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.

- anonymous

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- anonymous

Something not right, write word for word, verbatim the problem.

- amistre64

a surface would see m more reasonable than a curve; but maybe i havent drawn the curve out anough to see it

- amistre64

the tangent to the postion vector is just the derivatives of the components..

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## More answers

- amistre64

<3,-4t,6t> is te tangent vector at any given point

- 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

the line equation for that vector is:
x = Px +3
y = Py -4t
z = Pz +6t
right?

- 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

the line equation for the tangent vector i meant to say

- anonymous

@ changuanas I understood it as passing through the point (8,-8,11) ? The tangent that is.

- amistre64

when the vector from P(x,y,z) matches the same vector from the given point; they are in line right?

- anonymous

I believe so.

- amistre64

or... simply equate the vector between points maybe

- 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

my tangent vector is wrong, i didnt see the -t in the z spot

- anonymous

lol Hold on I'm solving something, think I got it!

- 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

V =:
x = 7 -3t
y = -10 +2t
z = 11 +t -2t^3
right?

- anonymous

Umm yes, granted I'm not sure how that's helping.. ?

- amistre64

im thinking on it lol

- 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

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

the tangent vector(t) = <3,-4t, 6t-1> right?

- anonymous

Nope. :), well yes but that isnt the answer yet. :P

- amistre64

so lets see if we can equate these vectors and get a result :) it may or maynot work; but im game lol

- amistre64

7-3t = 3
-3t = -4
t = 3/4 maybe?
..................

- amistre64

well, i see a possibility that has a scalar; so let apply that as well; say 's'

- amistre64

7 -3t = 3s
-10+2t = -4st
11+t - 2t^3 = 6st - s got no idea where im going yet :)

- 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

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

err
I g2g to class but I'll explain what I did when I get back... in 3 hours ):

- anonymous

Good luck, finals week. Good work, Amistre, i'm a bit lost, figure it out later.

- amistre64

11+(-13) -2(-13)^3 ?=? 6(46/3)(-13) - 46/3
4392 ?=? 92 - 46/3 ..... taht dint work for me :)

- amistre64

maybe determine the tangent plane the the given is on? :)

- 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

right, the position vector is determined by the vector function; and the tangent is then given by the derivative of the components right?

- amistre64

if it were the F(x,y,z) = 0 then the derivatives would give us a gradient/normal to the plane...

- amistre64

the line from a one point to another in R^3 is the vector from one point to the other right?

- 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

right; the backwards of what i stated is what that looks like ... if i read it right

- amistre64

or the backwards of what im thinking :)

- 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

that vector should be:
<8+3, -8-4t, 11 + 6t-1> right?

- anonymous

Yeah, I'm still thinking how to proceed.

- amistre64

i also did that to the point itself; so those 2 lines should at least be parallel... is my thought

- amistre64

that should also equal the vector from teh given to the point right? or at least some scalar of it?

- anonymous

It is just matter of putting this seeming complicated problem into simple Cal I: find the point of tangency.

- anonymous

froggy click this link
http://www.twiddla.com/537628

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