p cost , p sint , pt) for Osts at 78(t) = < g (t), h(t), f(t)) where g(t) = Pcost, h(t) = psint and f(t)=Pt We know that, Ate length (2) = ( (8'(+372 + (6 (0)7² + (8 (+)) ?/** 10 g' (t) = (g(t) = a (P Cost) = -P sinct) : . (g' (t)) = (-psint) 2 = P² sin²t f'(t) = (h(t)] = (P sint) = peost 2. Ch' (t))² = (P cost) 2 = p2 cost f'(tai (lt)) = (Pt) =P . (f' (t)) ² = (P)2 =p² From equation we get L = 821 / p²sin²t + p2265²t + p2 dt = 12 w p2 (sin?t + cos² t) + p 2 dt dt ES = f d 2p2 dt rap2 [t] = √2 [2a-o (as, p=1) = 252 al units. Gire equation : The points passes through the given equation are P(6, 1,-1), 8(0, -2,0) and R (1,-1,7) The points must satisfy the equation. 6a+b-c.=K C for point p) ; b=-- and a-b + TC = K (for point "R) - ③ Plug in the values of b) in the equations@x® 6a.-K/2-C=k ² 6a-c=k+1/2 60-c=3K 24 0 27 2 K-21k 2 43K-21k 86 22K 86 11K 43 w p2 p2 :24 axt by tez =k (17 SHOT ON REDMI Y3 AI DUAL CAMERA AS, 3K 11K and ca 86 43 Na²+6²+ c2 +-+*821 7 KN 1849 7 KV .0.316657 = 1.77 (approx) √0.316657 0.56272 IIK 721 9. + 4 + 7396 = K = 2 0-2b to =k & 77 -26= Ksor point a