An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each other. One airplane is 90 miles from the point and has a speed of 270 miles per hour. The other is 120 miles from the point and has a speed of 360 miles per hour. Exercise (a) At what rate is the distance between the planes changing? Step 1 Lets denote the distance in miles as shown in the following figure. The origin represents the point of convergence. 140 120 100 80 $ 60 y 40 20 X 20 40 60 80 100 The distances between the planes given by the formula, s=√x2 + y2. Thus, s² = x² + y² One plane is 90 miles away and moving towards the point of convergence at 270 miles per hour so x = 90 and The other plane is 120 miles away and moving towards the point of convergence at 360 miles per hour, so y = 120 and= [ X
An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each other. One airplane is 90 miles from the point and has a speed of 270 miles per hour. The other is 120 miles from the point and has a speed of 360 miles per hour. Exercise (a) At what rate is the distance between the planes changing? Step 1 Lets denote the distance in miles as shown in the following figure. The origin represents the point of convergence. 140 120 100 80 $ 60 y 40 20 X 20 40 60 80 100 The distances between the planes given by the formula, s=√x2 + y2. Thus, s² = x² + y² One plane is 90 miles away and moving towards the point of convergence at 270 miles per hour so x = 90 and The other plane is 120 miles away and moving towards the point of convergence at 360 miles per hour, so y = 120 and= [ X
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Exercise (a)
At what rate is the distance between the planes changing?
### Step 1
Let \( s \) denote the distance in miles as shown in the following figure. The origin represents the point of convergence.
#### Diagram Description
The diagram is a right-angled triangle with a vertical y-axis and a horizontal x-axis. The hypotenuse of the triangle, labeled \( s \), represents the distance between the two planes. The vertical leg is labeled \( y \), and the horizontal leg is labeled \( x \). Points on the axes are labeled at intervals of 20 miles.
#### Equation of Distance
The distance \( s \) between the planes is given by the formula:
\[ s = \sqrt{x^2 + y^2} \]
Thus,
\[ s^2 = x^2 + y^2 \]
One plane is 90 miles away and moving towards the point of convergence at 270 miles per hour so \( x = 90 \) and
\[ \frac{dx}{dt} = -270 \, \text{miles per hour} \]
The other plane is 120 miles away and moving towards the point of convergence at 360 miles per hour, so \( y = 120 \) and
\[ \frac{dy}{dt} = -360 \, \text{miles per hour} \]
#### Calculation
To find the rate at which the distance \( s \) between the planes is changing, substitute the given values into the derivative of the distance formula:
1. Differentiate both sides of \( s^2 = x^2 + y^2 \):
\[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]
2. Substitute the known values:
\[ s = \sqrt{90^2 + 120^2} = \sqrt{8100 + 14400} = 150 \, \text{miles}\]
\[ x = 90 \]
\[ \frac{dx}{dt} = -270 \]
\[ y = 120 \]
\[ \frac{dy}{dt} = -360 \]
3. Substitute these values into the differentiated equation:
\[ 2(150) \frac{ds}{dt} = 2(90)(-270) + 2(120)(-360) \]
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6e1a3f1-b538-4464-8269-7761edea80d0%2Fe25650d8-cf50-4a57-b2ec-d605d66f7fb5%2Fyyhc0dk_processed.png&w=3840&q=75)
Transcribed Image Text:### Exercise (a)
At what rate is the distance between the planes changing?
### Step 1
Let \( s \) denote the distance in miles as shown in the following figure. The origin represents the point of convergence.
#### Diagram Description
The diagram is a right-angled triangle with a vertical y-axis and a horizontal x-axis. The hypotenuse of the triangle, labeled \( s \), represents the distance between the two planes. The vertical leg is labeled \( y \), and the horizontal leg is labeled \( x \). Points on the axes are labeled at intervals of 20 miles.
#### Equation of Distance
The distance \( s \) between the planes is given by the formula:
\[ s = \sqrt{x^2 + y^2} \]
Thus,
\[ s^2 = x^2 + y^2 \]
One plane is 90 miles away and moving towards the point of convergence at 270 miles per hour so \( x = 90 \) and
\[ \frac{dx}{dt} = -270 \, \text{miles per hour} \]
The other plane is 120 miles away and moving towards the point of convergence at 360 miles per hour, so \( y = 120 \) and
\[ \frac{dy}{dt} = -360 \, \text{miles per hour} \]
#### Calculation
To find the rate at which the distance \( s \) between the planes is changing, substitute the given values into the derivative of the distance formula:
1. Differentiate both sides of \( s^2 = x^2 + y^2 \):
\[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]
2. Substitute the known values:
\[ s = \sqrt{90^2 + 120^2} = \sqrt{8100 + 14400} = 150 \, \text{miles}\]
\[ x = 90 \]
\[ \frac{dx}{dt} = -270 \]
\[ y = 120 \]
\[ \frac{dy}{dt} = -360 \]
3. Substitute these values into the differentiated equation:
\[ 2(150) \frac{ds}{dt} = 2(90)(-270) + 2(120)(-360) \]
\[
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