As for #33:
It's given that the parabola is y = x². Find the vectors parallel to the tangent line at the point (2,4). This is asking for the slope of the parabola y = x² at the point (2,4). As you know from Calculus I, the slope is given by the derivative. Our derivative here is 2x. Plugging 2 into x, we get a slope of 4. Slope is rise over run, so rise (y) = 4, run (x) = 1. Converting into vector notation, we have <1,4> or (i, 4j). Don't forget their "hats." We know that a unit vector is a vector whose length is 1. In order to convert a vector into a unit vector, we must divide that vector by its magnitude. If we have <1,4>, then we can use Pythagorean's theorem to find the magnitude: √(1²+4²). This gives us a magnitude of √(17). We must divide the given vectors by √(17) (or multiply by 1/√(17)) to find the unit vectors we're looking for. That gives us 1/√(17)×<1,4>, or <1/√(17), 4/√(17)>, or 1/√(17)×(i, 4j). Finally, we must note that the question asked us for any vectors that are PARALLEL to this tangent line. A vector parallel to a line could point in the same or opposite direction, hence we add a ±. The final answer comes to ±1/√(17)×(i + 4j).
P.S. This question was already answered in the Calculus III forum under the topic "HWK 9.2 EX 33."