I have tried to answer this question for you here:
If there is something about that explanation that you don't understand, please use that thread to say what the problem is and I'll try to help you.
As already stated by two people you need to make a tangent to the circle from the origin. One will have the maximum gradient; the other has the minimum.
Let the point where the max tangent hits the circle be P, and the centre of the circle be C, origin O.
Then OC^2 = CP^2 + PO^2 => OP = root 12.
So P is on a circle x^2 + y^2 = 12.
Put this with the other circle to get x + y = 4. Substitute y = 4 - x into the first circle and solve for x.
You'll get two answers. One is the max and one the min.
Find y and compute y/x
Let's start with
Why is that true?
In right angled triangle trig, using Pythagoras
Divide by H squared
But we can also define sine cosine and tangent for angles beyond 0-90 by considering a unit circle centred on the origin and letting a point rotate around (0,0) :
defining sin(θ) = y coordinate of the point and cos(θ) as the x coordinate.
As a result equation (1) continues to hold for all angles.
Now divide by sin^2(θ)
which is the same as
There's a third trig identity like these two which you can derive for yourself as an exercise
I'm not understanding you.
Let's say A = 8, C = 2 and E = 4. Suppose the multiplier is 7.
Then G = 56, H = 14 and K = 28.
Now suppose B = 1, D = 3 and F = 2 with multiplier 5.
Then L = 5, M = 15 and N = 10.
G > L (56 > 5) but H is not greater than M ( 14 < 15).
So I guess that was not what you meant. Please clarify.
4. Find the maximum value of y/x over all real numbers x and y that satisfy (x - 3)^2 + (y - 3)^2 = 6.
That locus is a circle, centre (3,3) and radius root 6.
Pick any point on the circle and join to the origin. The gradient of the line will be the value of y/x .
To maximise this value move the point around the circle until the line makes a tangent to the circle, at approximately (0.59, 3.38).
You can calculate the exact point using coordinate geometry.
OK. Here's the coordinate geometry method:
strategy: simplify whenever possible to keep the algebra simple (relatively) and work through the given information, getting all the coordinates and using product of gradients of perpendicular lines = -1.
Let A be (0 , 0) and B (b , 0). Also C (c , d). As AB = AC =>
D is the midpoint of BC so coordinates are
gradient of AC = d/c => gradient of ED = -c/d
=> equation of ED is
equation of AC is y = dx/c so equating the y values to get the x coordinate of E
So, gradient of BE is
and gradient of AF is
The product of these gradients is
So AF is perpendicular to BE.
There was a } missing in the above post which I have added for you. Unless the bit 'under the square root' is a single symbol you have to enclose all in curly brackets, plus the one to close off the fraction numerator.
What is the 'it' referred to above. I assume you were trying to add an image.
I'm using the diagram in post 7. I need to label one more point. H' is the point where AA' and C'B' cross.
As we only want ratios of areas it doesn't matter exactly what units I use, so I've simplified my working by assuming the easiest units.
Let's say the area of ABC is 81 units.
Then (using the scale factor 2/3) the area of A'B'C' must be 81 x2/3 x2/3 = 36
And let's say that AH = 9 units.
Then AG = 6 units, GH = 3 units.
(G is the centroid of both triangles, so I don't need G')
A'H' = 2/3 of 9 = 6
So A'G = 4 (and GH' = 2)
so A'H = A'G - HG = 4 - 3 = 1
So the little triangles like A'JK are 1/9 of ABC.
area AJK = 1/9 x 1/9 x 81 = 1
So the required are is 36 - 1 - 1 - 1 = 33
So k = 33/81 = 11/27
Welcome to the forum.
I'll show how the right hand side becomes the left. It amounts to the same thing.
If you has t^2 /2 and you wanted to differentiate with respect to t, that would be straight forward. You would get 2t / 2 = t
But you want to differentiate v^2 /2. It hasn't got a 't' in it, although we suspect that v is a function of t.
