Problem 94 - Calculus Volume 3

Question

Find \(d y / d x\) at the value of the parameter. $$ x=\sqrt{t}, \quad y=2 t+4, \quad t=9 $$

Step-by-Step Solution

Answer

The derivative \(d y / d x\) at \(t = 9\) is 12.

Step 1: Find derivatives with respect to t

First, determine the derivatives of the given parametric equations with respect to the parameter \(t\). For \(x = \sqrt{t}\), use the power rule to get \(\frac{d x}{d t} = \frac{1}{2} t^{-1/2}\). For \(y = 2t + 4\), the derivative is simple: \(\frac{d y}{d t} = 2\).

Step 2: Apply chain rule to find dy/dx

To find \(\frac{d y}{d x}\), use the chain rule for parametric equations: \(\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}\). Substitute the derivatives from Step 1: \(\frac{d y}{d x} = \frac{2}{\frac{1}{2} t^{-1/2}} = 4t^{1/2}\).

Step 3: Substitute t = 9 and solve

Replace \(t\) with 9 in the expression for \(\frac{d y}{d x}\) obtained in Step 2: \(\frac{d y}{d x} = 4(9)^{1/2} = 4 \times 3 = 12\). Thus, \(\frac{d y}{d x}\) evaluated at \(t = 9\) is 12.

Key Concepts

Chain Rule

When dealing with parametric equations, the chain rule plays a crucial role in connecting how two variables relate through a third parameter, like time \(t\). The chain rule states that if you have a function composed of another function, the derivative of the outer function is the derivative of the inner times the derivative of the outer evaluated at the inner function.
For parametric equations,

• To find \( \frac{dy}{dx} \), you actually calculate it as \( \frac{dy}{dt} \) divided by \( \frac{dx}{dt} \).
• This principle allows you to determine the rate of change of \(y\) with respect to \(x\) without having to explicitly solve \(y\) in terms of \(x\).

In our problem, we computed \( \frac{dy}{dt} = 2 \), and \( \frac{dx}{dt} = \frac{1}{2} t^{-1/2} \). Using the chain rule, \( \frac{dy}{dx} \) is efficiently obtained by dividing these two derivatives.

Derivative

Derivatives are the backbone of calculus and help us understand how functions change. When you take a derivative, you're essentially finding the slope of a function at any given point, which tells us how fast it is changing.
In the context of parametric equations:

• We take derivatives with respect to a parameter, usually \(t\), that controls the evolution of the system.
• The derivative \( \frac{dx}{dt} \) or \( \frac{dy}{dt} \) signifies the instantaneous rate of change of \(x\) or \(y\) with respect to \(t\).

For example, in this exercise, \(x = \sqrt{t}\) and \(y = 2t + 4\). We found that \( \frac{dx}{dt} = \frac{1}{2} t^{-1/2} \). It means as \(t\) changes, this is how \(x\) is adjusting.

Power Rule

The power rule is a fundamental tool in calculus used to quickly find the derivative of functions raised to a power. This rule simplifies the differentiation process for power functions.

• For any function \(x^n\), the derivative is \(nx^{n-1}\).

In parametric equations, such as \(x = t^{1/2}\), apply the power rule:

• The derivative \( \frac{dx}{dt} \) is \( \frac{1}{2} t^{-1/2} \), here \(1/2\) is brought down and the exponent is reduced by 1.

This simplification is crucial because it makes calculating derivatives of power functions straightforward.
Remember, the power rule is applicable to any real number exponent, not just whole numbers.

Calculus

Calculus is the mathematical study of change, encompassing differentiation and integration. It allows mathematicians and engineers to understand how quantities change over time. In addressing problems with parametric equations:

• Differentiation helps us find the derivative, which tells us how functions are changing at any specific moment.
• Using calculus, specifically the chain rule and power rule, we can untangle complex relationships between variables.

In the provided example, we leveraged calculus to derive \( \frac{dy}{dx} \) from the parametric form directly.
This demonstrates the power of calculus in providing a framework for systematically solving problems that involve changing quantities.