TrueTears question thread

[dejan91]

Lol you're right (at least I can't see any correct options). Usually I just look for an undefined gradient and sub the coordinates in to each one to see which one is undefined...but not working for this one. And btw...wtf kind of a differential is that??? So trippy

[TrueTears]

Lol you're right (at least I can't see any correct options). Usually I just look for an undefined gradient and sub the coordinates in to each one to see which one is undefined...but not working for this one. And btw...wtf kind of a differential is that??? So trippy

True, I subbed in (0,0) all undefined cept for E, but E is wrong clearly.

GG

[TonyHem]

gradient is 0 at x = 0, y = -1
A)1/-(-1)
B)0-1/-(-1) = 1
C)0-(-1)/(-1) = -1
D)= 0
E) = 0-(-1)^2 = 1

But yeah, that (0,0) thing, leaves E :S

really weird
i saw this question when it was coming out of my printer 20 mins ago lol
now i dont really wanna do it

[TrueTears]

Good paper to do, just some stupid questions (like always with trial exams)

[TrueTears]

can also be written as

Implicit differentiating we get:



so

But the domain they give is so both and satisfies.

Thus subbing in

How come answer is just Why don't they also include the negative half?

[TrueTears]

a) This I can do:







b) My working is as follows:









When









Now how do I get rid of the mod signs to get their form? I did the following:







But then how to get rid of the mods in the denominator? Since it still could be negative and you need the mods to ensure it's positive?

Next part says:





When



Solving these on the TI-89 gives:

Now since the question asks for the speed we require

However which one do you choose?

Now if you solve their equation which is

you get only and not

So which formula do you use and most importantly when you use my formula how do you dismiss as a solution?

[bem9]

the bags velocity can only be from [0, sqrt(8820)) as for any value greater than this it would require the bags velocity to be greater than sqrt(8820) and therefore at some stage its velocity would have to be equal to sqrt(8820). This is impossible as this value is also the terminal velocity. Therefore you dont need to mod the bottom

Also in your second question, since 123.044 is greater than the terminal velocity, it is rejected as in order to reach this velocity, the sand bag must also pass and at some stage equal sqrt(8820), which is impossible.

i hope my reasoning is correct here :S

[Over9000]

the bags velocity can only be from [0, sqrt(8820)) as for any value greater than this it would require the bags velocity to be greater than sqrt(8820) and therefore at some stage its velocity would have to be equal to sqrt(8820). This is impossible as this value is also the terminal velocity. Therefore you dont need to mod the bottom

Also in your second question, since 123.044 is greater than the terminal velocity, it is rejected as in order to reach this velocity, the sand bag must also pass and at some stage equal sqrt(8820), which is impossible.

i hope my reasoning is correct here :S

How do you know is the terminal velocity, what is ur working/reasoning for it?

[NE2000]

I think it's because when


Solving for v gives the value that bem9 gave. This is the velocity of the object when acceleration would be zero and hence is the terminal velocity.

For your second value, the is greater than 9g, which means the acceleration is actually against gravity which we can reject. If I were doing that question I wouldn't work out terminal velocity but I would state that it is necessary for the resistive force to be less than or equal to the weight force as acceleration can only be down. Or something like that.

EDIT: the reason the don't get your solution is, as you probably know, because of the modulus being absent. That should indicate to you that the difference is that you have a negative log which you changed to be positive. Intuitively you should be wary of that, but obviously you can't just reject it because of the fact that it was negative before the modulus

[Over9000]

I think it's because when


Solving for v gives the value that bem9 gave. This is the velocity of the object when acceleration would be zero and hence is the terminal velocity.

For your second value, the is greater than 9g, which means the acceleration is actually against gravity which we can reject. If I were doing that question I wouldn't work out terminal velocity but I would state that it is necessary for the resistive force to be less than or equal to the weight force as acceleration can only be down. Or something like that.

EDIT: the reason the don't get your solution is, as you probably know, because of the modulus being absent. That should indicate to you that the difference is that you have a negative log which you changed to be positive. Intuitively you should be wary of that, but obviously you can't just reject it because of the fact that it was negative before the modulus

Thanks for explaining NE2000 and bem9.

[TrueTears]

Cool thanks guys!