true/false. yes it is home work. no it is not graded. i cant figer it out.
Code:
integral sign
-x
A(x) = | f(t) dt
-2
a) A'(-1) = A'(3)
b) max of A ocourse @ x = 1
c) max of A' ocourse @ x = 1
a is false. think they are all false
Code:
-x
A(x) = | f(t) dt
-2
A'(x) = f(t) + C ???
does this possibly belong in the 'general programming' section? just a thought.
No...because it is not implicitly programming, it's mathematics.
a lot of programming involves math... so if i had a problem with qb code that involves math, ill post it here, and it will get moved, or someone else will say isnt this programming help?
Quote:a lot of programming involves math... so if i had a problem with qb code that involves math, ill post it here, and it will get moved, or someone else will say isnt this programming help?
It's labelled Calc. It's clearly math. Math is not programming.
Look that f(x) dont change if t varies, so it remains constant under the integral, you cant take off the integral sign.
Code:
-x
A(x) = | f(x) dt
-2
A(x) = (x - 2) f(x)
A'(x) = f(x) + (x - 2)f'(x)
But if you mean f(t) and not f(x), then
Code:
-x
A(x) = | f(t) dt
-2
A'(x) = f(x)
oh yeah it's f(t) dt. anyways here is more i dont get.
Code:
A rocket id fired directly upward with initial velocity of 0 and burns fuel at a rate to produce a constant acceleration of 50 ft/sec^2 for 0 < t < 5 with time t in seconds. An observer 400ft away from the launch pad visually fallows the flight of the rocket.
1. Express the angle of elevation x of the rocket as a function of t
2. The observer perceives the rocket to be rising fastest when dx/dt is largest. Determine the height of the rocket at the moment of perceived maximum velocity.
Code:
1. gravity in ft + given acc
|\ a(t) = -32.152231 + 50 = 17.847769
| \ hypot v(t) =17.847769t + 0
y | \ h(t) = 8.9238845t^2
| \ x(t) = tan^-1 (8.92t^2/400) ???
| x \ y = h(t), tan^-1 y/400 = x
---------
400
2. t @ 5 h(5) = 8.9238845(5)^2 = 223.097112
WHAT?
I don't understand that stuff with the "|"..
Quote:1. Express the angle of elevation x of the rocket as a function of t
2. The observer perceives the rocket to be rising fastest when dx/dt is largest. Determine the height of the rocket at the moment if perceived maximum velocity.
Dude, please put the whole question "as is" in your post. You fragmented it.
Ok, the 2nd point is really simple and its obvious.
x = distance
dx/dt = v = velocity [1st order derivative]
d2x/dt2 = dv/dt = acceleration [2nd order derivative]
dx/dt is maximum means the rocket covers very large distance per unit time.
Others: Could you please stop the petty bickering about where the question should be placed? The admins/mods should be worried about that and not us.
![[Image: integral.gif]](http://math.furman.edu/~tlewis/integral.gif)