# related rates

• Mar 22nd 2007, 01:52 PM
zachb
related rates
I need help with this peoblem:

A rope that is 30 feet long is tied to a weight and strung through a pulley that is 8 feet above the weight and is tied to the bumber of a car that is 2 feet from the ground. The car moves away from the weight at a rate of 5 feet/sec. How fast is the weight rising when it is 4 feet from the ground?

Forgive me if this makes no sense, but this is exactly how my professor worded the question and he thinks is makes perfect sense. :rolleyes:
• Mar 22nd 2007, 07:12 PM
Jhevon
Quote:

Originally Posted by zachb
I need help with this peoblem:

A rope that is 30 feet long is tied to a weight and strung through a pulley that is 8 feet above the weight and is tied to the bumber of a car that is 2 feet from the ground. The car moves away from the weight at a rate of 5 feet/sec. How fast is the weight rising when it is 4 feet from the ground?

Forgive me if this makes no sense, but this is exactly how my professor worded the question and he thinks is makes perfect sense. :rolleyes:

ok, i think this is the right interpretation for your problem (see the diagram below--ALWAYS DRAW A DIAGRAM WHEN DOING RELATED RATES!)

so obviously there is a right angled triangle here, so the formula we will be using is Pythagoras' formula.

notice that x is increasing at the same rate y is decreasing, so dx/dt = -dy/dt

also notice that when the weight is four feet from the ground, y=4, x = 26 (since its a 30 foot rope, they have to add up to 30) and by pythagoras' formula, z = 25.69

we are told the the car moves 5 feet/s so dz/dt is increasing at 5 ft/s

we are looking for dy/dt when y = 4, here goes:

x^2 = y^2 + z^2
=> 2x dx/dt = 2y dy/dt + 2z dz/dt
=> 2x (-dy/dt) = 2y dy/dt + 2z dz/dt
=> - 2x dy/dt - 2y dy/dt = 2z dz/dt
=> dy/dt (-2x - 2y) = 2z dz/dt
=> dy/dt = [2z dz/dt]/(-2x - 2y)
now plug in all the values

=> dy/dt = [2(25.69)(5)]/(-2(26) - 2(4)) = 256.9/(-52 - 8) = 256.9/(-60)
=> dy/dt = - 4.28 ft/s

note that dy/dt is negative since the length of y according to our diagram is decreasing.

so the rate is just 4.28 ft/s
• Mar 22nd 2007, 07:23 PM
ThePerfectHacker
No, this is the type of car I want.
(I will without jokes, drive that thing. I do not know why people today say that car is ugly. I really think this has a nice look to it).
• Mar 22nd 2007, 07:34 PM
Jhevon
i seem to have made an error in my thought process above. note that as the length y = 4, the length x will not be hypotenuse of the triangle, rather, only a part of x will be. i leave it to you to correct this error, you have the frame work of the solution

or maybe it will work as it is, i dont know:)

i'll think about it later if you dont figure it out
• Mar 22nd 2007, 08:07 PM
Jhevon
i think replacing y with y + 2 will do the trick, since this will be the height that maintains the right angled triangle. so when y = 4, the height that maintains the triangle is 6ft, which is y + 2.

what do you guys think (speaking to the geniuses of the forum)?
• Mar 22nd 2007, 11:09 PM
ecMathGeek
Quote:

Originally Posted by Jhevon
i think replacing y with y + 2 will do the trick, since this will be the height that maintains the right angled triangle. so when y = 4, the height that maintains the triangle is 6ft, which is y + 2.

what do you guys think (speaking to the geniuses of the forum)?

You're correct on this one. The weight is initially 2ft above the ground while the pully is initially 8 ft above the weight. Since you've said that y is the distance from the weight to the pully, if the weight is 4ft above the ground then y is 2ft higher than its initial 2ft above the ground and so the distance from the weight and the pully is 2ft shorter than its initial length: y = 8ft - 2ft = 6ft.

Since y = 6, a lot of the other numbers in the equation, such as x and z, also change.

The total distance of x + y is 30ft, so x = 24. Using the pythagorean theorem, z = sqrt(24^2 - 6^2) = sqrt(540) = 23.24ft

Quote:

Originally Posted by Jhevon
x^2 = y^2 + z^2
=> 2x dx/dt = 2y dy/dt + 2z dz/dt
=> 2x (-dy/dt) = 2y dy/dt + 2z dz/dt
=> - 2x dy/dt - 2y dy/dt = 2z dz/dt
=> dy/dt (-2x - 2y) = 2z dz/dt
=> dy/dt = [2z dz/dt]/(-2x - 2y)
now plug in all the values

dy/dt = [2(23.24)(5)]/(-2*24-2*6) = -3.87 ft/s

But the rate that the weight is rising from the ground is 3.87 ft/s

P.S. As a side note, the diagram is a little misleading. The pully is 8ft above the weight, not 8ft above the ground. Besides that, very nice diagram.
• Mar 23rd 2007, 06:16 AM
zachb
Thanks everyone for your help. This was actually a question I had on an exam yesterday and I posted here after the exam to see just how WRONG I was.

Quote:

ALWAYS DRAW A DIAGRAM WHEN DOING RELATED RATES!
Yes. I wasted a lot of time during the exam trying to draw the picture, but it just didn't make any sense to me. I eventually got the picture right, but I only had five minutes left.

By looking at the work you did (everything before you plugged in the values) I see that I was actually on the right track. Of course, I second guessed myself and started over and got the whole thing wrong.
• Mar 23rd 2007, 06:51 AM
Jhevon
Quote:

Originally Posted by ecMathGeek
P.S. As a side note, the diagram is a little misleading. The pully is 8ft above the weight, not 8ft above the ground. Besides that, very nice diagram.

Yeah, i expected the diagram to be misleading since it was based on my initial logic which was faulty--but not in the way you mentioned. i figured initially the weight was on the ground, so being 8ft above the weight was basically being 8 ft above the ground. then the car drives and raises it to 4ft, which is the instant we are worried about. but maybe that logic was faulty as well, i don't know :)
• Apr 9th 2007, 09:16 PM
qbkr21
Re:
Quote:

Originally Posted by zachb
I need help with this peoblem:

A rope that is 30 feet long is tied to a weight and strung through a pulley that is 8 feet above the weight and is tied to the bumber of a car that is 2 feet from the ground. The car moves away from the weight at a rate of 5 feet/sec. How fast is the weight rising when it is 4 feet from the ground?

Forgive me if this makes no sense, but this is exactly how my professor worded the question and he thinks is makes perfect sense. :rolleyes:

If your professor didn't draw the diagram for you on the test then he is a real moron. This problem takes more left-handed artistic ability, than sheer math skill. What a numskull...
• Apr 10th 2007, 05:50 AM
Jhevon
Quote:

Originally Posted by qbkr21
If your professor didn't draw the diagram for you on the test then he is a real moron. This problem takes more left-handed artistic ability, than sheer math skill. What a numskull...

I agree with you in that a question of this nature should not be used on a test (unless of course it was a test on related rates and it was the ONLY question--which would be cruel), however, it does not require as much artisitic skill as i made it seem. I was feeling jovial at the time and decided to place a car in the diagram and draw the pulley and all that stuff, but it really wasn't necessary. drawing a simple right-angled triangle with the proper dimensions is sufficient to think about the solution to this problem