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Find the parametric equation?
Find the parametric equation for the line passing through p0=r(2) in the direction of the tangent to C at p0. C is the curve parameterized by r(t)= 2ti+t^2j+lntk. p0 is 4i+4j+ln4k. I don't know what to do after that.
1 Answer
- 8 years agoFavorite Answer
It's been a while since I did this sort of calculus, but I think I remember. (I hate i,j,k notation so I will do the short hand).
We want a line in the direction of r(t) = <2t, t^2, ln(t)> at t = 2. Then we simply take the derivative with respect to t, r'(t) = <2, 2t, 1/t>. Then at t = 2, the direction of the parametric vector curve is r'(2) = <2, 2*2, 1/2> = <2, 4, 1/2>. Then to get the line tangent to that point [p0=r(2)] on the curve, use the point/vector r(2) as a starting point for the line equation:
T(u) = <4, 4, ln(2)> + u*<2,4,1/2> = <4+2u, 4+4u, ln(2)+u/2>
And that's the line passing through p0=r(2) tangent to the parametric curve. Remember, if its a line not a curve you're looking for, you only need i.) a starting point and ii.) a direction vector. Then the thing practically parametrizes itself.
