integration question :( (1 Viewer)

[Mc_Meaney]

OOps my bad.

A partical p is at the origin at time t=0 and moves so its velocity for t=> 0 is given by v=1/t+1

(i) what is the acceleration (already answered)

(ii) what is the displacement x from P at the origin when t=1

 

[SoulSearcher]

x = ∫ 1/t + 1 dt
= ln (t) + t + C, this is where we hit a little snag, because when t = 0, ln 0 is undefined, so there's a little problem there. Unless the actual equation is 1/(t+1), but this also changes the acceleration. Anyway, continuing on with 1/(t+1),
x = ∫ 1/(t+1) dt
= ln (t+1) + C
x = 0 when t = 0, so
0 = ln 1 + C, therefore C = 0, so x = ln (t+1)
substituting t = 1, x = ln 2

 

[Mc_Meaney]

equation is 1/(t+1) tar

 

[SoulSearcher]

So acceleration would be dv/dt = -1/(t+1)²
t = 1, -1/(1+1)² = -1/4 m/s²

 

[Mc_Meaney]

Confuzzled

sin²x

Solve for pi/4 pi/2 3pi/4 and pi.

How do i do this?

 

[SoulSearcher]

I'm assuming you sub in the values of pi/4, pi/2, 3pi/4 and pi, although that does not sound like solving it. So for pi/4, sin²x = sin²pi/4 = (1/rt2)² = 1/2. Same with the rest.

 

[SomaFairy]

Hey sorry about hijacking your thread but I thought that it would better if i asked my integration question here than starting up a new thread...

Find the equation of the tangent to the parabola y= 2x^2 at (1,2). Calculate its point of intersection with the x-axis and the volume of the solid formed when the area betw'n the parabola, the tangent and the x axis is revolved about the x axis.

I have worked out that the tangent is y=4x -2 and the intersection with the x axis is x=1/2. The volume should be 2pi/3 but i can't work that out.

 

[SoulSearcher]

To find points of intersection, solve simutaneously
y = 2x²
y = 4x-2
2x² = 4x - 2
x² - 2x + 1 = 0
(x-1)² = 0
thereforer the curves only touch when x = 1, therefore we only need to be concerned with the volume found when rotating y = 4x-2, as the parabola does not affect that volume. Use the bounds x = 1 to x = 1/2
y = 4x-2
y² = 16x² - 16x + 4
pi _1/2∫¹ 16x² - 16x + 4 dx
= pi[16x³/3 - 8x² + 4x]¹_1/2
= pi[(16/3 - 8 + 4) - (2/3 - 2 + 2)]
= pi(16/3 - 4 - 2/3)
= pi(14/3 - 4)
= 2pi/3 units³

 

[SomaFairy]

ooo Thats where i went wrong~! I thought that the limits were 0 and 1... opps

Thankz for solving it for me

 

[word.]

that can't be right... it's the area bounded by x-axis, parabola and tangent rotated around the x-axis, so the parabola does affect it.

draw up a picture if you can't visualise it
you need to find the volume of the area under the parabola, and subtract the volume under the line (which happens to be a cone of radius 2 height 0.5 -> volume = (pi * r^2 * h)/3 = 2pi/3)

if you want to check by integration:
y^2 = (4x - 2)^2
V = pi * int(0.5->1)(4x - 2)^2dx -> (4(1) - 2)^3/12 - (4(0.5) - 2)^3/12 = 8pi/12 = 2pi/3

y = 2x^2
y^2 = 4x^4

pi * int(0->1) 4x^4dx = 4pi/5

so the total volume is pi(4/5 - 2/3) = 2pi/15 units^3

 

[lilkiwifruit]

that can't be right... it's the area bounded by x-axis, parabola and tangent rotated around the x-axis, so the parabola does affect it.

draw up a picture if you can't visualise it
you need to find the volume of the area under the parabola, and subtract the volume under the line (which happens to be a cone of radius 2 height 0.5 -> volume = (pi * r^2 * h)/3 = 2pi/3)

if you want to check by integration:
y^2 = (4x - 2)^2
V = pi * int(0.5->1)(4x - 2)^2dx -> (4(1) - 2)^3/12 - (4(0.5) - 2)^3/12 = 8pi/12 = 2pi/3

y = 2x^2
y^2 = 4x^4

pi * int(0->1) 4x^4dx = 4pi/5

so the total volume is pi(4/5 - 2/3) = 2pi/15 units^3

lol I got 2pi/15 u^3 as well....