Related rates, my answer differs from the book, misprint or me?How fast is the length of his shadow on the building decreasing when he is 4 m from the buildingRelated Rates: rate of change of the distance between two objectsRelated Rates change in theta
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Related rates, my answer differs from the book, misprint or me?
How fast is the length of his shadow on the building decreasing when he is 4 m from the buildingRelated Rates: rate of change of the distance between two objectsRelated Rates change in theta
$begingroup$
Did my answer go wrong or does the book have a misprint?(there have been some inconstancies between the definitions used in the chapters and answer key, like two different authors, though only one is listed)
The problem: An airplane is flying 500 miles per hour horizontal one mile high over a radar station. Find the rate at which the distance is increasing when the plane is 2 miles from the station.
My answer is $1000/sqrt5$ or $200*sqrt5$
The book gives $250*sqrt3$
My method: triangle abc, a=1, b=2, and c is the hypotenuse, $db/dt=500$, and $c^2=1^2+b^2$, and I want $dc/dt$ at b=2
I took the derivative: $2c*fracdcdt=0+2b*fracdbdt$,
solved for $dc/dt$; $dc/dt=frac2b*db/dt2c$
Substitute the variables; $c=sqrt1+4$ and so $fracdcdt=frac2*2*5002*sqrt5=frac1000sqrt5$
related-rates
asked 2 days ago
Max PowerMax Power
1085
New contributor
Max Power is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Did my answer go wrong or does the book have a misprint?(there have been some inconstancies between the definitions used in the chapters and answer key, like two different authors, though only one is listed)
The problem: An airplane is flying 500 miles per hour horizontal one mile high over a radar station. Find the rate at which the distance is increasing when the plane is 2 miles from the station.
My answer is $1000/sqrt5$ or $200*sqrt5$
The book gives $250*sqrt3$
My method: triangle abc, a=1, b=2, and c is the hypotenuse, $db/dt=500$, and $c^2=1^2+b^2$, and I want $dc/dt$ at b=2
I took the derivative: $2c*fracdcdt=0+2b*fracdbdt$,
solved for $dc/dt$; $dc/dt=frac2b*db/dt2c$
Substitute the variables; $c=sqrt1+4$ and so $fracdcdt=frac2*2*5002*sqrt5=frac1000sqrt5$
related-rates
asked 2 days ago
Max PowerMax Power
1085
New contributor
Max Power is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Did my answer go wrong or does the book have a misprint?(there have been some inconstancies between the definitions used in the chapters and answer key, like two different authors, though only one is listed)
The problem: An airplane is flying 500 miles per hour horizontal one mile high over a radar station. Find the rate at which the distance is increasing when the plane is 2 miles from the station.
My answer is $1000/sqrt5$ or $200*sqrt5$
The book gives $250*sqrt3$
My method: triangle abc, a=1, b=2, and c is the hypotenuse, $db/dt=500$, and $c^2=1^2+b^2$, and I want $dc/dt$ at b=2
I took the derivative: $2c*fracdcdt=0+2b*fracdbdt$,
solved for $dc/dt$; $dc/dt=frac2b*db/dt2c$
Substitute the variables; $c=sqrt1+4$ and so $fracdcdt=frac2*2*5002*sqrt5=frac1000sqrt5$
related-rates
asked 2 days ago
Max PowerMax Power
1085
New contributor
Max Power is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Did my answer go wrong or does the book have a misprint?(there have been some inconstancies between the definitions used in the chapters and answer key, like two different authors, though only one is listed)
The problem: An airplane is flying 500 miles per hour horizontal one mile high over a radar station. Find the rate at which the distance is increasing when the plane is 2 miles from the station.
My answer is $1000/sqrt5$ or $200*sqrt5$
The book gives $250*sqrt3$
My method: triangle abc, a=1, b=2, and c is the hypotenuse, $db/dt=500$, and $c^2=1^2+b^2$, and I want $dc/dt$ at b=2
I took the derivative: $2c*fracdcdt=0+2b*fracdbdt$,
solved for $dc/dt$; $dc/dt=frac2b*db/dt2c$
Substitute the variables; $c=sqrt1+4$ and so $fracdcdt=frac2*2*5002*sqrt5=frac1000sqrt5$
related-rates
related-rates
asked 2 days ago
Max PowerMax Power
1085
New contributor
Max Power is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Max PowerMax Power
1085
New contributor
Max Power is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 days ago
Max Power
asked 2 days ago
Max PowerMax Power
1085
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Max Power is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago
Max PowerMax Power
1085
asked 2 days ago
Max PowerMax Power
1085
1085
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Max Power is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Max Power is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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2 Answers
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$begingroup$
It may be a misprint, but I don't think your answer is right either. The question asks for $dc/dt$ when the distance from the station is 2 - this means when c=2 not when b=2. Other than that your answer is correct; plugging a=1, b=$sqrt 3$, c=2 in, I get $250sqrt3$ (not $250/sqrt3$)
answered 2 days ago
Ben Ben
263
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Ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
$begingroup$
Ah semantics. Probably not the last time they get the better of me. And yes $250/sqrt3$, was a typo on my part, fixed now.
