Question

Suppose \(\mathrm{f}\) is a differentiable mapping of \(R^{1}\) into \(R^{3}\) such that \(|\mathbf{f}(t)|=1\) for every \(t\). Prove that \(f^{\prime}(t) \cdot f(t)=0\). Interpret this result geometrically.

Short answer

Answer: For the given function, we've proved that f'(t) · f(t) = 0, which implies that the vectors f(t) and f'(t) are orthogonal or perpendicular to each other. Geometrically, this means that the tangent vector f'(t) to the curve traced by f(t) in R^3 is always perpendicular to the position vector f(t). The curve can be thought of as always being tangent to the unit sphere centered at the origin, since the magnitude of f(t) is always 1.

Step-by-step solution

Find the derivative of the square of the magnitude of f(t)

Let's first find the derivative of |\mathbf{f}(t)|^2. As we know that the magnitude of f(t) is 1, we have |\mathbf{f}(t)|^2 = 1. Nobody is taking the derivative both sides. [\frac{d}{dt}(|\mathbf{f}(t)|^2) = \frac{d}{dt}(1)]

Use the properties of dot product

Now let's use the property |\mathbf{f}(t)|^2= \mathbf{f}(t) \cdot \mathbf{f}(t). Rewriting the left side of the equation from the previous step: [\frac{d}{dt}(\mathbf{f}(t) \cdot \mathbf{f}(t)) = 0]

Apply the product rule

Next, we apply the product rule for the derivative of a dot product: [\frac{d}{dt}(\mathbf{f}(t) \cdot \mathbf{f}(t)) = \mathbf{f}^{\prime}(t) \cdot \mathbf{f}(t) + \mathbf{f}(t) \cdot \mathbf{f}^{\prime}(t)] Notice that the dot product is commutative, so the last equation simplifies to: [2 \mathbf{f}^{\prime}(t) \cdot \mathbf{f}(t) = 0] Now divide by 2: [\mathbf{f}^{\prime}(t) \cdot \mathbf{f}(t) = 0] So, we have proved the required relation.

Geometrical Interpretation

Geometrically, the result \mathbf{f}^{\prime}(t) \cdot \mathbf{f}(t) = 0 implies that the vectors \mathbf{f}(t) and \mathbf{f}^{\prime}(t) are orthogonal, or perpendicular to each other. In the specific case of \mathbf{f}(t), this means that the tangent vector to the curve traced by \mathbf{f}(t) in R^3, i.e., \mathbf{f}^{\prime}(t), is always perpendicular to the position vector \mathbf{f}(t) itself. This can be thought of as the curve being always tangent to the unit sphere centered at the origin, as the magnitude of \mathbf{f}(t) is always 1.

Key concepts

Dot Product

The dot product, often called the scalar product, is a key concept in vector mathematics. It is calculated between two vectors and gives a single number, or scalar.
The formula for the dot product is:

• For vectors \( \mathbf{a} \) and \( \mathbf{b} \), it is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

It effectively measures how much one vector goes in the direction of another. This means if two vectors are pointing in nearly the same direction, the dot product is positive and large. If they point in opposite directions, it is negative.
And importantly, if they are perpendicular to each other, i.e., orthogonal, the dot product is zero. The property \( \mathbf{f}(t) \cdot \mathbf{f}(t) \) represents the squared magnitude of \( \mathbf{f}(t) \), and using this in derivative calculations as in the problem helps us relate the curve's tangent properties.

Orthogonality

Orthogonality means that two vectors are perpendicular to each other. This is significant because it implies a dot product of zero. When you can prove two vectors are orthogonal, like \( \mathbf{f}(t) \) and \( \mathbf{f}^{\prime}(t) \) in the exercise, you're really saying they meet at a right angle.
In this particular problem, we prove that the derivative of a unit vector \( \mathbf{f}(t) \), is orthogonal to the vector itself at every point \( t \). This orthogonality stems from the constraint \( |\mathbf{f}(t)| = 1 \), which implies that \( \mathbf{f}(t) \) is constantly a unit vector. When a unit vector moves along a surface (like a sphere), it moves in such a way that its rate of change or direction is perpendicular to its original position.

Geometric Interpretation

Geometrically, the statement \( \mathbf{f}^{\prime}(t) \cdot \mathbf{f}(t) = 0 \) means that the tangent vector to the curve defined by \( \mathbf{f}(t) \) is orthogonal to the radius vector \( \mathbf{f}(t) \) itself. This is a fascinating result when considered for a curve lying on a unit sphere.
Because if \( |\mathbf{f}(t)| = 1 \) always, then \( \mathbf{f}(t) \) is essentially a vector traced on the surface of this sphere.
The orthogonality to \( \mathbf{f}^{\prime}(t) \) tells us the path taken does not change the radius (since magnitude remains 1), but rather moves in a direction tangent to the sphere surface at every point.

• This can be visualized as the vector \( \mathbf{f}(t) \) rotating around the sphere, always keeping the same distance from the origin.
• The actual rotation is in directions that do not "veer" away from the shell of the sphere, keeping at a constant radius.

By understanding this geometric relationship, we can truly appreciate the spherical and continuous nature of the path described.