Author Topic: Mathematics Question Thread (Read 1303402 times) Tweet Share

jamonwindeyer · « **Reply #45 on:** March 21, 2016, 09:39:11 pm »

Quote from: imtrying on March 21, 2016, 04:26:50 pm

Hey
This is a Geometrical Applications of Calculus question I'm stuck on (I've attached a photo of the question and answer). I know the quotient and product rules, but for some reason I keep ending up with some ridiculous answer. A step by step explanation would be hugely helpful

Sure thing!!

$y = \frac{5-x}{(4x^2+1)^3}$

The quotient rule says:

$y'=\frac{u'v-v'u}{v^2}$

$u = 5-x \quad\quad u'=-1 \quad\quad v=(4x^2+1)^3 \quad\quad v'=3(4x^2+1)^2\times8x = 24x(4x^2+1)^2$

Note that our differentiation of that last term is done with the chain rule!

So the first derivative by substitution is:

$y'=\frac{(-1)(4x^2+1)^3-24x(4x^2+1)^2(5-x)}{(4x^2+1)^6}$

Through cancellation and then expansion, we obtain:

$y'=\frac{(-1)(4x^2+1)-24x(5-x)}{(4x^2+1)^4} \\ =\frac{20x^2-120x-1}{(4x^2+1)^4}$

Now for the second derivative, it works the exact same way!

$y'=\frac{(20x^2-120x-1)}{(4x^2+1)^{-4}}$

$y''= \frac{a'b-b'a}{b^2}$

$a=(20x^2-120x-1) \quad\quad a'=(40x-120) \quad\quad b=(4x^2+1)^{4} \quad\quad b'=32x(4x^2+1)^{3}$

$\therefore y''=\frac{(40x-120)(4x^2+1)^{4}-32x(20x^2-120x-1)(4x^2+1)^{3}}{(4x^2+1)^8} \\ =\frac{(40x-120)(4x^2+1)-32x(20x^2-120x-1)}{(4x^2+1)^5}$

Which through expansion and factorisation will turn into your solution for Part B!

I hope this helps!

Marlo365 · « **Reply #46 on:** March 28, 2016, 06:57:44 pm »

Hey Jake, so i got my 2u half yearly tomrrow and not sure what i'm supposed to be doing this very moment wether its past papers of relaxing haha, any advice would be great

jakesilove · « **Reply #47 on:** March 28, 2016, 07:37:43 pm »

Quote from: Marlo365 on March 28, 2016, 06:57:44 pm

Hey Jake, so i got my 2u half yearly tomrrow and not sure what i'm supposed to be doing this very moment wether its past papers of relaxing haha, any advice would be great

Hey Marlo!

I think that it's totally up to you! Generally, I would recommend that you just keep doing past papers until a set time. Say you want to get 8 hours of sleep: figure out when you need to stop, and GO TO BED!

If you're feeling nervous, do a few past papers. If you find that this is making you more nervous, call it quits for the night! I really don't have any particular suggestion as to what you should do: just whatever makes you feel the most comfortable. If you're totally indifferent, maybe have a crack at one more paper (who know if it will help you tomorrow!) but definitely don't stay up all night.

Absolute best of luck, I'm sure you'll do great!

Jake

imtrying · « **Reply #48 on:** March 29, 2016, 06:11:02 pm »

Loan Repayments Question
The smith family buys a car for $38 000, paying a 10% deposit and taking out a loan for the balance. if the loan is over 5 years with interest of 1.5% monthly, find the amount of each monthly (a) loan repayment the total amount that the (b) smith family paid for the car.
Answer:
(a) $592.00 (b) $39 319.89

liiz · « **Reply #49 on:** March 29, 2016, 06:16:09 pm »

Hello, needing some help to integrate this: x / 2x² - 3
Thankyou!

jamonwindeyer · « **Reply #50 on:** March 29, 2016, 06:37:49 pm »

Quote from: liiz on March 29, 2016, 06:16:09 pm

Hello, needing some help to integrate this: x / 2x² - 3
Thankyou!

Hey Liiz! Sure thing!

