May 29, 2020, 12:54:35 pm

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#### RuiAce

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« Reply #4170 on: April 19, 2019, 03:41:52 pm »
+3
Use the trapezoidal rules with 4 sub intervals to find an approximation to the volume of the solid formed when rotating y=sqrcosx about the x-axis from x=0 to x=pi/2.

So I found the A with x=0,pi/8,pi/4,3pi/8,pi/2 and it was equal to approx 1.4 (maybe someone could confirm this?). I just don't know how to find the volume from this area. Do I square it and then multiply by pi??
$\text{Recall that the volume is given by}\\ V =\pi \int_a^b y^2\,dx.$
$\text{So here, }y = \sqrt{\cos x}\implies \boxed{y^2 = \cos x}\text{. Hence your volume is}\\ \boxed{V = \pi \int_0^{\pi/2} \cos x\,dx}.\\ \text{This is the integral you would like to use on the trapezoidal rule.}$
The area is not helpful in finding the volume of solid of revolution because the $y^2$ is entirely under the integral. As opposed to something like $\pi \left( \int_a^b y\,dx \right)^2$.

#### spnmox

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« Reply #4171 on: April 19, 2019, 04:55:30 pm »
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Thank you!

#### therese07

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« Reply #4172 on: April 20, 2019, 06:58:47 pm »
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Hi!

So I was doing some revision on curve sketching, and I came across this question (attached below).

I understand you have to do the produce rule first (I could be wrong while doing this working out, so sorry in advance)

u = 2x +1           v = (x-2)^4
u' = 2                 v' = 4(x-2)^3

dy/dx = (2x +1) x 4(x-2)^3 + (x-2)^4 x 2
= 4(2x +1)(x-2)^3 + 2(x-2)^4

I don't know what do to, or what to factor out to get to the next step and to find stationary points, and sketch it.

2020: Bachelor of Arts/Bachelor of Law @ Macquarie University

#### jamonwindeyer

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« Reply #4173 on: April 20, 2019, 07:11:20 pm »
+1
Hi!

So I was doing some revision on curve sketching, and I came across this question (attached below).

I understand you have to do the produce rule first (I could be wrong while doing this working out, so sorry in advance)

u = 2x +1           v = (x-2)^4
u' = 2                 v' = 4(x-2)^3

dy/dx = (2x +1) x 4(x-2)^3 + (x-2)^4 x 2
= 4(2x +1)(x-2)^3 + 2(x-2)^4

I don't know what do to, or what to factor out to get to the next step and to find stationary points, and sketch it.

Smashed the first bit! If you pull out a factor of $\left(x-2\right)^3$, you'll get:

$\frac{dy}{dx}=\left(x-2\right)^3\left(4(2x+1)+2(x-2)\right)=\left(x-2\right)^3\left(10x\right)$

From there you can solve for the $x=2$ and $x=0$ stationary points, and determine their nature using a table (the second derivative test is probably a bit cumbersome here)

#### Thankunext

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« Reply #4174 on: April 22, 2019, 10:14:21 pm »
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Help!
For what value of k will the equation 6x*2 + (k+15)x +(4+k)=0 have roots which are
I. Equal but opposite in sign?
2. Reciprocals?

#### RuiAce

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« Reply #4175 on: April 22, 2019, 10:42:47 pm »
+3
Help!
For what value of k will the equation 6x*2 + (k+15)x +(4+k)=0 have roots which are
I. Equal but opposite in sign?
2. Reciprocals?

$\text{You can just use the sum and product of roots.}\\ \text{In the first case, suppose that the roots are negatives of each other, i.e. }\boxed{\beta = -\alpha}.$
$\text{But since }\alpha+\beta = -(k+15),\text{ we therefore have}\\ \alpha-\alpha = -k-15 \implies \boxed{0 = -k-15}.$
You can do something similar for the next question.

