Author Topic: Mathematics Question Thread (Read 1626830 times) Tweet Share

Calley123 · « **Reply #3420 on:** April 08, 2018, 04:51:56 pm »

Hey again,

How do I solve this trig equation between o and 2pi
sin3x+ sinx=0

Thank you

StupidProdigy · « **Reply #3421 on:** April 08, 2018, 06:25:20 pm »

Quote from: Calley123 on April 08, 2018, 04:51:56 pm

Hey again,

How do I solve this trig equation between o and 2pi
sin3x+ sinx=0

Thank you

Basically just apply the relevant compound angle formula first (sin(3x) becomes sin(2x+x)), then apply the sin double angle formula and cosine double angle formula. The overall goal is to get the argument in any of the trig functions to all be the same and in this case be 'x' by itself. See the pic attached for working. I didn't do the final step because I'm trusting you know where to go from there, hope it helps

RuiAce · « **Reply #3422 on:** April 08, 2018, 08:45:34 pm »

Quote from: Calley123 on April 08, 2018, 04:51:56 pm

Hey again,

How do I solve this trig equation between o and 2pi
sin3x+ sinx=0

Thank you

Answer provided above. The solutions will then just be $x=0,\frac\pi2,\pi,\frac{3\pi}2,2\pi $.

However please note that this is a 3U question, and for these situations in the future I will move it to the relevant thread with only a small warning.

Calley123 · « **Reply #3423 on:** April 09, 2018, 09:01:33 am »

Quote from: StupidProdigy on April 08, 2018, 06:25:20 pm

Basically just apply the relevant compound angle formula first (sin(3x) becomes sin(2x+x)), then apply the sin double angle formula and cosine double angle formula. The overall goal is to get the argument in any of the trig functions to all be the same and in this case be 'x' by itself. See the pic attached for working. I didn't do the final step because I'm trusting you know where to go from there, hope it helps

Thank you !!

Mate2425 · « **Reply #3424 on:** April 10, 2018, 03:53:24 pm »

Hi could someone please help me with this question:
f(x) = 2x^3 +9x^2 +12x +1
Q. By halving the interval twice find approximation to this. (Answer = 0)
Thank you!!

Also when they ask to use the method of halving the interval twice to find an approximation / estimate to cube root of 12 ........; why do they have the answer as only one of the numbers of where the root lies e.g After halving interval twice gives me 2.25< Cube root of 12 < 2.5 and Answer is only 2.25.

Any help in understanding why, would be of much appreciation!!

Mod Edit: Post merge, use the 'Modify' button to add to your last post if no one has responded yet

RuiAce · « **Reply #3425 on:** April 10, 2018, 10:23:42 pm »

Quote from: Mate2425 on April 10, 2018, 03:53:24 pm

Hi could someone please help me with this question:
f(x) = 2x^3 +9x^2 +12x +1
Q. By halving the interval twice find approximation to this. (Answer = 0)
Thank you!!
Also when they ask to use the method of halving the interval twice to find an approximation / estimate to cube root of 12 ........; why do they have the answer as only one of the numbers of where the root lies e.g After halving interval twice gives me 2.25< Cube root of 12 < 2.5 and Answer is only 2.25.

Any help in understanding why, would be of much appreciation!!

Mod Edit: Post merge, use the 'Modify' button to add to your last post if no one has responded yet

This is actually a 3U concept! We can help with that in our 3U thread if you like, just so everything is in one spot when people read it back later!

Click here!

Mod Edit: Added link

LaraC · « **Reply #3426 on:** April 18, 2018, 10:26:36 am »

Hello

Just not quite sure how to tackle this question:
The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the first 20 terms.

So I've done part 1: as in, got an equation for the sum of the first 4 terms = 42 (my equation is 2a + d = 21.....is that right?!), and I know how to find the sum of the first 20 terms once I have the middle bit....I'm just a little confused about the 2nd step.....how do I work out an equation for the 3rd and 7th term adding to 46?

Thanks!

Opengangs · « **Reply #3427 on:** April 18, 2018, 11:34:17 am »

Quote from: LaraC on April 18, 2018, 10:26:36 am

Hello

Just not quite sure how to tackle this question:
The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the first 20 terms.

So I've done part 1: as in, got an equation for the sum of the first 4 terms = 42 (my equation is 2a + d = 21.....is that right?!), and I know how to find the sum of the first 20 terms once I have the middle bit....I'm just a little confused about the 2nd step.....how do I work out an equation for the 3rd and 7th term adding to 46?

Thanks!

Hey, so we begin with the sum of the first 4 terms being 42.
$\frac{4}{2}\left[2a_1 + (4 - 1)d\right] = 42 \\ 2a_1 + 3d = 21$
$\text{Furthermore, } a_3 + a_7 = 46 \\ a_1 + (3 - 1)d + a_1 + (7 - 1)d = 46 \\ 2a_1 + 8d = 46 \\ 2a_1 = 46 - 8d$
$\text{Substitute } 2a_1 \text{ back into the first equation we formed: } \\ (46 - 8d) + 3d = 21 \\ 46 - 5d = 21 \\ -5d = -25 \\ d = 5$
$\text{So, } 2a_1 = 46 - 8(5) = 46 - 40 = 6 \\ a_1 = 3$

$\text{You now have all you need to find the sum of the first 20 terms.}$

LaraC · « **Reply #3428 on:** April 18, 2018, 12:33:47 pm »

Thanks opengangs!
Dur! Just saw where I went wrong with my first equation for that question!

