Author Topic: 3U Maths Question Thread (Read 1522336 times) Tweet Share

anotherworld2b · « **Reply #2280 on:** July 03, 2017, 01:47:06 am »

Hi I was hoping to get help with these 2 questions please. so far I found that for q 6 that d= b and a = 2b but I'm not sure how to find c and do from here
I wasn't sure how to do part b on q32

RuiAce · « **Reply #2281 on:** July 03, 2017, 08:59:59 am »

Quote from: anotherworld2b on July 03, 2017, 01:47:06 am

Hi I was hoping to get help with these 2 questions please. so far I found that for q 6 that d= b and a = 2b but I'm not sure how to find c and do from here
I wasn't sure how to do part b on q32

$\text{As correctly stated, }\boxed{d=b}\text{ and }\boxed{a=2b}\\ \text{So we may rewrite }y=\frac{b(2x+1)}{cx+b}$
$\text{Further information is given about the }\textbf{gradient}\\ \text{so we must find the relevant derivative.}\\ \text{From applying the quotient rule (and messy algebra), we have }\boxed{\frac{dy}{dx}=\frac{b(2b-c)}{(b+cx)^2}}$
$\text{We are told that when }x=0, \frac{dy}{dx}=3\\ \text{Hence upon substitution, }3=\frac{(2b-c)}{b} \implies \boxed{c=-b}$
$\text{Substituting back gives }y=\frac{b(2x+1)}{-bx+b } \implies \boxed{y=\frac{2x+1}{-x+1}}$

anotherworld2b · « **Reply #2282 on:** July 03, 2017, 10:20:05 am »

Oh ok

I'm a bit confused because in the answer has k in it? (1st image)
I also wanted to ask what do you do when there is a negative in front of the log? In my book the answer has no negative but I'm not sure how to get rid of it (2nd image)

Quote from: RuiAce on July 03, 2017, 08:59:59 am

$\text{As correctly stated, }\boxed{d=b}\text{ and }\boxed{a=2b}\\ \text{So we may rewrite }y=\frac{b(2x+1)}{cx+b}$
$\text{Further information is given about the }\textbf{gradient}\\ \text{so we must find the relevant derivative.}\\ \text{From applying the quotient rule (and messy algebra), we have }\boxed{\frac{dy}{dx}=\frac{b(2b-c)}{(b+cx)^2}}$
$\text{We are told that when }x=0, \frac{dy}{dx}=3\\ \text{Hence upon substitution, }3=\frac{(2b-c)}{b} \implies \boxed{c=-b}$
$\text{Substituting back gives }y=\frac{b(2x+1)}{-bx+b } \implies \boxed{y=\frac{2x+1}{-x+1}}$

RuiAce · « **Reply #2283 on:** July 03, 2017, 10:27:05 am »

Quote from: anotherworld2b on July 03, 2017, 10:20:05 am

Oh ok
I'm a bit confused because in the answer has k in it? (1st image)

My bad there, I missed the fact they wanted the values of a, b, c and d

Explanation: If you think about it, a, c and d were all in terms of b. But b was never a fixed number. b could've been 1, 2, -100, 28415239052 or whatever you wanted it to be. So they first let b = k, where k is just some arbitrary constant. That gives us a basis for everything.

From there, we can find values of a, c and d in terms of k as well.

Note that we can't just say b = b in the final answer. b has to equal something arbitrary, it just can't equal to itself or we have a redundancy.

K9810 · « **Reply #2284 on:** July 03, 2017, 12:27:24 pm »

Hey!
Not sure how to do part b and c

anotherworld2b · « **Reply #2285 on:** July 03, 2017, 12:59:54 pm »

I see thank you for your help
I just have a quick question
For sin2x = 1
The range would be 90 and 270
But is 270 degrees = 3pi/2?
so 135 degrees would be 3pi/4 right?
In my book the answer says its 5pi/4 for some reason?

Quote from: RuiAce on July 03, 2017, 10:27:05 am

My bad there, I missed the fact they wanted the values of a, b, c and d

Explanation: If you think about it, a, c and d were all in terms of b. But b was never a fixed number. b could've been 1, 2, -100, 28415239052 or whatever you wanted it to be. So they first let b = k, where k is just some arbitrary constant. That gives us a basis for everything.

From there, we can find values of a, c and d in terms of k as well.

Note that we can't just say b = b in the final answer. b has to equal something arbitrary, it just can't equal to itself or we have a redundancy.

RuiAce · « **Reply #2286 on:** July 03, 2017, 01:15:50 pm »

Quote from: anotherworld2b on July 03, 2017, 12:59:54 pm

I see thank you for your help
I just have a quick question
For sin2x = 1
The range would be 90 and 270
But is 270 degrees = 3pi/2?
so 135 degrees would be 3pi/4 right?
In my book the answer says its 5pi/4 for some reason?

Range is irrelevant here as we are talking about an equation. The function f(x) = sin(2x) has range [-1,1].