#### spnmox

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« Reply #4176 on: April 23, 2019, 09:29:44 pm »
0
2014 HSC Q7 Multiple Choice

How many solutions of the equation (sinx-1)(tanx+2)=0 lie between 0 and 2pi?

#### jamonwindeyer

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« Reply #4177 on: April 23, 2019, 09:35:02 pm »
+1
2014 HSC Q7 Multiple Choice

How many solutions of the equation (sinx-1)(tanx+2)=0 lie between 0 and 2pi?

Hey! So all you are doing here is solving each part of the equation separately, like a quadratic!

$\sin{x}-1=0\\\sin{x}=1\\x=\frac{\pi}{2}\\\tan{x}+2=0\\\tan{x}=-2$

And I don't have a calculator handy, but you'll get two answers for that second bit - So it would appear that the answer is three! However, $x=\frac{\pi}{2}$ isn't a solution! Why? If you substitute $x=\frac{\pi}{2}$ into $\tan{x}$, it breaks! Math error! The tangent ratio doesn't allow odd multiples of 90 degrees

Really tricky, but you have to ignore that solution. Which means only the two solutions from the $\tan$ part of the equation work - The answer is two

#### spnmox

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« Reply #4178 on: April 23, 2019, 10:09:00 pm »
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Hey! So all you are doing here is solving each part of the equation separately, like a quadratic!

$\sin{x}-1=0\\\sin{x}=1\\x=\frac{\pi}{2}\\\tan{x}+2=0\\\tan{x}=-2$

And I don't have a calculator handy, but you'll get two answers for that second bit - So it would appear that the answer is three! However, $x=\frac{\pi}{2}$ isn't a solution! Why? If you substitute $x=\frac{\pi}{2}$ into $\tan{x}$, it breaks! Math error! The tangent ratio doesn't allow odd multiples of 90 degrees

Really tricky, but you have to ignore that solution. Which means only the two solutions from the $\tan$ part of the equation work - The answer is two

oh, thank you so much! haha now it seems obvious

#### Karlamineeeee

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« Reply #4179 on: April 24, 2019, 01:01:59 pm »
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I hope this question is allowed here, I'm trying out the first series and sequence test in the ATAR notes' HSC mathematics topic tests.

For the first question, why do the answers calculate for the 15th term instead of the 30th?

#### RuiAce

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« Reply #4180 on: April 24, 2019, 01:40:48 pm »
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I hope this question is allowed here, I'm trying out the first series and sequence test in the ATAR notes' HSC mathematics topic tests.

For the first question, why do the answers calculate for the 15th term instead of the 30th?
I had a look. That's definitely some accidental error. Should certainly sub in $n=30$ in $T_n = 6(2)^{n-1}$

#### Karlamineeeee

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« Reply #4181 on: April 24, 2019, 07:30:00 pm »
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I had a look. That's definitely some accidental error. Should certainly sub in $n=30$ in $T_n = 6(2)^{n-1}$

Thank you so much!!

Also, for the final question of that test, where did the 500 value come from? Wasn't $1000 deposited? Sorry for asking so late!! DX #### RuiAce • ATAR Notes Lecturer • Honorary Moderator • Great Wonder of ATAR Notes • Posts: 8737 • "All models are wrong, but some are useful." • Respect: +2516 ##### Re: Mathematics Question Thread « Reply #4182 on: April 24, 2019, 10:31:58 pm » 0 Thank you so much!! Also, for the final question of that test, where did the 500 value come from? Wasn't$1000 deposited? Sorry for asking so late!! DX
Only just arrived back at home and yea, that's another accidental error I think. Should be 1000.

#### jamonwindeyer

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« Reply #4183 on: April 25, 2019, 12:45:36 am »
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Confirming both of those are errors, and both have corrections at this link: https://atarnotes.com/product-updates/. There's a full fixed solution to that second one too!

Apologies for the confusion caused - It's been fixed for future prints, but we must still be working through existing stock!