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September 29th, 2010, 02:40 PM   #1
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Multivariable Calculus: parametric curves problem

Consider the parametric curves C1: , t?R and C2 : , ,t?R. Find all lines that are tangent to both of the parametric curves. * Some clues would be highly appreciative. Thanks
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October 3rd, 2010, 07:06 PM   #2
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Re: Multivariable Calculus: parametric curves problem

The way I went about solving this problem was to eliminate the parameter t to get Cartesian equations for the two curves.

C1: x1 = y1 or y1 = x1^()

C2: y2 = x2

Now, we require

dy1/dx1 = dy2/dx2 = (y2 - y1)/(x2 - x1)

(a) dy1/dx1 = x1^(-)
(b) dy2/dx2 = x2
(c) (y2 - y1)/(x2 - x1) = (x2 - x1^())/(x2 - x1)

Setting (a) and (b) equal, we find

x1^(-) = x2
(d) x2 = x1^(-)

Next, setting (c) and (a) equal, and using (d) to eliminate x2 we have

((x1^(-)) - x1^())/(x1^(-) - x1) = x1^(-)

(x1^(-1) - x1^()) = x1^(-)(x1^(-) - x1)

(x1^(-1) - x1^()) = x1^(-1) -x1^()

Multiply through by 4x1.

1 - 4x1^(3/2) = 2 - 2x1^(3/2)

2x1^(3/2) = -1

x1^(3/2) = -1/(2)

Here, we see the only real solution for x1 occurs when the plus/minus is minus (which we will use hereafter) and is:

x1 = (1/2)^(2/3) = 2^(-?)
x2 = -x1^(-) = -2^(?)

y1 = -x1^() = -2^(-?)
y2 = x2 = 2^(?) = 2^(-4/3)

Let's check to see if the slopes given by (a), (b) and (c) are equivalent.

(a) -x1^(-) = -(2^(-?))^(-) = -(2^(?)) = -2^(-?)
(b) x2 = (-2^(?)) = -2^(-?)
(c) (y2 - y1)/(x2 - x1) = [2^(-4/3) - -2^(-?)]/[-2^(?) - 2^(-?)] = [2^(-4/3)(1 + 2)]/[-2^(-?)(2 + 1)] = -2^(-?)

Thus, the one tangent line to both curves is given by:

y - 2^(4/3) = -2^(-?)[x + 2^(?)] or y + 2^(-?) = -2^(-?)[x - 2^(-?)]

both reduce to (had to check them both to make sure )

y = -2^(-?)x - 2^(-4/3)
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October 4th, 2010, 04:17 AM   #3
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Tangent to C1: (y - t) = (1/(2t))(x - t)
Tangent to C2: (y - u) = u(x - 2u)
For these to coincide, u = 1/(2t) and t/2 = -u, so and
Hence the common tangent has equation (as per the previous post).
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October 4th, 2010, 10:06 AM   #4
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Re: Multivariable Calculus: parametric curves problem

Wow! What a difference knowing what you're doing makes!

I truly appreciate seeing how to do this problem parametrically. It makes sense now that I see it, but I just didn't know how.
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