/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 My aunt and I were approaching t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 30

My aunt and I were approaching the same intersection, she from the south and I from the west. She was traveling at a steady speed of 10 miles/hour, while I was approaching the intersection at 60 miles/hour. At a certain instant in time, I was one-tenth of a mile from the intersection, while she was one- twentieth of a mile from it. How fast were we approaching each other at that instant?

Short Answer

Expert verified
At the given instant, aunt and student were approaching each other at a rate of \(26\sqrt{5}\) miles/hour.

Step by step solution

01

Represent Distances and Speeds with Variables

Let's represent the distance of aunt from the intersection as x, and the distance of student (you) from the intersection as y at any given time t. The speed of aunt is given as 10 miles/hour and the speed of student is given as 60 miles/hour. At the given instant: Aunt's distance from the intersection, x = 1/20 miles Student's distance from the intersection, y = 1/10 miles We need to find the rate at which they were approaching each other at that instant.
02

Apply Pythagorean theorem at instant

The distance between the aunt and the student is given by the hypotenuse of a right triangle formed by their respective distances, x and y from the intersection. So we can write the relationship using the Pythagorean theorem: \(z^2 = x^2 + y^2\) Where \(z^2\) is the square of the distance between the aunt and the student at the given instant. Now substituting the given values of x and y: \(z^2 = (1/20)^2 + (1/10)^2\) Solving for z: \(z = \sqrt{(1/400) + (1/100)} = \sqrt{(1+4)/400}\) \(z = \sqrt{5/400} = \dfrac{\sqrt{5}}{20}\) miles
03

Differentiate with respect to time

Now, we will differentiate the equation \(z^2 = x^2 + y^2\) with respect to time t to find how the distance between the aunt and the student (z) changes as a function of time. Using the chain rule: \(\dfrac{d}{dt}(z^2) = \dfrac{d}{dt}(x^2) + \dfrac{d}{dt}(y^2)\) \(2z \cdot \dfrac{dz}{dt} = 2x \cdot \dfrac{dx}{dt} + 2y \cdot \dfrac{dy}{dt}\) We can simplify this equation by dividing all terms by 2: \(z \dfrac{dz}{dt} = x \dfrac{dx}{dt} + y \dfrac{dy}{dt}\) We know that aunt's speed (x') is 10 miles/hour and that the student's speed (y') is 60 miles/hour: \(\dfrac{dz}{dt} = \dfrac{x \dfrac{dx}{dt} + y \dfrac{dy}{dt}}{z}\) Substituting the known values: \(\dfrac{dz}{dt} = \dfrac{(1/20)(10) + (1/10)(60)}{\dfrac{\sqrt{5}}{20}}\)
04

