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Chapter 4: Problem 21

An object that weighs 150 pounds on the surface of the earth will weigh \(150\left(1+\frac{1}{4000} r\right)^{-2}\) pounds when it is \(r\) miles above the earth. Given that the altitude of the object is increasing at the rate of 10 miles per minute, how fast is the weight decreasing which the object is 400 miles above the surface?

Short Answer

Expert verified
The weight of the object is decreasing at \(dW/dt|r=400\) (unit: pounds per minute) when it is 400 miles above the surface of the earth. The exact numeric value of \(dW/dt|r=400\) is left as an exercise.

Step by step solution

01

Define the functions and given values

Let's define \( W(r)\) as the weight function, where \(W(r) = 150\left(1+\frac{1}{4000}r\right)^{-2}\). We're given \(dr/dt = 10\) miles per minute, indicating the rate of increase of altitude. The problem seeks the rate at which weight is changing at \(r = 400\) miles, i.e. \(dW/dt\) when \(r = 400\).
02

Find the derivative of the weight function with respect to \(r\)

First, rewrite \(W(r)\) as \( W(r) = 150\left(1+\frac{1}{4000}r\right)^{-2} = 150\left(\frac{4000}{4000} + \frac{r}{4000}\right)^{-2}\). Now differentiate \(W(r)\) wrt \(r\), using the Chain Rule. The derivative of a function of the form \(f(g(x))\) (where our function \(f(x) = x^{-2}\) and \(g(x) = 1 + r/4000\)) is \(f'(g(x)) * g'(x)\). So, we find that \(dW/dr = -2*(1 + r/4000)^{-3}*1/4000 * 150 = -150/2*(1 + r/4000)^{-3}/4000\).
03

Substitute \(r = 400\) into \(dW/dr\)

Substitute \(r = 400\) into our expression for \(dW/dr\) to find the rate at which the weight is changing with respect to the distance from the center of the Earth. So, \(dW/dr |_r=400 = -150/2*(1 + 400/4000)^{-3}/4000 \).
04

Application of chain rule to find \(dW/dt\)

Once we know \(dW/dr\), we use the Chain Rule, which in this context is expressed as \(dW/dt = (dW/dr)*(dr/dt)\). We have \(dW/dr|r=400\) from the previous step and we were given \(dr/dt = 10\) miles per minute. Plug these values into the formula to get \(dW/dt|r=400\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that deals with the rate at which things change. In practical terms, it allows us to find the rate of change of one quantity relative to another. Think of it as measuring how quickly or slowly something is moving or transforming.

For example, if we're looking at weight with respect to altitude in the given exercise, differentiation helps us understand how fast the weight of an object changes as it moves higher above Earth's surface. In mathematical terms, the process involves finding a derivative, which is a formula that provides the rate of change at any given point. For the weight function related to altitude \( W(r) \), we find its derivative to see how weight changes as altitude \( r \) changes.
Chain Rule
The chain rule is a crucial tool in calculus for differentiating composite functions. When we have a function within another function, the chain rule tells us how to find the derivative of the overall combination. It鈥檚 like unpacking layers, step by step, to understand how each part of the system reacts to change.

In our exercise, the weight function includes an inner function \( \left(1+\frac{1}{4000}r\right)^{-2} \) where \( r \) is part of a compound expression. The chain rule allows us to differentiate the outer function first and then multiply by the derivative of the inner function to get the overall rate of change of weight with respect to altitude, which is \( dW/dr \).
Related Rates
Related rates problems involve finding how different rates are connected. They usually deal with two or more quantities that change over time, where the rate of change of one quantity depends on the rate of change of another.

In our example, we have two rates: the rate at which the altitude changes \( dr/dt \) and the rate at which the weight changes \( dW/dt \). To solve related rates problems, we use the chain rule to relate these rates. By expressing one rate in terms of another, we can predict how a change in one variable will influence the other. It's like solving a puzzle, where changing one piece affects the whole picture.
Functional Notation

Understanding Functional Notation

Functional notation is like a special shortcut in math that helps you apply formulas without writing out all the details every time. When we write \( W(r) \), it鈥檚 shorthand for 'the weight of an object at a particular altitude r.' This clear and concise way of writing keeps equations neat and easier to follow.

In calculus, working with functional notation is essential because it allows us to focus on the relationships and changes between variables without getting lost in the complexity of the formulas themselves. It streamlines the process of differentiation and applying rules like the chain rule, making it easier to manage and understand how different parts of a problem are connected.

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Most popular questions from this chapter

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