Author Topic: 3U Maths Question Thread (Read 1493564 times) Tweet Share

owidjaja · « **Reply #3240 on:** February 08, 2018, 06:53:23 pm »

Quote from: RuiAce on February 08, 2018, 06:41:38 pm

Look at her second last line. It's the same thing except rearranged - her answer is just tidier than what they provided.

Either would be marked correct.

Ok, so I finally understand the previous question, but I don't think I understand the method. I attempted the next question but I got it wrong..

RuiAce · « **Reply #3241 on:** February 08, 2018, 06:56:38 pm »

Quote from: owidjaja on February 08, 2018, 06:53:23 pm

Ok, so I finally understand the previous question, but I don't think I understand the method. I attempted the next question but I got it wrong..

You need to expand everything in before you do the integration.
$\int (5u-20)u^{1/2}\,du = \int \left(5u^{3/2} - 20u^{1/2}\right)\,du$

arii · « **Reply #3242 on:** February 08, 2018, 07:35:40 pm »

Can someone teach me part (c) of this question please? I've already done (a) and (b) so I've written down the answers to save whoever time. Thanks in advance.

A particle P moves so that its acceleration is proportional to its displacement x from a fixed point O and opposite in direction. Initally, the particle is at the origin, moving with velocity 12 m/s, and the particle is stationary when x=4.
(a) Find a and v² as functions of x. (DONE)
a=-w²x=-9x
v² = w²(A²-x²) = 9(16-x²)
(b) Find x, v and a as functions of t. (DONE)
x=4sin3t, v=12cos3t, a=-36sin3t
(c) Find the displacement, acceleration and times when the particle is at rest.

owidjaja · « **Reply #3243 on:** February 08, 2018, 08:11:11 pm »

Hey,
So I'm not entirely sure how to do question b).

Thanks in advance.

Opengangs · « **Reply #3244 on:** February 08, 2018, 08:50:35 pm »

Quote from: arii on February 08, 2018, 07:35:40 pm

Can someone teach me part (c) of this question please? I've already done (a) and (b) so I've written down the answers to save whoever time. Thanks in advance.

A particle P moves so that its acceleration is proportional to its displacement x from a fixed point O and opposite in direction. Initally, the particle is at the origin, moving with velocity 12 m/s, and the particle is stationary when x=4.
(a) Find a and v² as functions of x. (DONE)
a=-w²x=-9x
v² = w²(A²-x²) = 9(16-x²)
(b) Find x, v and a as functions of t. (DONE)
x=4sin3t, v=12cos3t, a=-36sin3t
(c) Find the displacement, acceleration and times when the particle is at rest.

$\text{Anytime the particle is at rest, we say that the velocity is equal to 0.} \\ \text{Thus, using the result from part (i): } 0 = 9(16 - x^{2}) \\ \text{Rearranging the expression, we get: } \\ 144 - 9x^{2} = 0 \\ 9x^{2} = 144 \\ x^{2} = 16 \Rightarrow x = \pm 4$ $\text{Notice that the acceleration by definition is given by: } \boxed{a = -w^2x} \\ \text{Using the result from part (i): } a = -9(\pm 4) = \mp 36$ $\text{Finally, we just use the result from part (b) to find the times where the particle is at rest: } \\ \text{That is: } 12cos(3t) = 0 \Rightarrow cos(3t) = 0 \\ 3t = cos^{-1}0 \Rightarrow t = \frac{\pi}{6}, \frac{3\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6} etc. \text{ } t > 0$

legorgo18 · « **Reply #3245 on:** February 08, 2018, 08:55:06 pm »

Quote from: owidjaja on February 08, 2018, 08:11:11 pm

Hey,
So I'm not entirely sure how to do question b).

Thanks in advance.

Hey! As usual working out in link, https://imgur.com/a/z64E9

If you dont understand something or have any questions please ask!

RuiAce · « **Reply #3246 on:** February 09, 2018, 07:42:41 pm »

Quote from: arii on February 09, 2018, 07:30:41 pm

Hello.

I'm really stuck with this question. Also note, the fraction we're meant to deduce has the variable t in it, rather than the normal theta.

$x = A \tan \theta \iff \theta = \tan^{-1} \frac{x}{A}\\ \text{and hence }\boxed{\frac{d\theta}{dx} = \frac{A}{A^2+x^2}}$
$\text{Subbing }x = A \sin t\text{ in gives}\\ \begin{align*}\frac{d\theta}{dx}&= \frac{A}{A^2 + A^2\sin^2 t}\\ &= \frac{1}{A(1+\sin^2 t)}\\ \therefore \frac{dx}{d\theta}&= A(1+\sin^2 t)\end{align*}\\ \text{as required.}$
Note that the whole point of differentiating the inverse function was because there was no way we could relate $ \theta $ in terms of $t$ without a lot of handwaving. Whereas we already have something relating $x$ to $t$.

owidjaja · « **Reply #3247 on:** February 09, 2018, 08:18:54 pm »

Hey there,
I need help with the attached question.

