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(; ) \(11 text(, ) 0 text(, ) 2\)

we have [tex]x=1+10\sqrt{t} ; y=t^{3} -t ;z=t^{3} +t[/tex] and the point (11,0,2); replacing the value in x; 11=1+10[tex]\sqrt{t}[/tex] → 10[tex]\sqrt{t}[/tex]=10 → t=1; checking the value t=1

[tex]y=t^{3}-t=1^{3} -1=0\\ z=t^{3} +t=1^{3} +1=2[/tex],

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

[tex]\frac{dx}{dt}=1+10\sqrt{t} = 1+10t^{\frac{1}{2} }=\frac{10}{2}t^{\frac{-1}{2} }=\frac{5}{\sqrt{t} }\\\frac{dy}{dt}=t^{3}-t=3t^{2}-1\\\frac{dz}{dt}=t^{3}+t=3t^{2}+1\\[/tex],

we denote the tangent vector for all points as

[tex]s(t)^{'}[/tex], so [tex]s(t)^{'} = (\frac{5}{\sqrt{t} })i^{'}+(3t^{2}-1)j^{'} +(3t^{2}+1)k^{'}[/tex] and evaluating the point t=1 in the vector: [tex]s(1)^{'}=5i^{'}+2j^{'}+4k^{'}[/tex], finally the parametric equation line to curve is: (x,y,z) = (11,0,2) + s(5[tex]i^{'}[/tex]+2[tex]j^{'}[/tex]+4[tex]k^{'}[/tex] → x=11+5i, y=2j, z=2+4k