Problem 54 - Calculus - AP Edition

Question

Use the differential \(d y\) to approximate \(\Delta y\) when \(x\) changes as indicated. $$y=x \sqrt{8 x+1} ; \text { from } x=3 \text { to } x=3.05$$

Step-by-Step Solution

Answer

The approximate change in \( y \) is 0.37.

Step 1: Identify the Function and Change in x

The given function is \( y = x \sqrt{8x + 1} \). We need to approximate \( \Delta y \), which is the change in \( y \) when \( x \) changes from 3 to 3.05. Thus, \( \Delta x = 0.05 \).

Step 2: Differentiate the Function

Find the derivative \( \frac{dy}{dx} \) using the product and chain rules. Let \( u = x \) and \( v = \sqrt{8x + 1} \). Thus, the derivative is \( \frac{dy}{dx} = u'v + uv' \). First, compute \( u' = 1 \). Then, for \( v = (8x + 1)^{1/2} \), use the chain rule to get \( v' = \frac{1}{2}(8x + 1)^{-1/2} \cdot 8 = \frac{4}{\sqrt{8x + 1}} \). Therefore, \( \frac{dy}{dx} = \sqrt{8x + 1} + \frac{4x}{\sqrt{8x + 1}} \).

Step 3: Evaluate the Derivative at x = 3

Substitute \( x = 3 \) into the derivative. Calculate \( \sqrt{8 \times 3 + 1} = \sqrt{25} = 5 \). Substitute to get \( \frac{dy}{dx} = 5 + \frac{12}{5} = 5 + 2.4 = 7.4 \).

Step 4: Calculate dy Using the Approximation

The differential \( dy \) is given by \( dy = \frac{dy}{dx}\Delta x \). Substituting the rate of change and \( \Delta x \): \( dy = 7.4 \times 0.05 = 0.37 \).

Step 5: Conclusion

Therefore, the approximate change in \( y \), represented by \( \Delta y \), is approximately 0.37 when \( x \) changes from 3 to 3.05.

Key Concepts

Derivative of a function

In calculus, the derivative of a function measures how the function value changes as its input changes. In this context, the function is given by \( y = x \sqrt{8x + 1} \). To find the derivative, we must understand how each part of the function contributes to its overall rate of change.A derivative is often denoted as \( \frac{dy}{dx} \). This notation helps signify that it describes the change in \( y \) with a small change in \( x \) (hence the \( dx \)). The derivative gives us the slope of the tangent line to the curve at any point \( x \), explaining how steep the curve is at that point.For the function above, calculating the derivative requires using specific rules tailored for different types of operations, such as multiplication and composition, which we will cover next.

Product rule

The product rule is a fundamental tool in differential calculus that allows us to differentiate functions that are multiplied together. If you have two functions, \( u(x) \) and \( v(x) \), their product's derivative is given by:\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]This rule ensures we account for both functions' changes. In our exercise, we set \( u = x \) and \( v = \sqrt{8x + 1} \).

• First, find \( u' \), which for \( u = x \) is simply 1, since the derivative of \( x \) with respect to \( x \) is 1.
• Next, identify \( v(x) \), and differentiate it using the chain rule to obtain \( v' \).

Applying the product rule, \( \frac{dy}{dx} = u'v + uv' \) computes the overall slope of the given function, which will be refined further by the chain rule.

Chain rule

The chain rule is essential when you have a composition of functions, that is, a function within another function. It's used to differentiate such composite functions, and it's usually represented as: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]For our exercise, the chain rule is applied to the function \( v(x) = (8x + 1)^{1/2} \).

• The outer function is \( w = u^{1/2} \) and its derivative \( w' \) is \( \frac{1}{2}u^{-1/2} \).
• The inner function is \( u = 8x + 1 \) with its derivative \( u' = 8 \).

Combining these, the derivative of \( v(x) \) is \( v'(x) = \frac{4}{\sqrt{8x + 1}} \), which contributes to the product rule's calculation of the full derivative of the original expression. The chain rule is pivotal in ensuring we correctly account for changes within nested functions like \( v(x) \) when obtaining \( \frac{dy}{dx} \).