imported_loudsubz

Here it is

Airplane A is flying SOUTH at a speed of 120ft/s. It passes over a bridge 12 MIN before another plane B, which is flying east at the same height at a speed of 160 ft/s. How fast are the airplanes sperating 12 MIN after B passes over the bridge.

Answer is 189 ft/s but I cant seem to get that answer.

imported_loudsubz

anyone?

imported_360_vortex

sounds like a related rates question


i would work it out for you on paper but i have to sleep, perhaps i'll do it tmr at work

imported_gatherer

this is a simple right angle triangle question .. once plane is going south and other is travelling east.

So you have a right angle triangle. you can find the rate of change in there distance because you know the rate of change in their distances relative to a single point...

remember this is monday morning (be easy on me if wrong)

so one side of the triangle is 120ft/s and the other side is 160 ft/s and you need to find the hypotinuse of the triangle. pythagorus theory and you get 200 ft/s

now if it asks the distance between the 2 planes 12 minutes after the second one crosses the bridge thats a whole new issue .. (and a pain in the *** to figure out)

szuberi

at the time when plane B passes over the bridge, the distance b/w the two planes will be 86,400 feet. lol but i dont think it can be this simple to figure out...i dont know how to arrive at 189f/s.

imported_JoonyaSI

who measures planes in the air ?

imported_gatherer

the question is how fast are the planes separating... not what is the distance between them... so you have to work with the velocities not the distances between them.. you are working with the rate of change of their distance... not their distance...

of course you coudl figure out the equation that gives you their distance apart after x seconds then take the dirivative of that....but tahts the complex way

bbarbulo

iTrader: (2)

read this, it explains it

http://www.karlscalculus.org/pr8_2-2.html

imported_loudsubz

ok I figured the first one out

This is the one thats a little more confusing to me.

This is the question.

A conical tank with a vertex down has an angle of 60 degrees. Water flows from the tank (drains) at a rate of 5cm cubed/min. At what rate is the inner surface of the tank being exposed when the water is 6.0cm deep?

Any idea on how to go about this question?

I have no idea what to do on it....

eurovento

How are you supposed to figure that out without getting at least one other dimension of the tank other than its a 60 degree taper.

imported_loudsubz

thats what I was wondering and out of all the examples I looked at in books and online I havn't found anything related.

bbarbulo

iTrader: (2)

the height of the cone = the depth of the water. what more could you want?

imported_loudsubz

it only gives you the rate at which water is draining and the vertex angle of the cone

it could be any height and any radius

imported_Simich

the second question is a related rates question. the first is an optimization problem, and the simplicity of it done above isn't correct im pretty sure.

id have to look back into my notes because i already forget most of this haha.

imported_t_dot_SiR

both are rates! haha

imported_t_dot_SiR

for your 2nd problem, you need to use ratio of depth compare with the radius of the top surface of water... then you'll know the radius when depth is 6.... that way you can use the derivative with respect to t, then you will have all the necessary variables to plug in

imported_t_dot_SiR

the angle being 60 tells you the height and the radius

------------
| /
| /
| /
| /
| /
| /
| /
|a /
| /
|/

angle a is 60

opp side is the radius
adj side is the height

imported_t_dot_SiR

the diagram didn't work =( but you get the idea

bbarbulo

iTrader: (2)

the angle is fixed!

......./
....../
...../
..../
.../
../
./

makes no difference what the total height and diameter of the funnel is. you start at 0 and fan out at a 60 degree angle, just take the height of remaining water, and given the angle you have the diameter.

ok six centimeters height... so it's a basic right angle triangle with a height of 6 cms. the so the triangle dimensions are (equilateral triangle).... 60-60-60 angles with a height of 6 cms. so, split that triangle into two triangles and you get two right angle triangles 90-30-60... use sin60 = 6/hypotenuse and that gives you the hypotenuse and height, so you can figure out the radius of the cone using pythagores.

I dunno how you do the related rates part, but that's how you get over that whole cone thing you seem worried about.