Compute ds/dt for the curve x = t^2, y = t at t = 0.5.

Compute ds/dt for the curve x = t^2, y = t at t = 0.5.

• A. √3
• B. √5
• C. √2
• D. 2

Answer: (C)

ds/dt represents the instantaneous speed along a parametric curve as the parameter t changes. For a curve given by x(t) and y(t), the arc length element is ds = sqrt((dx/dt)^2 + (dy/dt)^2) dt, so the rate ds/dt is sqrt((dx/dt)^2 + (dy/dt)^2). Here, dx/dt = 2t and dy/dt = 1. So ds/dt = sqrt((2t)^2 + 1^2) = sqrt(4t^2 + 1). At t = 0.5, 4t^2 = 1, giving ds/dt = sqrt(2).

ds/dt represents the instantaneous speed along a parametric curve as the parameter t changes. For a curve given by x(t) and y(t), the arc length element is ds = sqrt((dx/dt)^2 + (dy/dt)^2) dt, so the rate ds/dt is sqrt((dx/dt)^2 + (dy/dt)^2).

Here, dx/dt = 2t and dy/dt = 1. So ds/dt = sqrt((2t)^2 + 1^2) = sqrt(4t^2 + 1). At t = 0.5, 4t^2 = 1, giving ds/dt = sqrt(2).