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

Step 1: Understanding the Concept:
This question involves the integral of a dot product of a vector with its own derivative. We can simplify the integrand using the product rule for differentiation of a dot product.
Step 2: Key Formula or Approach:
Consider the derivative of the square of the magnitude of a vector \(\vec{A}\), which is \(A^2 = \vec{A} \cdot \vec{A}\). Using the product rule for differentiation: \[ \frac{d}{dt}(\vec{A} \cdot \vec{A}) = \frac{d\vec{A}}{dt} \cdot \vec{A} + \vec{A} \cdot \frac{d\vec{A}}{dt} \] Since the dot product is commutative (\(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\)), this simplifies to: \[ \frac{d(A^2)}{dt} = 2 \vec{A} \cdot \frac{d\vec{A}}{dt} \] Therefore, the integrand can be written as: \[ \vec{A} \cdot \frac{d\vec{A}}{dt} = \frac{1}{2} \frac{d(A^2)}{dt} \] Step 3: Detailed Explanation:
The integral becomes: \[ \int_2^3 \vec{A} \cdot \frac{d\vec{A}}{dt} dt = \int_2^3 \frac{1}{2} \frac{d(A^2)}{dt} dt \] By the Fundamental Theorem of Calculus, integrating a derivative gives the function back: \[ = \frac{1}{2} [A^2]_2^3 = \frac{1}{2} (|\vec{A}(3)|^2 - |\vec{A}(2)|^2) \] Now, we calculate the magnitudes squared: \[ |\vec{A}(2)|^2 = (2)^2 + (-1)^2 + (2)^2 = 4 + 1 + 4 = 9 \] \[ |\vec{A}(3)|^2 = (4)^2 + (-2)^2 + (3)^2 = 16 + 4 + 9 = 29 \] Substitute these values back into the expression: \[ \frac{1}{2} (29 - 9) = \frac{1}{2} (20) = 10 \] Step 4: Final Answer:
The value of the integral is 10.