TrueTears question thread

[TrueTears]

It is the same as rotating the region defined by and about the line . You just translated it down by 4 but the volume is the same.

, differentiating with respect to y. Then solve for k and C.

yeah i did for that question, keep getting wrong answer ><

[TrueTears]

actually nvm, i think i got it

What about this question

Find the constants a and b if is a solution of the differential equation , i worked out and subbed it back in and everything, but the 4x cancels out the . What do i do?

[shinny]

Bit too late at night for calculus, but from what I can see at first glance, I think you forgot to product rule the x (don't forget the t outside the brackets) to get to the second derivative. If you don't do that, then yeh it'll cancel out I think.

Note: Only did a few mental calculations so there's a 50% chance this is wrong =P

[ell]

Find the constants a and b if is a solution of the differential equation , i worked out and subbed it back in and everything, but the 4x cancels out the . What do i do?

DEs are always messy so I'll omit some of the working







Substituting into given equation:



after some work:



Equate:
and

[TrueTears]

ah yes thanks ell, soz was doing this a bit late at night, i did what shinny said lol forgot the product rule grrrrrrr

Oh and also, When u are doing differential equations, how do u know when to sub let like this example:

Find the general solution of each of following differential equations

a)

so and

then after some working u get But then my book says Let and so the equation becomes

just wondering what's the point of letting , and when do u know when to do it?

And just this question:
A tank initially contains 200L of pure water. A salt solution containing 5 kg of salt per litre is added at the rate of 10 litres per minute and the mixed solution is drained simultaneously at the rate of 12 litres per minute. there is m kg of salt in the tank after t minutes. Find an expression for

[kj_]

just wondering what's the point of letting , and when do u know when to do it?

Notice that when you let , your equation appears a lot neater - nothing else as the answers are equivalent. 

And just this question:
A tank initially contains 200L of pure water. A salt solution containing 5 kg of salt per litre is added at the rate of 10 litres per minute and the mixed solution is drained simultaneously at the rate of 12 litres per minute. there is m kg of salt in the tank after t minutes. Find an expression for

The expression for is defined by Inflow - Outflow.
Now, inflow is defined by input concentration by volume rate, and thus IN = 5kg/L x 10L/min = 50 kg/min.
Outflow is defined by output concentration by volume rate, and this is where it gets tricky.
You'll know that concentration is defined by .
Your mass is unknown and defined as .
Your volume is . Why? You initially have 200L of water, and you're adding water at 10L/min. At the same time, water is flowing out at 12L/min, and therefore, each minute, you are losing 2 litres of water. This is a variable of time, and thus the expression stated above.
And finally, your outflow volume rate is 12L/min, and therefore:
OUT = x


Thus, your expression is

EDIT: Note that in the specialist math syllabus, you're not required to solve these type of differential equations. Actually, you're not taught how to  :buck2:

[TrueTears]

ahh thank you soo much kj_, yeah it was that output part that confused me, just didn't quite recognize that overall its losing 2L per min xD. thanks again

[TrueTears]

just 2 more new questions

1. The rate of decay of a substance is where is the mass of substance remaining. Show that the half life, the time in which the amount of the original substance remaining is halved, is

2.
How do u do Q 3. Just a bit confused about part a)

Sorry about Large picture >< not sure how to make it smaller.

Many thanks

[Mao]

1.





let be the amount of mass initially





when (half decayed)

[Mao]

2. (actually 3 )

this question isn't straight forward. the variable 'A' is the cross sectional area of cylinder, with the side as a chord on the circle. There is no simple formula for this, hence we construct a circle (radius 2 circle centered at origin), where a is the horizontal axis and b is the vertical axis. We hence have the upper half of the circle

imagine x starts from (-2,0) on our a axis, hence our a value would be , and its height would be

this is the length of the chord across the cylinder. Hence,



the rest should be easy enough =]

[TrueTears]

thanks mao

Kinematics q

1. The position of an object travelling in a horizontal line is x metres from a point O on the line at t seconds. The position is described by
Find the displacement of the particle in the 5th second.

so x(5) = -10. And displacement means 'the position of the object relative to the origin (O)'. But my book says its -6. How do u get that

[Flaming_Arrow]

[TrueTears]

wats difference b/w 5th second and in 5 seconds? is 5th second just between 4-5 seconds? ( soz for stupid Q ><)

[Flaming_Arrow]

when it says find the distance traveled in the 5th second of motion that means t5-t4, the distance traveled in the 5TH SECOND OF MOTION.
if it says find the distance traveled in the first 5 seconds, thats the total distance traveled which is t5-t0, which is just t5

[TrueTears]

ahhh thanks.