Okay, so this question takes a bit of experience to spot the trick for. The integral of this will be a logarithm.

Remember, the derivative of the log of some function is:

$\ln{f(x)} = \frac{f'(x)}{f(x)}$

For your question, we almost have the derivative of the bottom on top of the fraction, which is what is required. To make it work, we multiply by 4 on the numerator, then multiply the integral by a quarter, thus balancing everything out. You could also think of it as taking out a factor of 1/4:

$\int\frac{x}{2x^2-3}dx = \frac{1}{4}\int\frac{4x}{2x^2-3}dx \\\\ =\frac{1}{4} \ln{(2x^2-3)} + C \\ \text{where C is a constant of integration}$

This isn't super easy to spot, it might take some practice to see it quickly! Hope this helps

jamonwindeyer · « **Reply #51 on:** March 29, 2016, 06:44:01 pm »

Quote from: imtrying on March 29, 2016, 06:11:02 pm

Loan Repayments Question
The smith family buys a car for $38 000, paying a 10% deposit and taking out a loan for the balance. if the loan is over 5 years with interest of 1.5% monthly, find the amount of each monthly (a) loan repayment the total amount that the (b) smith family paid for the car.
Answer:
(a) $592.00 (b) $39 319.89

I didn't forget you imtrying! I'm just not at my desk at the moment, I need to write out the series for your question (I think), so I'll tackle it tonight once I have some desk space! Unless someone else would like to have a crack

imtrying · « **Reply #52 on:** March 29, 2016, 06:58:30 pm »

Quote from: jamonwindeyer on March 29, 2016, 06:44:01 pm

I didn't forget you imtrying! I'm just not at my desk at the moment, I need to write out the series for your question (I think), so I'll tackle it tonight once I have some desk space! Unless someone else would like to have a crack

All good

RuiAce · « **Reply #53 on:** March 30, 2016, 11:18:21 am »

Quote from: imtrying on March 29, 2016, 06:11:02 pm

Loan Repayments Question
The smith family buys a car for $38 000, paying a 10% deposit and taking out a loan for the balance. if the loan is over 5 years with interest of 1.5% monthly, find the amount of each monthly (a) loan repayment the total amount that the (b) smith family paid for the car.
Answer:
(a) $592.00 (b) $39 319.89

$\text{Let the monthly repayment be equal to }M\text{ and observe that we are compounding interest monthly.}\\\text{This means we consider 60 periods.}\\\text{From experimentation, I have also determined that the interest rate they gave was per annum.}\\\text{Monthly interest rate}=0.015/12=0.00125$
$\begin{align*}{A}_{0}&=34200\\{A}_{1}&={A}_{0}\left(1.00125\right)-M\\&=34200\left(1.00125\right)-M\\{A}_{2}&={A}_{1}\left(1.00125\right)-M\\&=34200{\left(1.00125\right)}^{2}-M\left(1.00125\right)-M\\{A}_{3}&={A}_{2}\left(1.00125\right)-M\\&=34200{\left(1.00125\right)}^{3}-M{\left(1.00125\right)}^{2}-M\left(1.00125\right)-M\\&\vdots\\{A}_{n}&=34200{\left(1.00125\right)}^{n}-M{\left(1.00125\right)}^{n-1}-M{\left(1.00125\right)}^{n-2}-\dots-M\\&=34200{\left(1.00125\right)}^{n}-M\left(1+1.00125+{1.00125}^{2}+\dots+{1.00125}^{n-1}\right)\\&=34200{\left(1.00125\right)}^{n}-M\left(\frac{{1.00125}^{n}-1}{0.00125}\right)\end{align*}$

$\text{We want }{A}_{60}=0\\\begin{align*}\Rightarrow 0&=34200{\left(1.00125\right)}^{60}-M\left(\frac{{1.00125}^{60}-1}{0.00125}\right)\\\Leftrightarrow M&=\frac{34200{\left(1.00125\right)}^{60}}{\left(\frac{{1.00125}^{60}-1}{0.00125}\right)}\\\Leftrightarrow M&=591.9981715...\end{align*}\\\text{So the monthly repayment to the nearest cent is }\$592$

$\text{The total amount of payments is simply}\\\text{Amount paid in the deposit + Amount paid every month for ALL 60 months}\\=3800 + 60*591.9981715...\\=39319.89029...\\\text{Therefore, to the nearest cent, }\$39319.89$

liiz · « **Reply #54 on:** March 30, 2016, 04:59:15 pm »

Quote from: jamonwindeyer on March 29, 2016, 06:37:49 pm

Hey Liiz! Sure thing!

