Author Topic: VCE Methods Question Thread! (Read 5722798 times) Tweet Share

zhe0001 · « **Reply #2130 on:** June 23, 2013, 05:02:52 pm »

Quote from: Eugenet17 on June 23, 2013, 04:59:06 pm

|A| is determinant

THANKS! why maths quest throughout whole exercise say calculate det (a) and only until chapter review say |A| its like they want to fuck you over

chemdeath · « **Reply #2131 on:** June 24, 2013, 04:43:45 pm »

Need help with a few questions on differentiation (would generally ask teacher but is away on holidays) anyway what better place but here!

1) differentiate (cos√ $pix$ )

2)differentiate (sin(X))^2 + (cos(x))^2)

3) Fish bowl fills at rate 0.03cm/s, height=16, and V= $pi$ (36+16h-h^2) find dh/dt

Thanks in advance!!!

clıppy · « **Reply #2132 on:** June 24, 2013, 04:58:52 pm »

Quote from: chemdeath on June 24, 2013, 04:43:45 pm

2)differentiate (sin(X))^2 + (cos(x))^2)

You need to do the chain rule to sin(x) and cos(x)
$u = sin(x)$
$\\ v = u^{2}$
$\\ \frac{du}{dx} = cos(x)$ $\frac{dv}{du} = 2u$
$\\ \frac{dy}{du} = 2(sin(x))$
$\\ \frac{dy}{dx} = 2cos(x)sin(x)$

Do the exact same thing with cos(x)

$u = cos(x)$
$\\ v = u^{2}$
$\\ \frac{du}{dx} = -sin(x)$ $\frac{dv}{du} = 2u$
$\\ \frac{dy}{du} = 2(cos(x))$
$\\ \frac{dy}{dx} = -2sin(x)cos(x)$

Now that you've differentiated both, just add them like normal

$\frac{dy}{dx} = 2cos(x)sin(x) + -2sin(x)cos(x)$
$\frac{dy}{dx} = 0$

Ancora_Imparo · « **Reply #2133 on:** June 24, 2013, 05:08:57 pm »

Quote from: clippy on June 24, 2013, 04:58:52 pm

You need to do the chain rule to sin(x) and cos(x)
$u = sin(x)$
$\\ v = u^{2}$
$\\ \frac{du}{dx} = cos(x)$ $\frac{dv}{du} = 2u$
$\\ \frac{dy}{du} = 2(sin(x))$
$\\ \frac{dy}{dx} = 2cos(x)sin(x)$

Do the exact same thing with cos(x)

$u = cos(x)$
$\\ v = u^{2}$
$\\ \frac{du}{dx} = -sin(x)$ $\frac{dv}{du} = 2u$
$\\ \frac{dy}{du} = 2(cos(x))$
$\\ \frac{dy}{dx} = -2sin(x)cos(x)$

Now that you've differentiated both, just add them like normal

$\frac{dy}{dx} = 2cos(x)sin(x) + -2sin(x)cos(x)$
$\frac{dy}{dx} = 0$

Or perhaps more simply: $y=sin^2(x)+cos^2(x)=1$ , thus $\frac{dy}{dx}=0$

chemdeath · « **Reply #2134 on:** June 24, 2013, 05:10:06 pm »

Quote from: clippy on June 24, 2013, 04:58:52 pm

Spoiler
You need to do the chain rule to sin(x) and cos(x)
$u = sin(x)$
$\\ v = u^{2}$
$\\ \frac{du}{dx} = cos(x)$ $\frac{dv}{du} = 2u$
$\\ \frac{dy}{du} = 2(sin(x))$
$\\ \frac{dy}{dx} = 2cos(x)sin(x)$

Do the exact same thing with cos(x)

$u = cos(x)$
$\\ v = u^{2}$
$\\ \frac{du}{dx} = -sin(x)$ $\frac{dv}{du} = 2u$
$\\ \frac{dy}{du} = 2(cos(x))$
$\\ \frac{dy}{dx} = -2sin(x)cos(x)$

Now that you've differentiated both, just add them like normal

$\frac{dy}{dx} = 2cos(x)sin(x) + -2sin(x)cos(x)$
$\frac{dy}{dx} = 0$

Thanks heaps, that makes sense I just didn't realize you y=use the chain rule!!
Now just these 2: $:-\$

Quote from: chemdeath on June 24, 2013, 04:43:45 pm

1) differentiate (cos√ $pix$ )

3) Fish bowl fills at rate 0.03cm/s, height=16, and V= $pi$ (36+16h-h^2) find dh/dt

Phy124 · « **Reply #2135 on:** June 24, 2013, 05:17:29 pm »

Quote from: chemdeath on June 24, 2013, 04:43:45 pm

Need help with a few questions on differentiation (would generally ask teacher but is away on holidays) anyway what better place but here!

