91Ó°ÊÓ

In vector calculus, the velocity vector plays a crucial role in describing how an object's position changes over time. When given a position function like \( \mathbf{r}(t) = \langle 3 \cos t, 3 \sin t, t^{2} \rangle \), the velocity vector \( \mathbf{v}(t) \) is found by taking the derivative of the position function with respect to the parameter \( t \). This derivative is taken component-wise. For clarity:

• The derivative of \( 3 \cos t \) is \( -3 \sin t \).
• The derivative of \( 3 \sin t \) is \( 3 \cos t \).
• The derivative of \( t^2 \) is \( 2t \).

By assembling these components, we derive the velocity vector: \( \mathbf{v}(t) = \langle -3 \sin t, 3 \cos t, 2t \rangle \). This vector signifies the rate and direction of change along each axis. It gives us insights into the object's motion at any instant \( t \).

The acceleration vector \( \mathbf{a}(t) \) is essential in understanding how the change in velocity alters over time. It's derived by taking the derivative of the velocity vector \( \mathbf{v}(t) = \langle -3 \sin t, 3 \cos t, 2t \rangle \) with respect to \( t \). Again, this involves component-wise differentiation:

• The derivative of \( -3 \sin t \) results in \( -3 \cos t \).
• The derivative of \( 3 \cos t \) is \( -3 \sin t \).
• The derivative of \( 2t \) is simply \( 2 \).

Thus, the resulting acceleration vector is \( \mathbf{a}(t) = \langle -3 \cos t, -3 \sin t, 2 \rangle \). This vector indicates how the velocity changes, encompassing both the magnitude and direction of the acceleration at any given moment in time.

Understanding the magnitude of a vector is integral in vector calculus to determine the vector's size, regardless of its direction. For the velocity vector \( \mathbf{v}(t) = \langle -3 \sin t, 3 \cos t, 2t \rangle \), the magnitude, which equates to speed, is calculated using the formula:\[ \| \mathbf{v}(t) \| = \sqrt{(-3 \sin t)^2 + (3 \cos t)^2 + (2t)^2} \]Breaking down this calculation involves squaring each component:

• \((-3 \sin t)^2 = 9 \sin^2 t\).
• \((3 \cos t)^2 = 9 \cos^2 t\).
• \((2t)^2 = 4t^2\).

The Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \) simplifies calculation:\[ \| \mathbf{v}(t) \| = \sqrt{9(\sin^2 t + \cos^2 t) + 4t^2} = \sqrt{9 + 4t^2} \]This result tells us the speed of the object, offering clear insights into how the object's velocity magnitude varies with time.