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Chapter 5: Problem 24

Suppose the demand \(x\) and the price \(p\) are related by the equation \(x^{2}+100 p^{2}=20,000\). Find \(d p / d x\) at the point where \(x=100\).

Short Answer

Expert verified
At \(x = 100\), \(\frac{dp}{dx} = \pm\frac{1}{10}\) depending on \(p = 10\) or \(p = -10\).

Step by step solution

01

Differentiate the equation with respect to x

The given equation is \(x^2 + 100p^2 = 20,000\). To find \(\frac{dp}{dx}\), we will differentiate both sides of the equation with respect to \(x\). Apply the chain rule to differentiate \(p\) with respect to \(x\).\\\[\frac{d}{dx}(x^2) + \frac{d}{dx}(100p^2) = \frac{d}{dx}(20,000)\] \\[2x + 100 \cdot 2p \cdot \frac{dp}{dx} = 0\] \\[2x + 200p \cdot \frac{dp}{dx} = 0\].
02

Solve for dp/dx

The differentiated equation is \(2x + 200p \cdot \frac{dp}{dx} = 0\). Solve for \(\frac{dp}{dx}\) by isolating it on one side. \\\[200p \cdot \frac{dp}{dx} = -2x\] \\[\frac{dp}{dx} = \frac{-2x}{200p}\] \\[\frac{dp}{dx} = \frac{-x}{100p}\].
03

Evaluate p at x = 100

Substitute \(x = 100\) into the original equation to solve for \(p\).\\\((100)^2 + 100p^2 = 20,000\) \\[10,000 + 100p^2 = 20,000\] \\[100p^2 = 10,000\] \\[p^2 = 100\] \\[p = \pm10\].
04

Calculate dp/dx at x = 100 and p = 10

Now that we have \(p = 10\) when \(x = 100\), substitute these values into \(\frac{dp}{dx} = \frac{-x}{100p}\). \\\[\frac{dp}{dx} = \frac{-100}{100 \times 10} = \frac{-1}{10}\].
05

Consider p = -10 and recalculate dp/dx

We also need to check \(p = -10\). Substitute \(x = 100\) and \(p = -10\) into \(\frac{dp}{dx} = \frac{-x}{100p}\). \\\[\frac{dp}{dx} = \frac{-100}{100 \times (-10)} = \frac{1}{10}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a crucial tool in calculus when you need to differentiate composite functions. Simply put, it allows you to find the derivative of a function that is composed of two or more simpler functions. When you have an equation where one variable, say \( p \), is a function of another variable \( x \), such as in the exercise \( 100p^2 \), applying the chain rule lets you differentiate \( p \) with respect to \( x \).In this particular problem:
  • The demand \( x \) and price \( p \) are linked through the equation \( x^2 + 100 p^2 = 20,000 \).
  • To find \( \frac{dp}{dx} \), we apply the chain rule.
  • Differentiating \( 100p^2 \) with respect to \( x \), we recognize the implicit function of \( p \).
This leads to the result \( 200p \cdot \frac{dp}{dx} \), which accounts for the chain rule's significance. The chain rule encapsulates the need to consider the rate of change of \( p \) as it pertains to \( x \), hence leading us seamlessly to solve the derivative provided in the solution.
Implicit Differentiation
Implicit differentiation is a method used to find the derivative of a variable in equations where that variable is not isolated. Many real-world problems involve equations that are not solved for one variable in terms of another. For the equation in the exercise,
  • \( x^2 + 100p^2 = 20,000 \) does not solve \( p \) explicitly in terms of \( x \).
  • We need to differentiate both sides of the equation with respect to \( x \), applying implicit differentiation.
By doing so, we find terms including \( \frac{dp}{dx} \). This allows us to retrieve the derivative \( \frac{-x}{100p} \) without directly solving for \( p \) as a function of \( x \). This reflects how implicit differentiation is vital for finding derivatives in equations where variables intermingle.
Economic Modeling
In economic modeling, variables such as price \( p \) and demand \( x \) are often interconnected through non-linear equations. This exercise demonstrates such a relationship, allowing us to understand how changes in one variable can influence another.In the equation \( x^2 + 100p^2 = 20,000 \):
  • This can represent a model of an economic scenario where demand \( x \) and price \( p \) adhere to a constant value.
  • Finding \( \frac{dp}{dx} \) provides insights into how price varies with slight changes in demand.
  • The derivative results \( \frac{-1}{10} \) and \( \frac{1}{10} \) at specific points help model this relationship under different conditions.
Economic models leveraging such calculus tools allow businesses to predict and respond to changes systematically in market scenarios.

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Most popular questions from this chapter

Biology Cherry and associates \(^{21}\) created a mathematical model that showed that the number \(y\) of eggs laid per female of the sugarcane grub was approximated by \(y=\) \(14.6 e^{-(x-116.6) / 60.9},\) where \(x\) is a measure of soil moisture. Sketch a graph. Find where the number of eggs is increasing and where they are decreasing. Check using your grapher.

Biology When predators are not stimulated to search very hard for prey, the rate of change \(R\) of the number \(N\) of prey attacked per unit of area with respect to the number \(N\) of prey is given by \(R(x)=k x(a x-N),\) where \(x\) is the number of predators, \(k\) is the per capita effectiveness of the predators, and \(a\) is the maximum per capita effectiveness of the predators over an extended period. \({ }^{75}\) Find the number of predators that minimizes \(R\).

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A water tank in the shape of a circular cone (see figure) has a radius of 4 yards and a height of 10 yards. If water is being poured into the tank at the rate of 5 cubic yards per minute, find the rate at which the water level is rising when the water level is at 2 yards. (Hint: \(\left.V=\frac{1}{3} \pi r^{2} h .\right)\) Use similar triangles to find \(r\) as a function of \(h\) and substitute this into the last equation. Then differentiate the latter expression.
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