please help me i have no clue how to solve..please?

Question

#1.At noon, Ship A is 100km west of ship B. Ship A is sailing south at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 P.M?


#2.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?


#3.The alititude of a triangle is increasing at a rate of 1cm/min while the area of the triangle is increasing at a rate of 2 cm2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100cm2?

its too hard for me to do those..please help me please i have noo cluee :(

Answer 1

Here's a post, taken from Yahoo! Answers with a problem very similar to #1. You should be able to get clues from this.

http://answers.yahoo.com/question/index?qid=20061016183251AAv7pcZ

To find out how to work the rest of these problems, google "related rates." There are many websites which give useful info which you can apply to these problems.

Good luck! If there are no more answers in the morning, I will try to help some more then.

I'm back!

1.) The best way to do this is to imagine the two ships forming a triangle with each other. The horizontal distance between them is constant, because they are both headed directly due north or south. The vertical distance between them at 4 p.m. is y = 4 h (25 + 35) km/h = 240 km.**

Extend y southward from ship B. Extend x eastward from ship A. The point at which these extensions intersect form a right angle with respect to the x and y coordinates of the two ships. The diagonal distance between the two ships is the hypotenuse of the triangle containing this right angle. It is the rate at which this diagonal is increasing that we are trying to find.

Now, since we have a right triangle existing between the respective positions of these two ships, we can use the Pythagorean Theorem do describe the diagonal distance between the two ships:

Let x be their horizontal separation, y be their vertical separation and s be the diagonal separation between the two ships. Then the diagonal between the two ships is given by this equation:

x² + y² = s²

Differentiate this equation to find rates of change:

2x dx/dt + 2y dy/dt = 2s ds/dt

What we are interested in finding is ds/dt. Notice that since x is constant, then dx/dt = 0. So the term containing that variable drops out of the equation. Then the equation becomes:

0 + 2y dy/dt = 2s ds/dt ----> 2y dy/dt = 2s ds/dt

So (y/s) dy/dt = ds/dt.

The two ships are receding from each other in the vertical direction by the sum of their speeds (35 + 25) km/h = 60 km/h because they are headed in opposite directions. So dy/dt = 60 km/h.

Now we need y and s. We know y = 240 km from ** above. We can use the Pythagorean Theorem as I said earlier to find s.

s = √[(100 km)² + (240 km)²]
s = 260 km

Now we can find ds/dt.

(y/s) dy/dt = ds/dt
(240 / 260) (60 km/h) = ds/dt
55.38 km/h = ds/dt.

So the diagonal distance between the two ships is changing at the rate of 55.38 km/h at 4 p.m.

2.) This problem is actually similar to #1. the horizontal distance between the two people does not change. So that means dx/dt = 0.

The man has walked this distance after 20 minutes (5 min + 15 min):

(60 s/min)(20 min)(4 ft/s) = 4800 ft.

The woman has walked:

(60 s/min)(15 min)(5 ft/s) = 4500 ft

So, the vertical distance between the two people 15 minutes after the woman sets out is (4800 + 4500) ft = 9300 ft.

Like problem #1, they are receding from each other, so dy/dt is the sum of their walking speeds: (4 + 5) ft/s = 9 ft/s.

We can again use the Pythagorean Theorem to find s fifteen minutes after the woman started walking.

s = √[9300 ft)² + (500 ft)²]
s = 9313.43 ft.

We differentiate as in problem #1.

x² + y² = s²
2x dx/dt + 2y dy/dt = 2s ds/dt
2x (0) + 2y dy/dt = 2s ds/dt
(y/s) dy/dt = ds/dt

Now, since we know y and s and dy/dt = (4 + 5) ft/s, we can find ds/dt.

(9300 / 9313.43)(9 ft/s) = ds/dt
8.99 ft/s = ds/dt.

So, the rate of increase of the diagonal distance between them is almost the same as the vertical rate of change, 8.99 ft/s.

3.) The area of any triangle is given by this formula:

A = ½ bh, where b is the base and h is the altitude of the triangle. Since the area is changing with respect to time, differentiate the above equation with respect to time:

dA/dt = ½ [b dh/dt + h db/dt] + bh (0)
dA/dt = ½ [b dh/dt + h db/dt] + 0
dA/dt = ½ [b dh/dt + h db/dt]

Since we are given h = 10 cm when A = 100 cm², then we can calculate what b must be at that moment given those values:

A = ½ bh
100 cm² = ½ b (10 cm)
(100 cm² / 10 cm) = ½ b
10 cm = ½ b
2 (10 cm) = b
20 cm = b

From the first part of the problem, we are given dA/dt = 2 cm²/min and dh/dt = 1 cm/min. We are given that h = 10 cm when A = 100 cm² and have calculated b from our area formula for a triangle. All we need to do now is find db/dt. From our differential equation above, we can find algebraically what db/dt must be in terms of the other values we have. Then we can simply plug those in and find the actual value for db/dt when A = 100 cm². We can use that formula to find db/dt because it is valid for all values of A.

dA/dt = ½ [b dh/dt + h db/dt]
2 dA/dt = [b dh/dt + h db/dt]
[2 dA/dt - b dh/dt] / h = db/dt

Now we actually plug in the values to find db/dt.

[2 (2cm²/min) - 20 cm (1 cm/min)] / 10 cm = db/dt
[4 cm²/min - 20 cm²/min] / 10 cm = db/dt
[-16 cm²/min] / 10 cm = db/dt
-1.6 cm/min = db/dt

So b is decreasing by 1.6 cm/min when A = 100 cm² and h = 10 cm..