Math Is Fun Forum

If you rearrange the equation into

you can see what you're doing a bit more, and just pick the coordinates of the centre and the size of the radius straight off. You were right with the answers though.

I'm not totally sure what you mean in part (b). Surely there could be any number of possible tangents to the circle? If you're asking for the points at which the x-axis intersects the circle, then this can be calculated by solving the simultaneous equation:

So all you need to do is substitute the value of y into the equation for the circle:

Therefore the circle intersects the x-axis at points (√5 + 1,0) and (-√5 + 1,0). I think.

Yeah, i agree with Mathsyperson. Think about it this way:

On time:

The only way for the work to be completed on time is for both supliers to be on time (0 days late). This gives the one possibility

0 and 0 = 1/2 × 2/3 = 1/3.

One day late:

There are three possible ways for the work to be completed one day late. Either both suppliers are a day late, or the first is on time and the second is a day late, or the first is a day late and the second is on time. This gives:

1 and 1 = 1/4 × 1/6 = 1/24
0 and 1 = 1/2 × 1/6 = 2/24
1 and 0 = 1/4 × 2/3 = 4/24

Therefore the overall probability is 1/24 + 2/24 + 4/24 = 7/24.

Two days late:

Using similar reasoning as before we have five possible ways in which the work can be completed two days late. These are:

2 and 2 = 1/4 × 1/6 = 1/24
1 and 2 = 1/4 × 1/6 = 1/24
2 and 1 = 1/4 × 1/6 = 1/24
0 and 2 = 1/2 × 1/6 = 2/24
2 and 0 = 1/4 × 2/3 = 4/24

This gives an overall probability of 1/24 + 1/24 + 1/24 + 2/24 + 4/24 = 9/24.

Composite Rule (sometimes called the Chain Rule)

If k is a function with rule of the form k(x) = g(f(x)), where f and g are smooth functions, then k is smooth and

      k'(x) = g'(f(x)).f'(x)

(or using Leibniz form:  If y = g(u), where u = f(x), then dy/dx = (dy/du)(du/dx) ) .

So, for instance, if you have k(x) = ln(x^2 + 1), then y = g(u), where g(u) = lnu and u = x^2 + 1. This gives k'(x) = (1/(x^2 + 1)).2x = 2x/(x^2 + 1) .

Hope that helps.

You need to calculate the derivative of each function...

1) d/dt(4/t +4lnt) = -4/t^2 + 4/t

2) d/dt(sin(t^2 + 1)) = 2cos(t^2 + 1)t

    d/dt(cos(2t - 3)) = -2sin(2t - 3)            (both using the composite rule)

3) d/dt(2(e^(-t/2))cos(2t)) = (-e^(-t/2))cos(2t) - 4(e^(-t/2))sin(2t)       (using the product rule)

4) I'm not sure about this one, it's a little bit unclear what you mean, but you get the general idea so far. Hopefully the rest are okay, but it's late so i'd check em!

Oh math, i've just realised you've specified t=1.2, so you'd have to use that value with the derived functions to get your answers.

Rather than struggle to do it in here, i've knocked up a basic proof thingy. I couldn't figure out how to post it as an image, so here's a link to the image instead:

http://img64.imageshack.us/img64/8128/f … of16xb.png

Hope that helps.

I'm a bit confused by the converge/diverge thing. I thought that a sequence diverges when r>1, so it doesn't tend to any one limit?

Okay cool thanks. It's really annoying me too now lol!