Problem 5. An observer is positioned 3km away from a rocket launch pad. How fast is the distance between the rocket and the observer increasing, when the rocket is 4km above the ground and is moving straight up at the speed of 300m/s?
Problem 5. An observer is positioned 3km away from a rocket launch pad. How fast is the distance between the rocket and the observer increasing, when the rocket is 4km above the ground and is moving straight up at the speed of 300m/s?
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem 5.** An observer is positioned \(3 \text{ km}\) away from a rocket launch pad. How fast is the distance between the rocket and the observer increasing, when the rocket is \(4 \text{ km}\) above the ground and is moving straight up at the speed of \(300 \text{ m/s}\)?
---
**Explanation:**
To solve this related rates problem, one needs to consider the position of the observer and the rocket as two vertices of a right triangle, with the horizontal distance from the observer to the launch pad forming one leg, the vertical distance from the launch pad to the rocket forming the other leg, and the hypotenuse representing the distance between the observer and the rocket.
Given:
1. Distance of observer from the launch pad: \(3 \text{ km}\)
2. Height of the rocket above the ground at the specific instance: \(4 \text{ km}\)
3. Vertical speed of the rocket: \(300 \text{ m/s}\) or \(0.3 \text{ km/s}\)
We need to find the rate at which the distance \(z\) between the observer and the rocket is increasing when the height \(y\) of the rocket is \(4 \text{ km}\).
---
**Steps Involving Pythagorean Theorem and Differentiation:**
1. Use the Pythagorean theorem to relate the distances:
\[
x^2 + y^2 = z^2
\]
where \(x = 3 \text{ km}\) (constant), \(y = 4 \text{ km}\), and \(z\) is the hypotenuse:
\[
3^2 + 4^2 = z^2 \implies 9 + 16 = 25 \implies z = 5 \text{ km}
\]
2. Differentiate both sides of the Pythagorean theorem equation with respect to time \(t\):
\[
\frac{d}{dt} (x^2 + y^2 = z^2) \implies 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}
\]
Since \(x\) is constant, \(\frac{dx}{dt} = 0\):
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Transcribed Image Text:**Problem 5.** An observer is positioned \(3 \text{ km}\) away from a rocket launch pad. How fast is the distance between the rocket and the observer increasing, when the rocket is \(4 \text{ km}\) above the ground and is moving straight up at the speed of \(300 \text{ m/s}\)?
---
**Explanation:**
To solve this related rates problem, one needs to consider the position of the observer and the rocket as two vertices of a right triangle, with the horizontal distance from the observer to the launch pad forming one leg, the vertical distance from the launch pad to the rocket forming the other leg, and the hypotenuse representing the distance between the observer and the rocket.
Given:
1. Distance of observer from the launch pad: \(3 \text{ km}\)
2. Height of the rocket above the ground at the specific instance: \(4 \text{ km}\)
3. Vertical speed of the rocket: \(300 \text{ m/s}\) or \(0.3 \text{ km/s}\)
We need to find the rate at which the distance \(z\) between the observer and the rocket is increasing when the height \(y\) of the rocket is \(4 \text{ km}\).
---
**Steps Involving Pythagorean Theorem and Differentiation:**
1. Use the Pythagorean theorem to relate the distances:
\[
x^2 + y^2 = z^2
\]
where \(x = 3 \text{ km}\) (constant), \(y = 4 \text{ km}\), and \(z\) is the hypotenuse:
\[
3^2 + 4^2 = z^2 \implies 9 + 16 = 25 \implies z = 5 \text{ km}
\]
2. Differentiate both sides of the Pythagorean theorem equation with respect to time \(t\):
\[
\frac{d}{dt} (x^2 + y^2 = z^2) \implies 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}
\]
Since \(x\) is constant, \(\frac{dx}{dt} = 0\):
\[
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