My Math Forum Related Rates Help Please

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 November 30th, 2008, 06:14 AM #1 Newbie   Joined: Nov 2008 Posts: 12 Thanks: 0 Related Rates Help Please I'm not asking these because I'm trying to cheat on my homework, I have a test coming up and I need to understand how to do these. 1. The minute hand of a certain clock is 4 in. long. Starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand? 2. Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 6 mi ^2/hr. How fast is the radius of the spill increasing when the area is 9 mi^2? Thanks for any help you could give me
 November 30th, 2008, 06:24 AM #2 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Related Rates Help Please This is the internet. There is no way for anyone to really know if this is or is not homework, regardless of what is seen in print [I'm tall, young, dark skinned, good looking, and rich, and you have to believe me because I said so.] A base guideline is that it is still "homework" until you have shown that you can at least start these problems ...one at a time. You could list forty as easily as four. It's simply a matter of showing your own goodwill if you wish. If someone simply gives you a full solution with no dialogue or consideration of your own effort, then you are no further ahead than you are now, having already seen some in your text and the classroom. It only shows that they can still do it easily, not you. So, if you can show what you have at least tried for just the first one ....
 November 30th, 2008, 06:27 AM #3 Newbie   Joined: Nov 2008 Posts: 12 Thanks: 0 Re: Related Rates Help Please I have done work but I can't really show it on here because I don't know how to do all the math symbols
November 30th, 2008, 06:36 AM   #4
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 Originally Posted by pranavpuck if your not going to help then don't bother posting
If you are not going to try by your own effort, then don't bother posting.

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 1. A man starts walking north at 4 ft/s from a point P. Five minutes later a woman starts walking south at 5 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?
Is the woman 500 feet directly due east when he started to walk, or when he had walked for the first five minutes?

Let's assume at the start. So, at 5 minutes after he has started to walk, you should have a point of his location [where on the y-axis], the woman still being 500' due east of his starting point.

You can then extend in the diagram, his position increasing by "s", and hers due south by (5/4)s [can you see why?]

You can then form a right triangle containing for one side the distance East of the woman [500], the northerly distance between them [1200 + s + (5/4)s], and can then find the direct distance between them, using Pythagoras, in terms of th one variable, "s". The 1200 is from your first calculation of his location when she starts to walk.

You can then apply your knowledge of calculus to find the rate of change of that distance in terms of ds/dt [which is given.]

November 30th, 2008, 06:40 AM   #5
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Originally Posted by Dave
Quote:
 Originally Posted by pranavpuck if your not going to help then don't bother posting
If you are not going to try by your own effort, then don't bother posting.
Dude please stop wasting my time with these worthless replies, and who spends their time on hanging out on math forums?

November 30th, 2008, 06:54 AM   #6
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 who spends their time on hanging out on math forums?
I do, I reckon that's makes me a mathnerd and proud of it. Someone should take an interest in something besides sports, celebrities, pop culture, and other mind-numbing things that do nothing to push us forward.

Quote:
 2. Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 6 mi ^2/hr. How fast is the radius of the spill increasing when the area is 9 mi^2? Thanks for any help you could give me
The area of a circle, as you probably know, is $A={\pi}r^{2}$

Now, we are given the area is 9. So, plug that in for A and solve for r. That is the radius when the area is 9.

Differentiate the area equation w.r.t time, plug in the knowns and solve for dr/dt.

You have dA/dt and r. All you need to plug in and find dr/dt. Does that help?.

 November 30th, 2008, 07:00 AM #7 Newbie   Joined: Nov 2008 Posts: 12 Thanks: 0 Re: Related Rates Help Please Yes it does, thank you!
November 30th, 2008, 07:02 AM   #8
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Originally Posted by galactus
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 who spends their time on hanging out on math forums?
I do, I reckon that's makes me a mathnerd and proud of it.
So do I. I have helped countless people for decades, and still do so in several forums. What I don't do is to perpetuate the quest for others to do their work. I recognise the difference between an honest request [I've done this so far and get stuck here] and NO effort ...a simple list of problems. I recognise the fact that if a person has seen examples in a textbook, has seen examples in a classroom, then his seeing one more example will not have the magic effect I think I can have over previous attempts by equally capable authors. I recognise that if he does not show some initial effort, then all is in vain ...and I just HATE wasting my time, even if it does show how clever I can be.

November 30th, 2008, 07:03 AM   #9
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 1. The minute hand of a certain clock is 4 in. long. Starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand?
Here's a start.

The area of a sector of a circle is $A=\frac{1}{2}r^{2}{\theta}$

The radius remains constant since that is the length of the hand (4 inches).

$\frac{d{\theta}}{dt}=\frac{\pi}{30}$. Do you see why?.

November 30th, 2008, 07:23 AM   #10
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Originally Posted by galactus
Quote:
 1. The minute hand of a certain clock is 4 in. long. Starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand?
Here's a start.

The area of a sector of a circle is $A=\frac{1}{2}r^{2}{\theta}$

The radius remains constant since that is the length of the hand (4 inches).

$\frac{d{\theta}}{dt}=\frac{\pi}{6}$. Do you see why?.
Oh, I see now, I've been using the area of a circle formula instead of the area of a sector of a circle formula.
Thanks for helping!

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