(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.
Homework Answers
We reparametrization the given curve in terms of arclength with
starting point (4,3,2).
![Sot 19+9+4 do Set V22 de = 522 st da = 122 []. = 122 (t-0) = t122 (+) = 4122 .: = S - $122 Now Substitute to hear (1) and get](http://img.homeworklib.com/questions/7df9d690-18b2-11ec-ab1a-c3d55b2de005.png?x-oss-process=image/resize,w_560)
Add Answer to:
(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,-1,0) reparametrize the curve r(t) = (-4+ 3t)i + (-1+2t)j...
(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)...
(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
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.
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.
U © © and Me if it) = (x+4?) 7+ (2+t')}+*k , evaluate S'iltydt Reparametrize the...
U © © and Me if it) = (x+4?) 7+ (2+t')}+*k , evaluate S'iltydt Reparametrize the curve with respect to arc length measured from the point where too in the direchin of increasing t. Fit) = (2+3t) i + (8+9+) 3 - 6t he 22 fenchon
(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
let two vectors be a(t) = e^t i + (sin 2t) j + t^3 k and b(t) = (e^-t , cos 3t, - 2 t^3) in euclidean three space R^3. Find d/dt [a(t) * b(t)].
let two vectors be a(t) = e^t i + (sin 2t) j + t^3 k and b(t) = (e^-t , cos 3t, - 2 t^3) in euclidean three space R^3. Find d/dt [a(t) * b(t)].
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 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) 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 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
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.
U © © and Me if it) = (x+4?) 7+ (2+t')}+*k , evaluate S'iltydt Reparametrize the curve with respect to arc length measured from the point where too in the direchin of increasing t. Fit) = (2+3t) i + (8+9+) 3 - 6t he 22 fenchon
(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 curvature of the space curve. r(t) = -5 i + (10 + 2t)j + (t? + 8) k Ov-2021 2052 or OK 2(+1312
Most questions answered within 3 hours.
-
Albinism is an autosomal recessive condition characterized by
absence of melanin pigment from the skin, eye...
asked 1 minute ago -
Suppose you're looking at a physics example online, and come
across this expression in the middle...
asked 9 minutes ago -
I was looking for help with a computer science c programming
class. not c++
This program...
asked 11 minutes ago -
Compile a list (7 or more) of other commands useful for
navigating or manipulating the UNIX/Linux...
asked 18 minutes ago -
How many grams of PbBr2 will precipitate when excess CrBr3
solution is added to 61.0 mL...
asked 21 minutes ago -
If I was given the address of 134.15.0.0/16 from my ISP and I
wanted to use...
asked 25 minutes ago -
What is the pH of the solution that results of dissolving 1.74g
of sodium hydroxide in...
asked 29 minutes ago -
Given a standardized normal distribution (with μ = 0 and a σ =
1), what is...
asked 1 hour ago -
Given the following information:
acetic acid
CH3COOH
Ka = 1.8×10-5
triethylamine
(C2H5)3N
Kb = 5.2×10-4
(1)...
asked 53 minutes ago -
Potassium permanganate(KMNO4)is has a solubility of 6.4 g/ 100 g
of water at 20ºC, and 250...
asked 50 minutes ago -
51.
As the marginal propensity to expend rises, the multiplier:
decreases.
is impossible to determine.
increases....
asked 56 minutes ago -
The Baldwin Company currently has the following balances on their
balance sheet:
Total
Liabilities
$69,309
Common...
asked 59 minutes ago