anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Something not right, write word for word, verbatim the problem.
amistre64
  • amistre64
a surface would see m more reasonable than a curve; but maybe i havent drawn the curve out anough to see it
amistre64
  • amistre64
the tangent to the postion vector is just the derivatives of the components..

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amistre64
  • amistre64
<3,-4t,6t> is te tangent vector at any given point
anonymous
  • anonymous
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?
amistre64
  • amistre64
the line equation for that vector is: x = Px +3 y = Py -4t z = Pz +6t right?
anonymous
  • anonymous
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.
amistre64
  • amistre64
the line equation for the tangent vector i meant to say
anonymous
  • anonymous
@ changuanas I understood it as passing through the point (8,-8,11) ? The tangent that is.
amistre64
  • amistre64
when the vector from P(x,y,z) matches the same vector from the given point; they are in line right?
anonymous
  • anonymous
I believe so.
amistre64
  • amistre64
or... simply equate the vector between points maybe
amistre64
  • amistre64
x = Px +8 y = Py -8 z = Pz +11 = the vector in line with P(x,y,z) and (8,-8,11) right?
amistre64
  • amistre64
my tangent vector is wrong, i didnt see the -t in the z spot
anonymous
  • anonymous
lol Hold on I'm solving something, think I got it!
amistre64
  • amistre64
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?
amistre64
  • amistre64
V =: x = 7 -3t y = -10 +2t z = 11 +t -2t^3 right?
anonymous
  • anonymous
Umm yes, granted I'm not sure how that's helping.. ?
amistre64
  • amistre64
im thinking on it lol
amistre64
  • amistre64
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
anonymous
  • anonymous
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.
amistre64
  • amistre64
the tangent vector(t) = <3,-4t, 6t-1> right?
anonymous
  • anonymous
Nope. :), well yes but that isnt the answer yet. :P
amistre64
  • amistre64
so lets see if we can equate these vectors and get a result :) it may or maynot work; but im game lol
amistre64
  • amistre64
7-3t = 3 -3t = -4 t = 3/4 maybe? ..................
amistre64
  • amistre64
well, i see a possibility that has a scalar; so let apply that as well; say 's'
amistre64
  • amistre64
7 -3t = 3s -10+2t = -4st 11+t - 2t^3 = 6st - s got no idea where im going yet :)
anonymous
  • anonymous
: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.
amistre64
  • amistre64
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....
anonymous
  • anonymous
err I g2g to class but I'll explain what I did when I get back... in 3 hours ):
anonymous
  • anonymous
Good luck, finals week. Good work, Amistre, i'm a bit lost, figure it out later.
amistre64
  • amistre64
11+(-13) -2(-13)^3 ?=? 6(46/3)(-13) - 46/3 4392 ?=? 92 - 46/3 ..... taht dint work for me :)
amistre64
  • amistre64
maybe determine the tangent plane the the given is on? :)
anonymous
  • anonymous
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.
amistre64
  • amistre64
right, the position vector is determined by the vector function; and the tangent is then given by the derivative of the components right?
amistre64
  • amistre64
if it were the F(x,y,z) = 0 then the derivatives would give us a gradient/normal to the plane...
amistre64
  • amistre64
the line from a one point to another in R^3 is the vector from one point to the other right?
anonymous
  • anonymous
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.
amistre64
  • amistre64
right; the backwards of what i stated is what that looks like ... if i read it right
amistre64
  • amistre64
or the backwards of what im thinking :)
anonymous
  • anonymous
A place to start i guess is build a vector from point (8, -8, 11) and P (xyz) That represents the tangent line.
amistre64
  • amistre64
that vector should be: <8+3, -8-4t, 11 + 6t-1> right?
anonymous
  • anonymous
Yeah, I'm still thinking how to proceed.
amistre64
  • amistre64
i also did that to the point itself; so those 2 lines should at least be parallel... is my thought
amistre64
  • amistre64
that should also equal the vector from teh given to the point right? or at least some scalar of it?
anonymous
  • anonymous
It is just matter of putting this seeming complicated problem into simple Cal I: find the point of tangency.
anonymous
  • anonymous
froggy click this link http://www.twiddla.com/537628

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