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We let the vector be the velocity vector. If the bee travels along a path in given by , then is given by

We are told that the bee is travelling along a path defined by the intersection of the two surfaces: and . Therefore, we can differentiate each relationship with respect to t to get a pair of equations describing the velocity of the bee at any point along the curve:

We are interested in analyzing the velocity at (1,1,0), so plug this point into the above equations:

Solving the two equations to get in terms of gives:

So at the point (1,1,0), we can write the velocity vector in terms of :

We know that the speed of the bee is 6 at the point (1,1,0), so we use the equation

Solving for gives

Taking the positive solution, (since the bee is travelling in the direction of increasing z, we can write the velocity vector

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