A curve given parametrically by (x,y,z)=(−1+3t, 2−2t^2, −t+2t^3). There is a unique point P on the curve with the property that the tangent line at P passes through the point (8,−8,11). The point P has coordinates... ? Hmm not quite sure where to start.

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A curve given parametrically by (x,y,z)=(−1+3t, 2−2t^2, −t+2t^3). There is a unique point P on the curve with the property that the tangent line at P passes through the point (8,−8,11). The point P has coordinates... ? Hmm not quite sure where to start.

Mathematics
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Something not right, write word for word, verbatim the problem.
a surface would see m more reasonable than a curve; but maybe i havent drawn the curve out anough to see it
the tangent to the postion vector is just the derivatives of the components..

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<3,-4t,6t> is te tangent vector at any given point
That is the correct wording of the problem. I don't think it would matter if its a curve or a surface, assuming it does actually intersect...? Hmm does that makes sense 3 is never going to pass through 8, am i right?
the line equation for that vector is: x = Px +3 y = Py -4t z = Pz +6t right?
OK, just clumsy wording, it is saying that at some point the tangent line passes through line (8, -8, 11) and that is not the coordinates of point P, I think.
the line equation for the tangent vector i meant to say
@ changuanas I understood it as passing through the point (8,-8,11) ? The tangent that is.
when the vector from P(x,y,z) matches the same vector from the given point; they are in line right?
I believe so.
or... simply equate the vector between points maybe
x = Px +8 y = Py -8 z = Pz +11 = the vector in line with P(x,y,z) and (8,-8,11) right?
my tangent vector is wrong, i didnt see the -t in the z spot
lol Hold on I'm solving something, think I got it!
the vector(x,y,z) from P(-1+3t, 2-2t, -2+2t^3) to (8,-8,11) is equal to; -1+3t +Vx = 8 2-2t + Vy = -8 -2+2t^3 + Vz = 11 right?
V =: x = 7 -3t y = -10 +2t z = 11 +t -2t^3 right?
Umm yes, granted I'm not sure how that's helping.. ?
im thinking on it lol
this gives us the vector from P to the given(g) when this vector matches the tangent vector, we might bea ble to relate them for an answer
They are describing the point P (x,y,z) where the tangent vector r (evaluated at that point), and the given curve r' is the tangent line evaluated at point P.
the tangent vector(t) = <3,-4t, 6t-1> right?
Nope. :), well yes but that isnt the answer yet. :P
so lets see if we can equate these vectors and get a result :) it may or maynot work; but im game lol
7-3t = 3 -3t = -4 t = 3/4 maybe? ..................
well, i see a possibility that has a scalar; so let apply that as well; say 's'
7 -3t = 3s -10+2t = -4st 11+t - 2t^3 = 6st - s got no idea where im going yet :)
:P Well I'm not really sure what I did... but I got the answer which was subbing in t as 1. Forgot what work I did to show that.
s = (7-3t)/3 -4(7-3/t)(t) = 10 +2t -28 +12 = 10 +2t -28 +12 -10 = 2t -26 = 2t t = -13; so..... s = 7-3(-13)/3 = 7+39/3 = 46/3 = 15' 1/3 gong for broke here....
err I g2g to class but I'll explain what I did when I get back... in 3 hours ):
Good luck, finals week. Good work, Amistre, i'm a bit lost, figure it out later.
11+(-13) -2(-13)^3 ?=? 6(46/3)(-13) - 46/3 4392 ?=? 92 - 46/3 ..... taht dint work for me :)
maybe determine the tangent plane the the given is on? :)
There are several relations: first the curve is given in parametric equations. That is usually formed from a point and a vector. So we can work backwards.
right, the position vector is determined by the vector function; and the tangent is then given by the derivative of the components right?
if it were the F(x,y,z) = 0 then the derivatives would give us a gradient/normal to the plane...
the line from a one point to another in R^3 is the vector from one point to the other right?
Another relation is the tangent line containing (8, -8, 11) is a vector let's call it r' This vector is the tangent to the unique point P. Then there is the tangent vector which is the integral of that tangent line. That is the tangent vector which shoots from the origin to point P.
right; the backwards of what i stated is what that looks like ... if i read it right
or the backwards of what im thinking :)
A place to start i guess is build a vector from point (8, -8, 11) and P (xyz) That represents the tangent line.
that vector should be: <8+3, -8-4t, 11 + 6t-1> right?
Yeah, I'm still thinking how to proceed.
i also did that to the point itself; so those 2 lines should at least be parallel... is my thought
that should also equal the vector from teh given to the point right? or at least some scalar of it?
It is just matter of putting this seeming complicated problem into simple Cal I: find the point of tangency.
froggy click this link http://www.twiddla.com/537628

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