Author Topic: VCE Methods Question Thread! (Read 5695820 times) Tweet Share

BubbleWrapMan · « **Reply #2100 on:** June 15, 2013, 09:06:06 pm »

The area of the triangle is $100\sin\left(\frac{\pi \theta}{180} \right)$ (where $\sin$ operates in radians), so use linear approximation with $f(\theta )=\sin\left(\frac{\pi \theta}{180} \right)$ and $h=1$ .

Zealous · « **Reply #2101 on:** June 15, 2013, 09:15:44 pm »

Quote from: Sanguinne on June 15, 2013, 08:59:38 pm

q4 exercise 8G MATHS quest
An isosceles triangle has its equal sides of length 10 cm with an included angle Ө as shown in the diagram at right. (I don't know how to upload stuff onto AN so it basically is an isosceles triangle with a theta for angle at the top as well has both its diagonal sides having the length of 10cm)

If Ө changes from 60 to 61, find correct to 2 decimal places:
a) the approximate area of the triangle when Ө=61
b) the approximate increse in the area, A, of the triangle.

My main problem is setting up this equation....

Oh Timmeh beat me.

But anyway, for a triangle with two lengths and the angle in between, the area can be found with:

$A=\frac{1}{2}sin(A)bc$ (where A is the angle in between the sides B and C)

That's where Timmeh got the equation from, just watch out, if you're calculator is in degrees, you can put the angle straight in (but you should be using rad for methods).

some working if you're interested:

Spoiler

$A=\frac{1}{2}sin(A)bc$

$A=\frac{1}{2}sin(60)\times100$

$A=50\times sin(\theta)$

differentiate:

$A'=50\times cos(\theta)$

and put it all back into the linear approximation formula:

$A=50\times sin(60)+(\frac{\pi}{180}\times 50cos(60))$

$A=43.74cm^2$

BubbleWrapMan · « **Reply #2102 on:** June 15, 2013, 09:27:28 pm »

It needs to be in radians, since the derivative is different.

Zealous · « **Reply #2103 on:** June 15, 2013, 09:33:51 pm »

Quote from: Timmeh on June 15, 2013, 09:27:28 pm

It needs to be in radians, since the derivative is different.

I'm not too sure what you mean -
I know that if derive sin/cos on a CAS and you are in degrees, you'll get a really weird result, before I derived by hand and didn't come across any issues.

BubbleWrapMan · « **Reply #2104 on:** June 15, 2013, 09:47:30 pm »

If $\sin$ is working in radians, the derivative of $\sin\left(\frac{\pi \theta }{180}\right)$ is $\frac{\pi}{180}\cos\left(\frac{\pi \theta }{180}\right)$ (the usual derivative formula only applies for radians), so if you're working in degrees, the derivative of $\sin(\theta)$ is $\frac{\pi}{180}\cos(\theta)$ .

Planck's constant · « **Reply #2105 on:** June 16, 2013, 12:17:58 am »

Quote from: sushi. on June 15, 2013, 09:33:51 pm

I'm not too sure what you mean -
I know that if derive sin/cos on a CAS and you are in degrees, you'll get a really weird result, before I derived by hand and didn't come across any issues.

What timmeh is trying to explain to you is that ALL the familiar rules of calculus for circular functions only apply if the angles are in radians.

For instance, when we say that the derivative of sinx is cosx, this is only true if the angle x is in radians.
If the angle x is not in radians, it must be converted to radians. Converting the angle from degrees to radians involves the multiplication of x by a constant, which introduces the requirement to apply the chain rule when we differentiate.

Sanguinne · « **Reply #2106 on:** June 16, 2013, 12:32:09 pm »

thanks everyone for their responses

helped clear everything up

Sanguinne · « **Reply #2107 on:** June 16, 2013, 02:54:19 pm »

Peter is quite poor at doing crossword puzzles and the probability of him completing one is 0.2. FInd the probability that he successfully completes the first three
puzzles that he tries given that he was successful in completing his first two;

b^3 · « **Reply #2108 on:** June 16, 2013, 03:19:22 pm »

Hint: Use the conditional probability formula $\begin{alignedat}{1}\text{Pr}\left(B|A\right) & =\frac{\text{Pr}\left(B\cap A\right)}{\text{Pr}\left(A\right)}\end{alignedat}$

zhe0001 · « **Reply #2109 on:** June 16, 2013, 05:25:12 pm »

Hey guys,

Please help me with this matrices question, Q10. b. I've been looking at it for like 10 minutes and still can't figure it out.

The question is attached.

P.S I've only covered addition, subtraction and multiplication of matrices.

P.S.S The answer is [1 -2 3 0] (a 2x2 matrix) -by the way how do you format a matrix question?

Thanks!

b^3 · « **Reply #2110 on:** June 16, 2013, 05:37:13 pm »

Normally you'd do this with an inverse matrix, but you can do it without.
Let $\begin{alignedat}{1}Q & =\begin{bmatrix}q_{1} & q_{2} \\ q_{3} & q_{4} \end{bmatrix}\end{alignedat}$ , then you can multiply the left hand side to get a matrix in terms of $q_{i}$ , which will give you 4 equations to solve to find the coefficients.

