Find the cartesian equation, x=2+3t/1+t y=3-2t/1+t?
回答 (5)
2011-02-11 10:32 am
✔ 最佳答案
Attempt to solve it(x',y'z')= (-3,-2t,1+6t )
since the above is the direction vector of the tangent T then I tried to express the parametric equation of the tangent in function of t which has given me
x=-3s-3
y=-2ts-2
z=(1+6t)s+2
after that I tried to solve xp=x by replacing x in the line equation by the curve equation but I can't solve that !!! I really don't know how to approach this exercise ...
2016-11-30 8:18 pm
x = (2 - 3t)/(a million+t) y = (3+2t)/(a million+t) i'm uncertain why it quite is going to likely be a line. Can this be solved for constants a,b,c ax + by capacity of = c a(2-3t) + b(3+2t) = c(a million+t) i think of it could 2a +3b = c -3a +2b = c 2a + 3b = -3a + 2b 5a +b = 0 enable a=a million then b = -5 and c = -13 x - 5y + 13 = 0
2011-02-11 12:26 pm
We need to eliminate t and leave an equation in just x, y and constants.
x(1 + t) = 2 + 3t
x + xt = 2 + 3t
xt - 3t = 2 - x
so, t = (2 - x)/(x - 3)
y(1 + t) = 3 - 2t
y + yt = 3 - 2t
yt + 2t = 3 - y
so, t = (3 - y)/(y + 2) = (2 - x)/(x - 3)
then, (3 - y)(x - 3) = (2 - x)(y + 2)
i.e. 3x - 9 -yx + 3y = 2y + 4 -xy - 2x
=> 3x + 3y - 9 = 2y - 2x + 4
i.e. 5x + y - 13 = 0
or y = - 5x + 13
then (3 - y)/(y + 2) =
x(1 + t) = 2 + 3t
x + xt = 2 + 3t
xt - 3t = 2 - x
so, t = (2 - x)/(x - 3)
y(1 + t) = 3 - 2t
y + yt = 3 - 2t
yt + 2t = 3 - y
so, t = (3 - y)/(y + 2) = (2 - x)/(x - 3)
then, (3 - y)(x - 3) = (2 - x)(y + 2)
i.e. 3x - 9 -yx + 3y = 2y + 4 -xy - 2x
=> 3x + 3y - 9 = 2y - 2x + 4
i.e. 5x + y - 13 = 0
or y = - 5x + 13
then (3 - y)/(y + 2) =
2011-02-11 10:35 am
x=2+3t/1+t whence t = (2 -- x) / (x -- 3)
y=3-2t/1+t? whence t = (3 -- y) /(y + 2)
equating (2 -- x) / (x -- 3) = (3 -- y) /(y + 2)
OR (2 -- x)(y + 2) = (3 -- y)(x -- 3) OR 5x + y = 13
y=3-2t/1+t? whence t = (3 -- y) /(y + 2)
equating (2 -- x) / (x -- 3) = (3 -- y) /(y + 2)
OR (2 -- x)(y + 2) = (3 -- y)(x -- 3) OR 5x + y = 13
2011-02-11 10:35 am
Easy to do- you need only eliminate t from these equations. To do this, start by isolating your x in the first equation (I assume that's a 2+3t/(1+t), right?), and after a little manipulation find that t=(x-2)/(5-x). Now just plug this in for t in your other equation and simply to find that y=(13-2x)/3. Hey, it's a straight line!
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