Question

If \(f(2)=2\hat i-\hat j+2\hat k\) and \(f(3)=4\hat i-2\hat j+3\hat k\), then the value of \(\int_2^3\left(f\cdot\dfrac{df}{dt}\right)dt\) is

• A. \(10\)
• B. \(11\)
• C. \(-10\)
• D. \(-11\)

Hint:

Use \(f\cdot f'= \frac{1}{2}\frac{d}{dt}(|f|^2)\) for integrals involving \(f\cdot \frac{df}{dt}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to evaluate a definite integral involving the dot product of a vector function and its derivative. This can be simplified using a property of differentiation of dot products.

Step 2: Key Formula or Approach:
Recall the product rule for the dot product of a vector function with itself:
\[ \frac{d}{dt}(f \cdot f) = f \cdot \frac{df}{dt} + \frac{df}{dt} \cdot f \] Since the dot product is commutative ($a \cdot b = b \cdot a$), this simplifies to:
\[ \frac{d}{dt}(|f|^2) = 2 \left( f \cdot \frac{df}{dt} \right) \] Therefore, the integrand can be rewritten as:
\[ f \cdot \frac{df}{dt} = \frac{1}{2} \frac{d}{dt}(|f|^2) \]

Step 3: Detailed Explanation:
Substitute the expression from Step 2 into the integral:
\[ \int_2^3 \left( f \cdot \frac{df}{dt} \right) dt = \int_2^3 \frac{1}{2} \frac{d}{dt}(|f(t)|^2) dt \] By the Fundamental Theorem of Calculus, the integral of a derivative gives the function evaluated at the limits:
\[ = \frac{1}{2} \left[ |f(t)|^2 \right]_2^3 \] \[ = \frac{1}{2} \left( |f(3)|^2 - |f(2)|^2 \right) \] Now, we calculate the squared magnitudes of $f(2)$ and $f(3)$.
For $f(2) = 2\hat{i} - \hat{j} + 2\hat{k}$:
\[ |f(2)|^2 = (2)^2 + (-1)^2 + (2)^2 = 4 + 1 + 4 = 9 \] For $f(3) = 4\hat{i} - 2\hat{j} + 3\hat{k}$:
\[ |f(3)|^2 = (4)^2 + (-2)^2 + (3)^2 = 16 + 4 + 9 = 29 \] Substitute these values back into the expression:
\[ \text{Value} = \frac{1}{2} (29 - 9) = \frac{1}{2} (20) = 10 \]

Step 4: Final Answer:
The value of the integral is 10.