Since you're solving an equation, they should specify a domain that they want you to solve it for. If it's 0≤x≤2π then x=5π/4 is one possible solution.

anotherworld2b · « **Reply #2287 on:** July 03, 2017, 01:24:46 pm »

Oh ok. I read the question again and the domain is as you said. But I'm not sure why it's 5pi/4?

Quote from: RuiAce on July 03, 2017, 01:15:50 pm

Range is irrelevant here as we are talking about an equation. The function f(x) = sin(2x) has range [-1,1].

Since you're solving an equation, they should specify a domain that they want you to solve it for. If it's 0≤x≤2π then x=5π/4 is one possible solution.

RuiAce · « **Reply #2288 on:** July 03, 2017, 01:47:57 pm »

Quote from: anotherworld2b on July 03, 2017, 01:24:46 pm

Oh ok. I read the question again and the domain is as you said. But I'm not sure why it's 5pi/4?

The question is meant to specify a domain.

You are meant to find all values of x within the domain given that are solutions to your equation.

x=5π/4 is a solution to sin(2x)=1

anotherworld2b · « **Reply #2289 on:** July 03, 2017, 02:31:04 pm »

why is 225 degrees?
I'm still confused isn't it 90 and 270?

Quote from: RuiAce on July 03, 2017, 01:47:57 pm

The question is meant to specify a domain.

You are meant to find all values of x within the domain given that are solutions to your equation.

x=5π/4 is a solution to sin(2x)=1

RuiAce · « **Reply #2290 on:** July 03, 2017, 02:45:36 pm »

Quote from: anotherworld2b on July 03, 2017, 02:31:04 pm

why is 225 degrees?
I'm still confused isn't it 90 and 270?

Firstly, 270 degrees should not be considered as sin(270deg) = -1, not +1
$\text{If the equation was for }0\le x \le 2\pi,\text{ we must adjust by considering }0\le 2x \le 4\pi\\ \sin 2x = 1\text{, over this domain, solves to give }2x = \frac{\pi}{2}, \frac{5\pi}{2}\\ \text{Leading to }x=\frac{\pi}{4}, \frac{5\pi}{4}$
Please provide the domain of interest in the future. An assumption was made here.

(Relevant corresponding degree values - 90 and 450 become 45 and 225)

anotherworld2b · « **Reply #2291 on:** July 03, 2017, 02:58:05 pm »

I see what I did wrong now

Thank you for your help
I will provide the domain for future questions
I was looking at another question but I wasn't sure what to do for part b

Quote from: RuiAce on July 03, 2017, 02:45:36 pm

Firstly, 270 degrees should not be considered as sin(270deg) = -1, not +1
$\text{If the equation was for }0\le x \le 2\pi,\text{ we must adjust by considering }0\le 2x \le 4\pi\\ \sin 2x = 1\text{, over this domain, solves to give }2x = \frac{\pi}{2}, \frac{5\pi}{2}\\ \text{Leading to }x=\frac{\pi}{4}, \frac{5\pi}{4}$
Please provide the domain of interest in the future. An assumption was made here.

(Relevant corresponding degree values - 90 and 450 become 45 and 225)

RuiAce · « **Reply #2292 on:** July 03, 2017, 03:03:26 pm »

Quote from: anotherworld2b on July 03, 2017, 02:58:05 pm

I see what I did wrong now
Thank you for your help
I will provide the domain for future questions
I was looking at another question but I wasn't sure what to do for part b

Looking at part a) confuses me a bit. in the WACE when they say log do they mean log₁₀ or ln?

jakesilove · « **Reply #2293 on:** July 03, 2017, 03:30:32 pm »

Quote from: anotherworld2b on July 03, 2017, 02:58:05 pm

I see what I did wrong now
Thank you for your help
I will provide the domain for future questions
I was looking at another question but I wasn't sure what to do for part b

Assuming that it is log base 10, as I believe that is how the Richter scale actually works;

$R=\log_{10}(\frac{I}{I_0})$
$5.4=\log_{10}(\frac{I}{I_0})$

This literally means that 10 to the power for 5.4 equals the value in the brackets. Therefore, we can rewrite it as;

$10^{5.4}=\frac{I}{I_0}$
$I=10^{5.4}I_0$

If it meant log base e (ie. natural logarithm), the 10 would be replaced with e.

anotherworld2b · « **Reply #2294 on:** July 03, 2017, 05:25:01 pm »

how do you know its 10 to the power of 5.4?

Quote from: jakesilove on July 03, 2017, 03:30:32 pm

Assuming that it is log base 10, as I believe that is how the Richter scale actually works;

$R=\log_{10}(\frac{I}{I_0})$
$5.4=\log_{10}(\frac{I}{I_0})$

This literally means that 10 to the power for 5.4 equals the value in the brackets. Therefore, we can rewrite it as;

$10^{5.4}=\frac{I}{I_0}$
$I=10^{5.4}I_0$

If it meant log base e (ie. natural logarithm), the 10 would be replaced with e.

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