Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x=3 ln(t), y=4t^{frac{1}{2}}, z=t^{3}, (0, 4, 1)

Joni Kenny 2021-02-12 Answered
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x=3 ln(t), y=4t12, z=t3, (0, 4, 1)
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Expert Answer

timbalemX
Answered 2021-02-13 Author has 108 answers

Step 1 Given: x=3 ln(t) ......(1) y=4t1/2 ......(2) z=t3 ......(3) (x0, y0, z0)=(0, 4, 1) For the specified points, Substitute 1 for z in equation 3. z=t3
1=t3
t=1 The function r(t) of these equations is, r(t)= < x, y, z >Substitute the values in the above equation. r(t)= < 3 ln(t),4t12, t3 > Step 2 Derivative of the vector r(t) is, r(t)= < 3t, 2t, 3t2 > Substitute 1 for t in the above vector. r(1)= < 3, 2, 3 > Step 3 The parametric equations for the tangent line are, X=x0 + (r(1))xt
Y=y0 + (r(1))yt
Z=z0 + (r(1))zt Substitute the values in the above equation. X=3t
Y=4 + 2t
Z=1 + 3t Thus, the parametric equations for the tangent line are X=3t, Y=4 + 2t, and z=1 + 3t.

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