Question

The vector function is given by \(\vec{f}(t) = t\hat{i} + t^2\hat{j} + 5\hat{k}\), then at point \(t = 1\) the slope is

• A. \(\hat{i} + 2\hat{j}\)
• B. \(\hat{i} + 3\hat{j}\)
• C. \(2\hat{i} + \hat{j} + \hat{k}\)
• D. \(\hat{i} + 3\hat{j} + 5\hat{k}\)

Hint:

When asked for the "slope" of a vector function, it almost always refers to the tangent vector, which is its first derivative. Remember to differentiate each component of the vector independently with respect to the parameter.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a vector function \(\vec{f}(t)\) that describes a curve in space, the "slope" at a particular point is represented by the tangent vector to the curve at that point. The tangent vector is found by taking the first derivative of the vector function with respect to the parameter 't'.
Step 2: Key Formula or Approach:
To find the derivative of a vector function \(\vec{f}(t) = f_x(t)\hat{i} + f_y(t)\hat{j} + f_z(t)\hat{k}\), we differentiate each component function with respect to 't': \[ \vec{f}'(t) = \frac{d\vec{f}}{dt} = \frac{df_x}{dt}\hat{i} + \frac{df_y}{dt}\hat{j} + \frac{df_z}{dt}\hat{k} \] Step 3: Detailed Explanation:
The given vector function is: \[ \vec{f}(t) = t\hat{i} + t^2\hat{j} + 5\hat{k} \] To find the tangent vector (slope), we differentiate \(\vec{f}(t)\) with respect to t: \[ \vec{f}'(t) = \frac{d}{dt}(t)\hat{i} + \frac{d}{dt}(t^2)\hat{j} + \frac{d}{dt}(5)\hat{k} \] Using the power rule for differentiation: \[ \vec{f}'(t) = (1)\hat{i} + (2t)\hat{j} + (0)\hat{k} \] \[ \vec{f}'(t) = \hat{i} + 2t\hat{j} \] Now, we need to find the tangent vector at the point t = 1. We substitute t = 1 into the derivative \(\vec{f}'(t)\): \[ \vec{f}'(1) = \hat{i} + 2(1)\hat{j} \] \[ \vec{f}'(1) = \hat{i} + 2\hat{j} \] Step 4: Final Answer:
The slope (tangent vector) at t = 1 is \(\hat{i} + 2\hat{j}\). So, option (A) is correct.