Harvey's Question Thread

[HarveyD]

Differential Equations QUestion attached

ANy help is appreciated

[Greatness]

Here: Re: wildareal's questions thread

[HarveyD]

thanks for that

[HarveyD]

Having problems with 2b) and 2cii)

Any help is appreciated

[Mao]

a. AB = ((v-3)t - 56) i + ((29+7v)t + 8 ) j

b. Solve (v-3)t - 56 = 0 and (29+7v)t + 8 = 0:

Rearrange the first to get t=56/(v-3). Substitute into second one to get 56(29+7v)=-8(v-3) , solve to get v=-4 at t=-8. This is obviously unphysical, so the particles will not collide for all v.

c. i. AB = -56 i + (50t + 8 ) j
c. ii. As time goes on, the j-component will become larger, thus increasing magnitude of AB. Therefore the minimum occurs at t=0.

[HarveyD]

ahh k
thanks!
for cii) when you're finding the min/max distance
would you also be able to differentiate the equation inside the square root and make it equal to zero, to find the time when it is a max/min

Edit: also do you have any idea on how I could do the attached question :/

[b^3]

To find the tangent to the path you need to find the velocity vector, i..e integrate differentiate (whoops) the displacement vector which is given by the points here so x=6ti+(t²+4)j
From their just find the unit vector by diving the vector for velocity by the magnitude (root(x^2+y^2), you know what I mean) then sub in t-4 to get your answer.

[b^3]

To procrastinate from doing phsyics I'll show you the working.
so v=6i+2tj
now the mag of this is root(36+4t^2)=2root(9+t^2)
so now the vector we are looking for is given by (1/2root(9+t^2))*(6i+2tj)
so let t=4 ,(1/2root(25))*6i+8j
5/2*(6i+8j)
1/10*(6i+8j)

Let me know if it works, because now that I think about it, I may have done something slightly wrong.

[HarveyD]

so let t=4 ,(1/2root(25))*6i+8j
5/2*(6i+8j)

not sure how you got to the 5/2 * (6i + 8j)
Because the one above it: 1/10 (6i + 8j) is the correct answer lol

thanks

[b^3]

so let t=4 ,(1/2root(25))*6i+8j
5/2*(6i+8j)

not sure how you got to the 5/2 * (6i + 8j)
Because the one above it: 1/10 (6i + 8j) is the correct answer lol

thanks

Sorry 1/2root(25) is not 5/2 its 1/5*2, I could tell I did something wrong, but I didn't think it was that. Silly mistakes sorry.