Question

What is the velocity vector of a particle traveling to the right along the hyperbola y=x^-1 with a constant speed 5cm/s when the particle's location is (2,1/2)?

Answer 1

:  The gradient of the curve 1/x at x=2 is m = -¼ 
We may choose any length of line to represent the direction of the slope (direction vector) at that point. We could choose a line for which x = 2 and then y would have to be -½ so that the gradient is still = -½/2 = -¼. It is simply convenient to choose a unit length for x, making y = -¼ The length of the resultant of x and y is √(1²+¼²) = √(17/16) = √(17)/4 which is a direction vector. If we had taken the direction vector to be (2, ½) then we would have a resultant direction vector of √17/2. It doesn't really matter what length the direction vector is - it's job is only to show the direction. So their choice of 1 is quite arbitrary but convenient, since it is easy to work with units – that's why we use units! 
Now, we know that the magnitude of the velocity vector must be 5 and the magnitude of our direction vector at the moment is √(17)/4. We therefore need to multiply this direction vector by 20/√(17) to get 5 – just try it : √(17)/4 × 20/√(17) = 5. 

We could equally well have done this with (2, ½) and would have got 2½ for lambda.