Mixing
At what rate is the angle $theta$ changing when 10 ft. of rope is out?
$begingroup$
A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft. above the bow. The rope is hauled in a rate of 2 ft/s.
At what rate is the angle $theta$ changing when 10 ft. of rope is out?
I understand related rates problems, but the trig and angle part of the question is confusing me. How do I solve this?
calculus
edited Jun 15 '16 at 22:53
N. F. Taussig
45.2k103358
asked Jun 14 '16 at 18:47
user10984587user10984587
1376
$endgroup$
|
show 2 more comments
$begingroup$
A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft. above the bow. The rope is hauled in a rate of 2 ft/s.
At what rate is the angle $theta$ changing when 10 ft. of rope is out?
I understand related rates problems, but the trig and angle part of the question is confusing me. How do I solve this?
calculus
edited Jun 15 '16 at 22:53
N. F. Taussig
45.2k103358
asked Jun 14 '16 at 18:47
user10984587user10984587
1376
$endgroup$
$begingroup$
Draw a picture. How are the 6, 10 and $theta$ related? Which derivative corresponds to 2 ft/s?
$endgroup$
– John Douma
Jun 14 '16 at 18:57
$begingroup$
The height of the dock is fixed. So the length of the rope $r = 6sectheta,frac{dr}{dt} = 2 ft/s = 6 sectheta tantheta frac{dtheta}{dt}$
$endgroup$
– Doug M
Jun 14 '16 at 19:11
$begingroup$
Thanks for all the help! I got $dtheta/dt = (1/3)(costheta cottheta) = (1/3)(6/10)(6/8) = 3/20 theta$.
$endgroup$
– user10984587
Jun 14 '16 at 19:17
$begingroup$
Is $theta$ the angle the rope forms with the water or the angle the rope forms with the side of the dock?
$endgroup$
– N. F. Taussig
Jun 15 '16 at 10:18
$begingroup$
@N.F.Taussig Here's a picture for a better explanation: Page 7. I believe the answer to your question is the side of the dock.
$endgroup$
– user10984587
Jun 15 '16 at 22:40
|
show 2 more comments
$begingroup$
A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft. above the bow. The rope is hauled in a rate of 2 ft/s.
At what rate is the angle $theta$ changing when 10 ft. of rope is out?
I understand related rates problems, but the trig and angle part of the question is confusing me. How do I solve this?
calculus
edited Jun 15 '16 at 22:53
N. F. Taussig
45.2k103358
asked Jun 14 '16 at 18:47
user10984587user10984587
1376
$endgroup$
A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft. above the bow. The rope is hauled in a rate of 2 ft/s.
At what rate is the angle $theta$ changing when 10 ft. of rope is out?
I understand related rates problems, but the trig and angle part of the question is confusing me. How do I solve this?
calculus
calculus
edited Jun 15 '16 at 22:53
N. F. Taussig
45.2k103358
asked Jun 14 '16 at 18:47
user10984587user10984587
1376
edited Jun 15 '16 at 22:53
N. F. Taussig
45.2k103358
asked Jun 14 '16 at 18:47
user10984587user10984587
1376
edited Jun 15 '16 at 22:53
N. F. Taussig
45.2k103358
edited Jun 15 '16 at 22:53
N. F. Taussig
45.2k103358
edited Jun 15 '16 at 22:53
N. F. Taussig
45.2k103358
45.2k103358
asked Jun 14 '16 at 18:47
user10984587user10984587
1376
asked Jun 14 '16 at 18:47
user10984587user10984587
1376
asked Jun 14 '16 at 18:47
user10984587user10984587
1376
1376
$begingroup$
Draw a picture. How are the 6, 10 and $theta$ related? Which derivative corresponds to 2 ft/s?
$endgroup$
– John Douma
Jun 14 '16 at 18:57
$begingroup$
The height of the dock is fixed. So the length of the rope $r = 6sectheta,frac{dr}{dt} = 2 ft/s = 6 sectheta tantheta frac{dtheta}{dt}$
$endgroup$
– Doug M
Jun 14 '16 at 19:11
$begingroup$
Thanks for all the help! I got $dtheta/dt = (1/3)(costheta cottheta) = (1/3)(6/10)(6/8) = 3/20 theta$.
