We are given two vectors:
- $\vec{P} = f(t)\hat{i} + g(t)\hat{j} + \hat{k}$
- $\vec{Q} = -f''(t)\hat{i} + f'(t)\hat{j} + \hat{k}$
The problem states that $\vec{P}$ is always parallel to $\vec{Q}$. This means that $\vec{P}$ is a scalar multiple of $\vec{Q}$. Let this scalar multiple be $k$. So, $\vec{P} = k\vec{Q}$.
Equating the components of the vectors:
- $f(t) = k(-f''(t))$
- $g(t) = k(f'(t))$
- $1 = k(1)$
Determining the Scalar Multiple (k)
From the third component equality, $1 = k(1)$, we directly find the scalar $k=1$.
Vector P Condition
Substituting $k=1$ back into the first two component equations gives us the relationships:
- $f(t) = -f''(t)$
- $g(t) = f'(t)$
The first relationship, $f(t) = -f''(t)$, can be rewritten as $f''(t) + f(t) = 0$. This is a standard second-order linear homogeneous differential equation, characteristic of simple harmonic motion. The solutions for $f(t)$ are typically of the form $f(t) = A \cos(\omega t) + B \sin(\omega t)$, where A, B, and $\omega$ are constants.
Magnitude Calculation
The magnitude of vector $\vec{P}$ is calculated as:
$|\vec{P}| = \sqrt{(f(t))^2 + (g(t))^2 + (1)^2}$
Now, substitute the derived conditions $f(t) = -f''(t)$ and $g(t) = f'(t)$ into the magnitude formula:
$|\vec{P}| = \sqrt{(-f''(t))^2 + (f'(t))^2 + 1}$
$|\vec{P}| = \sqrt{(f''(t))^2 + (f'(t))^2 + 1}$
Verifying Constant Magnitude
Let's consider the general solution $f(t) = A \cos(t) + B \sin(t)$ (assuming $\omega=1$ without loss of generality, as any $\omega$ leads to a constant result). Its derivatives are:
- $f'(t) = -A \sin(t) + B \cos(t)$
- $f''(t) = -A \cos(t) - B \sin(t)$
Now, let's compute $(f'(t))^2 + (f''(t))^2$:
$(f'(t))^2 = (-A \sin(t) + B \cos(t))^2 = A^2\sin^2(t) - 2AB\sin(t)\cos(t) + B^2\cos^2(t)$
$(f''(t))^2 = (-A \cos(t) - B \sin(t))^2 = A^2\cos^2(t) + 2AB\sin(t)\cos(t) + B^2\sin^2(t)$
Adding these two:
$(f'(t))^2 + (f''(t))^2 = A^2(\sin^2(t) + \cos^2(t)) + B^2(\cos^2(t) + \sin^2(t)) = A^2(1) + B^2(1) = A^2 + B^2$
Substitute this back into the magnitude expression for $\vec{P}$:
$|\vec{P}| = \sqrt{(A^2 + B^2) + 1}$
Since $A$ and $B$ are constants determined by the specific function $f(t)$, the value $A^2 + B^2 + 1$ is a constant. Therefore, the magnitude of $\vec{P}$ is constant.
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