i have some problems with these attached file , need help as soon as possible need to make sure i did them correctly:p

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- May 22nd 2007, 09:18 AMcarlasaderA level Maths
i have some problems with these attached file , need help as soon as possible need to make sure i did them correctly:p

- May 22nd 2007, 10:14 AMjanvdl
$\displaystyle \cos 2x = 0.5 $

$\displaystyle 2x = 60^{\circ}$

$\displaystyle 2x = 60^{\circ} \pm 360 k$

$\displaystyle x = 30^{\circ} \pm 180 k$

$\displaystyle \mbox{Or } x = 150^{\circ} \pm 180k

$ - May 22nd 2007, 10:21 AMjanvdl
1. cos 2x = 0.5

2x = 60 degrees

2x = 60 degrees \pm 360 k

x = 30 degrees \pm 180 k

Or x = 150 degrees \pm 180k

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2. Look at the RHS, where you have $\displaystyle cos^2 \theta $

Remember that $\displaystyle cos^2 = 1 - sin^2 $

Therefore:

$\displaystyle 4 sin^2 \theta - 2 sin \theta = 4 - 4 sin^2 \theta - 1$

$\displaystyle 8 sin^2 \theta - 2 sin \theta - 3 = 0$

By solving for theta we find:

$\displaystyle sin \theta = 0.75 \ OR \ sin \theta = -0.5 $ - May 22nd 2007, 10:26 AMjanvdl
3. $\displaystyle A = \pi R^2 $

But Perimeter $\displaystyle P = 2 \pi R $

$\displaystyle 20 = 2 \pi R $

$\displaystyle \frac{10}{\pi} = R $

Set R into A

$\displaystyle A = \pi (\frac{10}{\pi})^2 $

$\displaystyle A = \frac{100}{\pi} $

Now my area formula does not seem at all correct. See if you can figure out where i made a mistake.

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Now lets ASSUME we did get the right formula.

To maximise, we must find the derivative and equal it to 0

So dA/dr = 0

$\displaystyle 25 - (r - 5)^2 = 0 $

$\displaystyle 25 - r^2 + 10r - 25 = 0 $

$\displaystyle r^2 + 10r = 0 $

$\displaystyle r(r + 10) = 0 $

Therefore $\displaystyle r = 0 \ or \ r = -10 $

Set both 0 and -10 into the ORIGINAL equation

If r = 0

Area = 0

If r = -10

Area = -200 - May 22nd 2007, 10:53 AMjanvdl
For the Architect Sum:

a) $\displaystyle 4x + 4y + x \pi $

b) $\displaystyle 4xy + \frac{1}{2} x^2 \pi $

c) $\displaystyle 100 = 4x + 4y + 2x \pi $

$\displaystyle y = 25 - (x + \frac{x \pi}{2}) $

Set y into the Area formula

Then we will eventually get: $\displaystyle A = 50x - 2x^2 + \frac{1}{2} x^2 \pi $

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Why cant i get these sums right??

Right lets assume the given values again

$\displaystyle \frac{dA}{dx} = 100 - 4x + x \pi = 0 $

$\displaystyle 100 = 4x - x \pi $

$\displaystyle 100 = x(4 - \pi) $

$\displaystyle x = \frac{100}{(4 - \pi)} $

Now just set $\displaystyle x $ into the ORIGINAL equation to find $\displaystyle y $ - May 22nd 2007, 11:11 AMjanvdl
For the Graph sum:

a) To find where they intersect we must set y = y

$\displaystyle x^2 - 2x + 2 = -x + 4 $

$\displaystyle x^2 - x - 2 = 0 $

$\displaystyle x = 2 \ or \ x = -1 $

Set these values into an original equation.

Then the points are: y = -(2) + 4

And y = -(-1) + 4

So point P = (-1 ; 5)

Point Q = (2 ; 2)

b) For MPQ to be right-angled, the product of the two formulas must equal -1

I do not get the product = -1, so i am assuming that they are not right angled. - May 22nd 2007, 11:13 AMjanvdl
I will see if i can answer the rest tomorrow. I hope that the things that i could get right myself helped you.

:) - May 22nd 2007, 01:28 PMJhevon
we have M(1 , 1), P(-1 , 5) and Q(2 , 2)

The slope of the line connecting P and Q is given by:

$\displaystyle m_{1} = \frac {2 - 5}{2 + 1} = \frac {-3}{3} = -1 $

The slope of the line connecting M and Q is given by:

$\displaystyle m_{2} = \frac {2 - 1}{2 - 1} = 1 $

since 1 is the negative reciprocal of -1, the line connecting P and Q is perpendicular to the line connecting M and Q. Thus this forms a right angle at Q. If we draw a line connecting M and P therefore, we form a right-triangle - May 22nd 2007, 02:06 PMJhevon
Let's find the arc length first of all.

