Question

At \( t = 2 \), the slope of the vector function \( \vec{f}(t) = 2\hat{i} + 3\hat{j} + 5t^2\hat{k} \) is

• A. \( 20\hat{k} \)
• B. \( 10\hat{k} \)
• C. \( 5\hat{k} \)
• D. \( 12\hat{k} \)

Hint:

For vector functions \( \vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k} \), the velocity vector is \( \vec{v}(t) = \vec{r}'(t) = x'(t)\hat{i} + y'(t)\hat{j} + z'(t)\hat{k} \), which is tangent to the path of motion. The term "slope" in this context is another way of asking for this tangent/velocity vector.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The "slope" of a vector function at a particular point refers to the tangent vector at that point. The tangent vector is found by taking the derivative of the vector function with respect to its parameter (in this case, \( t \)).
Step 2: Key Formula or Approach:
The tangent vector (or slope) is given by the derivative \( \frac{d\vec{f}}{dt} \).
We need to:
1. Differentiate the vector function \( \vec{f}(t) \) with respect to \( t \).
2. Evaluate the resulting derivative vector at \( t = 2 \).

Step 3: Detailed Explanation:
The given vector function is: \[ \vec{f}(t) = 2\hat{i} + 3\hat{j} + 5t^2\hat{k} \] We differentiate each component of the vector function with respect to \( t \): \[ \frac{d\vec{f}}{dt} = \frac{d}{dt}(2\hat{i}) + \frac{d}{dt}(3\hat{j}) + \frac{d}{dt}(5t^2\hat{k}) \] The derivatives of the constant components are zero: \[ \frac{d\vec{f}}{dt} = 0\hat{i} + 0\hat{j} + (2 \cdot 5t^{2-1})\hat{k} \] \[ \frac{d\vec{f}}{dt} = 10t\hat{k} \] This is the general expression for the tangent vector. Now, we evaluate it at \( t = 2 \): \[ \left.\frac{d\vec{f}}{dt}\right|_{t=2} = 10(2)\hat{k} = 20\hat{k} \] Step 4: Final Answer:
The slope of the vector function at \( t=2 \) is \( 20\hat{k} \). Therefore, the correct option is (i).