Math / Geometry

QuestionExample 3 A plane flying north at 640 miles per hour passes over a certain town at noon, and a second plane going east at 600 miles per hour is directly over the same town 15 minutes later. If the planes are flying at the same altitude, how fast will they be separating at $1: 15$ p.m.?

Studdy Solution

STEP 1

What is this asking? How fast are two planes moving apart from each other at a specific time, given their speeds and directions? Watch out! Don't forget to convert minutes to hours, and remember we're looking for the rate of change of the distance between the planes!

STEP 2

1. Set up the variables and the relationship between them.
2. Differentiate with respect to time.
3. Plug in the known values.

STEP 3

Let $t$ be the time in hours after noon.
Let $x$ be the distance the eastbound plane has traveled, and $y$ be the distance the northbound plane has traveled.
Let $s$ be the distance between the two planes.

STEP 4

We can use the Pythagorean theorem to relate these variables: $x^2 + y^2 = s^2$ .
This is because the planes' paths form a right triangle, with $x$ and $y$ as the legs and $s$ as the hypotenuse.

STEP 5

We know that the northbound plane is traveling at 640 mph, so $y = 640t$ .
The eastbound plane starts 15 minutes (or $\frac{1}{4}$ hour) later, so its distance traveled is $x = 600(t - \frac{1}{4})$ .

STEP 6

Now, we differentiate both sides of the Pythagorean equation with respect to time $t$ : $\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(s^2)$ $2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 2s \cdot \frac{ds}{dt}$ Remember, $\frac{dx}{dt}$ is the speed of the eastbound plane, and $\frac{dy}{dt}$ is the speed of the northbound plane!
And $\frac{ds}{dt}$ is what we want to find: the rate at which the distance between the planes is changing.

STEP 7

We can divide both sides of the equation by $2$ to simplify: $x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = s \cdot \frac{ds}{dt}$

STEP 8

We want to find $\frac{ds}{dt}$ at $t = 1.25$ (1:15 pm).
First, let's find $x$ and $y$ at this time: $x = 600(1.25 - 0.25) = 600(1) = \textbf{600}$ $y = 640(1.25) = \textbf{800}$

STEP 9

Now we can find $s$ using the Pythagorean theorem: $s = \sqrt{x^2 + y^2} = \sqrt{600^2 + 800^2} = \sqrt{360000 + 640000} = \sqrt{1000000} = \textbf{1000}$

STEP 10

We know $\frac{dx}{dt} = 600$ and $\frac{dy}{dt} = 640$ .
Now we can plug everything into our differentiated equation: $(600)(600) + (800)(640) = (1000) \cdot \frac{ds}{dt}$ $360000 + 512000 = 1000 \cdot \frac{ds}{dt}$ $872000 = 1000 \cdot \frac{ds}{dt}$

STEP 11

Finally, we solve for $\frac{ds}{dt}$ : $\frac{ds}{dt} = \frac{872000}{1000} = \textbf{872}$

STEP 12

At 1:15 pm, the planes are separating at a rate of 872 miles per hour.

Was this helpful?

QuestionExample 3 A plane flying north at 640 miles per hour passes over a certain town at noon, and a second plane going east at 600 miles per hour is directly over the same town 15 minutes later. If the planes are flying at the same altitude, how fast will they be separating at $1: 15$ p.m.?

Studdy Solution

STEP 1

What is this asking? How fast are two planes moving apart from each other at a specific time, given their speeds and directions? Watch out! Don't forget to convert minutes to hours, and remember we're looking for the rate of change of the distance between the planes!

STEP 2

1. Set up the variables and the relationship between them.
2. Differentiate with respect to time.
3. Plug in the known values.

STEP 3

Let $t$ be the time in hours after noon.
Let $x$ be the distance the eastbound plane has traveled, and $y$ be the distance the northbound plane has traveled.
Let $s$ be the distance between the two planes.

STEP 4

We can use the Pythagorean theorem to relate these variables: $x^2 + y^2 = s^2$ .
This is because the planes' paths form a right triangle, with $x$ and $y$ as the legs and $s$ as the hypotenuse.

STEP 5

We know that the northbound plane is traveling at 640 mph, so $y = 640t$ .
The eastbound plane starts 15 minutes (or $\frac{1}{4}$ hour) later, so its distance traveled is $x = 600(t - \frac{1}{4})$ .

