For x = t^2, y = t, compute the speed ds/dt at t = 2.

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Multiple Choice

For x = t^2, y = t, compute the speed ds/dt at t = 2.

Explanation:
The speed along a parametric path is the magnitude of the velocity vector, so ds/dt = sqrt( (dx/dt)^2 + (dy/dt)^2 ). Here dx/dt = 2t and dy/dt = 1. At t = 2, these are dx/dt = 4 and dy/dt = 1. The speed is sqrt(4^2 + 1^2) = sqrt(16 + 1) = sqrt(17). So the correct value is sqrt(17). The other options would arise if the y-derivative contributed differently (for example, if dy/dt were 0 or 2), which isn’t the case here.

The speed along a parametric path is the magnitude of the velocity vector, so ds/dt = sqrt( (dx/dt)^2 + (dy/dt)^2 ). Here dx/dt = 2t and dy/dt = 1. At t = 2, these are dx/dt = 4 and dy/dt = 1. The speed is sqrt(4^2 + 1^2) = sqrt(16 + 1) = sqrt(17). So the correct value is sqrt(17). The other options would arise if the y-derivative contributed differently (for example, if dy/dt were 0 or 2), which isn’t the case here.

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