Question

If $\overline{r}(t) = ti - t^{2}j + (t - 1)k$ and $\overline{s}(t) = 2t^{2}i + 6tk$ then the value of $\int_{0}^{2}\overline{r} \cdot \overline{s} dt$ is}

• A. 4
• B. 6
• C. 8
• D. 12

Hint:

Convert vector multiplication into a scalar expression first, then integrate like a standard calculus problem.

Answer 1

The Correct Option is D

Step 1: Concept
The dot product of two vectors is a scalar function, which can then be integrated over the specified limits.

Step 2: Meaning
Compute $\overline{r} \cdot \overline{s} = (t)(2t^{2}) + (-t^{2})(0) + (t - 1)(6t) = 2t^{3} + 6t^{2} - 6t$.

Step 3: Analysis
Integrate: $\int_{0}^{2} (2t^{3} + 6t^{2} - 6t) dt = [\frac{2t^{4}}{4} + \frac{6t^{3}}{3} - \frac{6t^{2}}{2}]_{0}^{2}$.

Step 4: Conclusion
$[\frac{t^{4}}{2} + 2t^{3} - 3t^{2}]_{0}^{2} = (16/2 + 2(8) - 3(4)) = 8 + 16 - 12 = 12$. Final Answer: (D)