Question

For the following exercises, find the equation of the tangent line to the given curve. Graph both the function and its tangent line. Find \(\frac{d y}{d x}, \frac{d x}{d y}\), and \(\frac{d^{2} x}{d y^{2}}\) of \(y=\left(2+e^{-t}\right), x=1-\sin (t)\)

Short answer

Find derivatives: \(\frac{dy}{dx} = \frac{e^{-t}}{\cos(t)}\), \(\frac{dx}{dy} = \frac{\cos(t)}{e^{-t}}\), \(\frac{d^2x}{dy^2} = e^{2t}(\sin(t) - \cos(t))\).

Step-by-step solution

Find dy/dt and dx/dt

Given the parametric equations are \(y=2+e^{-t}\) and \(x=1-\sin(t)\). First, differentiate both the expressions with respect to \(t\). The derivative of \(y\) is \(\frac{dy}{dt} = -e^{-t}\). For \(x\), the derivative is \(\frac{dx}{dt} = -\cos(t)\).

Find dy/dx using the chain rule

Since \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) are known, use the formula for parametric derivatives \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). Thus, \(\frac{dy}{dx} = \frac{-e^{-t}}{-\cos(t)} = \frac{e^{-t}}{\cos(t)}\).

Find dx/dy

To find \(\frac{dx}{dy}\), take the reciprocal of \(\frac{dy}{dx}\). Thus, \(\frac{dx}{dy} = \frac{\cos(t)}{e^{-t}}\).

Find d²x/dy²

First, find the derivative of \(\frac{dx}{dy}\) with respect to \(t\) to obtain \(\frac{d}{dt}\left(\frac{\cos(t)}{e^{-t}}\right)\). Use the quotient rule: \(\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \cdot du/dt - u \cdot dv/dt}{v^2}\), where \(u = \cos(t)\) and \(v = e^{-t}\). Compute: \(-e^{-t} \sin(t) + \cos(t) e^{-t}\) leads to \(e^{-t}(\cos(t) - \sin(t))\). Thus \(\frac{d^2x}{dy^2} = \frac{e^t(\sin(t) - \cos(t))}{(-e^{-t})^2} = e^{2t}(\sin(t) - \cos(t))\).

Key concepts

Parametric Derivatives

When dealing with curves, often the expressions for x and y coordinates are given in terms of a third variable, usually called the parameter \( t \). This kind of representation is very useful for describing motions or curves in space rather than just on a Cartesian plane.

This is where parametric derivatives come into play. When we have parametric equations \( x = f(t) \) and \( y = g(t) \), we can find the derivative \( \frac{dy}{dx} \) by differentiating each with respect to \( t \). The derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) give us the rates of change of \( y \) and \( x \) with respect to time or the parameter \( t \). By using these, the formula for finding the derivative of \( y \) with respect to \( x \) is:

• \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \)

This key formula allows us to find slopes and examine the behavior of the curves effectively.

Chain Rule

The chain rule is an essential concept in calculus for finding derivatives of composite functions. Composite functions are functions formed by combining one function inside another, such as \( h(x) = f(g(x)) \). The chain rule states that the derivative of such a function is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function.

Applying this idea in the context of parametric equations, we consider \( y \) and \( x \) both as separate functions of \( t \), and we want to find \( \frac{dy}{dx} \). Using chain rule principles, we interpret this as:

• \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)

This approach helps us understand how changes in \( t \) affect changes in \( y \) and \( x \), and it ensures we obtain the correct slope of the tangent at any point on our parametric curve.

Quotient Rule

The quotient rule is particularly handy when finding derivatives of ratios or fractions of functions. Suppose you have a function that is the quotient of two other functions, specifically \( h(t) = \frac{u(t)}{v(t)} \). To find its derivative, we use the quotient rule:

• \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v(t)\frac{du}{dt} - u(t)\frac{dv}{dt}}{(v(t))^2} \)

In our parametric context, especially when determining \( \frac{d^2x}{dy^2} \), we apply this rule when differentiating \( \frac{dx}{dy} \) with respect to \( t \). This is necessary because \( dx/dy \) itself is a quotient, and using the quotient rule allows us to find a higher-order derivative like \( \frac{d^2x}{dy^2} \) accurately.

Parametric Equations

Parametric equations are a powerful way of expressing mathematical curves without a direct dependency of y upon x or vice versa. Instead, both coordinates are dependent on a third parameter \( t \). Using parametric equations, we can construct paths and motions that might not be easily expressed otherwise.

For example, consider the parametric equations \( x = 1 - \sin(t) \) and \( y = 2 + e^{-t} \). Here, \( t \) can represent time, and as it varies, the values \( x \) and \( y \) trace out a curve.

• This approach offers a more flexible representation, allowing us to study the geometry and dynamics of curves more comprehensively.

Parametric equations thus provide insight into intricate curves by transforming them into understandable algebraic forms.