Related Rates Problem Involving Airplanes
I took a test yesterday, and would like to know how to answer this specific question on the exam:
One airplane flew over an airport at the rate of $300$ mi/hr. Ten minutes later another airplane flew over the airport at $240$ mi/hr. If the first airplane was flying west and the second flying south (both at the same altitude), determine the rate at which they were separating $20$ minutes after the second plane flew over the airport.
I know that the pythagorean theorem should be used: $x^2 + y^2 = z^2$; but I don't know what to use for the $x, y$, and $z$ values.
$dy/dt = 240$ mi/hr
$dx/dt = 300$ mi/hr
calculus
asked Jan 29 '15 at 23:21
AudreyAudrey
448
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
I took a test yesterday, and would like to know how to answer this specific question on the exam:
One airplane flew over an airport at the rate of $300$ mi/hr. Ten minutes later another airplane flew over the airport at $240$ mi/hr. If the first airplane was flying west and the second flying south (both at the same altitude), determine the rate at which they were separating $20$ minutes after the second plane flew over the airport.
I know that the pythagorean theorem should be used: $x^2 + y^2 = z^2$; but I don't know what to use for the $x, y$, and $z$ values.
$dy/dt = 240$ mi/hr
$dx/dt = 300$ mi/hr
calculus
asked Jan 29 '15 at 23:21
AudreyAudrey
448
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
– Pranav Marathe
Jan 29 '15 at 23:31
1
To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
– apnorton
Jan 29 '15 at 23:46
1
@anorton Thank you for your help. I added some background information.
– Audrey
Jan 30 '15 at 0:03
@Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
– Loreno Heer
Jan 30 '15 at 0:22
add a comment |
I took a test yesterday, and would like to know how to answer this specific question on the exam:
One airplane flew over an airport at the rate of $300$ mi/hr. Ten minutes later another airplane flew over the airport at $240$ mi/hr. If the first airplane was flying west and the second flying south (both at the same altitude), determine the rate at which they were separating $20$ minutes after the second plane flew over the airport.
I know that the pythagorean theorem should be used: $x^2 + y^2 = z^2$; but I don't know what to use for the $x, y$, and $z$ values.
$dy/dt = 240$ mi/hr
$dx/dt = 300$ mi/hr
calculus
asked Jan 29 '15 at 23:21
AudreyAudrey
448
I took a test yesterday, and would like to know how to answer this specific question on the exam:
One airplane flew over an airport at the rate of $300$ mi/hr. Ten minutes later another airplane flew over the airport at $240$ mi/hr. If the first airplane was flying west and the second flying south (both at the same altitude), determine the rate at which they were separating $20$ minutes after the second plane flew over the airport.
I know that the pythagorean theorem should be used: $x^2 + y^2 = z^2$; but I don't know what to use for the $x, y$, and $z$ values.
$dy/dt = 240$ mi/hr
$dx/dt = 300$ mi/hr
calculus
calculus
asked Jan 29 '15 at 23:21
AudreyAudrey
448
asked Jan 29 '15 at 23:21
AudreyAudrey
448
edited Jan 30 '15 at 0:03
Audrey
asked Jan 29 '15 at 23:21
AudreyAudrey
448
asked Jan 29 '15 at 23:21
AudreyAudrey
448
asked Jan 29 '15 at 23:21
AudreyAudrey
448
448
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
– Pranav Marathe
Jan 29 '15 at 23:31
1
To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
– apnorton
Jan 29 '15 at 23:46
1
@anorton Thank you for your help. I added some background information.
– Audrey
Jan 30 '15 at 0:03
@Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
– Loreno Heer
Jan 30 '15 at 0:22
add a comment |
1
Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
– Pranav Marathe
Jan 29 '15 at 23:31
1
To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
– apnorton
Jan 29 '15 at 23:46
1
@anorton Thank you for your help. I added some background information.
– Audrey
Jan 30 '15 at 0:03
@Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
– Loreno Heer
Jan 30 '15 at 0:22
1
1
Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
– Pranav Marathe
Jan 29 '15 at 23:31
Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
– Pranav Marathe
Jan 29 '15 at 23:31
1
1
To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
– apnorton
Jan 29 '15 at 23:46
To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
– apnorton
Jan 29 '15 at 23:46
1
1
@anorton Thank you for your help. I added some background information.
– Audrey
Jan 30 '15 at 0:03
@anorton Thank you for your help. I added some background information.
– Audrey
Jan 30 '15 at 0:03
@Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
– Loreno Heer
Jan 30 '15 at 0:22
@Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
– Loreno Heer
Jan 30 '15 at 0:22
add a comment |
2 Answers
2
active
oldest
votes
Paths of the planes
$p_1(t) = (t cdot 300, 0)$
$p_2(t) = (0,(t-1/6) cdot (240))$
$s(t) = p_1(t) - p_2(t)$
$left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$
answered Jan 29 '15 at 23:31
Loreno HeerLoreno Heer
3,32411534
add a comment |
You have $x^2 + y^2 = z^2$
your goal is to find $frac {dz}{dt}$
$2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$
Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$
$x$ is the distance the west-bound airplane travels in 30 minutes.
