Problem 1 - Essential Calculus

Question

Find \(d y / d x\). $$ x=t^{2}, y=5-4 t $$

Step-by-Step Solution

Answer

The derivative of \(y\) with respect to \(x\), \(dy/dx\), is equal to \(-2 / t\).

Step 1: Calculate \(dx/dt\)

First, need to differentiate \(x = t^2\) with respect to \(t\). Using the power rule for differentiation, we get \(dx/dt = 2t\).

Step 2: Calculate \(dy/dt\)

Next, differentiate \(y = 5 - 4t\) with respect to \(t\). Here we apply the power rule again, resulting in \(dy/dt = -4\).

Step 3: Express \(dy/dx\) as \((dy/dt) / (dx/dt)\)

Finally, express \(dy/dx\) as \((dy/dt) / (dx/dt)\). Substituting the previously calculated values, \((dy/dt) / (dx/dt) = -4 / (2t) = -2 / t\).

Key Concepts

Parametric Equations

In mathematics, parametric equations are a way of expressing a set of related quantities as explicit functions of an independent variable, known as a parameter. In practical terms, instead of expressing the relationship between the variables directly, we use a parameter like \( t \) to express both \( x \) and \( y \).
For example, in our exercise, \( x \) and \( y \) are given as \( x = t^2 \) and \( y = 5 - 4t \). Here, \( t \) is the parameter. This can be very useful when dealing with curves in the plane because it allows us to easily express more complex relationships and motion along paths. The parameter \( t \) can represent time or any other quantity that the variables \( x \) and \( y \) depend upon.

• Expression of the relationship using a third variable
• Flexibility in dealing with curves and paths
• Utility in representing direction and motion

Parametric equations provide a powerful foundation for solving more complex calculus problems, as they allow for more detailed analysis of trajectories and shapes.

Chain Rule

The chain rule is a fundamental technique for differentiation in calculus, used when dealing with composite functions. It provides a way to find the derivative of a function that is composed of multiple functions. This rule is particularly invaluable in parametric equations, where two or more different functions are expressed in terms of a third variable.
In our exercise, we want to differentiate \( y \) with respect to \( x \), not directly but through their shared parameter \( t \).

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

By breaking down the process, we can easily solve complex derivatives that involve multiple interconnected functions. In this case, calculating \( \frac{dy/dx} \) involves first finding \( \frac{dy}{dt} = -4 \) and \( \frac{dx}{dt} = 2t \), then computing the result as \( \frac{-4}{2t} = \frac{-2}{t} \).

Calculus

Calculus is the mathematical study that focuses on change, represented via derivatives and integrals. It allows us to explore how one quantity changes in relation to another and understand the behavior of functions graphically and analytically. In calculus, differentiation is the process of finding the derivative, which measures how a function changes as its input changes.
In the context of our exercise, we are performing implicit differentiation with parametric equations. This process allows us to explore how one quantity, \( y \), changes with respect to another quantity, \( x \), through an intermediary parameter, \( t \).
Doing this effectively:

• Identifies slopes and rates of change
• Helps in graphing complex curves
• Aids in solving real-world motion problems

Ultimately, such calculus techniques are crucial for deeper scientific explorations and practical applications, from engineering to economics.