Question

A vector given by \[ \vec{P}=f(t)\hat{i}+g(t)\hat{j}+\hat{k} \] moves in such a way that it is always parallel to the vector \[ \vec{Q}=-f^{\prime\prime}(t)\hat{i}+f^{\prime}(t)\hat{j}+\hat{k}. \] The magnitude of \(\vec{P}\) is:

• A. a linear function of time
• B. a quadratic function of time
• C. a cubic function of time
• D. constant

Hint:

The transformation pair $f(t) = A\sin t + B\cos t$ and $g(t) = A\cos t - B\sin t$ traces out a uniform circular path in 2D space, satisfying $f^2 + g^2 = R^2$. Adding a constant third component out-of-plane turns the total motion into a uniform helix, which maintains a completely fixed distance from the central axis.

Answer 1

The Correct Option is D

Concept: Two non-zero vectors $\vec{P}$ and $\vec{Q}$ are collinear or parallel if and only if their corresponding directional component parameters are directly proportional. This vector condition translates into a system of independent differential equations. Step 1: Set up the component proportionality system.
Because $\vec{P} \parallel \vec{Q}$, we can write $\vec{P} = \kappa(t) \cdot \vec{Q}$ for some scalar function $\kappa(t)$. Equating components:
• From $\hat{k}$: $1 = \kappa(t) \cdot 1 \implies \kappa(t) = 1$
• From $\hat{i}$: $f(t) = 1 \cdot (-f''(t)) \implies f''(t) + f(t) = 0$
• From $\hat{j}$: $g(t) = 1 \cdot f'(t) \implies g(t) = f'(t)$

Step 2: Solve the resulting linear differential equation.
The relationship $f''(t) + f(t) = 0$ represents a classic second-order homogeneous linear differential equation with constant coefficients. Its characteristic auxiliary equation is $r^2 + 1 = 0 \implies r = \pm i$. The general solution is: $$f(t) = A\sin t + B\cos t$$ where $A$ and $B$ are arbitrary real integration constants.

Step 3: Determine the functional form of $g(t)$.
Using the third component tracking rule $g(t) = f'(t)$, differentiate our solution for $f(t)$: $$g(t) = \frac{d}{dt}(A\sin t + B\cos t) = A\cos t - B\sin t$$

Step 4: Calculate the magnitude of vector $\vec{P}$.
The magnitude of vector $\vec{P}$ is given by the standard Euclidean norm formula: $$|\vec{P}| = \sqrt{f(t)^2 + g(t)^2 + 1^2}$$ Let us evaluate the sum of the squares of the trigonometric components: $$f(t)^2 + g(t)^2 = (A\sin t + B\cos t)^2 + (A\cos t - B\sin t)^2$$ Expanding the binomial expressions: $$= A^2\sin^2 t + B^2\cos^2 t + 2AB\sin t\cos t + A^2\cos^2 t + B^2\sin^2 t - 2AB\sin t\cos t$$ Cancel out the cross-product terms and apply the Pythagorean identity $\sin^2 t + \cos^2 t = 1$: $$= A^2(\sin^2 t + \cos^2 t) + B^2(\cos^2 t + \sin^2 t) = A^2 + B^2$$ Substitute this constant sum back into our absolute magnitude metric: $$|\vec{P}| = \sqrt{A^2 + B^2 + 1}$$ Since $A$ and $B$ are numerical constants, the absolute length value $|\vec{P}|$ is independent of the temporal variable $t$, making it perfectly constant.