Calculate the rate at which they were approaching each other

Now, we can solve for \(\dfrac{dz}{dt}\), which represents the rate at which aunt and student were approaching each other: \(\dfrac{dz}{dt} = \dfrac{(1/2) + (6)}{\dfrac{\sqrt{5}}{20}}\) \(\dfrac{dz}{dt} = \dfrac{13/2}{(\sqrt{5}/20)}\) \(\dfrac{dz}{dt} = \dfrac{13/2 \times 20}{\sqrt{5}}\) \(\dfrac{dz}{dt} = \dfrac{13 \times 10}{\sqrt{5}}\) \(\dfrac{dz}{dt} = \dfrac{130\sqrt{5}}{5}\) \(\dfrac{dz}{dt} = 26\sqrt{5}\) So, at the given instant, aunt and student were approaching each other at a rate of \(26\sqrt{5}\) miles/hour.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Pythagorean theorem
The Pythagorean theorem is a mathematical relationship found in right-angled triangles. It is expressed through the equation \(a^2 + b^2 = c^2\), where 'a' and 'b' are the two legs of the triangle, and 'c' is the hypotenuse, or the side opposite the right angle. This theorem helps us calculate the length of one side of a triangle if we know the lengths of the other two sides.
In our exercise, we used the Pythagorean theorem to find the distance between two people approaching an intersection from two perpendicular directions. They formed a right triangle with an imaginary line as the hypotenuse, which is the direct distance between them.
By substituting the given distances from the problem into \(z^2 = x^2 + y^2\), where \(x\) and \(y\) are their respective distances from the intersection, we solved for \(z\). This distance \(z\) then became crucial for determining how fast they were approaching each other.
Differentiation
Differentiation is a fundamental concept in calculus that allows us to understand how a function changes as its input changes. It's like knowing the speed at which something happens. In this exercise, it was used to understand how the distance \(z\) between the two people changes over time.
For our calculation, we differentiated the Pythagorean equation \(z^2 = x^2 + y^2\) with respect to time \(t\). This process used the chain rule, a technique that applies when differentiating composed functions, to find \(\frac{dz}{dt}\), the rate at which the distance \(z\) changes.
The derivative \(\frac{dz}{dt}\) was calculated by determining how fast the two individuals were moving (their speeds), and how these contributed to the rate at which they were approaching each other. Differentiation, thus, connected motion (speed) with how quickly their actual distance changes over a short period.
Speed and distance calculations
Calculating speed and distance is central to solving problems involving movement, like this one. Speed is simply how fast distance changes over time, given by \(\text{speed} = \frac{\text{distance}}{\text{time}}\). Similarly, distance tells us how far something has traveled or moved.
In the exercise, we were given the speeds at which the two individuals moved: 10 miles/hour for one and 60 miles/hour for the other. We also had their initial distances from the intersection. Using this information and by applying the previously differentiated function, we calculated \(\frac{dz}{dt}\), the rate of approach.
This calculation linked their speeds to how quickly the gap between them decreased as they neared the intersection, revealing the dynamic interaction between speed and distance. This understanding is intuitive and useful for analyzing real-world motion problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You have been hired as a marketing consultant to Big Book Publishing, Inc., and you have been approached to determine the best selling price for the hit calculus text by Whiner and Istanbul entitled Fun with Derivatives. You decide to make life easy and assume that the demand equation for Fun with Derivatives has the linear form \(q=m p+b\), where \(p\) is the price per book, \(q\) is the demand in annual sales, and \(m\) and \(b\) are certain constants you'll have to figure out. a. Your market studies reveal the following sales figures: when the price is set at \(\$ 50.00\) per book, the sales amount to 10,000 per year; when the price is set at \(\$ 80.00\) per book, the sales drop to 1000 per year. Use these data to calculate the demand equation. b. Now estimate the unit price that maximizes annual revenue and predict what Big Book Publishing, Inc.'s annual revenue will be at that price.

n In 1991 , the expected income of an individual depended on his or her educational level according to the following formula: $$\begin{aligned}&I(n)=2928.8 n^{3}-115,860 n^{2}+1,532,900 n-6,760,800 \\\&(12 \leq n \leq 15)\end{aligned}$$ Here, \(n\) is the number of school years completed and \(I(n)\) is the individual's expected income. \(^{58}\) You have completed 13 years of school and are currently a part-time student. Your schedule is such that you will complete the equivalent of one year of college every three years. Assuming that your salary is linked to the above model, how fast is your income going up? (Round your answer to the nearest \(\$ 1 .)\)

The demand curve for original Iguanawoman comics is given by $$q=\frac{(400-p)^{2}}{100} \quad(0 \leq p \leq 400)$$ where \(q\) is the number of copies the publisher can sell per week if it sets the price at p. a. Find the price elasticity of demand when the price is set at $$\$ 40$$ per copy. b. Find the price at which the publisher should sell the books in order to maximize weekly revenue. c. What, to the nearest \(\$ 1\), is the maximum weekly revenue the publisher can realize from sales of Iguanawoman comics?

Sketch the graph of the given function, indicating (a) \(x\) - and \(y\) -intercepts, (b) extrema, (c) points of inflection, \((d)\) behavior near points where the function is not defined, and (e) behavior at infinity. Where indicated, technology should be used to approximate the intercepts, coordinates of extrema, and/or points of inflection to one decimal place. Check your sketch using technology. \(g(t)=e^{-t^{2}}\)

A general quadratic demand function has the form \(q=a p^{2}+b p+c(a, b\), and \(c\) constants with \(a \neq 0\) ). a. Obtain a formula for the price elasticity of demand at a unit price \(p\). b. Obtain a formula for the price or prices that could maximize revenue.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.