Mixing
Related Rates Calculus Trigonometric Problem
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I've been stuck on this related rates problem for a while now and I just can't figure out how to even approach it. The problem goes something like this:
The above diagram shows two objects moving at different speeds. Both objects are 0.5 miles from the origin. The blue object is moving at 50mph. The straight-line distance between the blue object and the red object is increasing at a rate of 35 mph. Find the speed of the red object.
I tried to solve it using Pythagorean Theorem and finding the derivative of the top side of the triangle. Anyway, I ended up getting a negative number, and even ignoring the sign, the answer I got was obviously wrong. I know the speed of the red object is obviously greater than the blue object because the distance between the objects is increasing. I just really don't know how to calculate the magnitude of said number. There's also a variation of this problem where the straight-line distance is changing at a rate of -35mph but I think that should be doable once I understand how to go about solving the original. Any responses would be appreciated!
calculus
edited 2 hours ago
Parcly Taxel
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asked 2 hours ago
Joe Tec
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up vote
5
down vote
favorite
I've been stuck on this related rates problem for a while now and I just can't figure out how to even approach it. The problem goes something like this:
The above diagram shows two objects moving at different speeds. Both objects are 0.5 miles from the origin. The blue object is moving at 50mph. The straight-line distance between the blue object and the red object is increasing at a rate of 35 mph. Find the speed of the red object.
I tried to solve it using Pythagorean Theorem and finding the derivative of the top side of the triangle. Anyway, I ended up getting a negative number, and even ignoring the sign, the answer I got was obviously wrong. I know the speed of the red object is obviously greater than the blue object because the distance between the objects is increasing. I just really don't know how to calculate the magnitude of said number. There's also a variation of this problem where the straight-line distance is changing at a rate of -35mph but I think that should be doable once I understand how to go about solving the original. Any responses would be appreciated!
calculus
edited 2 hours ago
Parcly Taxel
38k137097
asked 2 hours ago
Joe Tec
261
New contributor
Joe Tec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I've been stuck on this related rates problem for a while now and I just can't figure out how to even approach it. The problem goes something like this:
The above diagram shows two objects moving at different speeds. Both objects are 0.5 miles from the origin. The blue object is moving at 50mph. The straight-line distance between the blue object and the red object is increasing at a rate of 35 mph. Find the speed of the red object.
I tried to solve it using Pythagorean Theorem and finding the derivative of the top side of the triangle. Anyway, I ended up getting a negative number, and even ignoring the sign, the answer I got was obviously wrong. I know the speed of the red object is obviously greater than the blue object because the distance between the objects is increasing. I just really don't know how to calculate the magnitude of said number. There's also a variation of this problem where the straight-line distance is changing at a rate of -35mph but I think that should be doable once I understand how to go about solving the original. Any responses would be appreciated!
calculus
edited 2 hours ago
Parcly Taxel
38k137097
asked 2 hours ago
Joe Tec
261
New contributor
Joe Tec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I've been stuck on this related rates problem for a while now and I just can't figure out how to even approach it. The problem goes something like this:
The above diagram shows two objects moving at different speeds. Both objects are 0.5 miles from the origin. The blue object is moving at 50mph. The straight-line distance between the blue object and the red object is increasing at a rate of 35 mph. Find the speed of the red object.
I tried to solve it using Pythagorean Theorem and finding the derivative of the top side of the triangle. Anyway, I ended up getting a negative number, and even ignoring the sign, the answer I got was obviously wrong. I know the speed of the red object is obviously greater than the blue object because the distance between the objects is increasing. I just really don't know how to calculate the magnitude of said number. There's also a variation of this problem where the straight-line distance is changing at a rate of -35mph but I think that should be doable once I understand how to go about solving the original. Any responses would be appreciated!
calculus
calculus
edited 2 hours ago
Parcly Taxel
38k137097
asked 2 hours ago
Joe Tec
261
New contributor
Joe Tec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Parcly Taxel
38k137097
asked 2 hours ago
Joe Tec
261
New contributor
Joe Tec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Parcly Taxel
38k137097
edited 2 hours ago
Parcly Taxel
38k137097
edited 2 hours ago
Parcly Taxel
38k137097
38k137097
asked 2 hours ago
Joe Tec
261
New contributor
Joe Tec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Joe Tec
261
asked 2 hours ago
Joe Tec
261
261
New contributor
Joe Tec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Joe Tec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Joe Tec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
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1 Answer
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Let $a$ be the distance of the blue object from the origin, $b$ be that of the blue object and $c=sqrta^2+b^2$ be the distance between the objects. All of these are functions of time $t$.
