Is that e^(2y) in (y/4)+((e2y-e-2y)/16)?

]]>Welcome to the forum.

I have added [img].....[/img] to your http addresses so that the images are imbedded in the page.

These graphs ought to have scales indicated. I'm guessing that each square represents 1 unit.

Q1. Draw on the horizontal line y = -1. Everywhere that it crosses the curve is a value.

Q2. When x = -2 what is y? You can estimate the right value but, much better, look at the end points of that line and work out exactly what the y coordinate is.

Q3. Firstly look at the y coordinate when x = 7. Now apply the function to that. ie. Call that answer the new x coordinate and work out the new y.

Q4. This time, you're given the y coordinate (y = 4) and they want you to say what the x coordinate is.

Hope that helps. If you wish, I'll check your answers for you when you post back.

Bob

]]>Could anyone please help me with the following?

This is question number 6 of Further Problems III on page 143 of K A Stroud's Further Engineering Mathematics.

given

and

and

prove that

Here's my attempt but I end up missing out the middle term of the answer. Can anyone see where I've gone wrong?

Starting with the generic formulae:

and

we have:

and

and

and

therefore

and

therefore

=

=

and similarly therefore

=

=

so now we subtract these two expressions to give

which simplifes to

=

which finally simplifies to

which is not the desired result, I am missing the

term.

So either the book is wrong, or I am wrong or the missing term evaluates to zero. I strongly suspect I am wrong. Can anyone help me see where I made a mistake?

Thanks,

Mitch.

The variables are separable so should be straightforward.

Next find a particular solution to the given equation

The general solution to the question is Ans 1 + Ans 2.

]]>Thumbs up, guys!

Hmm, when will I reach that level too.

And, oh! Lest i forget, thanks guest for sharing info.

]]>I was idly scrolling through old, unanswered posts and I found this. Not sure how much simplification is expected but how about this:

Bob

]]>I've found it is usually a good thing to agree with bobbym.

no doubt about it.

]]>Example.

Primes 3 5 7 = coordinate 2,2 (5-3,7-5)

I'm still tweeking this, but here is the newer code.

```
var canvas = document.getElementById('myCanvas');
var context = canvas.getContext('2d');
var primes = [3];
var index = 0;
// takes an odd number and returns true if prime, false if composite
var isPrime = function(num) {
var divisor = num - 1,
testing = true;
while (divisor > 2) {
if (num % divisor === 0) {
return false;
} else {
divisor -= 1;
}
}
return true;
};
// takes an arrawy of primes, looks at the largest, then looks at every other number until it finds a prime and adds it to the primes array
var addPrime = function (primes) {
var topPrime = +primes.slice(-1),
candidate = topPrime + 2,
primeSought = true;
while (primeSought) {
if (isPrime(candidate)) {
primeSought = false;
primes.push(candidate);
} else {
candidate += 2;
}
}
};
var i = 0;
for (i = 0; i < 3; i += 1) {
addPrime(primes);
}
console.log("done");
function draw() {
context.fillStyle = 'rgba(0,0,0,0.3)';
context.fillRect(primes[index+1]-primes[index], primes[index+2]-primes[index+1], 2, 2);
}
function update() {
index += 3;
addPrime(primes);
addPrime(primes);
addPrime(primes);
}
function loop() {
update();
draw();
window.requestAnimationFrame(loop);
}
loop();
```

Welcome to the forum.

Q1. Because the x^2 and y^2 terms have equal positive coefficients, that is the equation of a circle. So work it round to the standard form for a circle.

http://www.mathsisfun.com/algebra/circle-equations.html

add 1 + 1 to each side.

So the centre is at (1,1) and the radius is root 2. So you need to work out the coordinate of the 'rightmost' point on the circumference.

Q2. If you call the angles of the lines to the x axis A and B

tan(A) = 4/3 and tan(B) = 5/12

The angle of the bisector will be (A+B)/2.

You can work this out using the following tan formulas

and

http://www.mathsisfun.com/algebra/trigo … ities.html

Q3.

Let A = (1,2), B = (0,1), and C = (5,0). There exists a point Q and a constant k such that for any point P, PA^2 + PB^2 + PC^2 = 3PQ^2 + k.

So call P the point (x,y) and Q the point (h,j) and form the equation:

Simplify carefully and you should be able to find h, j, and k.

(Because this must be true for all x and all y you can set equal the x terms to get one equation; the y terms to get a second and the non x,y terms to get a third. You will find all the squared terms all cancel out so you can set the x terms equal to get h, the y terms to get j and finally the non-x,y terms to get k.)

Bob

]]>Hey! No tears. Fun is what we're here for.

Looks like no one has told you that

So if 5sinθ=2cosθ divide by cos and also by 5, and you'll have sin/cos = tan = a fraction straight away.

So then part b falls into place fairly easily. Solve for 2x and thus for x. Make sure you find all the answers.

Q2. Re-write as sin(2x) / cos(2x) - 5sin(2x) = 0 and then is should be possible for you to finish.

If you post your answers, I'll check them for you.

Bob

]]>For 1), what is the significance level?

]]>