1) differentiate (cos√ $pix$ )

Hint: Use the chain rule.

Spoiler

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

$y = \cos(u)$

$\frac{dy}{du} = -\sin(u) = -\sin(\sqrt{\pi x})$

$u = \sqrt{\pi x}$

$\frac{du}{dx} = \frac{\pi}{2\sqrt{\pi x}} = \frac{\sqrt{\pi}}{2\sqrt{x}}$

$\Rightarrow \frac{dy}{dx} = -\sin(\sqrt{\pi x}) \times \frac{\sqrt{\pi}}{2\sqrt{x}} = -\frac{\sqrt{\pi}}{2\sqrt{x}} \sin(\sqrt{\pi x})$

Quote from: chemdeath on June 24, 2013, 04:43:45 pm

2)differentiate (sin(X))^2 + (cos(x))^2)

Hint: What identity could simplify this?

Spoiler

$\sin^2(x) + \cos^2(x) = 1$

$\frac{d}{dx}(1) = 0$

$\therefore \frac{d}{dx}(\sin^2(x) + \cos^2(x)) = 0$

Quote from: chemdeath on June 24, 2013, 04:43:45 pm

3) Fish bowl fills at rate 0.03cm/s, height=16, and V= $pi$ (36+16h-h^2) find dh/dt

Thanks in advance!!!

Hint: $\frac{dh}{dt}=\frac{dh}{dV}\frac{dV}{dt}$

Spoiler

I assume you mean rate 0.03cm³/s, in which case:

$\frac{dh}{dt}=\frac{dh}{dV}\frac{dV}{dt}$

$\frac{dV}{dt} = 0.03$

$\frac{dV}{dh} = \pi(16-2h)$

$\Rightarrow \frac{dh}{dV} = \frac{1}{\pi(16-2h)}$

When $h=16$

$\frac{dh}{dV} = \frac{1}{\pi(16-2(16))} = -\frac{1}{16\pi}$

$\frac{dh}{dt} = -\frac{1}{16\pi}\times 0.03 = -\frac{3}{1600\pi}$ cm/s

(It appears as though the equation you gave could be wrong? Unless I'm missing something

)

chemdeath · « **Reply #2136 on:** June 24, 2013, 05:35:22 pm »

Quote from: 2/cos(c) on June 24, 2013, 05:17:29 pm

Hint: Use the chain rule.
Spoiler
$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

$y = \cos(u)$

$\frac{dy}{du} = -\sin(u) = -\sin(\sqrt{\pi x})$

$u = \sqrt{\pi x}$

$\frac{du}{dx} = \frac{\pi}{2\sqrt{\pi x}} = \frac{\sqrt{\pi}}{2\sqrt{x}}$

$\Rightarrow \frac{dy}{dx} = -\sin(\sqrt{\pi x}) \times \frac{\sqrt{\pi}}{2\sqrt{x}} = -\frac{\sqrt{\pi}}{2\sqrt{x}} \sin(\sqrt{\pi x})$
Hint: What identity could simplify this?
Spoiler
$\sin^2(x) + \cos^2(x) = 1$

$\frac{d}{dx}(1) = 0$

$\therefore \frac{d}{dx}(\sin^2(x) + \cos^2(x)) = 0$
Hint: $\frac{dh}{dt}=\frac{dh}{dV}\frac{dV}{dt}$

Spoiler
I assume you mean rate 0.03cm³/s, in which case:

$\frac{dh}{dt}=\frac{dh}{dV}\frac{dV}{dt}$

$\frac{dV}{dt} = 0.03$

$\frac{dV}{dh} = \pi(16-2h)$

$\Rightarrow \frac{dh}{dV} = \frac{1}{\pi(16-2h)}$

When $h=16$

$\frac{dh}{dV} = \frac{1}{\pi(16-2(16))} = -\frac{1}{16\pi}$

$\frac{dh}{dt} = -\frac{1}{16\pi}\times 0.03 = -\frac{3}{1600\pi}$ cm/s

(It appears as though the equation you gave could be wrong? Unless I'm missing something )

Thank you so much!!!!

and for 3) I dunno that's it i think.....?