STEP 6

STEP 7

We can divide both sides of the equation by $2$ to simplify: $x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = s \cdot \frac{ds}{dt}$

STEP 8

We want to find $\frac{ds}{dt}$ at $t = 1.25$ (1:15 pm).
First, let's find $x$ and $y$ at this time: $x = 600(1.25 - 0.25) = 600(1) = \textbf{600}$ $y = 640(1.25) = \textbf{800}$

STEP 9

Now we can find $s$ using the Pythagorean theorem: $s = \sqrt{x^2 + y^2} = \sqrt{600^2 + 800^2} = \sqrt{360000 + 640000} = \sqrt{1000000} = \textbf{1000}$

STEP 10

We know $\frac{dx}{dt} = 600$ and $\frac{dy}{dt} = 640$ .
Now we can plug everything into our differentiated equation: $(600)(600) + (800)(640) = (1000) \cdot \frac{ds}{dt}$ $360000 + 512000 = 1000 \cdot \frac{ds}{dt}$ $872000 = 1000 \cdot \frac{ds}{dt}$

STEP 11

Finally, we solve for $\frac{ds}{dt}$ : $\frac{ds}{dt} = \frac{872000}{1000} = \textbf{872}$

STEP 12

At 1:15 pm, the planes are separating at a rate of 872 miles per hour.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

What is this asking? How fast are two planes moving apart from each other at a specific time, given their speeds and directions? Watch out! Don't forget to convert minutes to hours, and remember we're looking for the rate of change of the *distance* between the planes!

STEP 2

1. Set up the variables and the relationship between them.2. Differentiate with respect to time.3. Plug in the known values.

STEP 3

Let ttt be the time in hours after noon.Let xxx be the distance the eastbound plane has traveled, and yyy be the distance the northbound plane has traveled.Let sss be the distance between the two planes.

STEP 4

We can use the Pythagorean theorem to relate these variables: x2+y2=s2x^2 + y^2 = s^2x2+y2=s2.This is because the planes' paths form a right triangle, with xxx and yyy as the legs and sss as the hypotenuse.

STEP 5

We know that the northbound plane is traveling at 640 mph, so y=640ty = 640ty=640t.The eastbound plane starts 15 minutes (or 14\frac{1}{4}41​ hour) later, so its distance traveled is x=600(t−14)x = 600(t - \frac{1}{4})x=600(t−41​).

STEP 6

STEP 7

We can divide both sides of the equation by 222 to simplify: x⋅dxdt+y⋅dydt=s⋅dsdt x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = s \cdot \frac{ds}{dt} x⋅dtdx​+y⋅dtdy​=s⋅dtds​

STEP 8

We want to find dsdt\frac{ds}{dt}dtds​ at t=1.25t = 1.25t=1.25 (1:15 pm).First, let's find xxx and yyy at this time: x=600(1.25−0.25)=600(1)=600 x = 600(1.25 - 0.25) = 600(1) = \textbf{600} x=600(1.25−0.25)=600(1)=600 y=640(1.25)=800 y = 640(1.25) = \textbf{800} y=640(1.25)=800

STEP 9

Now we can find sss using the Pythagorean theorem: s=x2+y2=6002+8002=360000+640000=1000000=1000 s = \sqrt{x^2 + y^2} = \sqrt{600^2 + 800^2} = \sqrt{360000 + 640000} = \sqrt{1000000} = \textbf{1000} s=x2+y2​=6002+8002​=360000+640000​=1000000​=1000

STEP 10

STEP 11

Finally, we solve for dsdt\frac{ds}{dt}dtds​: dsdt=8720001000=872 \frac{ds}{dt} = \frac{872000}{1000} = \textbf{872} dtds​=1000872000​=872

STEP 12

At 1:15 pm, the planes are separating at a rate of **872** miles per hour.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? How fast are two planes moving apart from each other at a specific time, given their speeds and directions? Watch out! Don't forget to convert minutes to hours, and remember we're looking for the rate of change of the distance between the planes!

1. Set up the variables and the relationship between them.
2. Differentiate with respect to time.
3. Plug in the known values.

Let $t$ be the time in hours after noon.
Let $x$ be the distance the eastbound plane has traveled, and $y$ be the distance the northbound plane has traveled.
Let $s$ be the distance between the two planes.

We can use the Pythagorean theorem to relate these variables: $x^2 + y^2 = s^2$ .
This is because the planes' paths form a right triangle, with $x$ and $y$ as the legs and $s$ as the hypotenuse.

We know that the northbound plane is traveling at 640 mph, so $y = 640t$ .
The eastbound plane starts 15 minutes (or $\frac{1}{4}$ hour) later, so its distance traveled is $x = 600(t - \frac{1}{4})$ .

We can divide both sides of the equation by $2$ to simplify: $x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = s \cdot \frac{ds}{dt}$

We want to find $\frac{ds}{dt}$ at $t = 1.25$ (1:15 pm).
First, let's find $x$ and $y$ at this time: $x = 600(1.25 - 0.25) = 600(1) = \textbf{600}$ $y = 640(1.25) = \textbf{800}$

Now we can find $s$ using the Pythagorean theorem: $s = \sqrt{x^2 + y^2} = \sqrt{600^2 + 800^2} = \sqrt{360000 + 640000} = \sqrt{1000000} = \textbf{1000}$

Finally, we solve for $\frac{ds}{dt}$ : $\frac{ds}{dt} = \frac{872000}{1000} = \textbf{872}$

At 1:15 pm, the planes are separating at a rate of 872 miles per hour.