Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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?

Answer 6

the line equation for that vector is:

x = Px +3
y = Py -4t
z = Pz +6t

right?

Answer 7

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.

Answer 8

the line equation for the tangent vector i meant to say

Answer 9

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

Answer 10

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

Answer 11

I believe so.

Answer 12

or... simply equate the vector between points maybe

Answer 13

x = Px +8
y = Py -8
z = Pz +11 = the vector in line with P(x,y,z) and (8,-8,11) right?

Answer 14

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

Answer 15

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

Answer 16

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?

Answer 17

V =:

x = 7 -3t
y = -10 +2t
z = 11 +t -2t^3

right?

Answer 18

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

Answer 19

im thinking on it lol

Answer 20

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

Answer 21

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.

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

7-3t = 3

-3t = -4

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

Answer 26

well, i see a possibility that <t> has a scalar; so let apply that as well; say 's'

Answer 27

7 -3t = 3s

-10+2t = -4st

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

Answer 28

: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.

Answer 29

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....

Answer 30

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

Answer 31

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

Answer 32

11+(-13) -2(-13)^3 ?=? 6(46/3)(-13) - 46/3

4392 ?=? 92 - 46/3 ..... taht dint work for me :)

Answer 33

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

Answer 34

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.

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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.

Answer 39

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

Answer 40

or the backwards of what im thinking :)

Answer 41

A place to start i guess is build a vector from point (8, -8, 11) and P (xyz) That represents the tangent line.

Answer 42

that vector should be:

<8+3, -8-4t, 11 + 6t-1> right?

Answer 43

Yeah, I'm still thinking how to proceed.

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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