Okay, so this question takes a bit of experience to spot the trick for. The integral of this will be a logarithm.

Remember, the derivative of the log of some function is:

$\ln{f(x)} = \frac{f'(x)}{f(x)}$

For your question, we almost have the derivative of the bottom on top of the fraction, which is what is required. To make it work, we multiply by 4 on the numerator, then multiply the integral by a quarter, thus balancing everything out. You could also think of it as taking out a factor of 1/4:

$\int\frac{x}{2x^2-3}dx = \frac{1}{4}\int\frac{4x}{2x^2-3}dx \\\\ =\frac{1}{4} \ln{(2x^2-3)} + C \\ \text{where C is a constant of integration}$

This isn't super easy to spot, it might take some practice to see it quickly! Hope this helps

Ah okay, yep I understand! Thanks so much

SimonaB · « **Reply #55 on:** March 30, 2016, 09:54:08 pm »

I've got a question that was on my exam and I got it wrong after literally 10 attempts.
The question is: Find the area bounded by y=sin x and y=cos 2x for π/6 ≤ x ≤ 5π/6
Thanks!

jamonwindeyer · « **Reply #56 on:** March 30, 2016, 10:50:49 pm »

Quote from: SimonaB on March 30, 2016, 09:54:08 pm

I've got a question that was on my exam and I got it wrong after literally 10 attempts.
The question is: Find the area bounded by y=sin x and y=cos 2x for π/6 ≤ x ≤ 5π/6
Thanks!

Hey Simona! Welcome to the forums!!

Let's have a look.

The first step for an area question of this nature is to draw a quick sketch, just to get a better picture of what we are dealing with. Wolfram Alpha helped me out here, but you don't need any great level of accuracy, we just want the position of the curves with relation to each other. Note also that this sketch from Wolfram Alpha has the axes centred at the first point of intersection, totally irrelevant to this question, just ignore the coordinates!

Okay! So it is clear from a quick sketch that the sinx curve is above the cos2x curve in this domain.

There is a really neat trick here. Even when the curves go above or below the x axis, any weird combination, we can always find the area by finding the area under the upper curve and subtracting the area under the lower curve, in the given domain. It's nice, and eliminates the need for weird sums of absolute values of integrals. Literally, the integral of the upper curve subtract the integral of the lower curve, works every time

So, the working would be:

$A = \int^\frac{5\pi}{6}_\frac{\pi}{6}\sin{x}-\cos{2x}dx \\ = [-\cos{x}-\frac{1}{2}\sin{2x}]^\frac{5\pi}{6}_\frac{\pi}{6} \\ = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4} \\ =\frac{3\sqrt{3}}{2}u^2$

I hope this helps!!

SimonaB · « **Reply #57 on:** March 30, 2016, 11:12:37 pm »

Thanks heaps! I see my mistake now..

Maz · « **Reply #58 on:** April 03, 2016, 04:47:10 pm »

hey
can u please help me with how to differentiate this...i really don't know how to...
i'd really appreciate it

RuiAce · « **Reply #59 on:** April 03, 2016, 04:56:02 pm »

Quote from: mq123 on April 03, 2016, 04:47:10 pm

hey
can u please help me with how to differentiate this...i really don't know how to...
i'd really appreciate it

Can't see your image. If your integral was something along the lines of this then in NSW it belongs under MX2.

$\int_{1}^{1+{x}^{2}}{\frac{1}{\sqrt{1+{e}^{t}}}\,dt}$

Also, you said differentiate? Did you mean you wanted to put d/dx in front of that integral? (or d/dt inside it?)

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