Nato · « **Reply #2137 on:** June 24, 2013, 05:51:17 pm »

Hey guys, i need help with a question (methods 1/2, but seems relevant to 3/4)

i am having a bit of trouble finding the exact value of $\cos(\frac{-21\pi}{3})$

btw Latex: how do you put the negative symbol to the left of the fraction?

p.s hey 2/cos(c), i totally get what it mean hahahahaahaha

clıppy · « **Reply #2138 on:** June 24, 2013, 05:55:38 pm »

Quote from: Student0001 on June 24, 2013, 05:51:17 pm

Hey guys, i need help with a question (methods 1/2, but seems relevant to 3/4)

i am having a bit of trouble finding the exact value of $\cos(\frac{-21\pi}{3})$

$(\frac{-21\pi}{3}) = -7\pi$
Well if you look at your unit circle $-7\pi$ has the same position as $\pi$ and we know that $cos(\pi) = -1$ so $cos(-7\pi) = -1$

Phy124 · « **Reply #2139 on:** June 24, 2013, 07:00:47 pm »

Quote from: Student0001 on June 24, 2013, 05:51:17 pm

btw Latex: how do you put the negative symbol to the left of the fraction?

Code: [Select]

-\frac{a}{b}
$\Rightarrow -\frac{a}{b}$

Code: [Select]

\frac{-a}{b}
$\Rightarrow \frac{-a}{b}$

Quote from: Student0001 on June 24, 2013, 05:51:17 pm

p.s hey 2/cos(c), i totally get what it mean hahahahaahaha

Quote from: Loch Ness on June 24, 2013, 06:05:45 pm

Hello...was wondering if you could help me with a question.

There is a parabola (restricted domain x needs to be equal or greater than 0)
It goes through (0,30) =A and (8,10) =B
There is also an angle at the top of A= 10degrees.

The equation given is y=ax^2+bx+c
The question is to find a (you can't find b or c before finding a), given that a= (-2tan(170)-5) / 16
Please help! Thank you!

This is a little odd. To me it is intuitive to find b and c before a.

For example;

$y = ax^2 + bx + c$

$\frac{dy}{dx}=2ax + b$

At the point A, $x=0$ and the slope , $\frac{dy}{dx}$ , is $-\tan(10^{\circ})$ , hence:

$2a(0)+b = -\tan(10^{\circ})$

$\Rightarrow b = -\tan(10^{\circ})$

At the point A, $x = 0$ and $y = 30$ and hence

$30 = a(0)^2 -\tan(10^{\circ})(0) + c$

$\Rightarrow c = 30$

At the point B, $x=8$ and $y=10$ , hence

$10 = a(8)^2 -\tan(10^{\circ})(8) + 30$

$\Rightarrow a = \frac{1}{16}(2\tan(10^{\circ})-5) = \frac{1}{16}(-2\tan(170^{\circ})-5)$
However, I suppose if you're given a (I don't really understand why they gave it) you can just do the above in reverse and you should be all sweet

12AM · « **Reply #2140 on:** June 24, 2013, 08:02:24 pm »

A hyperbola has the equation (considering the right hand side of the graph) of

$y=\frac{25}{x+1}-2$

With an x-intercept at (11.5, 0), a y-intercept of (0, 23), whilst going through the point (4,3)

A rectangle within the hyperbola has it's corner touching the hyperbola at all times.
Find an equation for the height of the rectangle, in terms of x.

No idea where to even start with this.

Help?

Phy124 · « **Reply #2141 on:** June 24, 2013, 09:07:30 pm »

Quote from: 12AM on June 24, 2013, 08:02:24 pm

A hyperbola has the equation (considering the right hand side of the graph) of

$y=\frac{25}{x+1}-2$

With an x-intercept at (11.5, 0), a y-intercept of (0, 23), whilst going through the point (4,3)

A rectangle within the hyperbola has it's corner touching the hyperbola at all times.
Find an equation for the height of the rectangle, in terms of x.

No idea where to even start with this. Help?

If the corner of the rectangle is touching a point on the hyperbola at all times, then the height of the rectangle would be the y value of the hyperbola at that point x.

Therefore the equation for the height of the rectangle is simply the equation of the hyperbola, $\frac{25}{x+1}-2$ , no?

SocialRhubarb · « **Reply #2142 on:** June 24, 2013, 10:15:12 pm »

Where's the other corner though?

BubbleWrapMan · « **Reply #2143 on:** June 25, 2013, 05:45:18 pm »

I'd say there isn't enough information given...

nspire · « **Reply #2144 on:** June 27, 2013, 05:48:07 pm »

Hello! I'm having trouble working out this question, so can someone please help me?
I really appreciate it!!

A body starts from O and moves in a straight line. After t seconds (t ≥ 0) its velocity
(v cm/s) is given by v = 2t − 3. Find the distance travelled in the first 3 seconds.

Thank you~

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Author Topic: VCE Methods Question Thread! (Read 5722798 times) Tweet Share

zhe0001

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