Problem 325 - Calculus Volume 1

Question

The volume of a right circular cone of radius \(x\) and height \(y\) is given by \(V=\frac{1}{3} \pi x^{2} y .\) Suppose that the volume of the cone is 85\(\pi \mathrm{cm}^{3} .\) Find \(\frac{d y}{d x}\) when \(x=4\) and \(y=16\)

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -8 \)

Step 1: Understand the Given Information

The volume of a cone is given by the formula \( V = \frac{1}{3} \pi x^2 y \). We know the volume is \( 85\pi \) cm\(^3\), therefore, \( \frac{1}{3} \pi x^2 y = 85\pi \). We need to find \( \frac{dy}{dx} \) when \( x = 4 \) and \( y = 16 \).

Step 2: Simplify the Equation

First, eliminate \( \pi \) from the equation since it appears on both sides: \( \frac{1}{3} x^2 y = 85 \). Next, multiply through by 3 to clear the fraction: \( x^2 y = 255 \).

Step 3: Differentiate with respect to \( x \)

Differentiate both sides of \( x^2 y = 255 \) with respect to \( x \), applying the product rule to the left side: \( \frac{d}{dx}(x^2 y) = 2x y + x^2 \frac{dy}{dx} = 0 \) because the derivative of a constant is zero.

Step 4: Solve for \( \frac{dy}{dx} \)

Rearrange the differentiated equation to solve for \( \frac{dy}{dx} \): \( x^2 \frac{dy}{dx} = -2xy \). Thus, \( \frac{dy}{dx} = \frac{-2xy}{x^2} = \frac{-2y}{x} \).

Step 5: Substitute \( x = 4 \) and \( y = 16 \) into the Derivative

Substitute the given values into the equation: \( \frac{dy}{dx} = \frac{-2(16)}{4} = \frac{-32}{4} = -8 \).

Key Concepts

Volume of a Cone

The volume of a cone is a fundamental concept in geometry. Cones are three-dimensional shapes with a circular base that narrows to a point called the apex. The volume of a cone is calculated using the formula:

• \( V = \frac{1}{3} \pi x^2 y \)

where \( V \) is the volume, \( x \) is the radius of the base, and \( y \) is the height. The factor \( \frac{1}{3} \) shows the relationship of the cone's volume as a third of what a cylinder with the same base and height would have. In this exercise context, we use the volume formula to understand how changes in radius and height affect the cone's total volume. Calculating the derivative helps us determine how these variations interplay under fixed parameters.

Product Rule

When dealing with differentiation of equations involving products of variables, the product rule is essential. For a function \( f \cdot g \), where both \( f \) and \( g \) depend on \( x \), the derivative is given by:

• \( \frac{d}{dx}(f \cdot g) = \frac{df}{dx} \cdot g + f \cdot \frac{dg}{dx} \)

This rule is useful in our exercise where we have the function \( x^2 \cdot y \) in the volume equation of the cone. By applying the product rule, we differentiate \( x^2 \) and \( y \) separately before combining the results. This is necessary for solving the derivative \( \frac{dy}{dx} \) which indicates how the height of the cone changes with respect to its radius.

Chain Rule

The chain rule helps us differentiate composite functions by breaking them down into simpler parts. If we have an expression like \( h(g(x)) \), the chain rule states:

• \( \frac{dh}{dx} = \frac{dh}{dg} \cdot \frac{dg}{dx} \)

In real-world applications such as this exercise, it allows us to handle situations where one variable depends indirectly on another. Although the main focus here is the product rule, understanding the chain rule supports broader comprehension of differentiation techniques and how different rules can interconnect for solving more complex derivative problems.

Derivatives

Derivatives measure how a function changes as its input changes. They're a cornerstone of calculus, often represented as \( \frac{dy}{dx} \) or \( f'(x) \). The calculation of derivatives provides the rate at which one quantity changes with respect to another. This exercise involves finding the derivative \( \frac{dy}{dx} \), which indicates how the height \( y \) of a cone changes as its radius \( x \) changes, given a constant volume. By differentiating the volume equation, we obtain relations like \( \frac{-2y}{x} \), which specifies the precise rate of change. Mastery of derivatives and their applications is vital for tackling dynamic changes in mathematical and real-world scenarios.