Help! Maths Problem...

jcb33 · 29 Mar 2007 at 00:21

Hi all, me and my room mates have a problem with maths, and we have sat all day trying to solve it, if anyone can shed some light on how to do this, please please do so!!!!

The question we are stuck on is 4, D):

http://img63.imageshack.us/img63/656/pdftf7.jpg

Also could anyone help us confirm, we have 3 different answers for 4, A) 0.5, 0.71 and 1.66, which is right? Thanks!

jcb33 · 29 Mar 2007 at 01:13

Urgh, more time, and we still cannot solve D

jcb33 · 29 Mar 2007 at 03:06

Well we have a) 0.71, b) 7.81, c) 5, we just dont know how to do D

jcb33 · 29 Mar 2007 at 03:11

Zefan said:
You'll have to use the equation and the distances between the points together, not sure really what you'll have to do with them but that's basically it.

Just use Billy's graphical solution it gives you the exact answer.

Had a go in paint with all the stuff up the right way and worked it out to be about 6.2, 3.3 so it seems good to me

But how can we get that without the graph, thats what we need to do *sighs*

jcb33 · 29 Mar 2007 at 03:16

touch said:
No idea, but i'v wasted 8 sheets of A4 trying

(and i dont see how your answer for A can be a single number? )

We just have to find 1 line, not the intersect point of 2?

jcb33 · 29 Mar 2007 at 09:14

Sorry all, was fried last night, answer for A is Y = 3/7x + 0.71

jcb33 · 29 Mar 2007 at 13:21

DaveF said:
You want to do this all parametrically.

Equation of the line is x = 3+7t, y = 2+3t. (Then 3x-7y = 9+21t-14-21t = -5 is a cartesian equation of the line, if you care).

Separation at start = sqrt(5^2+6^2) = sqrt(61).

Separation at collision is 5 (sum of the radii).

For the last part: using the parametric equation, separation in x is (8-(3+7t)) = 5 - 7t; separation in y is 8-(2+3t) = 6 - 3t.

If S is the total separation, then S^2 = (5-7t)^2+(6-3t)^2 = 25-70t+49t^2+36-36t+9t^2 = 58t^2 - 106t + 61.

The balls collide when S^2 = 25: i.e. 58t^2-106t+36 = 0 or 29t^2-53t+18 = 0

t=(53+/- sqrt(53^2-4*18*29))/56 = (53 +/- 26.851)/58. Since we want the first point where S^2 = 25, we take t = (53 - 26.851)/58 = 0.451

Substituting back, we get x = 3+7*0.451 = 6.157, y = 2+3*0.451= 3.353

Check: distance^2 = (x-8)^2+(y-8)^2 = 3.397 + 21.594 = 24.991 (close enough to 25 considering we've only been working to 3dp).

Thats to complicated for my brain, I understand using a diagram that I can get x and y, and that using the following equation i can find 25, but im not sure what 25 is for, or how to find x and y with no diagram!

(8-(0.71)) = 7.29

(8-(3/7x) = 3/7x

(8-6.16)^2+(7.29 – 3/7*6.16)^2

25

(8-x)^2+(8-(3/7x + 0.71))^2 = 25

jcb33 · 29 Mar 2007 at 13:25

This is what I have so far:

d. What are the coordinates of player 1 at this point? [8]

You have the coordinates of player 2 (8,8) we need to take these, along with m1 to work out the value of x:

From the use of a graph I found x to be 6.16, and y to be 3.35

We can work out If x is correct using the following equation, if the answer works out to be 25, then x is correct
(8-x)^2+(8-(3/7x + 0.71))^2

(8-(0.71)) = 7.29

(8-(3/7x) = 3/7x

(8-6.16)^2+(7.29 – 3/7*6.16)^2 = 25

(8-x)^2+(8-(3/7x + 0.71))^2 = 25

The coordinates of player 1 at this point are (6.16, 3.35)

But Idealy I need to know how to fix x and y with no graph use!