Find parametric equations for the tangent line to the curve with the given parametric equation s at the specified point.x=t2+1,y=4t,z=et2−t;(2,4,1)

Question

Cruz Marshall · Accepted Answer

Frist we need to find the tangent vector r&#039;(t)We haver(t)=t2+1i+4tj+et2−tkDifferentiate to get the tangent vectorr′(t)=2ti+2tj+(2t−1)et2−tkNote that the point (2,4,1) corresponds to t=1Since as we put t=1 in r(t), we get (2,4,1) as the position.r′(1)=2⋅1i+21j+(2⋅1−1)e12−1kr&#039;(1)=2i+2j+kRemember that: Equation of a line passing through a point with position vector a, and parallel to the vector b is r(t)=a+t bHence equation of the tangent line at (2,4,1) isr(t)=⟨2,4,1⟩+t⟨2,2,1⟩r(t)=⟨2+2t,4+2t,1+t⟩Hence the parametric equations of the line arex=2+2t, y=4+2t, z=1+tResult:The required equation for tangent vector is x=2+2t, y=4+2t, z=1+t.

Find parametric equations for the tangent line to the curve with the given parametric equation s at the specified point. x=t^(2)+1, y=4sqrt(t), z=e^(t^(2)-t);(2,4,1)

Open question

Answer & Explanation

New Questions in Integral Calculus