Problem 62 - Thomas Calculus

Question

Find \(d y / d t\) when \(x=1\) if \(y=x^{2}+7 x-5\) and \(d x / d t=1 / 3\)

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dt} \) when \( x = 1 \) is 3.

Step 1: Differentiate y with respect to x

Start by finding the derivative of the given function \(y = x^2 + 7x - 5\) with respect to \(x\). Use the power rule for differentiation: \( \frac{d}{dx}(x^n) = nx^{n-1} \). Computing the derivative gives:\[ \frac{dy}{dx} = 2x + 7 \]

Step 2: Apply the Chain Rule to find dy/dt

To find \( \frac{dy}{dt} \), apply the chain rule. The chain rule states:\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]We already know \( \frac{dy}{dx} = 2x + 7 \) from Step 1 and \( \frac{dx}{dt} = \frac{1}{3} \) is given in the problem. Substitute these into the chain rule formula:\[ \frac{dy}{dt} = (2x + 7) \left( \frac{1}{3} \right) \]

Step 3: Evaluate at x = 1

Insert \(x = 1\) into the expression obtained in Step 2:\[ \frac{dy}{dt} = (2(1) + 7) \left( \frac{1}{3} \right) = (2 + 7) \left( \frac{1}{3} \right) = 9 \times \frac{1}{3} = 3 \]Thus, \( \frac{dy}{dt} = 3 \) when \( x = 1 \).

Key Concepts

Understanding the Chain Rule

In calculus, the chain rule is essential for differentiating composite functions. It allows you to find the derivative of a function based on a sequence of other functions. The chain rule is expressed as:

• For a function that composes two functions, say, \(f(g(x))\), the derivative is given by \(f'(g(x)) \cdot g'(x)\).
• This idea extends to more complex situations where functions are nested within each other.

In our exercise, we used the chain rule to relate \(\frac{dy}{dt}\) and \(\frac{dy}{dx}\) via \(\frac{dx}{dt}\). It simplifies finding how changes in \(x\) affect \(y\) considering an external parameter \(t\). This demonstrates the chain rule's power in connecting derivatives across different variables and functions. Since \(\frac{dx}{dt} = \frac{1}{3}\) is provided, combining it with \(\frac{dy}{dx} = 2x+7\) achieves the objective of finding \(\frac{dy}{dt}\). This is done by multiplying the two derivatives together.

Introduction to Differentiation

Differentiation is a fundamental concept in calculus. It involves calculating the derivative, which represents the rate of change of a function's value with respect to changes in its input variable. Here are some basics:

• A derivative, \(f'(x)\), gives the slope of a function at any given point \(x\).
• The derivative tells you how quickly a function is increasing or decreasing at any point.
• Common rules include the power rule, product rule, quotient rule, and chain rule.

In our exercise, we applied the power rule to differentiate \(y = x^2 + 7x - 5\). The power rule indicates that for \(x^n\), the derivative \(\frac{d}{dx}(x^n) = n\cdot x^{n-1}\). Therefore:

• The derivative of \(x^2\) is \(2x\).
• The derivative of \(7x\) is \(7\).
• The derivative of a constant \(-5\) is \(0\).

Thus, the expression \(2x + 7\) represents \(\frac{dy}{dx}\). This step executed the necessary differentiation to proceed with finding \(\frac{dy}{dt}\).

Effective Calculus Problem Solving

Solving calculus problems efficiently often involves a systematic approach. Alright, let's break it down!

• Start by understanding the problem's requirements and known variables. Here, we need \(\frac{dy}{dt}\) when \(x=1\).
• Identify the correct tools and methods, such as differentiation rules or the chain rule, to address these requirements.
• Carefully calculate derivatives: First, find \(\frac{dy}{dx}\) through differentiation. Then use the chain rule to derive \(\frac{dy}{dt}\).

Incorporating these strategies, it becomes easier to solve problems like the one presented. Systematically follow each step:

• First, differentiate with respect to \(x\).
• Next, apply applicable calculus rules like the chain rule.
• Finally, substitute known values and evaluate the expressions.

Using these steps ensures accurate results, such as finding \(\frac{dy}{dt} = 3\) when \(x = 1\). By breaking the problem into manageable parts, complex problems become solvable with ease.