Recall:

$\displaystyle s = r \theta $

where s is the arc length and $\displaystyle \theta$ is the angle subtending the arc in radians

We are told that the perimeter is 20 cm this means that:

$\displaystyle P = r + r + s = 2r + r \theta = r(2 + \theta)$, where $\displaystyle P$ is the perimeter

$\displaystyle \Rightarrow r(2 + \theta) = 20 \Rightarrow 2 + \theta = \frac {20}{r} \Rightarrow \theta = \frac {20}{r} - 2$

Now, the area A of the sector will be given by:

$\displaystyle A = \frac {1}{2} r^2 \theta$

$\displaystyle = \frac {1}{2} r^2 ( \frac {20}{r} - 2) $

$\displaystyle = r^2 ( \frac {10}{r} - 1)$

$\displaystyle = 10r - r^2$

By completing the square, you will get the desired result

Quote:

Now lets ASSUME we did get the right formula.

To maximise, we must find the derivative and equal it to 0

So dA/dr = 0

$\displaystyle 25 - (r - 5)^2 = 0 $

$\displaystyle 25 - r^2 + 10r - 25 = 0 $

$\displaystyle r^2 + 10r = 0 $

$\displaystyle r(r + 10) = 0 $

Therefore $\displaystyle r = 0 \ or \ r = -10 $

Set both 0 and -10 into the ORIGINAL equation

If r = 0

Area = 0

If r = -10

Area = -200

We wish to maximize A, so we must find A' and set it equal to zero, a value for r that causes this to happen is the value we are looking for.

From above, $\displaystyle A = 25 - (r - 5)^2$

$\displaystyle \Rightarrow A = 10r - r^2 $

$\displaystyle \Rightarrow A' = 10 - 2r $

Set $\displaystyle A' = 0 $ we get:

$\displaystyle 10 - 2r = 0 $

$\displaystyle \Rightarrow r = 5 $

Note, the formula for a is a downward opening parabola, all we needed to do was find the value of r that gives the vertex. we didn't even need calculus here, we could have found this value by using:

$\displaystyle r = \frac {-b}{2a} $, where $\displaystyle a$ is the coefficient of $\displaystyle r^2$ and $\displaystyle b$ is the coefficient of $\displaystyle r$. - May 22nd 2007, 02:19 PMJhevon
- May 22nd 2007, 02:52 PMJhevon
I assume you actually know how to draw the graphs, draw them both on the same pair of axis. see the graphs below.

a)

Note that we have the roots of $\displaystyle x^2 + e^x - 4 = 0$ where the graphs intersect (do you see why?). from the graphs we see that the intersections occur such that one is for a positive x-value and the other is for a negative x-value, thus we will have a positive and negative root.

b)

I can't really see the power of e very well, i think it is $\displaystyle x_{n}$ so let's run with that.

$\displaystyle x_{n+1} = - \sqrt {4 - e^{x_{n}}}$

$\displaystyle \Rightarrow x_{1} = - \sqrt {4 - e^{x_{0}}} = - \sqrt {4 - e^{-2}} = -1.96588$

$\displaystyle \Rightarrow x_{2} = - \sqrt {4 - e^{x_{1}}} = - \sqrt {4 - e^{-1.96588}} = -1.96468$

$\displaystyle \Rightarrow x_{3} = - \sqrt {4 - e^{x_{2}}} = - \sqrt {4 - e^{-1.96468}} = -1.96463$

so the approx for the negative root to three decimal places is -1.965

c)

the approximation will go towards the negative root, since within any reasonable range $\displaystyle \sqrt {4 - e^{x_{n}}$ will give a positive result. so with the minus on the outside, we will get a negative result

EDIT: So i believe i have taken care of all the problems janvdl didn't do or the ones he expressed doubt about, for the others, i think you're in good hands with him on the job, so i won't even check them - May 23rd 2007, 11:06 AMjanvdl
- May 23rd 2007, 11:22 AMtopsquark
An angle in radians is defined to be

$\displaystyle \theta = \frac{s}{r}$

where s is the length of the arc the angle subtends in a cirlce of radius r.

For conversion purposes, you can use this formula to determine that there are $\displaystyle \pi$ radians in 180 degrees.

-Dan