Problem 67 - Calculus Early Transcendentals

Question

If \(f^{\prime}\) is continuous and \(f^{\prime}(t) \neq 0\) for \(a \leqslant t \leqslant b,\) show that the parametric curve \(x=f(t), y=g(t), a \leq t \leq b,\) can be put in the form \(y=F(x) .\left[\) Hint. Show that \(f^{-1}\) exists. \right\(]\)

Step-by-Step Solution

Answer

The curve can be written as \(y = g(f^{-1}(x))\).

Step 1: Understanding the Problem

We need to show that the parametric curve defined by \(x = f(t)\) and \(y = g(t)\) can be expressed as \(y = F(x)\) where \(f'(t) eq 0\) for all \(a \leq t \leq b\). This hints at the possibility that \(f\) is invertible within this domain.

Step 2: Exploring Invertibility

For a function \(f\) to be invertible, it must be one-to-one. Since \(f'(t) eq 0\) for \(a \leq t \leq b\), by the Mean Value Theorem, \(f\) is either strictly increasing or strictly decreasing on this interval. This ensures each \(x\) value corresponds to a unique \(t\). Hence, \(f\) is invertible.

Step 3: Using Continuity and Differentiability

Given that \(f'(t)\) is continuous, by the Inverse Function Theorem, the inverse function \(f^{-1}\) exists and is also differentiable on the interval \([f(a), f(b)]\). This means we can express \(t\) as a function of \(x\), that is, \(t = f^{-1}(x)\).

Step 4: Expressing y in terms of x

Since we can write \(t = f^{-1}(x)\), substitute in the expression for \(y = g(t)\), we have \(y = g(f^{-1}(x))\). Thus, the parametric curve can be represented in the Cartesian form \(y = F(x)\) where \(F(x) = g(f^{-1}(x))\).

Step 5: Confirming the Solution

By showing that \(f\) is strictly monotonic and its inverse exists, we have successfully transformed the parametric representation into a functional form, verifying the problem's requirement. This confirms that \(y = F(x) = g(f^{-1}(x))\).

Key Concepts

Inverse Function

An inverse function essentially reverses the operation of the original function such that if you apply an inverse function to the output of the original function, you get back the original input. In mathematical terms, if we have a function \( f \), then its inverse, denoted \( f^{-1} \), has the property that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).
For a function to have an inverse, it must be bijective, meaning it should be both one-to-one (injective) and onto (surjective).
In the context of the given exercise, we are interested in proving that the parametric curve \( x = f(t) \) is invertible so that we can express \( t \) as a function of \( x \).

• Since \( f'(t) eq 0 \) over the interval \( a \leq t \leq b \), \( f \) is strictly monotonic, being either strictly increasing or decreasing.
• A strictly monotonic function on an interval ensures that distinct inputs yield distinct outputs, satisfying the injective requirement.
• This means that every \( x \) within the range of \( f \) corresponds uniquely to a \( t \), thus making \( f \) invertible, allowing us to find \( f^{-1} \).

Differentiability

Differentiability refers to the existence of a derivative for a function at every point in its domain. If a function is differentiable, it means you can compute its derivative with respect to its variable at each point.
In our exercise, differentiability plays a crucial role because:

• The problem states that \( f'(t) \) is continuous, indicating that \( f(t) \) is differentiable on the closed interval \([a, b]\).
• This continuous derivative leads to crucial conclusions via the Inverse Function Theorem.

The Inverse Function Theorem helps us by stating that if a function \( f \) has a non-zero continuous derivative, then its inverse function \( f^{-1}(x) \) is differentiable wherever \( f \) is invertible.
This means we can express our parameter \( t \) in terms of \( x \) using \( f^{-1}(x) \), supporting us in turning the parametric curve into a Cartesian form.

Mean Value Theorem

The Mean Value Theorem (MVT) provides a bridge connecting differentiability and monotonicity. It states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in the interval \((a, b)\) such that:\[f'(c) = \frac{f(b) - f(a)}{b - a}.\]
This theorem confirms that because \( f'(t) eq 0 \) throughout the interval, no two distinct \( t \)-values in the interval can lead to the same \( x \)-value. This is crucial in proving \( f \) is strictly monotonic over \([a, b]\). As a result:

• The function is either increasing or decreasing consistently throughout the interval, ensuring one-to-one mapping.
• This guarantees \( f \) is invertible, aligning with the verification of the solution.

The importance is that the MVT proves that our assumptions about \( f \)'s behavior lead to the correct conclusion that we can invert the function.

Continuous Functions

A continuous function is one where small changes in the input will result in small changes in the output, without any jumps or breaks. For a function to be continuous, it must fulfil three conditions:

• The function is defined at every point within the interval.
• The limit of the function as it approaches from either side of each point is equal and finite.
• The value of the function at each point must agree with the limit.

In our particular exercise, continuity is vital because:

• We rely on \( f'(t) \)'s continuity to ensure the behavior of \( f \) across the interval \([a, b]\).
• This continuous nature of \( f'(t) \) supports the derivative's consistent behavior, helping argue the smoothness required for differentiability.
• It also underpins the Mean Value Theorem, fostering the conditions needed for invertibility.

Through these properties, continuity assures that no jumps or abrupt changes disrupt the smooth transformation from parametric to functional representation.