Question - Vector Function Derivative: Velocity Calculation
mathphysicsvelocity calculationvector function derivativederivative of vector functioncalculus operation
Solution:
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.
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