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(1 point) Starting from the point (-4,-1,0) reparametrize the curve r(t) = (-4+ 3t)i + (-1+2t)j...
(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.
(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.
(1 point) Starting from the point (1,1, -3) reparametrize the curve r(t) = (1 – 1t)...
(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
Reparametrize the curve with respect to arc length measured from the point where t = 0...
Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter your answer in terms of s.) r(t) = 2t i + (2 − 3t) j + (8 + 4t) k r(t(s)) =
(1 point) Given the acceleration vector a(t) = (-4 cos (2t))i + (-4 sin (2t))j +...
(1 point) Given the acceleration vector a(t) = (-4 cos (2t))i + (-4 sin (2t))j + (-3t) k , an initial velocity of v (0) =i+ k, and an initial position of r (0)=i+j+ k, compute: A. The velocity vector v (t) = j+ . B. The position vector r(t) = j+ k
Match each given vector equation with the corresponding curve. y4 0 b a (0, 1,0) (1,0,0 , 1,0 d C 2 A (0,0. 2 y- r...
Match each given vector equation with the corresponding curve. y4 0 b a (0, 1,0) (1,0,0 , 1,0 d C 2 A (0,0. 2 y- r(t)= (, ? r(t) (sin (t),t) r (t) (t, cos (2t), sin (2t)) ? v r (t) (1 +t,3t,-t) r (t) (t)i-cos (t)j+sin (t) k =COS r(t)=i+tj+k r(t) i+tj+2k r(t)= (1,cos (t).2sin (t)
Match each given vector equation with the corresponding curve. y4 0 b a (0, 1,0) (1,0,0 , 1,0 d C 2 A...
(1 point) Given R(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tkR(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tk Find the derivative R′(t)R′(t) and norm of the derivative. R′(t)=R′(t)= ∥R′(t)∥=‖R′(t)‖= Then...
(1 point)
Given
R(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tkR(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tk
Find the derivative R′(t)R′(t) and norm of the derivative.
R′(t)=R′(t)= ∥R′(t)∥=‖R′(t)‖=
Then find the unit tangent vector T(t)T(t) and the principal
unit normal vector N(t)N(t)
T(t)=T(t)= N(t)=N(t)=
(1 point) Given R(t) = cos(36) i + e sin(3t) 3 + 3e"k Find the derivative R') and norm of the derivative. R'(t) = R' (t) Then find the unit tangent vector T(t) and the principal unit normal vector N() T(0) N() Note: Yn can can on the hom
e.) What is the equation of the tangent plane to the function z = x2 +...
e.) What is the equation of the tangent plane to the function z = x2 + 4y2 at the point with x = 2, y = -1? [8 points) f.) For the curve through space F(t) =< sin(3t), cos(3t), 2t>, what is the unit tangent vector at t = 7/2? [8 points] g.) Starting from t= 0, reparameterize the curve r(t) = (1 - 2t) î +(-4+ 2t)ſ+(-3 – t)k in terms of arclength. [8 points]
Find the distance of the point (3,4,-4) from the line r(t) = (1 + 3t, -1...
Find the distance of the point (3,4,-4) from the line r(t) = (1 + 3t, -1 + 2t, 5 – 3t). Answer:
Find the curvature of the space curve. r(t) = -5 i + (10 + 2t)j +...
Find the curvature of the space curve. r(t) = -5 i + (10 + 2t)j + (t? + 8) k Ov-2021 2052 or OK 2(+1312
(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 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 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 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 point) Given the acceleration vector a(t) = (-4 cos (2t))i + (-4 sin (2t))j + (-3t) k , an initial velocity of v (0) =i+ k, and an initial position of r (0)=i+j+ k, compute: A. The velocity vector v (t) = j+ . B. The position vector r(t) = j+ k
Match each given vector equation with the corresponding curve. y4 0 b a (0, 1,0) (1,0,0 , 1,0 d C 2 A (0,0. 2 y- r(t)= (, ? r(t) (sin (t),t) r (t) (t, cos (2t), sin (2t)) ? v r (t) (1 +t,3t,-t) r (t) (t)i-cos (t)j+sin (t) k =COS r(t)=i+tj+k r(t) i+tj+2k r(t)= (1,cos (t).2sin (t)
Match each given vector equation with the corresponding curve. y4 0 b a (0, 1,0) (1,0,0 , 1,0 d C 2 A...
(1 point)
Given
R(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tkR(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tk
Find the derivative R′(t)R′(t) and norm of the derivative.
R′(t)=R′(t)= ∥R′(t)∥=‖R′(t)‖=
Then find the unit tangent vector T(t)T(t) and the principal
unit normal vector N(t)N(t)
T(t)=T(t)= N(t)=N(t)=
(1 point) Given R(t) = cos(36) i + e sin(3t) 3 + 3e"k Find the derivative R') and norm of the derivative. R'(t) = R' (t) Then find the unit tangent vector T(t) and the principal unit normal vector N() T(0) N() Note: Yn can can on the hom
e.) What is the equation of the tangent plane to the function z = x2 + 4y2 at the point with x = 2, y = -1? [8 points) f.) For the curve through space F(t) =< sin(3t), cos(3t), 2t>, what is the unit tangent vector at t = 7/2? [8 points] g.) Starting from t= 0, reparameterize the curve r(t) = (1 - 2t) î +(-4+ 2t)ſ+(-3 – t)k in terms of arclength. [8 points]
Find the distance of the point (3,4,-4) from the line r(t) = (1 + 3t, -1 + 2t, 5 – 3t). Answer:
Find the curvature of the space curve. r(t) = -5 i + (10 + 2t)j + (t? + 8) k Ov-2021 2052 or OK 2(+1312
(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 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)
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