If x'(t) = 3t and y'(t) = 2, the speed is which expression?

If x'(t) = 3t and y'(t) = 2, the speed is which expression?

• A. sqrt((3t)^2 + 2^2)
• B. 3t + 2
• C. (3t)^2 + 2^2
• D. sqrt(9t^2 - 4)

Answer: (A)

Speed is the magnitude of the velocity vector in the plane, which for parametric coordinates x(t), y(t) is the length of (x′(t), y′(t)). That magnitude is sqrt[(x′(t))^2 + (y′(t))^2]. With x′(t) = 3t and y′(t) = 2, the speed becomes sqrt[(3t)^2 + 2^2] = sqrt(9t^2 + 4). This is the correct expression because it measures how fast the position is changing in any direction. The other forms don’t represent a speed: adding the derivatives gives a scalar that’s not the magnitude, squaring without the square root omits the actual length, and using a minus sign inside the square root would be invalid for real speeds.

Speed is the magnitude of the velocity vector in the plane, which for parametric coordinates x(t), y(t) is the length of (x′(t), y′(t)). That magnitude is sqrt[(x′(t))^2 + (y′(t))^2]. With x′(t) = 3t and y′(t) = 2, the speed becomes sqrt[(3t)^2 + 2^2] = sqrt(9t^2 + 4). This is the correct expression because it measures how fast the position is changing in any direction. The other forms don’t represent a speed: adding the derivatives gives a scalar that’s not the magnitude, squaring without the square root omits the actual length, and using a minus sign inside the square root would be invalid for real speeds.