$endgroup$
– user10984587
Jun 14 '16 at 19:17
$begingroup$
Is $theta$ the angle the rope forms with the water or the angle the rope forms with the side of the dock?
$endgroup$
– N. F. Taussig
Jun 15 '16 at 10:18
$begingroup$
@N.F.Taussig Here's a picture for a better explanation: Page 7. I believe the answer to your question is the side of the dock.
$endgroup$
– user10984587
Jun 15 '16 at 22:40
|
show 2 more comments
$begingroup$
Draw a picture. How are the 6, 10 and $theta$ related? Which derivative corresponds to 2 ft/s?
$endgroup$
– John Douma
Jun 14 '16 at 18:57
$begingroup$
The height of the dock is fixed. So the length of the rope $r = 6sectheta,frac{dr}{dt} = 2 ft/s = 6 sectheta tantheta frac{dtheta}{dt}$
$endgroup$
– Doug M
Jun 14 '16 at 19:11
$begingroup$
Thanks for all the help! I got $dtheta/dt = (1/3)(costheta cottheta) = (1/3)(6/10)(6/8) = 3/20 theta$.
$endgroup$
– user10984587
Jun 14 '16 at 19:17
$begingroup$
Is $theta$ the angle the rope forms with the water or the angle the rope forms with the side of the dock?
$endgroup$
– N. F. Taussig
Jun 15 '16 at 10:18
$begingroup$
@N.F.Taussig Here's a picture for a better explanation: Page 7. I believe the answer to your question is the side of the dock.
$endgroup$
– user10984587
Jun 15 '16 at 22:40
$begingroup$
Draw a picture. How are the 6, 10 and $theta$ related? Which derivative corresponds to 2 ft/s?
$endgroup$
– John Douma
Jun 14 '16 at 18:57
$begingroup$
Draw a picture. How are the 6, 10 and $theta$ related? Which derivative corresponds to 2 ft/s?
$endgroup$
– John Douma
Jun 14 '16 at 18:57
$begingroup$
The height of the dock is fixed. So the length of the rope $r = 6sectheta,frac{dr}{dt} = 2 ft/s = 6 sectheta tantheta frac{dtheta}{dt}$
$endgroup$
– Doug M
Jun 14 '16 at 19:11
$begingroup$
The height of the dock is fixed. So the length of the rope $r = 6sectheta,frac{dr}{dt} = 2 ft/s = 6 sectheta tantheta frac{dtheta}{dt}$
$endgroup$
– Doug M
Jun 14 '16 at 19:11
$begingroup$
Thanks for all the help! I got $dtheta/dt = (1/3)(costheta cottheta) = (1/3)(6/10)(6/8) = 3/20 theta$.
$endgroup$
– user10984587
Jun 14 '16 at 19:17
$begingroup$
Thanks for all the help! I got $dtheta/dt = (1/3)(costheta cottheta) = (1/3)(6/10)(6/8) = 3/20 theta$.
$endgroup$
– user10984587
Jun 14 '16 at 19:17
$begingroup$
Is $theta$ the angle the rope forms with the water or the angle the rope forms with the side of the dock?
$endgroup$
– N. F. Taussig
Jun 15 '16 at 10:18
$begingroup$
Is $theta$ the angle the rope forms with the water or the angle the rope forms with the side of the dock?
$endgroup$
– N. F. Taussig
Jun 15 '16 at 10:18
$begingroup$
@N.F.Taussig Here's a picture for a better explanation: Page 7. I believe the answer to your question is the side of the dock.
$endgroup$
– user10984587
Jun 15 '16 at 22:40
$begingroup$
@N.F.Taussig Here's a picture for a better explanation: Page 7. I believe the answer to your question is the side of the dock.