$endgroup$
– Max Power
2 days ago
add a comment |
$begingroup$
You approach is good, you just mixed up $b$ and $c$. When plane is two miles from the station, $c=2$, $b=sqrt3$. So I think the answer should be $250sqrt3$.
answered 2 days ago
VasyaVasya
4,0681618
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It may be a misprint, but I don't think your answer is right either. The question asks for $dc/dt$ when the distance from the station is 2 - this means when c=2 not when b=2. Other than that your answer is correct; plugging a=1, b=$sqrt 3$, c=2 in, I get $250sqrt3$ (not $250/sqrt3$)
answered 2 days ago
Ben Ben
263
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Ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
1
$begingroup$
Ah semantics. Probably not the last time they get the better of me. And yes $250/sqrt3$, was a typo on my part, fixed now.
$endgroup$
– Max Power
2 days ago
add a comment |
$begingroup$
It may be a misprint, but I don't think your answer is right either. The question asks for $dc/dt$ when the distance from the station is 2 - this means when c=2 not when b=2. Other than that your answer is correct; plugging a=1, b=$sqrt 3$, c=2 in, I get $250sqrt3$ (not $250/sqrt3$)
answered 2 days ago
Ben Ben
263
New contributor
Ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Ah semantics. Probably not the last time they get the better of me. And yes $250/sqrt3$, was a typo on my part, fixed now.
$endgroup$
– Max Power
2 days ago
add a comment |
$begingroup$
It may be a misprint, but I don't think your answer is right either. The question asks for $dc/dt$ when the distance from the station is 2 - this means when c=2 not when b=2. Other than that your answer is correct; plugging a=1, b=$sqrt 3$, c=2 in, I get $250sqrt3$ (not $250/sqrt3$)
answered 2 days ago
Ben Ben
263
New contributor
Ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
It may be a misprint, but I don't think your answer is right either. The question asks for $dc/dt$ when the distance from the station is 2 - this means when c=2 not when b=2. Other than that your answer is correct; plugging a=1, b=$sqrt 3$, c=2 in, I get $250sqrt3$ (not $250/sqrt3$)
answered 2 days ago
Ben Ben
263
New contributor
Ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Ben Ben
263
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answered 2 days ago
Ben Ben
263
answered 2 days ago
Ben Ben
263
263
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New contributor
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Check out our Code of Conduct.
Ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
$begingroup$
Ah semantics. Probably not the last time they get the better of me. And yes $250/sqrt3$, was a typo on my part, fixed now.
$endgroup$
– Max Power
2 days ago
add a comment |
1
$begingroup$
Ah semantics. Probably not the last time they get the better of me. And yes $250/sqrt3$, was a typo on my part, fixed now.
$endgroup$
– Max Power
2 days ago
1
1
$begingroup$
Ah semantics. Probably not the last time they get the better of me. And yes $250/sqrt3$, was a typo on my part, fixed now.
$endgroup$
– Max Power
2 days ago
$begingroup$
Ah semantics. Probably not the last time they get the better of me. And yes $250/sqrt3$, was a typo on my part, fixed now.
$endgroup$
– Max Power
2 days ago
add a comment |
$begingroup$
You approach is good, you just mixed up $b$ and $c$. When plane is two miles from the station, $c=2$, $b=sqrt3$. So I think the answer should be $250sqrt3$.
answered 2 days ago
VasyaVasya
4,0681618
$endgroup$
add a comment |
$begingroup$
You approach is good, you just mixed up $b$ and $c$. When plane is two miles from the station, $c=2$, $b=sqrt3$. So I think the answer should be $250sqrt3$.
answered 2 days ago
VasyaVasya
4,0681618
$endgroup$
add a comment |
$begingroup$
You approach is good, you just mixed up $b$ and $c$. When plane is two miles from the station, $c=2$, $b=sqrt3$. So I think the answer should be $250sqrt3$.
answered 2 days ago
VasyaVasya
4,0681618
$endgroup$
You approach is good, you just mixed up $b$ and $c$. When plane is two miles from the station, $c=2$, $b=sqrt3$. So I think the answer should be $250sqrt3$.
answered 2 days ago
VasyaVasya
4,0681618
answered 2 days ago
VasyaVasya
4,0681618
answered 2 days ago
VasyaVasya
4,0681618
answered 2 days ago
VasyaVasya
4,0681618
4,0681618
add a comment |
add a comment |
Max Power is a new contributor. Be nice, and check out our Code of Conduct.
Max Power is a new contributor. Be nice, and check out our Code of Conduct.
Max Power is a new contributor. Be nice, and check out our Code of Conduct.
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