Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = 1+10 sqrt(t) text(, ) y = t^3 - t text(, ) z=t^3 + t text(; ) \(11 text(, ) 0 text(, ) 2\)

x=11+5i, y=2j, z=2+4K

Step-by-step explanation:

we have and the point (11,0,2); replacing the value in x; 11=1+10  → 10=10 → t=1; checking the value t=1

,

that is correct, let´s go to find the tangent vector:

,

we denote the tangent vector for all points as

, so and evaluating the point t=1 in the vector: , finally the parametric equation line to curve is: (x,y,z) = (11,0,2) + s(5+2+4 → x=11+5i, y=2j, z=2+4k

answer:  

123455421

step-by-step explanation: easy

answer: 0.1,0.85

step-by-step explanation: