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anonymous
 5 years ago
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
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Something not right, write word for word, verbatim the problem.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0a surface would see m more reasonable than a curve; but maybe i havent drawn the curve out anough to see it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the tangent to the postion vector is just the derivatives of the components..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0<3,4t,6t> is te tangent vector at any given point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That 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
 5 years ago
Best ResponseYou've already chosen the best response.0the line equation for that vector is: x = Px +3 y = Py 4t z = Pz +6t right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, 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
 5 years ago
Best ResponseYou've already chosen the best response.0the line equation for the tangent vector i meant to say

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@ changuanas I understood it as passing through the point (8,8,11) ? The tangent that is.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when the vector from P(x,y,z) matches the same vector from the given point; they are in line right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or... simply equate the vector between points maybe

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x = Px +8 y = Py 8 z = Pz +11 = the vector in line with P(x,y,z) and (8,8,11) right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0my tangent vector is wrong, i didnt see the t in the z spot

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol Hold on I'm solving something, think I got it!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the vector(x,y,z) from P(1+3t, 22t, 2+2t^3) to (8,8,11) is equal to; 1+3t +Vx = 8 22t + Vy = 8 2+2t^3 + Vz = 11 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0V =: x = 7 3t y = 10 +2t z = 11 +t 2t^3 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Umm yes, granted I'm not sure how that's helping.. ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0im thinking on it lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this 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
 5 years ago
Best ResponseYou've already chosen the best response.0They 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
 5 years ago
Best ResponseYou've already chosen the best response.0the tangent vector(t) = <3,4t, 6t1> right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nope. :), well yes but that isnt the answer yet. :P

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so lets see if we can equate these vectors and get a result :) it may or maynot work; but im game lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.073t = 3 3t = 4 t = 3/4 maybe? ..................

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, i see a possibility that <t> has a scalar; so let apply that as well; say 's'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.07 3t = 3s 10+2t = 4st 11+t  2t^3 = 6st  s got no idea where im going yet :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0: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
 5 years ago
Best ResponseYou've already chosen the best response.0s = (73t)/3 4(73/t)(t) = 10 +2t 28 +12 = 10 +2t 28 +12 10 = 2t 26 = 2t t = 13; so..... s = 73(13)/3 = 7+39/3 = 46/3 = 15' 1/3 gong for broke here....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0err I g2g to class but I'll explain what I did when I get back... in 3 hours ):

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good luck, finals week. Good work, Amistre, i'm a bit lost, figure it out later.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.011+(13) 2(13)^3 ?=? 6(46/3)(13)  46/3 4392 ?=? 92  46/3 ..... taht dint work for me :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0maybe determine the tangent plane the the given is on? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There 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
 5 years ago
Best ResponseYou've already chosen the best response.0right, the position vector is determined by the vector function; and the tangent is then given by the derivative of the components right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if it were the F(x,y,z) = 0 then the derivatives would give us a gradient/normal to the plane...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the line from a one point to another in R^3 is the vector from one point to the other right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Another 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
 5 years ago
Best ResponseYou've already chosen the best response.0right; the backwards of what i stated is what that looks like ... if i read it right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or the backwards of what im thinking :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A place to start i guess is build a vector from point (8, 8, 11) and P (xyz) That represents the tangent line.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that vector should be: <8+3, 84t, 11 + 6t1> right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I'm still thinking how to proceed.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i also did that to the point itself; so those 2 lines should at least be parallel... is my thought

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that should also equal the vector from teh given to the point right? or at least some scalar of it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is just matter of putting this seeming complicated problem into simple Cal I: find the point of tangency.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0froggy click this link http://www.twiddla.com/537628
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