$endgroup$
– user10984587
Jun 15 '16 at 22:40
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
You have the relation $costheta=dfrac{6}{r}$ so $-sin(theta)dfrac{dtheta}{dt}=-dfrac{6}{r^2}dfrac{dr}{dt}$, or
begin{equation}
sin(theta)frac{dtheta}{dt}=frac{6}{r^2}frac{dr}{dt}
end{equation}
When $r=10$ we know that $dfrac{dr}{dt}=-2$ and $sin(theta)=dfrac{6}{10}=dfrac{3}{5}$ so
begin{equation}
dfrac{dtheta}{dt}=frac{5}{3}cdotfrac{6}{100}(-2)=-frac{1}{5} text{ per seconds}
end{equation}
answered Jun 19 '16 at 5:41
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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$begingroup$
You have the relation $costheta=dfrac{6}{r}$ so $-sin(theta)dfrac{dtheta}{dt}=-dfrac{6}{r^2}dfrac{dr}{dt}$, or
begin{equation}
sin(theta)frac{dtheta}{dt}=frac{6}{r^2}frac{dr}{dt}
end{equation}
When $r=10$ we know that $dfrac{dr}{dt}=-2$ and $sin(theta)=dfrac{6}{10}=dfrac{3}{5}$ so
begin{equation}
dfrac{dtheta}{dt}=frac{5}{3}cdotfrac{6}{100}(-2)=-frac{1}{5} text{ per seconds}
end{equation}
answered Jun 19 '16 at 5:41
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
You have the relation $costheta=dfrac{6}{r}$ so $-sin(theta)dfrac{dtheta}{dt}=-dfrac{6}{r^2}dfrac{dr}{dt}$, or
begin{equation}
sin(theta)frac{dtheta}{dt}=frac{6}{r^2}frac{dr}{dt}
end{equation}
When $r=10$ we know that $dfrac{dr}{dt}=-2$ and $sin(theta)=dfrac{6}{10}=dfrac{3}{5}$ so
begin{equation}
dfrac{dtheta}{dt}=frac{5}{3}cdotfrac{6}{100}(-2)=-frac{1}{5} text{ per seconds}
end{equation}
answered Jun 19 '16 at 5:41
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
You have the relation $costheta=dfrac{6}{r}$ so $-sin(theta)dfrac{dtheta}{dt}=-dfrac{6}{r^2}dfrac{dr}{dt}$, or
begin{equation}
sin(theta)frac{dtheta}{dt}=frac{6}{r^2}frac{dr}{dt}
end{equation}
When $r=10$ we know that $dfrac{dr}{dt}=-2$ and $sin(theta)=dfrac{6}{10}=dfrac{3}{5}$ so
begin{equation}
dfrac{dtheta}{dt}=frac{5}{3}cdotfrac{6}{100}(-2)=-frac{1}{5} text{ per seconds}
end{equation}
answered Jun 19 '16 at 5:41
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
You have the relation $costheta=dfrac{6}{r}$ so $-sin(theta)dfrac{dtheta}{dt}=-dfrac{6}{r^2}dfrac{dr}{dt}$, or
begin{equation}
sin(theta)frac{dtheta}{dt}=frac{6}{r^2}frac{dr}{dt}
end{equation}
When $r=10$ we know that $dfrac{dr}{dt}=-2$ and $sin(theta)=dfrac{6}{10}=dfrac{3}{5}$ so
begin{equation}
dfrac{dtheta}{dt}=frac{5}{3}cdotfrac{6}{100}(-2)=-frac{1}{5} text{ per seconds}
end{equation}
answered Jun 19 '16 at 5:41
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jun 19 '16 at 5:41
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jun 19 '16 at 5:41
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jun 19 '16 at 5:41
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
add a comment |
add a comment |
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$begingroup$
Draw a picture. How are the 6, 10 and $theta$ related? Which derivative corresponds to 2 ft/s?
$endgroup$
– John Douma
Jun 14 '16 at 18:57
$begingroup$
The height of the dock is fixed. So the length of the rope $r = 6sectheta,frac{dr}{dt} = 2 ft/s = 6 sectheta tantheta frac{dtheta}{dt}$
$endgroup$
– Doug M
Jun 14 '16 at 19:11
$begingroup$
Thanks for all the help! I got $dtheta/dt = (1/3)(costheta cottheta) = (1/3)(6/10)(6/8) = 3/20 theta$.
$endgroup$
– user10984587
Jun 14 '16 at 19:17
$begingroup$
Is $theta$ the angle the rope forms with the water or the angle the rope forms with the side of the dock?
$endgroup$
– N. F. Taussig
Jun 15 '16 at 10:18
$begingroup$
@N.F.Taussig Here's a picture for a better explanation: Page 7. I believe the answer to your question is the side of the dock.
$endgroup$
– user10984587
Jun 15 '16 at 22:40