|
If a rocket is rising vertically at 880 ft/sec when it is 4,000 ft up, how fast is the camera - to - rocket distance changing at that instant?
how do i start this off? i drew this, but im not sure if its correctly drawn:
i just wanted to start it off, i mean it would equal to 4,000 right?
----
Suppose a liquid is to be cleared of sediments by pouring it through a cone - shaped filter. Assume the height of the cone is 16 in. and the radius at the base of the cone is 4 in. If the liquid is flowing out of the cone at a constant rate of 2 cubic inches per minute, how fast is the depth of the liquid decreasing when the level is 8 in. deep?
just from reading that, i think it would be best to get the cone formula? cone: A = (PIr^2h) / 3 after this, what would be the next step? thanks guys.
|
don't you need to know how far you are from the base of the launching rocket?
|
|
|
United States24497 Posts
I don't mind people occasionally getting help, but why are you using tl as your primary source for learning everything that is math? It is getting to be too much :-/
|
because you guys help me more than the internet does first time, its my primary source and basically my #1 place to go first. sometimes the sites i go to via google gives example that arent related or hard to understand.
|
If he wants help and people wants helping it's not a problem imo
+ it's his blog
|
well, don't you have a book that teaches you this shit? o__O
when i took calc, everything was in the book lol
for #1, i'm guessing the answer is 700ft/s?
|
I'd help him except I spent all my free time watching the TSL and don't have anymore....LOL.
|
United States24497 Posts
The distance between the camera and the rocket:
r = sqrt(y^2+3000^2)
differentiate:
dr/dt = .5 (y^2+3000^2)^(-.5) * 2y dy/dt
Plug in 880 for dy/dt and 4000 for y? I haven't plugged it in yet.
Edit: would give you 704 ft/s
|
i love related rates. they're so fun. you just need to find a relationship and then take the derivative and plug in a number.
"If a rocket is rising vertically at 880 ft/sec when it is 4,000 ft up, how fast is the camera - to - rocket distance changing at that instant?"
so using your triangle, d^2=h^2+90,000 (pythagorean theorem)
you want to find dd/dt when h=4,000, and you're given dh/dt to be 880.
derive and plug in h=4,000, and then solve for dd/dt
"
Suppose a liquid is to be cleared of sediments by pouring it through a cone - shaped filter. Assume the height of the cone is 16 in. and the radius at the base of the cone is 4 in. If the liquid is flowing out of the cone at a constant rate of 2 cubic inches per minute, how fast is the depth of the liquid decreasing when the level is 8 in. deep?
"
again, use the formula for the volume of the cone. derive it. plug in what you know and find dh/dt, where h is the height of the cone. you're given dv/dt and you want to find dh/dt when h=8
also note, you can find the radius when h=8 by using similar triangles
|
I got 2*4*88=704 on paper not sure it's right though, I did it pretty fast
edit: yup, seems right. Did it just like micronesia
|
Use pythagorean theorem
x^2 + y^2 = d^2
Differentiate implicitly wrt t to get something like
x dx/dt + y dy/dt = d dd/dt
dy/dt = given, d = can figure out easily, dx/dt = 0, dd/dt = answer
|
United States24497 Posts
On June 02 2008 09:08 Purind wrote:
Use pythagorean theorem
x^2 + y^2 = d^2
Differentiate implicitly wrt t to get something like
x dx/dt + y dy/dt = d dd/dt
dy/dt = given, d = can figure out easily, dx/dt = 0, dd/dt = answer
your alphabetical collision is XD
|
Use the volume of a cone formula.
V = something.
it will be in terms of r and h. You can find a relation between r and h. It'll be like... h = 4r or something like that (draw one view of the cone so that it looks like a triangle, you should be able to figure out what the equation is from that). Plug that relationship into the volume equation so that the only unknown is h.
Differentiate wrt t to get
dV/dt = something dh/dt
dh/dt should be your answer
|
On June 02 2008 09:10 micronesia wrote:Show nested quote +On June 02 2008 09:08 Purind wrote:
Use pythagorean theorem
x^2 + y^2 = d^2
Differentiate implicitly wrt t to get something like
x dx/dt + y dy/dt = d dd/dt
dy/dt = given, d = can figure out easily, dx/dt = 0, dd/dt = answer your alphabetical collision is XD
Shut up, I don't read or think!!
|
On June 02 2008 08:55 micronesia wrote:
The distance between the camera and the rocket:
r = sqrt(y^2+3000^2)
differentiate:
dr/dt = .5 (y^2+3000^2)^(-.5) * 2y dy/dt
Plug in 880 for dy/dt and 4000 for y? I haven't plugged it in yet.
Edit: would give you 704 ft/s
i get the dr/dt part, but getting the r = sqrt(y^2+3000^2) part i got lost, why y^2 which i dont know, i know its the pythagorean formula, but that confused me, to get "r" dont i need 2 numbers? i guess its calculus and you dont huh?
i have a question guys when solving these kinds of stuff, im still looking at it, which one am i solving for; meaning, which side? its asking how fast the camera is, so we're solving for the hypotenuse right? thanks again. and ydg can you explain the second one like write it out?
so..
A = (PIr^2h) / 3
derive this? PI = 0 r^2 would be 2r and h would be dy/dx? the 3 going on top would be 3^-1, am i on the right track?
|
United States24497 Posts
On June 02 2008 09:30 Raithed wrote:Show nested quote +On June 02 2008 08:55 micronesia wrote:
The distance between the camera and the rocket:
r = sqrt(y^2+3000^2)
differentiate:
dr/dt = .5 (y^2+3000^2)^(-.5) * 2y dy/dt
Plug in 880 for dy/dt and 4000 for y? I haven't plugged it in yet.
Edit: would give you 704 ft/s i get the dr/dt part, but getting the r = sqrt(y^2+3000^2) part i got lost, why y^2 which i dont know, i know its the pythagorean formula, but that confused me, to get "r" dont i need 2 numbers? i guess its calculus and you dont huh?
I'm not sure if I understand your question. I kept 3000 as a number because it is always 3000. I kept y as a variable because, even though we know the value at some point in time, we need to keep it as a variable in order to determine the overall relationship between y and r, and dy and dr. Once we have that relationship, then we can plug in the actual value of 4000.
|
for the first one, r is the hypotenuse and y is the height and 3000 is the given distance between the launch and the rocket.
okay second one..
derive, plug in, solve
|
ydg, does the 16/4 always have to be h/r? and which do i derive for? the V = 1/3PI(h/4)^2 or h = PI(h^3/48)? im guessing the latter?
|
similar triangles, so the height/base of one triangle is equal to the height/base of another
and uh, lol, it's just two equations that are equal to each other, v=that, which =the last thing. derive the v=pi h^3/48
|
|
|
|
|
|
EPT![[image loading]](http://i28.tinypic.com/2a7dgfc.jpg)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21V%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2%20h.gif)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21%5Cfrac%7B16%7D%7B4%7D%3D%5Cfrac%7Bh%7D%7Br%7D%3B%20r%3D%5Cfrac%7Bh%7D%7B4%7D.gif)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21V%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%28%5Cfrac%7Bh%7D%7B4%7D%29%5E2%20h%3D%5Cpi%20%5Cfrac%7Bh%5E3%7D%7B48%7D.gif)
