Problem 325 - Calculus Volume 3

Question

True or False? Justify your answer with a proof or a counterexample. Given \(x=f(t)\) and \(y=g(t),\) if \(\frac{d x}{d y}=\frac{d y}{d x},\) then \(f(t)=g(t)+\mathrm{C},\) where \(\mathrm{C}\) is a constant.

Step-by-Step Solution

Answer

True. If the derivatives are zero, both functions are constants, complying with the given relationship.

Step 1: Understand the given derivative relationship

We are given that \( \frac{d x}{d y} = \frac{d y}{d x} \). This means that the derivative of \( x \) with respect to \( y \) is equal to the derivative of \( y \) with respect to \( x \).

Step 2: Analyze the mathematical implications

The expression \( \frac{d x}{d y} = \frac{d y}{d x} \) implies that \( \left( \frac{d x}{d y} \right)\left( \frac{d y}{d x} \right) = 1 \). By default, if \( a = b \), then \( a \times b = a^2 = b^2 \), which would only hold true if both \( a \) and \( b \) are either 0 or \( \pm 1 \).

Step 3: Consider the special values of derivatives

For derivatives, this scenario implies both derivatives are 0 since for other cases like \( 1 \) and \( -1 \), they do not comply with the equality \( a = b \). So, the derivatives \( \frac{d x}{d y} = 0 \) and \( \frac{d y}{d x} = 0 \) stand in consideration.

Step 4: Interpret \( \frac{d x}{d y} = 0 \) and \( \frac{d y}{d x} = 0 \)

If \( \frac{d x}{d y} = 0 \), this implies \( x \) is constant with respect to \( y \), and similarly, \( y \) is constant in terms of \( x \). This implies both \( f(t) \) and \( g(t) \) are constant functions.

Step 5: Evaluate if \( f(t) = g(t) + C \) holds

Given both functions are constant, \( f(t) = C_1 \) and \( g(t) = C_2 \). Hence, \( C_1 = C_2 + C \), which aligns with the equation \( f(t) = g(t) + C \). Thus, it can be true under such a condition.

Key Concepts

Derivative Relationship

In the exercise, we observe a derivative relationship between two parametric equations: \(x=f(t)\) and \(y=g(t)\). We are given that \(\frac{d x}{d y} = \frac{d y}{d x}\). This implies that the rate of change of \(x\) with respect to \(y\) is equal to the rate of change of \(y\) with respect to \(x\). When such a scenario occurs, there is an interesting mathematical implication. We need to explore what conditions allow the derivatives to be equal in this peculiar manner. Generally, for the derivatives of two functions to be exactly equal, specific constraints on the functions and their relationship are necessary. In parametric terms, derivatives interrelating like this can signify a profound symmetry or a constant nature of the functions involved, paving the way to the solutions discussed further.

Constant Functions

A constant function is a pivotal element in this proof. A function is called constant if its derivative is zero across its domain. If \( f'(t) = 0 \), it means \( f(t) = C_1 \), where \( C_1 \) is a constant. Similarly, if \( g'(t) = 0 \), it means \( g(t) = C_2 \). When analyzing \( \frac{d x}{d y} = 0 \) and \( \frac{d y}{d x} = 0 \), we realize both these derivatives suggest that \(x\) and \(y\) do not change as \(t\) varies. Hence, \(f(t)\) and \(g(t)\) are constant functions. Even though \(f(t)\) and \(g(t)\) may have different constant values, the relationship \(f(t) = g(t) + C\) is feasible since it defines a fixed offset between the two constants.

Mathematical Implications

The mathematical implications of \( \frac{d x}{d y} = \frac{d y}{d x} \) derive from symmetrical relationships between derivatives. We have previously established that the equation holds if both derivatives are zero, which points toward constant functions. Consider the case where these derivatives are not zero.

• If they are non-zero, the initial equality won't hold as it would mean \(a \times b = 1\) with no general solution matching \(a = b\) other than zero.
• The derivatives equal to each other directly imply a state of equality that can only uphold across constant functions, aligning with the given equation \(f(t) = g(t) + C\).

This equality unrolls a simple, yet important realization regarding constant intervals and dependencies in parametric equations.

Proofs and Counterexamples

To prove the statement, the idea of using counterexamples helps cement the understanding when exceptions can invalidate a statement. However, in this case, the proof revolves around showing constancy using derivatives.Here’s the logical flow:- Assume \( \frac{d x}{d y} = \frac{d y}{d x} \).- Explore by setting these derivatives zero, leading to constant results \(f(t) = C_1\) and \(g(t) = C_2\).- Apply it mathematically: \(f(t)-g(t) = C\), illustrating \(C = C_1 - C_2\), proving the exercise's true, with a constant \(C\).By carefully confirming this is the only satisfying input circumstances under which the claimed mathematical relationship holds, we ensure complete validity of the statement via constant functions proof.