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Hi, im having problems with the following problem. Any help would be great, thanks!
Find the two points on the curve y = x^4 - 2x^2 - x that have a common tangent line.
Find the derivative:
The curve of the derivative has an infinite number of coordinates which share the same 'y' values.
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True, but that wouldn't necessarily mean that they'd have a common tangent. For that, the tangents would need to have the same y-intercept as well as the same gradient.
This is quite a tricky and interesting problem, actually. I'll come back to this in a while.
Why did the vector cross the road?
It wanted to be normal.
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So equation of a tangent to the curve at any point x is:
replacing y from original equation gives: which gives intercept asSo say we have two points which have a common tangent, (a,b) and (c,d) then the x coordinates will have to simultaneously satisfy:
andSo we "just" have to solve these for a and c where a is not equal to b.
Ok, then I "cheat" and use mathematica:
eqns = {4a^3 - 4a - 1 == 4c^3 - 4c - 1, -3a^4 + 2a^2 == -3c^4 - 2c^2}
Solve[eqns, {a, c}]
Whicih gives six solutions including:
and
The others are similarly unpleasant.
This solution gives a is approx. 0.853 and c is approx 0.247 and if you look at the graph of the function at these points:
This looks reasonable.
Last edited by gnitsuk (2006-10-26 02:22:37)
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Wow. That was tricky. From the bit where you needed to solve those simultaneous equations, you could do it by writing the second equation in the form (pa²+q)²+r and then writing that in terms of a, then substituting into the first equation and solving for c.
But that would involve a ton of work and Mathematica is a far easier option.
Why did the vector cross the road?
It wanted to be normal.
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