Step 1: Understanding the Concept:
Two vectors \( \vec{A} \) and \( \vec{B} \) are parallel if \( \vec{A} = c \vec{B} \) for some scalar \( c \). In this specific problem, both vectors \( \vec{P} \) and \( \vec{Q} \) have their \( \hat{k} \) component equal to 1. This implies that the scalar multiple \( c \) must be equal to 1, meaning the vectors themselves are identical in every component.
Step 2: Key Formula or Approach:
1. Equate corresponding components of \( \vec{P} \) and \( \vec{Q} \).
2. Solve the resulting differential equations for \( f(t) \) and \( g(t) \).
3. Calculate the magnitude \( |\vec{P}| = \sqrt{f(t)^2 + g(t)^2 + 1} \).
Step 3: Detailed Explanation:
Since \( \vec{P} \parallel \vec{Q} \) and the coefficients of \( \hat{k} \) are both 1:
\[ f(t) = -f''(t) \quad \text{and} \quad g(t) = f'(t) \]
The first equation is \( f''(t) + f(t) = 0 \), which is the standard second-order differential equation for simple harmonic motion.
The general solution is:
\[ f(t) = A \cos t + B \sin t \]
Now, let's find \( g(t) \) using the second relationship \( g(t) = f'(t) \):
\[ g(t) = \frac{d}{dt} (A \cos t + B \sin t) = -A \sin t + B \cos t \]
Now we calculate the squared magnitude of \( \vec{P} \):
\[ |\vec{P}|^2 = [f(t)]^2 + [g(t)]^2 + 1^2 \]
Substitute the expressions for \( f(t) \) and \( g(t) \):
\[ |\vec{P}|^2 = (A \cos t + B \sin t)^2 + (-A \sin t + B \cos t)^2 + 1 \]
Expand the squares:
\[ |\vec{P}|^2 = (A^2 \cos^2 t + B^2 \sin^2 t + 2AB \sin t \cos t) + (A^2 \sin^2 t + B^2 \cos^2 t - 2AB \sin t \cos t) + 1 \]
Notice that the cross terms \( 2AB \sin t \cos t \) cancel each other out:
\[ |\vec{P}|^2 = A^2(\cos^2 t + \sin^2 t) + B^2(\sin^2 t + \cos^2 t) + 1 \]
Using the fundamental identity \( \sin^2 t + \cos^2 t = 1 \):
\[ |\vec{P}|^2 = A^2(1) + B^2(1) + 1 = A^2 + B^2 + 1 \]
Since \( A, B, \) and \( 1 \) are all constants, the entire expression \( |\vec{P}|^2 \) is independent of time \( t \).
Thus, the magnitude \( |\vec{P}| = \sqrt{A^2 + B^2 + 1} \) is a constant.
Step 4: Final Answer:
The magnitude of the vector remains unchanged as time progresses. Therefore, the magnitude is constant, which matches option (D).