Problem 5 - Essential Calculus

Question

In Exercises 5 and \(6,\) use the information to evaluate and compare \(\Delta y\) and \(d y\). $$ y=\frac{1}{2} x^{3} \quad x=2 \quad \Delta x=d x=0.1 $$

Step-by-Step Solution

Answer

For the given function \(y=\frac{1}{2} x^{3}\), when \(x=2\), \(Δx=0.1\), the differential \(dy\) is \(0.6\) and the increment \(Δy\) is \(0.6415\). Hence, \(Δy\) > \(dy\) for the given values.

Step 1: Differentiate the Given Function

Given that \(y=\frac{1}{2} x^{3}\). The first derivative of \(y\) with respect to \(x\) is \(y' = \frac{3}{2}x^{2}\).

Step 2: Compute dy

Now calculate the approximate change in \(y\), denoted by \(dy\). From the formula for differentials \(dy = y' dx\), we find, \(dy = \frac{3}{2}*2^{2}*0.1 = 0.6\).

Step 3: Compute Δy

The increment in \(y\) (Δy) is computed by the formula \(Δy = f(x+Δx) - f(x)\). Substituting the given values into \(y=\frac{1}{2} x^{3}\) results in \(Δy = \frac{1}{2} *(2.1)^{3} - \frac{1}{2} *2^{3} = 0.6415\).

Step 4: Compare Δy and dy

Both Δy and dy represent approximations of how \(y\) changes as \(x\) increases by \(0.1\). However, Δy is a more accurate approximation than dy, as it uses the actual function \(y=\frac{1}{2} x^{3}\). In this case, we can see that Δy = 0.6415 is slightly greater than dy = 0.6

Key Concepts

Differentials

Understanding differentials is like learning about small changes in mathematics. Let's say you have a function and you want to see how a tiny change in the input (which we call \(dx\)) affects the output (which we call \(dy\)).

This is important in calculus because it helps us predict how a function behaves. For our exercise, we have a function \(y = \frac{1}{2} x^3\). When we find the differential \(dy\), it tells us how much \(y\) changes in response to a tiny change in \(x\).

The formula for \(dy\) is given by the derivative of the function with respect to \(x\) multiplied by \(dx\):

• \(dy = y' \cdot dx\)

Think of \(dy\) like making a prediction. In our example, for a small change \(dx = 0.1\) when \(x = 2\), the differential is \(0.6\). While \(dy\) isn't perfect, it's a quick way to estimate change.

Derivatives

Derivatives are one of the most fundamental ideas in calculus. They measure the rate at which something changes. It's like asking "how fast" something is changing at a particular instant. When you see \(y'\), it represents the derivative of \(y\) with respect to \(x\).

In our exercise, we're given \(y = \frac{1}{2}x^3\). To find how \(y\) changes, we differentiate it:

• The derivative \(y' = \frac{3}{2}x^2\) tells us the rate of change of \(y\) at any \(x\).

This means if you know \(x\), you can calculate how fast \(y\) changes when \(x\) increases by just a little bit.
The derivative helps us understand the behavior of curves as it reveals spots where the function increases, decreases, or remains flat. In physics, it's often used to describe velocity or acceleration, giving a deeper understanding of motion.

Function Approximation

Function approximation is about estimating the value of a function using simpler methods. It's crucial when you want a quick, close answer without performing lengthy calculations.

In this exercise, we have two ways to approximate change:

• \(\Delta y\), which uses the actual formula for more precision.
• \(dy\), which uses the slope, or the derivative, for a faster estimate.

The goal is to see how close our predictions are to the real change.
Let's break it down:- \(\Delta y\) represents the true difference using \(f(x + \Delta x) - f(x)\). In our example, this is \(0.6415\).
- \(dy = 0.6\) quickly estimates this change.

Even though \(dy\) is simpler, it might not be as accurate because it only considers the tangent's slope. As you can see, \(\Delta y\) is slightly closer to true values because it factors in more than just instantaneous change.