Question

The value of \( \int_2^3 \vec{A} \cdot \frac{d\vec{A}}{dt} dt \) if \( \vec{A}(2) = 2\hat{i} - \hat{j} + 2\hat{k} \) and \( \vec{A}(3) = 4\hat{i} - 2\hat{j} + 3\hat{k} \) is

• A. 8
• B. 9
• C. 10
• D. 11

Hint:

The identity \( \vec{A} \cdot \frac{d\vec{A}}{dt} = \frac{1}{2} \frac{d}{dt}(A^2) = A \frac{dA}{dt} \) is extremely useful in vector calculus and mechanics (where it relates to the rate of change of kinetic energy). Recognizing this identity immediately converts a complex integral into a simple evaluation at the endpoints.

Answer 1

The Correct Option is C

Objective: Evaluate the integral of the dot product of a vector with its time derivative.
Methodology: Utilize the product rule for dot products and the Fundamental Theorem of Calculus.
Derivation:Consider the square of the magnitude of vector \(\vec{A}\), defined as \(A^2 = \vec{A} \cdot \vec{A}\). Applying the product rule for differentiation with respect to time \(t\):\[ \frac{d}{dt}(\vec{A} \cdot \vec{A}) = \frac{d\vec{A}}{dt} \cdot \vec{A} + \vec{A} \cdot \frac{d\vec{A}}{dt} \]Due to the commutative property of the dot product, this simplifies to:\[ \frac{d(A^2)}{dt} = 2 \vec{A} \cdot \frac{d\vec{A}}{dt} \]Rearranging, we express the integrand in a more integrable form:\[ \vec{A} \cdot \frac{d\vec{A}}{dt} = \frac{1}{2} \frac{d(A^2)}{dt} \]The integral is transformed as follows:\[ \int_2^3 \vec{A} \cdot \frac{d\vec{A}}{dt} dt = \int_2^3 \frac{1}{2} \frac{d(A^2)}{dt} dt \]Applying the Fundamental Theorem of Calculus:\[ = \frac{1}{2} [A^2]_2^3 = \frac{1}{2} (|\vec{A}(3)|^2 - |\vec{A}(2)|^2) \]Calculate the squared magnitudes at the integration limits:\[ |\vec{A}(2)|^2 = (2)^2 + (-1)^2 + (2)^2 = 4 + 1 + 4 = 9 \]\[ |\vec{A}(3)|^2 = (4)^2 + (-2)^2 + (3)^2 = 16 + 4 + 9 = 29 \]Substitute these values to find the final result:\[ \frac{1}{2} (29 - 9) = \frac{1}{2} (20) = 10 \]Conclusion: The integral evaluates to 10.