Calculus - Related Rates - Tilted Forum Project Discussion Community

threewingedfury · 10-16-2005, 10:16 AM

Ok.. I'm new to this place.. but I see that you guys do help with calc.. which is good since people at NCSU don't like helping others.. but anyways.. I'm having a serious problem with related rates. I don't know if its the fact that I can't put 2 and 2 together or I can't draw a picture that makes sense. And on top of all that, I hate calculus. So without furthur adue.. these are the problems I'm struggling with:

1. At 2 P.M., ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 6 P.M.?

2. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 2.3 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

3. A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 m from the dock?

4. Water is leaking out of an inverted conical tank at a rate of 10500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

And no I'm not asking anyone to do my homework, but if someone could just show me the method that would help me get the idea of how to do it, I'd appreciate it.

Slavakion · 10-16-2005, 02:16 PM

Related rate problems are (almost always?) solved by using a formula from geometry. The first three problems that you posted can be solved by drawing a triangle. There are two main parts.

Fill in the geometrical formula with given lengths/distances.
Find the derivative of the formula and fill in with given speeds or changes.

This is a rushed post, I know, but it should get you started. If nobody posts by the time I get back, I'll try to explain in more detail.

n0nsensical · 10-16-2005, 09:13 PM

Yeah, it's just a matter of finding a formula for what you're looking for and plugging in the unknowns.

Like number 1, it asks for the rate of change of the distance between them.
The distance D between them is
D(t) = Sqrt(x(t)^2+y(t)^2)
where x(t) is the distance between them in x and y(t) is the distance between them in y. We'll call +/- x East/West respectively and +/- y North/South respectively, assume 2 PM is t=0, and t is in hours.

The rate of change of the distance between them is
d/dt(D) = dD/dt = (2x(t)x'(t)+2y(t)y'(t))/(2sqrt(x(t)^2+y(t)^2))
(so ugly in ASCII form, I know)
( d/dt(sqrt(u)) = 1/2 * 1/sqrt(u) * du/dt = du/dt / (2sqrt(u)) and our u(t) = x(t)^2+y(t)^2 thus u'(t) = 2x(t)x'(t)+2y(t)y'(t) )

So then all you need to know are the unknowns of dD/dt: x, y, x', y', and t which is 4.
x(t) = 150-35t km
y(t) = 25t km
x'(t) = -35 km/h
y'(t) = 25 km/h
(Really x' and y' are given and we integrate to get x and y, but in this case it's so mentally intuitive we have no need to show it explicitly)
Plug everything in for dD/dt|t=4 and you get the answer. I hope I did this right.

Same procedure for the others with different functions.

10-16-2005, 10:16 AM	#1 (permalink)
threewingedfury Upright	Calculus - Related Rates Ok.. I'm new to this place.. but I see that you guys do help with calc.. which is good since people at NCSU don't like helping others.. but anyways.. I'm having a serious problem with related rates. I don't know if its the fact that I can't put 2 and 2 together or I can't draw a picture that makes sense. And on top of all that, I hate calculus. So without furthur adue.. these are the problems I'm struggling with: 1. At 2 P.M., ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 6 P.M.? 2. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 2.3 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building? 3. A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 m from the dock? 4. Water is leaking out of an inverted conical tank at a rate of 10500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. And no I'm not asking anyone to do my homework, but if someone could just show me the method that would help me get the idea of how to do it, I'd appreciate it.

10-16-2005, 02:16 PM	#2 (permalink)
Slavakion Mjollnir Incarnate Location: Lost in thought	Related rate problems are (almost always?) solved by using a formula from geometry. The first three problems that you posted can be solved by drawing a triangle. There are two main parts. Fill in the geometrical formula with given lengths/distances. Find the derivative of the formula and fill in with given speeds or changes. This is a rushed post, I know, but it should get you started. If nobody posts by the time I get back, I'll try to explain in more detail.

10-16-2005, 09:13 PM	#3 (permalink)
n0nsensical Junkie Location: San Francisco	Yeah, it's just a matter of finding a formula for what you're looking for and plugging in the unknowns. Like number 1, it asks for the rate of change of the distance between them. The distance D between them is D(t) = Sqrt(x(t)^2+y(t)^2) where x(t) is the distance between them in x and y(t) is the distance between them in y. We'll call +/- x East/West respectively and +/- y North/South respectively, assume 2 PM is t=0, and t is in hours. The rate of change of the distance between them is d/dt(D) = dD/dt = (2x(t)x'(t)+2y(t)y'(t))/(2sqrt(x(t)^2+y(t)^2)) (so ugly in ASCII form, I know) ( d/dt(sqrt(u)) = 1/2 * 1/sqrt(u) * du/dt = du/dt / (2sqrt(u)) and our u(t) = x(t)^2+y(t)^2 thus u'(t) = 2x(t)x'(t)+2y(t)y'(t) ) So then all you need to know are the unknowns of dD/dt: x, y, x', y', and t which is 4. x(t) = 150-35t km y(t) = 25t km x'(t) = -35 km/h y'(t) = 25 km/h (Really x' and y' are given and we integrate to get x and y, but in this case it's so mentally intuitive we have no need to show it explicitly) Plug everything in for dD/dt\|t=4 and you get the answer. I hope I did this right. Same procedure for the others with different functions.