Vector Function Derivative: Velocity Calculation
The provided image contains a vector function labeled as number 4:
\[ \mathbf{r}(t) = 4i + 2\cos(t)j + 3\sin(t)k \]
To solve this question, you would typically need more context, such as finding the derivative of the vector function for velocity, the second derivative for acceleration, or evaluating at a specific point t. However, since the question isn't provided other than the expression for the vector function, I will show you how to find the derivative, which is a common operation on vector functions in calculus.
The derivative of a vector function is found by differentiating each of the component functions with respect to t. Here's how you would find the derivative of the given vector function \(\mathbf{r}(t)\):
\[ \frac{d}{dt}\mathbf{r}(t) = \frac{d}{dt}(4i) + \frac{d}{dt}(2\cos(t)j) + \frac{d}{dt}(3\sin(t)k) \]
For each of the components:
1. The derivative of a constant (4i) is 0.
2. The derivative of \(2\cos(t)j\) with respect to t is \(-2\sin(t)j\).
3. The derivative of \(3\sin(t)k\) with respect to t is \(3\cos(t)k\).
Putting it all together, we get:
\[ \mathbf{r}'(t) = 0i - 2\sin(t)j + 3\cos(t)k \]
So the derivative of the vector function \(\mathbf{r}(t) = 4i + 2\cos(t)j + 3\sin(t)k\) is:
\[ \mathbf{r}'(t) = -2\sin(t)j + 3\cos(t)k \]
This vector function represents the velocity of a particle moving along the path defined by \(\mathbf{r}(t)\) at any time t.
