Spec '10 - Help forum

[QuantumJG]

and  

[Chavi]

How would I do the following?

Find the cartesian equation [and the restriction on domain + range] of the path of a particle whose position vector at any time t is given by


Many thanks

[kamil9876]

if (it never does anyway as you can check by plugging in x=-1 into the second equation)

Now find you have in terms of so plug into the equation for to get it in terms of

[Chavi]

Can I get help with some vectors:

2 ships are observed from station O. Ship A at 1pm and ship B at 2pm. Relative to O the respective position + velocity vectors of the ships are given by:

A: r= 2i + 6j and v= 5i + 4j
B: r=11i + 6j and v= 2i + 7j

time is measured in hours. Prove that if the ships maintain their velocities they will collide, and find the time they collide.

Thanks

[hyperblade01]

Bit rusty...

Alright I'm assuming those position vectors are at 1pm and 2pm for A and B respectively

Antidiff the velocity vectors:

A r(t) = 5ti + 4tj + c
B r(t) = 2ti + 7tj + c

Let t be the time past 1pm...

A r(0) = 5(0)i + 4(0)j + c
 2i+6j = c

r(t) = (5t+2)i + (4t+6)j

Do the same for B  but this time sub in t=1 to get r(t) = (2t+9)i + (7t-1)j

Now when it says they collide, their positions must be the same, so let the i components equal and the j components equal

5t+2 = 2t+9           4t+6=7t-1

Solve both equations and you should get t = 7/3 for both, which means they will indeed be in the same position at a particular time.

So they hit 7/3hrs after 1pm..which is 3:20pm

[Chavi]

The adjacent sides of a parallelogram are 9cm and 11cm. One of its angles is 67º. Find the length of the longer diagonal, correct to two decimal places.
I don't understand how/what they want you to find, if all the opposite angles/sides of a parallelogram are equal.

[fady_22]

Find the length of the DIAGONALS: i.e. the lines drawn from opposite vertices to make the parallelogram two triangles.

You know that there are two angles of 67 (opposite angles) and two angles of 113 (again, opposite angles) so to find the length of the diagonals, use the cos rule:

Sub in the known values:


Do this again, but use 113 as the angle:

Therefore, the longer diagonal is 16.71 cm long.

[Juddinator]

Sketch the graph of: x^2 +y^2 +3x-4y=6

I can get the equation which is (x+3/2)^2 + (y-2)^2 = 49/4    but I cannot get the x and y intercepts even when I let both equal 0 :|

[superflya]

Sketch the graph of: x^2 +y^2 +3x-4y=6

I can get the equation which is (x+3/2)^2 + (y-2)^2 = 49/4    but I cannot get the x and y intercepts even when I let both equal 0 :|

there are no axial intercepts.

[Juddinator]

Sketch the graph of: x^2 +y^2 +3x-4y=6

I can get the equation which is (x+3/2)^2 + (y-2)^2 = 49/4    but I cannot get the x and y intercepts even when I let both equal 0 :|

there are no axial intercepts.

ahh touché... thanks man

[superflya]

no prob, whenever u cant solve for y or x after letting one of the variables equal 0, usually means there are no intercepts.

[brightsky]

there are no axial intercepts.

Not true. Assuming that I'm interpreting stuff correctly.

Let's work with to make matters simpler.

To find y-intercepts, substitute .





By the quadratic formula:

or

So your y-intercepts are:

and


Do that for the x-intercepts, and you'll find that the x-intercepts are and

[superflya]

there are no axial intercepts.

Not true. Assuming that I'm interpreting stuff correctly.

Let's work with to make matters simpler.

To find y-intercepts, substitute .





By the quadratic formula:

or

So your y-intercepts are:

and


Do that for the x-intercepts.

LOL i didnt work it out -.-  tis correct brightsky

[superflya]

for the x ints





so your 2 x ints are &

[Juddinator]

Sweet thanks superfly and brightsky. It makes so much more sense to use the quadratic formula when you have your a, b and c. Cheers (Y)