Question

Let \(x=f(t), y=g(t),\) where \(t\) varies over an open interval \(I \subseteq E^{1},\) define a curve in \(E^{2}\) parametrically. Prove that if \(f\) and \(g\) have derivatives on \(I\) and \(f^{\prime} \neq 0,\) then the function \(h=f^{-1}\) has a derivative on \(f[I]\), and the slope of the tangent to the curve at \(t_{0}\) equals \(g^{\prime}\left(t_{0}\right) / f^{\prime}\left(t_{0}\right)\). [Hint: The word "curve" implies that \(f\) and \(g\) are continuous on \(I\) (Chapter \(4, \S 10)\), so Theorems 1 and 3 apply, and \(h=f^{-1}\) is a function. Also, \(y=g(h(x)) .\) Use Theorem 3 of \(\S 1 .]\)

Short answer

Slope of tangent at \(t_0\) is \(g'(t_0)/f'(t_0)\).

Step-by-step solution

Understand the Parametric Functions and Inverses

We have parametric equations defined by two functions: \(x = f(t)\) and \(y = g(t)\). The function \(f\) is assumed to be invertible, meaning there exists a function \(h = f^{-1}\) such that \(h(f(t)) = t\) over the interval \(I\). In this setup, \(h(f(t)) = t\), and hence \(y = g(h(x))\). Our task is to show that \(h\) has a derivative on \(f[I]\) and to find the slope of the tangent line to the curve at a specific \(t_0\).

Use the Derivatives of Inverse Functions

By the properties of inverses, if \(f\) has a derivative at \(t\) and \(f'(t) eq 0\), then the inverse function \(h = f^{-1}\) is differentiable at \(x = f(t)\). The derivative of \(h\) at a point \(x\) is given by \(h'(x) = \frac{1}{f'(h(x))}\). Since \(h = f^{-1}\), we specifically have \(h'(f(t)) = \frac{1}{f'(t)}\).

Differentiate the Composition Function

We express \(y\) as a composite function: \(y = g(h(x))\). By the chain rule, we differentiate \(y\) with respect to \(x\). \(\frac{dy}{dx} = g'(h(x)) \cdot h'(x)\). Substituting the expression for \(h'(x)\), we obtain \(\frac{dy}{dx} = g'(h(x)) \cdot \frac{1}{f'(h(x))}\).

Evaluate at Point \(t_0\)

For the specific point \(t_0\) from the interval \(I\), we evaluate the derivative: \(x = f(t_0)\) and \(y = g(t_0)\). We find that \(\frac{dy}{dx} = \frac{g'(t_0)}{f'(t_0)}\) which gives the slope of the tangent to the parametric curve at the point \((x_0, y_0)\), where \(x_0 = f(t_0)\) and \(y_0 = g(t_0)\).

Key concepts

Inverse Functions

Understanding inverse functions is crucial when working with parametric equations in calculus. An inverse function essentially reverses the effect of the original function. So, if we have a function \(f\) and its inverse \(h = f^{-1}\), then applying \(f\) followed by \(h\) returns the original input: \(h(f(t)) = t\).
This property is instrumental when dealing with parametric equations, such as \(x = f(t)\) and \(y = g(t)\). Here, \(f\) is assumed to be invertible, allowing us to express \(y\) as a function of \(x\). Specifically, \(y = g(h(x))\) expresses \(y\) in terms of \(x\), utilizing the inverse function \(h\).
When an inverse function exists, it allows us to switch between the variable representations effortlessly. This is especially helpful in solving problems that require us to analyse the relationship between \(x\) and \(y\) without directly involving \(t\).

Tangent Slope

In calculus, the slope of the tangent line is a fundamental concept, representing how a curve behaves at a specific point. For parametrically defined curves, calculating this slope involves differentiating the equations used to define the curve.
Suppose we have parametric equations like \(x = f(t)\) and \(y = g(t)\). To find the slope of the tangent at a particular parameter \(t = t_0\), we focus on finding \(\frac{dy}{dx}\).
We remember that the slope of a curve is given by the ratio of the change in \(y\) to the change in \(x\). By finding \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), we calculate the slope \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}\).
This expression \(\frac{g'(t_0)}{f'(t_0)}\) at the point \(t_0\) helps us determine how steep the curve is at that specific spot.

Derivatives

Differentiation, a core concept in calculus, finds the rate at which a function changes. This process is vital for understanding how parametric equations behave.
For parametric equations, where \(x = f(t)\) and \(y = g(t)\), we seek derivatives \(f'(t)\) and \(g'(t)\) to explore the dynamic nature of the curve.
One powerful aspect of differentiation is its application to inverses. If \(f\) is invertible and smooth (meaning differentiable and \(f' eq 0\)), then the inverse function \(h = f^{-1}\) also has a derivative. The formula for the derivative of an inverse function \(h\) at a point is \(h'(x) = \frac{1}{f'(h(x))}\).
Using these derivatives in parametric curves, we apply the chain rule to express \(y\) as \(y = g(h(x))\). Differentiation yields \(\frac{dy}{dx} = g'(h(x)) \cdot h'(x)\), which simplifies using the inverse derivative rules.

Composite Functions

Composite functions are built by applying one function to the results of another. These are particularly useful when handling parametric equations.
For instance, with \(x = f(t)\) and \(y = g(t)\), by recognizing \(h = f^{-1}\), we allow expression of \(y\) as \(y = g(h(x))\). This showcases how evaluating the outputs of one function through another creates a composite function.
The magic happens when differentiating composite functions. The chain rule is the perfect tool for this: it tells us how to differentiate a function of a function.
When we need the derivative \(\frac{dy}{dx}\) in parametric terms, applying the chain rule gives us \(\frac{dy}{dx} = g'(h(x)) \cdot h'(x)\). By substituting \(h'(x)\) with \(\frac{1}{f'(h(x))}\), the differentiation becomes complete, highlighting the synergy between derivatives and composite functions.