Spoiler

$\begin{alignedat}{1}Q\times\begin{bmatrix}3 & 4 \\ 1 & 2 \end{bmatrix} & =\begin{bmatrix}1 & 0 \\ 9 & 12 \end{bmatrix} \\ \begin{bmatrix}q_{1} & q_{2} \\ q_{3} & q_{4} \end{bmatrix}\times\begin{bmatrix}3 & 4 \\ 1 & 2 \end{bmatrix} & =\begin{bmatrix}1 & 0 \\ 9 & 12 \end{bmatrix} \\ \begin{bmatrix}3q_{1}+q_{2} & 4q_{1}+2q_{2} \\ 3q_{3}+q_{4} & 4q_{3}+2q_{3} \end{bmatrix} & =\begin{bmatrix}1 & 0 \\ 9 & 12 \end{bmatrix} \\ 3q_{1}+q_{2} & =1....[1] \\ 4q_{1}+2q_{2} & =0 \\ \implies q_{2} & =-2q_{1}....[2] \\ \text{substitute into [1]} \\ 3q_{1}-2q_{1} & =1 \\ \implies q_{1} & =1 \\ \implies q_{2} & =-2 \\ 3q_{3}+q_{4} & =9 \\ q_{4} & =9-3q_{3}....[3] \\ 4q_{3}+2q_{3} & =12 \\ 4q_{3}+18-6q_{3} & =12 \\ -2q_{3} & =-6 \\ \implies q_{3} & =3 \\ \implies q_{4} & =9-3\left(3\right) \\ & =0 \\ \therefore Q & =\begin{bmatrix}1 & -2 \\ 3 & 0 \end{bmatrix} \end{alignedat} $

This is the long way of doing it though....

Sanguinne · « **Reply #2111 on:** June 16, 2013, 06:33:46 pm »

Quote

Peter is quite poor at doing crossword puzzles and the probability of him completing one is 0.2. FInd the probability that he successfully completes the first three
puzzles that he tries given that he was successful in completing his first two;

Hint: Use the conditional probability formula

I know I have to use conditional probability but Im not what the intercept is in this particular case

BubbleWrapMan · « **Reply #2112 on:** June 16, 2013, 06:38:21 pm »

I don't think it's conditional. The "probability of him completing one is 0.2" so it doesn't matter if he completed the previous ones.

b^3 · « **Reply #2113 on:** June 16, 2013, 06:57:01 pm »

$\begin{alignedat}{1}\text{Pr}\left(S|SS\right) & =\frac{\text{Pr}\left(SSS\right)}{\text{Pr}\left(SS\right)} \\ & =\frac{0.2^{3}}{0.2^{2}} \\ & =0.2 \end{alignedat}$
Which gives the same result as what Tim was getting at, I guess the above is a little pointless then.

zhe0001 · « **Reply #2114 on:** June 16, 2013, 07:22:46 pm »

Quote from: b^3 on June 16, 2013, 05:37:13 pm

Normally you'd do this with an inverse matrix, but you can do it without.
Let $\begin{alignedat}{1}Q & =\begin{bmatrix}q_{1} & q_{2} \\ q_{3} & q_{4} \end{bmatrix}\end{alignedat}$ , then you can multiply the left hand side to get a matrix in terms of $q_{i}$ , which will give you 4 equations to solve to find the coefficients.
Spoiler
$\begin{alignedat}{1}Q\times\begin{bmatrix}3 & 4 \\ 1 & 2 \end{bmatrix} & =\begin{bmatrix}1 & 0 \\ 9 & 12 \end{bmatrix} \\ \begin{bmatrix}q_{1} & q_{2} \\ q_{3} & q_{4} \end{bmatrix}\times\begin{bmatrix}3 & 4 \\ 1 & 2 \end{bmatrix} & =\begin{bmatrix}1 & 0 \\ 9 & 12 \end{bmatrix} \\ \begin{bmatrix}3q_{1}+q_{2} & 4q_{1}+2q_{2} \\ 3q_{3}+q_{4} & 4q_{3}+2q_{3} \end{bmatrix} & =\begin{bmatrix}1 & 0 \\ 9 & 12 \end{bmatrix} \\ 3q_{1}+q_{2} & =1....[1] \\ 4q_{1}+2q_{2} & =0 \\ \implies q_{2} & =-2q_{1}....[2] \\ \text{substitute into [1]} \\ 3q_{1}-2q_{1} & =1 \\ \implies q_{1} & =1 \\ \implies q_{2} & =-2 \\ 3q_{3}+q_{4} & =9 \\ q_{4} & =9-3q_{3}....[3] \\ 4q_{3}+2q_{3} & =12 \\ 4q_{3}+18-6q_{3} & =12 \\ -2q_{3} & =-6 \\ \implies q_{3} & =3 \\ \implies q_{4} & =9-3\left(3\right) \\ & =0 \\ \therefore Q & =\begin{bmatrix}1 & -2 \\ 3 & 0 \end{bmatrix} \end{alignedat} $
This is the long way of doing it though....

Thank you so much! Thanks for doing the long way as well!

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