Find parametric equations for the tangent line to the curve with the parametric equations x=t2+3, y=ln⁡(t2+3), z=t at the point (2, ln⁡4, 1)

Question

Find parametric equations for the tangent line to the curve with the parametric equations  x=t2+3, y=ln⁡(t2+3), z=t  at the point  (2, ln⁡4, 1)

Mary Moen · Accepted Answer

Step 1Given:The parametric equations, x=t2+3, y=ln⁡(t2+3), z=t and the point (2, ln⁡(4), 1)To find:The parametric equations for the tangent line to the curve with given parametric equations at the given point.Step 2Let,x=t2+3, y=ln⁡(t2+3), z=t and teh point is (2, ln⁡(4), 1)⇒2=t2+3, ln⁡(4)=ln⁡(t2+3), 1=tNow we will be lookin for the vector parallel to the tangent, and calculate r′(1)Let,r(t)=⟨t2+3, ln⁡(t2+3), t⟩⇒r′(t)=⟨2t2t2+3, 2t(t2+3), 1⟩r′(t)=⟨tt2+3, 2t(t2+3), 1⟩Step 3Put t=1⇒r′(1)=⟨11+3, 24, 1⟩r′(1)=⟨12, 12, 1⟩We can use our given starting point (2, ln⁡(4), 1) and the vector r′(1) to define our line.x=2+12t, y=ln⁡(4)+12t, z=1+tResult:The parametric equations of the required tangent line is,x=2+12t, y=ln⁡(4)+12t, z=1+t

Find parametric equations for the tangent line to the curve

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