Author Topic: Spesh Qs (Read 11952 times) Tweet Share

TrueTears · « **Reply #15 on:** July 28, 2009, 06:43:27 pm »

Haha I just did this Q /0, yeah their both right, I think the Q asked you to find the equation of the tangents? Both expressions for the gradient function give the same gradient at those points.

/0 · « **Reply #16 on:** July 28, 2009, 08:44:23 pm »

Quote from: TrueTears on July 28, 2009, 06:43:27 pm

Haha I just did this Q /0, yeah their both right, I think the Q asked you to find the equation of the tangents? Both expressions for the gradient function give the same gradient at those points.

Yeah that's right, but what I'm more worried about is that the expressions are not the same for other points.

i.e. for $x=2$ , solving for y you get $y=-1 \pm 2\sqrt{2}$

Let's take the positive value of y...

With $\frac{y^2-x^2y}{x^3+2xy}$ the gradient is $\frac{-2}{51}(113\sqrt{2}-151)$

With $\frac{2(1-xy)}{x^2+2xy}$ the gradient is $\frac{4}{93}(11\sqrt{2}-26)$

... which leads me to believe that somehow, one of these expressions is wrong.

TrueTears · « **Reply #17 on:** July 28, 2009, 08:48:17 pm »

Yeah that JUST happened to me on Kilbaha exam Q 1... the expressions don't work for some points but they do for others... I'm gonna ask kamil about it later and see what he says on it.

/0 · « **Reply #18 on:** July 28, 2009, 08:53:50 pm »

hmm interesting
Also another question, this time from MAV 2007 Exam 2, picture attached

This might be due to my lack of understanding with solids of revolutions, but I was under the impression that A would be the correct expression, but it is not.

Flaming_Arrow · « **Reply #19 on:** July 28, 2009, 08:57:41 pm »

is it B?

kamil9876 · « **Reply #20 on:** July 28, 2009, 09:01:06 pm »

Quote from: Flaming_Arrow on July 28, 2009, 08:57:41 pm

is it B?

Yep

/0 · « **Reply #21 on:** July 28, 2009, 09:03:00 pm »

ok don't worry i think i see why now

Damo17 · « **Reply #22 on:** July 28, 2009, 09:39:28 pm »

Quote from: /0 on July 28, 2009, 08:53:50 pm

hmm interesting
Also another question, this time from MAV 2007 Exam 2, picture attached

This might be due to my lack of understanding with solids of revolutions, but I was under the impression that A would be the correct expression, but it is not.

it can't be A because it is not in the form $\pi \int_a^b ((y_1)^2 - (y_2)^2) dx$ . where $y_1$ is upper function and $y_2$ is lower function.

you can not combine to two functions.
$\pi \int_a^b ((y_1)^2 - (y_2)^2) dx \ne \pi \int_a^b (y_1 - y_2)^2 dx$

Mao · « **Reply #23 on:** July 28, 2009, 09:44:49 pm »

Replying to the question relating implicit diff, $xy + \frac{y^2}{x} = 2$ , the solutions should be the same. It appears you have made a small error in solving the relation for y when x=2, as well as an error in implicit differentiation.

implicit differentiation:
$\implies x \frac{dy}{dx} + y + \frac{2y}{x}\frac{dy}{dx} - \frac{y^2}{x^2} = 0$

$\implies \frac{dy}{dx}\cdot \left( \frac{2y + x^2}{x}\right) = \frac{y^2 - x^2 y}{x^2}$

$\implies \frac{dy}{dx} = \frac{y^2 - x^2 y }{x^3 + 2xy}$

The other method:

$x^2 y + y^2 = 2x$

implicit diff:

$2xy + x^2 \frac{dy}{dx} + 2y\frac{dy}{dx} = 2$

$\implies \frac{dy}{dx} \cdot (x^2 + 2y) = 2 - 2xy$

$\implies \frac{dy}{dx} = \frac{2(1-xy)}{x^2 + 2y}$

When x=2, $2y + \frac{1}{2}y^2 = 2 \implies y = -2 \pm 2\sqrt{2}$

for the positive value, $\frac{dy}{dx}_1 =-2 + \frac{5}{2\sqrt{2}}$
$\frac{dy}{dx}_2 = -2 + \frac{5}{2\sqrt{2}}$

/0 · « **Reply #24 on:** August 09, 2009, 01:25:18 am »

Apparently the answer for this is 100N but i get 41.2N... what does everyone else get?

(highlight to uncover text)

unknown id · « **Reply #25 on:** August 09, 2009, 01:40:50 am »

I get 100N.

$F = ma$

$P - 10gsin30^\circ - \frac{\sqrt 3}{5}gcos30^\circ = 10 \times 2.16$

TrueTears · « **Reply #26 on:** August 09, 2009, 01:42:40 am »

Yeah I get 100.

EDIT: too fast unknown id lol was just about to post that.

TonyHem · « **Reply #27 on:** August 09, 2009, 01:48:42 am »

He forgot the 10

/0 · « **Reply #28 on:** August 09, 2009, 03:10:57 am »

thanks

/0 · « **Reply #29 on:** August 09, 2009, 09:09:29 pm »

Is it true/false that if you take the average of the three coordinates of a triangle you will get the coordinates of the centre of the circumcircle?
I've heard that it can be used to find the centroid but i'm not sure if that's the same thing.

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