Question

If \(y=x^{4}+2 x\), find the values of \(\Delta y\) and \(d y\) in each case. (a) \(x=2\) and \(d x=\Delta x=1\) (b) \(x=2\) and \(d x=\Delta x=0.005\)

Short answer

(a) \(\Delta y = 67\), \(dy = 34\); (b) \(\Delta y \approx 0.0902\), \(dy = 0.17\).

Step-by-step solution

Find the original function value when x=2

Substitute \(x=2\) into the function \(y = x^4 + 2x\): \[ y = (2)^4 + 2(2) = 16 + 4 = 20 \]

Calculate y for changed x-values in case (a)

For \(x=2\) and \(\Delta x = 1\), calculate new \(x\): \[ x + \Delta x = 2 + 1 = 3 \] Now find the new \(y\): \[ y = (3)^4 + 2(3) = 81 + 6 = 87 \]

Calculate \(\Delta y\) for case (a)

Find \(\Delta y\) using the values from Step 1 and Step 2: \[ \Delta y = y_{new} - y_{original} = 87 - 20 = 67 \]

Calculate the derivative of y with respect to x

Find \(\frac{dy}{dx}\) from the function \(y = x^4 + 2x\): \[ \frac{dy}{dx} = 4x^3 + 2 \]

Calculate dy for case (a)

Using \(x=2\), \(dx=1\), and \( \frac{dy}{dx} = 4x^3 + 2 \), calculate \(dy\): \[ \frac{dy}{dx}|_{x=2} = 4(2)^3 + 2 = 32 + 2 = 34 \] Then \[ dy = \frac{dy}{dx} \times dx = 34 \times 1 = 34 \]

Calculate y for changed x-values in case (b)

For \(x=2\) and \(\Delta x = 0.005\), calculate new \(x\): \[ x + \Delta x = 2 + 0.005 = 2.005 \] Now find the new \(y\): \[ y = (2.005)^4 + 2(2.005) \approx 16.080160125 + 4.01 \approx 20.090160125 \]

Calculate \(\Delta y\) for case (b)

Find \(\Delta y\) using the values from Step 1 and Step 6: \[ \Delta y = y_{new} - y_{original} \approx 20.090160125 - 20 = 0.090160125 \]

Calculate dy for case (b)

Using \(x=2\), \(dx=0.005\), and \( \frac{dy}{dx} = 34 \) (from Step 5): \[ dy = \frac{dy}{dx} \times dx = 34 \times 0.005 = 0.17 \]

Key concepts

Derivatives

In the realm of differential calculus, derivatives take center stage in analyzing how functions change. Essentially, a derivative is a tool that measures the sensitivity or rate of change of a function with respect to one of its variables. Think of it as a way to describe how steep or flat the graph of a function is at any given point.

For instance, in our exercise, the function given is \(y = x^4 + 2x\). The derivative of this function, \(\frac{dy}{dx}\), tells us how \(y\) changes for a small change in \(x\). By applying differentiation rules, we find that the derivative is \(4x^3 + 2\). You'll often see derivatives used extensively in physics for velocity and acceleration calculations because they inherently deal with changes.

When we computed \(dy\) for cases (a) and (b), we effectively used the derivative function to estimate how much \(y\) would change for each small change \(dx\). This step is crucial because it illustrates the practical utility of derivatives in predicting changes in variables accurately.

Functions

A function is much like a machine with an input and an output. You insert a value (input), apply the function (machine process), and retrieve a result (output). In our example, we have the polynomial function \(y = x^4 + 2x\). This equation defines how each input \(x\) corresponds to an output \(y\).

Understanding functions is crucial because they model relationships between different quantities. Functions are not limited to mathematics; they depict real-world phenomena like the trajectory of a launched object or the growth of a population. They can be linear, quadratic, polynomial, etc., each with different characteristics.

In this problem, we evaluated the function at various points to understand how small changes impact \(y\), helping us uncover responses in relationships modeled by the function. The initial step was evaluating the function at \(x=2\), which gave us a starting point (\(y=20\)). Later, we saw how modifying \(x\) led to a new \(y\) value.

Delta notation

The delta notation (\(\Delta\)) is commonly used to denote changes or differences in values. In the exercise, \(\Delta y\) represents the change in the function's output when \(x\) changes by a certain amount, denoted as \(\Delta x\).

Specifically, \(\Delta y = y_{new} - y_{original}\). This notation is handy in calculating actual increases or decreases in a function due to changing variables. For case (a), we calculated \(\Delta y = 87 - 20 = 67\), highlighting a significant change for a large \(\Delta x\) of 1. In contrast, case (b) had a much smaller \(\Delta y\) of approximately 0.090 when \(\Delta x = 0.005\).

Delta notation is essential in many scientific fields. It helps in quantifying change processes, be that physics, engineering, finance, or any domain requiring analysis of variations in quantities.

Change in variables

Examining changes in variables is a foundational activity in calculus. It involves observing how tweaking one variable influences another. This concept underpins the exercise and allows students to gauge the effect of different \(\Delta x\) values on the function's output \(y\).

Starting with an initial variable value (e.g., \(x=2\)), we introduce a small change (\(\Delta x\)), which yields a new \(x\) and a recalculated \(y\). Through this, we directly observe how the function behaves around specific points, crucial for predictions in many scientific fields.

Importantly, such exercises lambent how calculus moves beyond static evaluations into dynamic analysis. By exploring variable changes, one ushers in sophisticated real-world application capabilities, like optimizing output or controlling processes in industrial environments.