Thanks in advance.

chelseam · « **Reply #3248 on:** February 09, 2018, 10:07:27 pm »

Quote from: owidjaja on February 09, 2018, 08:18:54 pm

Hey there,
I need help with the attached question.

Thanks in advance.

Here's my working out! Please let me know if there's anything in the working out that you want to clarify

owidjaja · « **Reply #3249 on:** February 10, 2018, 02:11:02 pm »

Quote from: chelseam on February 09, 2018, 10:07:27 pm

Here's my working out! Please let me know if there's anything in the working out that you want to clarify

Hey!
Just a bit confused as to why is there a 2 outside the integral symbol?

RuiAce · « **Reply #3250 on:** February 10, 2018, 02:20:50 pm »

Quote from: owidjaja on February 10, 2018, 02:11:02 pm

Hey!
Just a bit confused as to why is there a 2 outside the integral symbol?

Try sketching the graph (hand-drawn or Desmos). She was exploiting symmetry.

Since the area below the x-axis is actually the same as the area above the x-axis (in magnitude), if you want the combined area you can just do 2 times one of them.

The question could also have done without exploiting symmetry by doing $ A =\left| \int_{-\sqrt2}^0 x(x^2-2)^4 \,dx \right| + \int_0^{\sqrt2} x(x^2-2)^4\,dx $

Sine · « **Reply #3251 on:** February 10, 2018, 02:21:56 pm »

Quote from: owidjaja on February 10, 2018, 02:11:02 pm

Hey!
Just a bit confused as to why is there a 2 outside the integral symbol?

if you sketch the funciton it appears to have rotational symmetry about the origin so chelseam changed the terminals from -sqrt(2) to sqrt(2) to 0 to sqrt(2) which excludes half of the required area since the area will be the same you need to multiply it by 2 for the total it''s a common technique to make evaluating an integral easier and quicker.

EDIT: beaten by RuiAce

fireives1967 · « **Reply #3252 on:** February 11, 2018, 01:11:21 am »

I need help on both of these please! For 10, I understand how to find the area both under and above y = 1/2, but I'm not sure how to come up with an integral to take away the top half

For (b) I'm just entirely confused $:-\$

Highly Appreciated!

Edit: sorry I forgot to include (a), since it's meant to be a reference for (b) (I've already done a)

clovvy · « **Reply #3253 on:** February 11, 2018, 07:20:55 am »

Quote from: fireives1967 on February 11, 2018, 01:11:21 am

I need help on both of these please! For 10, I understand how to find the area both under and above y = 1/2, but I'm not sure how to come up with an integral to take away the top half

For (b) I'm just entirely confused $:-\$

Highly Appreciated!

Edit: sorry I forgot to include (a), since it's meant to be a reference for (b) (I've already done a)

I think for part B you have to do the integral again but this time in respect to y so instead of dx it will be dy

RuiAce · « **Reply #3254 on:** February 11, 2018, 11:06:17 am »

Quote from: fireives1967 on February 11, 2018, 01:11:21 am

I need help on both of these please! For 10, I understand how to find the area both under and above y = 1/2, but I'm not sure how to come up with an integral to take away the top half

For (b) I'm just entirely confused $:-\$

Highly Appreciated!

Edit: sorry I forgot to include (a), since it's meant to be a reference for (b) (I've already done a)

$\text{Q10 is just another classic area-between-two-curves}\\ \text{where as you identified, }y=\frac12\text{ is the bottom curve.}\\ A = \int_{\pi/6}^{5\pi/6} \left(\sin x - \frac12 \right)\,dx$
All that you really care about is the area above $ y=\frac12$ below $y=\sin x$
________________________________________________________
$\text{In a), you would've found the area between two curves}\\ \text{using }\boxed{A = \int_{a}^b (y_2 - y_1)\,dx}$
Where in your case, $y_2 = \cos x$ and $y_1 = \sin x$. Or you could've computed $\int_{\pi/6}^{\pi/4} \cos x\,dx $ and $\int_{\pi/6}^{\pi/4} \sin x\,dx $, before subtracting the second into the first at the end.
$\text{In a similar way, the volume of the solid of revolution}\\ \text{formed from rotating a region bound BETWEEN two curves}\\ \text{is }\boxed{V = \pi\int_a^b (y_2^2 - y_1^2)\,dx }$
$\text{So in your case}\\ V = \pi \int_{\pi/6}^{\pi/4}(\cos^2 x - \sin^2 x)\,dx$
To evaluate this integral, you can start by just quoting the double angle formula $ \cos^2 x - \sin^2 x = \cos 2x $

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