Let's solve the problem step by step! Given: \[ \vec{r}(t) = 7t^2 \vec{i} - 3t^2 \vec{j} + \frac{4}{t} \vec{k} \] We are asked to **find the tangent vector**, which is the derivative \(\vec{r}'(t)\), and then evaluate it at \(t=1\). --- ### 1. Differentiate Each Component #### **i-component:** \[ \frac{d}{dt}(7t^2) = 14t \] #### **j-component:** \[ \frac{d}{dt}(-3t^2) = -6t \] #### **k-component:** \[ \frac{d}{dt}\left(\frac{4}{t}\right) = 4 \cdot \frac{d}{dt}(t^{-1}) = 4 \cdot (-1)t^{-2} = -\frac{4}{t^2} \] So the derivative is: \[ \vec{r}'(t) = 14t\,\vec{i} - 6t\,\vec{j} - \frac{4}{t^2}\,\vec{k} \] --- ### 2. Plug in \( t = 1 \): - \(14t \to 14 \cdot 1 = 14\) - \(-6t \to -6 \cdot 1 = -6\) - \(-\frac{4}{t^2} \to -\frac{4}{1^2} = -4\) So, \[ \vec{r}'(1) = 14\,\vec{i} - 6\,\vec{j} - 4\,\vec{k} \] --- ## **Final Answer** \[ \boxed{ \vec{r}'(1) = 14\,\vec{i} - 6\,\vec{j} - 4\,\vec{k} } \] The values you should enter are: - **14** for \(\vec{i}\), - **-6** for \(\vec{j}\), - **-4** for \(\vec{k}\). --- **Explanation:** The tangent vector to the curve at \(t=1\) is just the derivative evaluated at \(t=1\). If you have any more questions, let me know!

Let's carefully rework the problem step-by-step to ensure accuracy. --- ### Given: \[ \vec{r}(t) = 7t^2 \,\vec{i} - 3t^2 \,\vec{j} + \frac{4}{t} \,\vec{k} \] --- ### Step 1: Find the derivative \(\vec{r}'(t)\) Differentiate each component with respect to \(t\): 1. **i-component:** \[ \frac{d}{dt}(7t^2) = 14t \] 2. **j-component:** \[ \frac{d}{dt}(-3t^2) = -6t \] 3. **k-component:** \[ \frac{d}{dt}\left(\frac{4}{t}\right) = 4 \cdot \frac{d}{dt}(t^{-1}) = 4 \cdot (-1)t^{-2} = -\frac{4}{t^2} \] --- ### Step 2: Write the derivative vector: \[ \boxed{ \vec{r}'(t) = 14t\,\vec{i} - 6t\,\vec{j} - \frac{4}{t^2}\,\vec{k} } \] --- ### Step 3: Evaluate at \(t=1\): - \(14t \to 14 \times 1 = 14\) - \(-6t \to -6 \times 1 = -6\) - \(-\frac{4}{t^2} \to -\frac{4}{1^2} = -4\) ### **Final answer:** \[ \boxed{ \vec{r}'(1) = 14\,\vec{i} - 6\,\vec{j} - 4\,\vec{k} } \] --- ### **Summary:** - The derivative (tangent vector) at \(t=1\) is **(14, -6, -4)**. - The components are: | Component | Value | |--------------|--------| | \(i\) | **14** | | \(j\) | **-6** | | \(k\) | **-4** | --- **Note:** If your question was asking for something different, such as the magnitude or a different derivative, please clarify. But based on the provided information, this is the correct and accurate computation.

The derivative of a vector function at a given point represents the tangent vector to the curve at that point, indicating the direction and rate of change of the position vector with respect to the parameter. It is obtained by differentiating each component of the vector function with respect to the parameter.