Problem 90 - University Calculus: Early Transcendentals

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 value of \( \frac{dy}{dt} \) when \(x=1\) is 3.

Step 1: Find the Expression for dy/dx

Given the function \(y = x^2 + 7x - 5\), differentiate it with respect to \(x\) to find \( \frac{dy}{dx} \). Using the power rule, the derivative of \(x^2\) is \(2x\), and the derivative of \(7x\) is \(7\). Therefore, \[ \frac{dy}{dx} = 2x + 7. \]

Step 2: Use the Chain Rule to Find dy/dt

We can find \( \frac{dy}{dt} \) by using the chain rule, which states: \[ \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}. \]Using the result from Step 1: \[ \frac{dy}{dt} = (2x + 7) \times \frac{dx}{dt}. \]

Step 3: Substitute Given Values

Substitute the given values \(x = 1\) and \(\frac{dx}{dt} = \frac{1}{3}\) into the expression for \( \frac{dy}{dt} \). First calculate \( \frac{dy}{dx} = 2(1) + 7 = 9. \)Then substitute in the expression: \[ \frac{dy}{dt} = 9 \times \frac{1}{3} = 3. \]

Key Concepts

Chain Rule

The chain rule is a fundamental component of differentiation in calculus, especially when dealing with composite functions or interconnected rates of change. It helps you derive the derivative of a function based on how the individual parts (components) of the function change over time. Here's how it works:

• The chain rule involves differentiating the outer function and then multiplying it by the derivative of the inner function. Simply put, it's like peeling the layers of an onion, taking one step at a time.
• If you've got functions combined like in our "chain," this rule is your go-to tool! You'll often see it in the formula: \[\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}.\]

In our exercise, the function for `y` in terms of `x` is differentiated first with respect to `x` to get \(\frac{dy}{dx}\). Then, we apply the chain rule to connect it back to \(\frac{dy}{dt}\) by multiplying \(\frac{dy}{dx}\) by \(\frac{dx}{dt}\). The result is that the impact of changes in `x` over time directly influences `y`. The equation ensures you can track how both values interact as time shifts.

Rate of Change

The rate of change can be a powerful concept as it describes how one quantity changes relative to another. In calculus, it's often seen as a derivative, which gives us a precise measure of how a function changes. Let's break it down:

• For the expression given in our example, \(\frac{dx}{dt} = \frac{1}{3}\) indicates the rate at which `x` changes concerning time `t`. Simply put, for every 3 units of time, `x` increases by 1 unit.
• The derivative \(\frac{dy}{dx}\) tells us how `y` changes concerning `x`. In solving our exercise, this was found by differentiating `y` with respect to `x`, resulting in \(2x + 7\).
• Finally, the combination of these rates of change through the chain rule gives \(\frac{dy}{dt}\) — the overall rate of change of `y` relative to `t`. This holistic value informs us how `y` responds to the passage of time, accounting for all given parameters.

By understanding rate of change, we gain insights into the dynamics of every function and how each variable plays a crucial role in the overall behavior.

Derivative Calculation

Derivative calculation is like finding the 'instantaneous speed' of an object's movement along a function. It's a core skill in calculus, allowing us to determine how one quantity changes with respect to another. Here's how you can tackle derivative calculation:

• Start by identifying the function you need to differentiate. In our exercise, the function was \(y = x^2 + 7x - 5\).
• Differentiation rules, like the power rule, can help. The power rule states that if you have \(x^n\), the derivative is \(nx^{n-1}\). For \(y = x^2 + 7x - 5\), the derivative with respect to `x` is \(2x + 7\).
• Once you find \(\frac{dy}{dx}\), you can use it to explore how changing `x` affects `y`. In our case, substituting \(x = 1\) yields \(\frac{dy}{dx} = 9\).

Incorporating the derivative calculation outcomes into broader solutions like \(\frac{dy}{dt}\) with given rates \(\frac{dx}{dt}\) emphasizes the impact of derivative calculations. This step determines how meaningful these derivatives are in real-world applications, offering a microscopic view of changes as they happen.