Problem 45 - Thomas Calculus

Question

Let \(v\) be a differentiable vector function of \(t .\) Show that if \(v \cdot(d v / d t)=0\) for all \(t,\) then \(|v|\) is constant.

Step-by-Step Solution

Answer

The magnitude \(|v|\) is constant because \(v \cdot \frac{dv}{dt} = 0\) implies \(\frac{d}{dt}(|v|^2) = 0\).

Step 1: Understand the relationship of dot product and magnitude

We are given that the dot product of the vector function \(v(t)\) with its derivative is zero, i.e., \(v \cdot \frac{dv}{dt} = 0\) for all \(t\). This indicates that the vector \(v(t)\) and its derivative \(\frac{dv}{dt}\) are orthogonal (the angle between them is 90 degrees).

Step 2: Recall the expression for the magnitude squared

The magnitude (or length) of a vector \(v(t)\) is \(|v|\), and its square is given by \(v \cdot v\). Therefore, the magnitude squared is \(|v|^2 = v \cdot v\).

Step 3: Differentiate the magnitude squared

Let's differentiate \(|v|^2\) with respect to \(t\), which is \(\frac{d}{dt} (v \cdot v)\). Using the product rule for differentiation of dot products, we get:\[\frac{d}{dt} (v \cdot v) = 2v \cdot \frac{dv}{dt}.\]

Step 4: Use the given condition

Substitute the given condition \(v \cdot \frac{dv}{dt} = 0\) into the differentiated expression:\[2v \cdot \frac{dv}{dt} = 2 \times 0 = 0.\]So, \(\frac{d}{dt}(|v|^2) = 0.\)

Step 5: Interpret the derivative result

The result \(\frac{d}{dt}(|v|^2) = 0\) implies that \(|v|^2\) is constant with respect to \(t\). This means the magnitude \(|v|\) itself is constant, as the square of a constant is also constant.

Key Concepts

Dot Product

The dot product, also called the scalar product, is a fundamental operation in vector calculus. It's defined for two vectors, say \(\mathbf{a}\) and \(\mathbf{b}\), as \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\) for vectors in three dimensions. This can be generalized to more dimensions depending on the problem.

• The dot product produces a scalar result, unlike vector cross product which produces another vector.
• When the dot product is zero, the vectors are orthogonal (i.e., perpendicular).

In our specific problem, we are given a condition \(v \cdot \frac{dv}{dt} = 0\), indicating that the vector \(v(t)\) is perpendicular to its own derivative at any point \(t\). This orthogonal relationship plays a crucial role in deducing the constancy of the vector’s magnitude, as orthogonality affects how vectors scale and interact through their geometrical properties.

Differentiable Vector Functions

Differentiable vector functions are vectors whose components are differentiable with respect to a variable, often time \(t\). These functions are a powerful tool for modeling phenomena that change smoothly in space and time.

• Differentiability implies continuity. Thus, for any vector function \(v(t)\), not only are \(v(t)\) and \(\frac{dv}{dt}\) defined, but they also exhibit smooth changes.
• The derivative \(\frac{dv}{dt}\) represents the rate of change of the vector function. Geometrically, it serves as the tangent vector indicating the direction in which \(v(t)\) changes.

In the given problem, \(v(t)\) being differentiable ensures that we can correctly apply calculus operations like differentiation, allowing the analysis of the magnitude and orientation of the vector over time. This makes it possible to explore deeper properties like the invariance of the magnitude when the derivative and the original vector are orthogonal.

Magnitude of a Vector

The magnitude of a vector, sometimes referred to as the vector's "length," indicates its size but not its direction. For a vector \(\mathbf{v}\) in three-dimensional space with components \(v_1, v_2, v_3\), its magnitude is given by the expression \(|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\).

• The magnitude is always a non-negative value.
• When you have repeated vectors dotted with themselves, such as \(v \cdot v\), this is equivalent to the magnitude squared: \(|v|^2 = v \cdot v\).

In our discussion, differentiating the magnitude squared \(|v|^2\) and finding it to be zero across time reveals that \(|v(t)|\) does not change with time. This constancy of ||v|| stems from the given orthogonality condition, embodying the intuitive idea that when a vector only changes direction but not length, its magnitude remains intact. This is a fundamental property exploited across physics and engineering disciplines.