Find the domain for the following: r(t) = (3t, squareroot 1 - t^2, 4) a) (-infin

Question

Find the domain for the following: r(t) = (3t, squareroot 1 - t^2, 4) a) (-infinity, infinity) b) (-infinity, -1) union [1, infinity) c) [-1, 1] d) (-1, 1) e) (-infinity, -1) union (1, infinity) Find the limit of the following: lim _t rightarrow 0 [(cos (t))i+(2 sin (t)/t)j+(t^2)k] a) 0 b) 3 c) i d) i+2j e) None When t = 3pi/2 find the unit tangent vector of the curve defined by r(t) = (squareroot t)i+ (t)j a) 1/4 i+j b) 2/squareroot 5 i+1/squareroot j c) 1/squareroot 17 i+ 4/squareroot 17 j d) 1/squareroot 5 i+ 2/squareroot 5 j

Explanation / Answer

21) Thedomain for 3t is all real numbers,

the domain for sqrt(1-t^2) is [-1, 1]

Therefore, the domain of r(t) is the intersection of domains. Hence, the domain of r(t) is [-1, 1].

the correct option is c) [-1, 1].

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