$y$ is the distance the south-bound airplane travels in 20 minutes.
Use the Pythagorean theorem above to find $z.$
distance = speed $times$ time
You will need to convert units because the speed is in mph, and time is in minutes.
And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.
answered Feb 15 '18 at 22:42
Doug MDoug M
44.2k31854
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Paths of the planes
$p_1(t) = (t cdot 300, 0)$
$p_2(t) = (0,(t-1/6) cdot (240))$
$s(t) = p_1(t) - p_2(t)$
$left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$
answered Jan 29 '15 at 23:31
Loreno HeerLoreno Heer
3,32411534
add a comment |
Paths of the planes
$p_1(t) = (t cdot 300, 0)$
$p_2(t) = (0,(t-1/6) cdot (240))$
$s(t) = p_1(t) - p_2(t)$
$left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$
answered Jan 29 '15 at 23:31
Loreno HeerLoreno Heer
3,32411534
add a comment |
Paths of the planes
$p_1(t) = (t cdot 300, 0)$
$p_2(t) = (0,(t-1/6) cdot (240))$
$s(t) = p_1(t) - p_2(t)$
$left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$
answered Jan 29 '15 at 23:31
Loreno HeerLoreno Heer
3,32411534
Paths of the planes
$p_1(t) = (t cdot 300, 0)$
$p_2(t) = (0,(t-1/6) cdot (240))$
$s(t) = p_1(t) - p_2(t)$
$left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$
answered Jan 29 '15 at 23:31
Loreno HeerLoreno Heer
3,32411534
edited Jan 29 '15 at 23:43
answered Jan 29 '15 at 23:31
Loreno HeerLoreno Heer
3,32411534
answered Jan 29 '15 at 23:31
Loreno HeerLoreno Heer
3,32411534
answered Jan 29 '15 at 23:31
Loreno HeerLoreno Heer
3,32411534
3,32411534
add a comment |
add a comment |
You have $x^2 + y^2 = z^2$
your goal is to find $frac {dz}{dt}$
$2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$
Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$
$x$ is the distance the west-bound airplane travels in 30 minutes.
$y$ is the distance the south-bound airplane travels in 20 minutes.
Use the Pythagorean theorem above to find $z.$
distance = speed $times$ time
You will need to convert units because the speed is in mph, and time is in minutes.
And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.
answered Feb 15 '18 at 22:42
Doug MDoug M
44.2k31854
add a comment |
You have $x^2 + y^2 = z^2$
your goal is to find $frac {dz}{dt}$
$2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$
Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$
$x$ is the distance the west-bound airplane travels in 30 minutes.
$y$ is the distance the south-bound airplane travels in 20 minutes.
Use the Pythagorean theorem above to find $z.$
distance = speed $times$ time
You will need to convert units because the speed is in mph, and time is in minutes.
And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.
answered Feb 15 '18 at 22:42
Doug MDoug M
44.2k31854
add a comment |
You have $x^2 + y^2 = z^2$
your goal is to find $frac {dz}{dt}$
$2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$
Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$
$x$ is the distance the west-bound airplane travels in 30 minutes.
$y$ is the distance the south-bound airplane travels in 20 minutes.
Use the Pythagorean theorem above to find $z.$
distance = speed $times$ time
You will need to convert units because the speed is in mph, and time is in minutes.
And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.
answered Feb 15 '18 at 22:42
Doug MDoug M
44.2k31854
You have $x^2 + y^2 = z^2$
your goal is to find $frac {dz}{dt}$
$2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$
Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$
$x$ is the distance the west-bound airplane travels in 30 minutes.
$y$ is the distance the south-bound airplane travels in 20 minutes.
Use the Pythagorean theorem above to find $z.$
distance = speed $times$ time
You will need to convert units because the speed is in mph, and time is in minutes.
And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.
answered Feb 15 '18 at 22:42
Doug MDoug M
44.2k31854
edited Feb 15 '18 at 22:48
answered Feb 15 '18 at 22:42
Doug MDoug M
44.2k31854
answered Feb 15 '18 at 22:42
Doug MDoug M
44.2k31854
answered Feb 15 '18 at 22:42
Doug MDoug M
44.2k31854
44.2k31854
add a comment |
add a comment |
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1
Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
– Pranav Marathe
Jan 29 '15 at 23:31
1
To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
– apnorton
Jan 29 '15 at 23:46
1
@anorton Thank you for your help. I added some background information.
– Audrey
Jan 30 '15 at 0:03
@Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
– Loreno Heer
Jan 30 '15 at 0:22