$$fracpartial cpartial t=fracpartial cpartial afracpartial apartial t+fracpartial cpartial bfracpartial bpartial t=frac asqrta^2+b^2cdot(-50)+frac bsqrta^2+b^2cdot x=35$$
We substitute $a=b=0.5$:
$$frac0.5sqrt0.25+0.25=sqrt0.5$$
$$sqrt0.5(-50+x)=35$$
$$x=frac35sqrt0.5+50=99.4975$$
Hence the red object is moving at around 99.5 mph.
answered 2 hours ago
Parcly Taxel
38k137097
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let $a$ be the distance of the blue object from the origin, $b$ be that of the blue object and $c=sqrta^2+b^2$ be the distance between the objects. All of these are functions of time $t$.
$$fracpartial cpartial t=fracpartial cpartial afracpartial apartial t+fracpartial cpartial bfracpartial bpartial t=frac asqrta^2+b^2cdot(-50)+frac bsqrta^2+b^2cdot x=35$$
We substitute $a=b=0.5$:
$$frac0.5sqrt0.25+0.25=sqrt0.5$$
$$sqrt0.5(-50+x)=35$$
$$x=frac35sqrt0.5+50=99.4975$$
Hence the red object is moving at around 99.5 mph.
answered 2 hours ago
Parcly Taxel
38k137097
add a comment |Â
up vote
3
down vote
Let $a$ be the distance of the blue object from the origin, $b$ be that of the blue object and $c=sqrta^2+b^2$ be the distance between the objects. All of these are functions of time $t$.
$$fracpartial cpartial t=fracpartial cpartial afracpartial apartial t+fracpartial cpartial bfracpartial bpartial t=frac asqrta^2+b^2cdot(-50)+frac bsqrta^2+b^2cdot x=35$$
We substitute $a=b=0.5$:
$$frac0.5sqrt0.25+0.25=sqrt0.5$$
$$sqrt0.5(-50+x)=35$$
$$x=frac35sqrt0.5+50=99.4975$$
Hence the red object is moving at around 99.5 mph.
answered 2 hours ago
Parcly Taxel
38k137097
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $a$ be the distance of the blue object from the origin, $b$ be that of the blue object and $c=sqrta^2+b^2$ be the distance between the objects. All of these are functions of time $t$.
$$fracpartial cpartial t=fracpartial cpartial afracpartial apartial t+fracpartial cpartial bfracpartial bpartial t=frac asqrta^2+b^2cdot(-50)+frac bsqrta^2+b^2cdot x=35$$
We substitute $a=b=0.5$:
$$frac0.5sqrt0.25+0.25=sqrt0.5$$
$$sqrt0.5(-50+x)=35$$
$$x=frac35sqrt0.5+50=99.4975$$
Hence the red object is moving at around 99.5 mph.
answered 2 hours ago
Parcly Taxel
38k137097
Let $a$ be the distance of the blue object from the origin, $b$ be that of the blue object and $c=sqrta^2+b^2$ be the distance between the objects. All of these are functions of time $t$.
$$fracpartial cpartial t=fracpartial cpartial afracpartial apartial t+fracpartial cpartial bfracpartial bpartial t=frac asqrta^2+b^2cdot(-50)+frac bsqrta^2+b^2cdot x=35$$
We substitute $a=b=0.5$:
$$frac0.5sqrt0.25+0.25=sqrt0.5$$
$$sqrt0.5(-50+x)=35$$
$$x=frac35sqrt0.5+50=99.4975$$
Hence the red object is moving at around 99.5 mph.
answered 2 hours ago
Parcly Taxel
38k137097
answered 2 hours ago
Parcly Taxel
38k137097
answered 2 hours ago
Parcly Taxel
38k137097
answered 2 hours ago
Parcly Taxel
38k137097
38k137097
add a comment |Â
add a comment |Â
Joe Tec is a new contributor. Be nice, and check out our Code of Conduct.
Joe Tec is a new contributor. Be nice, and check out our Code of Conduct.
Joe Tec is a new contributor. Be nice, and check out our Code of Conduct.
Joe Tec is a new contributor. Be nice, and check out our Code of Conduct.
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