Author Topic: TyErd's questions (Read 43282 times) Tweet Share

Whatlol · « **Reply #375 on:** July 02, 2010, 03:46:53 pm »

Hmm sorry i cant help i honestly dont know how to do it )=

moekamo · « **Reply #376 on:** July 02, 2010, 04:01:29 pm »

$\frac{d\theta}{dt} = \frac{dx}{dt} \frac{d\theta}{dx}$ where x is the horizontal distance from the pbservation point to under the plane.

since $\tan \theta = 1000 /x \implies x = 1000/\tan\theta \implies \frac{dx}{d\theta} = \frac{\tan \theta \cdot 0 - 1000\sec^2 \theta}{\tan ^2 \theta} = -1000/\sin ^2 \theta$

when theta is 30 degrees, $\frac{dx}{d \theta} = -4000 \implies \frac{d \theta}{dx} = -\frac{1}{4000} \implies \frac{d \theta}{dt} = -\frac{1}{4000}\times \frac{480}{3.6} = -1/30$ so decreasing at 1/30 rad/sec

the next part is to find dh/dt where h is the hypotenuse, can you try it?

TyErd · « **Reply #377 on:** July 02, 2010, 04:25:43 pm »

Okays

$\frac{dh}{dt} = \frac {dh}{d\theta} \frac {d\theta}{dt}$

$h= \frac{1000}{sin\theta}, \frac{dh}{d\theta} = \frac{-1000cos\theta}{sin^2\theta}<br />$

$\frac{d\theta}{dt}$ was found to be $-\frac{1}{30}$

$So \frac{dh}{dt} = \frac{-1000cos30}{sin^2(30)} \times -\frac{1}{30}$

For some reason i think thats wrong

brightsky · « **Reply #378 on:** July 02, 2010, 04:26:39 pm »

Let me continue on...

$\frac{dh}{dt} = \frac{dh}{d\theta} \times \frac{d\theta}{dt}$ , where h is the length of the hypotenuse.

We know that $\frac{d\theta}{dt} = -\frac{1}{30}$

$\because \sin(\theta) = \frac{1000}{h}$

$\therefore h = \frac{1000}{\sin(\theta)}$

$\frac{dh}{d\theta} = -1000 \cot(\theta) \csc(\theta)$

When $x = \frac{\pi}{6}$ , $\frac{dh}{d\theta} = -2000 \sqrt{3}$

Hence:

$\frac{dh}{dt} = -2000\sqrt{3} \cdot -\frac{1}{30} = \frac{2000\sqrt{3}}{30} = \frac{200\sqrt{3}}{3}$

brightsky · « **Reply #379 on:** July 02, 2010, 04:30:36 pm »

Quote from: TyErd on July 02, 2010, 04:25:43 pm

Okays

$\frac{dh}{dt} = \frac {dh}{d\theta} \frac {d\theta}{dt}$

$h= \frac{1000}{sin\theta}, \frac{dh}{d\theta} = \frac{-1000cos\theta}{sin^2\theta}<br />$

$\frac{d\theta}{dt}$ was found to be $-\frac{1}{30}$

$So \frac{dh}{dt} = \frac{-1000cos30}{sin^2(30)} \times -\frac{1}{30}$

For some reason i think thats wrong

Change 30 degrees into radians. :p

TyErd · « **Reply #380 on:** July 02, 2010, 04:30:54 pm »

Yay I got it right, I just realised instead of 30 i have to sub in $\frac {\pi}{6}$

TyErd · « **Reply #381 on:** July 02, 2010, 04:31:18 pm »

damn Beaten lol

TyErd · « **Reply #382 on:** July 02, 2010, 04:31:56 pm »

Thanks everyone!

TyErd · « **Reply #383 on:** July 02, 2010, 04:36:53 pm »

I have another problem: ABCD is a trapezium with AB=CD, with vertices on the circle and with the centre O. AD is a diameter of the circle. The radius of the circle is 4units.

a) Find BC in terms of $\theta$ .
b) Find the area of the trapezium in terms of $\theta$ and hence find the maximum area.

image attached

brightsky · « **Reply #384 on:** July 02, 2010, 04:42:42 pm »

$\angle COD = \angle BOA$ (They subtend the same size arc).

Hence $\angle BOA = \theta$

So $\angle BOC = 180 - 2\theta$

$\therefore \angle OBC = \angle OCB = \theta$

By the sine rule:

$\frac{4}{\sin\theta} = \frac{BC}{\sin(180 - 2\theta)}$

Solve that to get your answer.

TyErd · « **Reply #385 on:** July 02, 2010, 05:15:34 pm »

error in latex btw

umm why is $\angle OBC$ and $\angle OCB$ equal to $\theta .$

brightsky · « **Reply #386 on:** July 02, 2010, 05:24:48 pm »

Because $\triangle OBC$ is an isosceles triangle. Hence its base angles are equal, hence $\angle OBC = \angle OCB$ .

So because $\angle BOC = 180 - 2\theta$

Then $\angle OBC = \angle OCB = \frac{180 - (180 - 2\theta)}{2} = \frac{2\theta}{2} = \theta$

Continuing on from $\frac{4}{\sin(\theta)} = \frac{BC}{\sin(180 - 2\theta)}$

$BC = \frac{4 \sin(180 - 2\theta)}{\sin(\theta)}$

$BC = \frac{4 \sin(2\theta)}{\sin (\theta)}$

$BC = 8\cos(\theta)$

TyErd · « **Reply #387 on:** July 02, 2010, 05:26:45 pm »

okaays thankyou very much I get that part now, what about the second part?

brightsky · « **Reply #388 on:** July 02, 2010, 05:34:48 pm »

Let $A$ be the area of the trapezium.

Then $A = \frac{1}{2} (BC + AD) h$ , where h is the height of the trapezium.

We already know BC and AD in terms of $\theta$ . $h$ is the height of $\triangle OBC$ which can be found by Pythagoras' theorem.

Once we have the equation, finding max area would be straightforward.

TyErd · « **Reply #389 on:** July 02, 2010, 05:46:00 pm »

Okay I got $A= \frac{1}{2} \times (\frac{4sin(2\theta)}{sin(\theta)} + 8 ) \times 4cos(90-\theta)$

not sure if that is correct though....

but then I have to differentiate it make it equal zero find $\theta$ and then sub back into the equation correct?

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