application of calculus help please. (1 Viewer)

[Timothy.Siu]

just not sure how to approach the question, in other words i have no clue how to do it...

The acceleration of a body P is given by a=18x(x^2+1) where x cm is the displacement at time t sec. Initially P starts from the origin with velocity 3cm/s

i) Show that v=3(x^2+1)
ii) Find x in terms of t

from 98 baulko trials only q4 so it shouldn't be too hard

 

[Trebla]

just not sure how to approach the question, in other words i have no clue how to do it...

The acceleration of a body P is given by a=18x(x^2+1) where x cm is the displacement at time t sec. Initially P starts from the origin with velocity 3cm/s

i) Show that v=3(x^2+1)
ii) Find x in terms of t

from 98 baulko trials only q4 so it shouldn't be too hard

i) a = 18x(x² + 1)
d(v²/2)/dx = 18x(x² + 1)
v² = 9 ∫ 4x(x² + 1) dx
= 9(x² + 1)² + c (using reverse chain rule or alternatively use substitution u = x² + 1)
At x = 0, v = 3
9 = 9 + c
=> c = 0
.: v² = 9(x² + 1)²
=> v = 3(x² + 1)
Note we take the positive root to satisfy x = 0 when v = 3. The negative case does not satisfy this condition.

ii) v = 3(x² + 1)
dx/dt = 3(x² + 1)
dt/dx = 1 / 3(x² + 1)
t = (1/3)tan^-1x + c
At t = 0, x = 0 => c = 0
t = (1/3)tan^-1x
.: x = tan 3t

 

[Timothy.Siu]

ah i see, thanks heaps, forgot about that identity of acceleration

 

[lolokay]

when i have dv/dt = f(x) I just multiply out by .dx to make it possible to integrate the RHS, and go from there (dv/dt .dx = dv .dx/dt = v.dv)

do we really need to write it as d(v²/2)/dx / is there really any difference?

 

[Timothy.Siu]

when i have dv/dt = f(x) I just multiply out by .dx to make it possible to integrate the RHS, and go from there (dv/dt .dx = dv .dx/dt = v.dv)

do we really need to write it as d(v²/2)/dx / is there really any difference?

i guess not but i'm still not sure how you do it, sorry i'm sorta new to this

soo u end up with v.dv =f(x).dx? and then what do you do

 

[Timothy.Siu]

also,

in b) i dont get the diagram..so i cant do the first part, but the second part is fine

 

[Trebla]

when i have dv/dt = f(x) I just multiply out by .dx to make it possible to integrate the RHS, and go from there (dv/dt .dx = dv .dx/dt = v.dv)

do we really need to write it as d(v²/2)/dx / is there really any difference?

Be careful there. There are some technical issues with doing it the way you suggested.

The equivalent to that is simply writing acceleration as v.dv/dx which is the same as d(v²/2)/dx. You're allowed to quote the derivative formulae for acceleration (dv/dt, d(v²/2)/dx and v.dv/dx) without deriving them and this way is preferred in the HSC exams.

 

[Trebla]

also,

in b) i dont get the diagram..so i cant do the first part, but the second part is fine




also if anyone has time can someone do the last question, for some reason i keep getting a different answer to the answers....i think the limits are -1 to root3

Let r be the radius of the cone made up of the water.
tan 60° = r/h
r = h√3
r² = 3h²
Volume of a cone is: πr²h/3 = πh³ cubic centres

 

[Timothy.Siu]

yup, thanks heaps

 

[lolokay]

i guess not but i'm still not sure how you do it, sorry i'm sorta new to this

soo u end up with v.dv =f(x).dx? and then what do you do

you would integrate both sides, so you get 1/2 v² = F(x) + C