So you have to use the chain rule:
dy/dx = dy/du . du/dx
In this problem y = v^2 /2. x = t. and u = v so the chain rule becomes
d(v^2/2)/dt = d(v^2/2)/dv . dv/dt
What this amounts to is this. Differentiate with respect to v and then times by dv/dt.
d(v^2/2)/dt = v.dv/dt
hope that helps,
date: Thursday, 23rd July. Time spent on maths work: 1 hour 30 mins.
Topic: plotting points on a graph.
OK with (3,4) and 7,2) but I don't know what to do with (-3,4) and (-1,-5)
Topic: practise at times tables
I can get 20 multiplications correct in 103 seconds. Target for tomorrow is under 90 seconds.
percentages. I am trying to change fractions into percentages but I'm not sure if I'm doing them right.
Here's my answers to 5 questions: .......................etc etc
How do you know that JD=x?
From a point you can make two tangents to a circle. The lengths of these two tangents (from point to circle) are equal.
How did you get the base as 20-2x?
Using the same rule AJ = AH = 10 - x so the base is AB = 20 - 2x
In triangle DAM, DM = FH = 6 and DA = 10, therefore AM = 8.
AH = 10 - x and also x + 8 => 10 - x = x + 8 => x = 1
Therefore AH = 9.
So the large cone is 9x as big as the small cone.
If the height of the small cone = h, then 9 times small height = large height => 9h = 6 + h => h = 3/4
So small cone
and large cone
Q1. For points A, B, and C, find the vector AB and the vector BC.
If AB = scalar x BC, then you know they are parallel, and as they have B in common, they must all be in one line.
Q2. The equation of the line of travel is
r = i + 4j + t(2i - 3j)
The parameter t will be the time in seconds so you can put t = 3 for the first part and solve for t in the second.
Q3. When k = 0, a point on the line is i - j.
Choose any other value for k and you'll get a second point.
then the gradient = (y2-y1)/(x2-x1) and substitution into y = mx + c will give you c.
Q4. The locus of points for b, is a circle (one) radius 2, centred on (0,0).
So if you add another 3i+4j to this you'll create a new circle (two) translated by that amount. If you draw a diagram the greatest and least should be obvious. (I think. I'm going to try this myself. )
OK. Done that now. Draw and extend the line from (0,0) to (3,4) and mark the two points where it cuts circle two. Any point on circle two is a possible a + b, but the two points you've just marked will give the greatest and least distances from (0,0) ie. will maximise and minimise |a+b|
note: If distances are in say metres, velocities in m/s, and time in seconds then distance travelled d = t.v.
And speed = |v|
And you can make bold with square brackets b to turn on and square brackets /b to turn off.
Would you advice I get a BSc degree or a BEd instead?
If you're asking this question then the answer is probably a BSc as it keeps your options open. But if you know you want to be a teacher then you'll save time (and in the UK money) by going straight for the BEd.
I really enjoy teaching and I think I'm good at it, but, in UK state schools, there are a lot of children who don't want to be there and who especially don't want to do maths. I found it very hard to deal with this. The nice thing about helping on MIF is my 'pupils' want to 'be there'.
Because the Earth's axis is tilted relative to the plane of the orbit around the Sun, the area that is illuminated includes a circle around the pole which is always in daylight during the summer. That is the definition of the Artic circle. The downside is that during the winter the opposite happens and the Sun never rises above the horizon.
The latitude for the circle is about 67 degrees. So on mid-summer's day everywhere inside that latitude will experience the midnight Sun. We were sailing North into the region but after mid-summer. So we probably didn't get the effect until we were a bit further North. We got all the way to Nordkapp at 71 degrees. At the equinoxes the daytime is 12 hours everywhere on Earth. We choose the timing of the holiday to make sure we saw midnight Sun. The downside is you won't see the aurora because you need a dark sky for that.