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By David Bachman

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Not if we want to define a linear operator. The result of ω ∧ ν is not just an area, it is a signed area; it can either be positive or negative. We will see a geometric interpretation of this soon. 3 Multiplying 1-forms ω ∧ ν(V1 , V2 ) = 39 ω(V1 ) ω(V2 ) . 5. Let ω and ν be the following 1-forms: ω( dx, dy ) = 2dx − 3dy ν( dx, dy ) = dx + dy. 1. Let V1 = −1, 2 and V2 = 1, 1 . Compute ω(V1 ), ν(V1 ), ω(V2 ) and ν(V2 ). 2. Use your answers to the previous question to compute ω ∧ ν(V1 , V2 ). 3. Find a constant c such that ω ∧ ν = c dx ∧ dy.

2 Hence a parameterization for this is (t, 1 − t ), where −1 ≤ t ≤ 1. This figure is also the graph of the polar equation r = 1, 0 ≤ θ ≤ π, hence the parameterization (cos t, sin t), where 0 ≤ t ≤ π. 3. Sketch and find parameterizations for the curves described by: 1. The graph of the polar equation r = cos θ. 2. The graph of y = sin x. 4. Find a parameterization for the line segment which connects the point (1, 1) to the point (2, 5). Parameterized curves may be familiar from a second semester calculus class.

So far all we have done is to take this pair of vectors and return another pair of vectors. But do we know of a way to take these vectors and get a number? Actually, we know several, but the most useful one turns out to be the area of the parallelogram that the vectors span. This is precisely what we define to be the value of ω ∧ ν(V1 , V2 ) (see Fig. 4). z ν V1 V2 ν(V1 ) ω ω(V1 ) x y Fig. 4. The product of ω and ν. Example 14. Let ω = 2dx − 3dy + dz and ν = dx + 2dy − dz be two 1forms on Tp R3 for some fixed p ∈ R3 .

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