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Jan26-12, 02:52 AM
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P: 10,380
Tangent Vectors on a 2D Graph

Quote Quote by ZdravkoBG View Post
Well, the general equations would be y=mx+b
However, at this point in the class, we are only using parametrics for lines.

In other words, the only lines we have built so far (involving vectors) have been in this format:

x = a + bt
y = a + bt
It is easier to use the first equation. But for both methods, you know one point of the line: (0,2), and need the
the slope m or the tangent vector of the line: For that, evaluate f''(x) = 3e^(3x) + 2cos(2x) + 3 at x=0.