Questions: Consider the following position function. Find (a) the velocity and the speed of the object and (b) the acceleration of the object. r(t)=⟨10 cos t, 10 sin t⟩ for 0 ≤ t ≤ 2 π (a) v(t)= .

Solution Steps

The velocity function is the derivative of the position function \( r(t) \) with respect to time \( t \). Given the position function:

\[ r(t) = \langle 10 \cos t, 10 \sin t \rangle \]

we differentiate each component with respect to \( t \):

\[ v(t) = \frac{d}{dt} \langle 10 \cos t, 10 \sin t \rangle = \langle -10 \sin t, 10 \cos t \rangle \]

The speed of the object is the magnitude of the velocity vector \( v(t) \). The magnitude is calculated as follows:

\[ \text{Speed} = \| v(t) \| = \sqrt{(-10 \sin t)^2 + (10 \cos t)^2} \]

Simplifying the expression:

\[ = \sqrt{100 \sin^2 t + 100 \cos^2 t} = \sqrt{100 (\sin^2 t + \cos^2 t)} = \sqrt{100} = 10 \]

The acceleration function is the derivative of the velocity function \( v(t) \) with respect to time \( t \). We differentiate each component of \( v(t) \):

\[ a(t) = \frac{d}{dt} \langle -10 \sin t, 10 \cos t \rangle = \langle -10 \cos t, -10 \sin t \rangle \]

Final Answer