Problem 55 - AP Calculus with 8 Practice Tests 15th

Question

A pair of equations that represent a curve parametrically is given. Choose the alternative that is the derivative \(\frac{d y}{d x}\). \(x=1-e^{-t} \quad\) and \(\quad y=t+e^{-t}\) (A) \(\frac{e^{-1}}{1-e^{-t}}\) (B) \(e^{-t}+1\) (C) \(e^{t}-e^{-2 t}\) (D) \(e^{t}-1\)

Step-by-Step Solution

Answer

The correct answer is (D) \( e^t - 1 \).

Step 1: Understanding Parametric Derivatives

To find the derivative \( \frac{d y}{d x} \) for parametric equations, we need to determine \( \frac{d y}{d t} \) and \( \frac{d x}{d t} \), then use the formula \( \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \).

Step 2: Compute \(\frac{dx}{dt}\)

Given \( x = 1 - e^{-t} \), differentiate with respect to \( t \):\[\frac{dx}{dt} = \frac{d}{dt}[1 - e^{-t}] = 0 + e^{-t} = e^{-t}\]

Step 3: Compute \(\frac{dy}{dt}\)

Given \( y = t + e^{-t} \), differentiate with respect to \( t \):\[\frac{dy}{dt} = \frac{d}{dt}[t + e^{-t}] = 1 - e^{-t}\]

Step 4: Apply \(\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)

Now, substitute \( \frac{dy}{dt} = 1 - e^{-t} \) and \( \frac{dx}{dt} = e^{-t} \) into the formula:\[\frac{d y}{d x} = \frac{1 - e^{-t}}{e^{-t}}\]

Step 5: Simplify the Derivative Expression

Simplify the expression:\[\frac{d y}{d x} = \frac{1}{e^{-t}} - \frac{e^{-t}}{e^{-t}} = e^{t} - 1\]This matches option (D).

Key Concepts

Understanding Derivatives

The concept of derivatives is fundamental in calculus. Derivatives represent the rate of change of a function with respect to a variable. For instance, if you have a function of time, its derivative will show how the function changes at any given moment. In the context of parametric equations, derivatives can help us find the slope of the tangent line to the curve described by the equations.

When dealing with parametric equations, you usually have a pair of functions, one for the x-coordinates and one for the y-coordinates, both in terms of a parameter such as \( t \). To find the derivative \( \frac{dy}{dx} \), you need to calculate how both x and y change concerning \( t \) and then use these rates to find how y changes with x. This involves finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), and then taking the ratio \( \frac{dy/dt}{dx/dt} \).

This ratio reveals the actual derivative of y with respect to x, providing insight into how the y-value changes as the x-value increases on the parametric curve. By understanding this concept, we can determine the slope of curves, which is vital for graph sketching and solving real-world rate problems.

Differentiation in Parametric Form

Differentiation is the process of finding a derivative, and this involves using specific rules and techniques. In parametric form, differentiation helps find the rate of change of one variable concerning another when both are given as functions of a parameter. This is a crucial skill in calculus and is immensely helpful in various applications, such as physics and engineering.

To differentiate parametric equations, follow these steps:

• Find the derivative \( \frac{dx}{dt} \). For the equation \( x = 1 - e^{-t} \), differentiate to get \( \frac{dx}{dt} = e^{-t} \).
• Find the derivative \( \frac{dy}{dt} \). For the equation \( y = t + e^{-t} \), differentiate to get \( \frac{dy}{dt} = 1 - e^{-t} \).
• Compute \( \frac{dy}{dx} \) using \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).

For example, substituting the derivatives in our equation gives us \( \frac{dy}{dx} = \frac{1 - e^{-t}}{e^{-t}} \), which simplifies to \( e^t - 1 \). This result shows how the y-component changes as compared to the x-component along the curve.

Foundations of Calculus

Calculus is a branch of mathematics that studies continuous change, encapsulating concepts like differentiation and integration. It's a powerful tool used to solve complex problems in fields such as physics, engineering, and economics. Understanding the foundational concepts of calculus allows you to model real-world phenomena accurately.

In this exercise, we applied calculus through differentiation to handle parametric equations and extract meaningful derivatives. This usage is central to calculus, showcasing its utility in analyzing dynamic systems whose variables change with respect to each other. The main ideas in this context include:

• Parametric equations, which define variables as functions of a parameter.
• Utilizing derivatives to find the slope or rate of change of variables.
• Applying differentiation techniques to solve real-world problems like motion analysis and optimization.

By mastering these concepts and techniques, you gain the ability to understand and predict how quantities evolve over time, a principle that sits at the heart of calculus and its applications in various disciplines.