Sorry, could I also ask about this qu....(I'm v dumb!

) :
The 20th term of an arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. Find the sum of the first 20 terms.

Its kind of similar, but do I need to form an equation that goes something like (sum of 10 terms - sum of 5 terms = 235)....or how do I go about it?

Opengangs · « **Reply #3429 on:** April 18, 2018, 12:41:17 pm »

Quote from: LaraC on April 18, 2018, 12:33:47 pm

Thanks opengangs!
Dur! Just saw where I went wrong with my first equation for that question!

Sorry, could I also ask about this qu....(I'm v dumb! ) :
The 20th term of an arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. Find the sum of the first 20 terms.

Its kind of similar, but do I need to form an equation that goes something like (sum of 10 terms - sum of 5 terms = 235)....or how do I go about it?

Hey, LaraC!

That's a very good question - don't feel dumb!

So, basically yep! You will need to form an equation where $ S_{10} - S_5 = 235 $ - that'd be the right approach!

[Hint: Use the fact that $ a_n = a_1 + (n - 1)d $ and use the formula for the sum to n terms]

Let me know if you get stuck!

Calley123 · « **Reply #3430 on:** April 20, 2018, 12:10:13 pm »

Hey,
It would be great if someone could help me with these questions. They are a bit tricky.

Thanks

RuiAce · « **Reply #3431 on:** April 20, 2018, 01:08:32 pm »

Quote from: Calley123 on April 20, 2018, 12:10:13 pm

Hey,
It would be great if someone could help me with these questions. They are a bit tricky.

Thanks

$\text{Using the product rule,}\\ \begin{align*}\frac{d}{dx} \left(x^2\log_e x\right) &= x^2\, \frac{1}{x}+2x \log_e x\\ &= x+2x\log_e x\\ &= x(1 + 2\log_e x).\end{align*}$
$\text{which also implies that }\frac{d}{dx} \left(2 x^2\log_e x\right)= 2x(1+2\log_e x).$
___________________________________________
$\text{Now the integral can be made much easier by recognising that}\\ \textbf{the derivative and the integral are just inverse operations.}$
$\text{That is to say, }\textbf{because } \boxed{\frac{d}{dx} \left(2 x^2\log_e x\right)= 2x(1+2\log_e x)}\\ \text{an immediate }\textbf{direct consequence}\text{ is }\boxed{\int 2x(1+2\log_e x) \,dx = 2x^2\log_e x + C}$
\begin{align*}\therefore \int_1^3 2x(1+2\log_ex)\,dx &= \left[2x^2\log_e x\right]_1^3\\ &= 18\log_e 3 - 0 \\ &= 18\log_e 3\end{align*}

Calley123 · « **Reply #3432 on:** April 20, 2018, 03:33:17 pm »

Quote from: RuiAce on April 20, 2018, 01:08:32 pm

$\text{Using the product rule,}\\ \begin{align*}\frac{d}{dx} \left(x^2\log_e x\right) &= x^2\, \frac{1}{x}+2x \log_e x\\ &= x+2x\log_e x\\ &= x(1 + 2\log_e x).\end{align*}$
$\text{which also implies that }\frac{d}{dx} \left(2 x^2\log_e x\right)= 2x(1+2\log_e x).$
___________________________________________
$\text{Now the integral can be made much easier by recognising that}\\ \textbf{the derivative and the integral are just inverse operations.}$
$\text{That is to say, }\textbf{because } \boxed{\frac{d}{dx} \left(2 x^2\log_e x\right)= 2x(1+2\log_e x)}\\ \text{an immediate }\textbf{direct consequence}\text{ is }\boxed{\int 2x(1+2\log_e x) \,dx = 2x^2\log_e x + C}$
\begin{align*}\therefore \int_1^3 2x(1+2\log_ex)\,dx &= \left[2x^2\log_e x\right]_1^3\\ &= 18\log_e 3 - 0 \\ &= 18\log_e 3\end{align*}

Thanks heaps

Calley123 · « **Reply #3433 on:** April 20, 2018, 03:36:22 pm »

Hey again,
I thought I attached this question before but obviously not.
Help please...

Cheers

RuiAce · « **Reply #3434 on:** April 20, 2018, 06:08:48 pm »

Quote from: Calley123 on April 20, 2018, 03:36:22 pm

Hey again,
I thought I attached this question before but obviously not.
Help please...

Cheers

$\text{If we're rotating about the }y\text{-axis, then we need to make }x\text{ the subject.}\\ y = e^x \implies x = \log_e y$
$\text{So we need to consider}\\ V = \pi \int_3^5 (\ln y)^2\,dy.$
$\text{The relevant Simpson's rule approximation is}\\ V \approx \pi \frac{5-3}{3\times 4}\left[ (\ln 3)^2 + 4(\ln 3.5)^2 + 2 (\ln 4)^2 + 4 (\ln 4.5)^2 + (\ln 5)^2 \right]\\ \text{which you can put into your calculator.}$

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