If velocity components are x'(t) = t and y'(t) = t^2, the speed is sqrt(t^2 + t^4).

If velocity components are x'(t) = t and y'(t) = t^2, the speed is sqrt(t^2 + t^4).

• A. sqrt(t^2 + t^4)
• B. t + t^2
• C. t^2 + t^4
• D. sqrt(t^4 - t^2)

Answer: (A)

Speed is the magnitude of the velocity vector. In the plane, if x′(t) and y′(t) are the velocity components, the speed is sqrt[(x′(t))^2 + (y′(t))^2]. Here x′(t) = t and y′(t) = t^2, so the speed is sqrt(t^2 + (t^2)^2) = sqrt(t^2 + t^4). This can be written as |t| sqrt(1 + t^2), which is nonnegative as speed must be. The other forms don’t represent the magnitude: t + t^2 is just a sum; t^2 + t^4 is the square of the speed; sqrt(t^4 - t^2) uses a difference under the root and isn’t correct.

Speed is the magnitude of the velocity vector. In the plane, if x′(t) and y′(t) are the velocity components, the speed is sqrt[(x′(t))^2 + (y′(t))^2]. Here x′(t) = t and y′(t) = t^2, so the speed is sqrt(t^2 + (t^2)^2) = sqrt(t^2 + t^4). This can be written as |t| sqrt(1 + t^2), which is nonnegative as speed must be. The other forms don’t represent the magnitude: t + t^2 is just a sum; t^2 + t^4 is the square of the speed; sqrt(t^4 - t^2) uses a difference under the root and isn’t correct.