Problem 52 - Calculus - AP Edition

Question

Use the differential \(d y\) to approximate \(\Delta y\) when \(x\) changes as indicated. $$y=\sqrt{x^{2}+8} ; \text { from } x=1 \text { to } x=0.97$$

Step-by-Step Solution

Answer

\( \Delta y \approx -0.01 \)

Step 1: Identify the Given Information

We need to approximate \( \Delta y = f(x + \Delta x) - f(x) \) using the differential \( dy \) where \( y = \sqrt{x^2 + 8} \). The initial value of \( x \) is 1 and it changes to 0.97, making \( \Delta x = -0.03 \).

Step 2: Find the Derivative

First, find the derivative \( \frac{dy}{dx} \). For \( y = \sqrt{x^2 + 8} \), let \( u = x^2 + 8 \) so \( y = u^{1/2} \). Then \( \frac{dy}{du} = \frac{1}{2}u^{-1/2} \) and \( \frac{du}{dx} = 2x \). Hence, \( \frac{dy}{dx} = \frac{1}{2}(x^2 + 8)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 8}} \).

Step 3: Calculate the Differential dy

The differential \( dy \) is given by \( \frac{dy}{dx}\cdot dx \). We substitute \( x = 1 \) into \( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 8}} \) which gives \( \frac{1}{\sqrt{9}} = \frac{1}{3} \). Then \( dy = \frac{1}{3} \cdot (-0.03) = -0.01 \).

Step 4: Approximate \( \Delta y \)

Since \( dy \approx \Delta y \), we approximate \( \Delta y \) as \( \Delta y \approx -0.01 \).

Key Concepts

Calculus

Calculus is a branch of mathematics focused on understanding changes. It plays a crucial role in finding rates of change in various phenomena. It's like the mathematical study of motion and allows us to model

• growth or decay
• motion
• accumulation

Calulus is divided into two major parts: differentiation and integration.
They are interconnected and each has great importance. In simple terms, differentiation deals with motion rates, whereas integration adds up all small changes.
Differential approximation is one of calculus's useful tools. This allows us to estimate changes in a function's value over a small interval, using its derivative. For instance, if we slightly alter input values, we can predict how the output will adjust, based on the calculated derivative.

Differentiation

Differentiation is a key concept within calculus. Its purpose is to find the derivative of a function. The derivative represents how a function's output values change as we tweak the input values. In simple terms, the derivative shows the rate at which the function's value is altering.
To differentiate a function means to calculate this rate of wiggle. It gives us an expression involving derivatives to describe the workings of the original function.
In our exercise, we differentiated the function \( y = \sqrt{x^2 + 8} \). The differentiation process involved expressing the function in terms of a simpler expression, \( u^{1/2} \). We then calculated \( \frac{dy}{dx} \) using formulas to get \( \frac{x}{\sqrt{x^2 + 8}} \). This represents the original function's rate of change, vital for understanding how small changes in the input affect the output, enabling differential approximation.

Derivative Calculation

The calculation of a derivative involves several steps, often beginning with rewriting the function for easier manipulation. This was exemplified in our exercise with the function \( y = \sqrt{x^2 + 8} \).Let's unpack this process:

• First, rewrite the function: We let \( u = x^2 + 8 \), making \( y = u^{1/2} \).
• Determine inner derivatives: Calculate \( \frac{dy}{du} \) and \( \frac{du}{dx} \). For example, we had \( \frac{1}{2}u^{-1/2} \) for one and \( 2x \) for the other.
• Apply chain rule: Combine to find overall derivative \( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 8}} \).

What we achieved was finding how much the function \( y \) will change in response to shifts in \( x \). This derivative lets us estimate changes in the \( y \) value for small \( x \) changes, as done when approximating \( \Delta y \). Such comprehensive steps of derivative calculation are essential in mathematics and practical applications.