Author Topic: 3U Maths Question Thread (Read 1505007 times) Tweet Share

RuiAce · « **Reply #3420 on:** May 10, 2018, 10:11:26 pm »

Quote from: K98100 on May 10, 2018, 09:55:06 pm

Hey,
For what values is y=1/(x^2+2x-2) concave up and concave down?
Thanks

$\text{The relevant second derivative is}\\ \frac{d^2y}{dx^2} = \frac{6(x^2+2x+2)}{(x^2+2x-2)^3}$
$\text{The curve is concave up when }\frac{d^2y}{dx^2} > 0\\ \text{so we are considering }\frac{6(x^2+2x+2)}{(x^2+2x-2)^3} > 0.$
$\text{It can be shown that }6(x^2+2x+2)\text{ is a positive definite quadratic.}\\ \text{Therefore, we may divide it out without affecting the inequality:}\\ \frac{1}{(x^2+2x-2)^3} > 0.$
$\text{We can then use the square-of-the-denominator trick}\\ \text{and multiply by }(x^2+2x-2)^2\text{, twice, to obtain}\\ \boxed{x^2 + 2x - 2 > 0}.$
$\text{This is now just a quadratic inequality}\\ \text{whose solutions are }x< -1-\sqrt{3}, x >-1+\sqrt{3}.\\ \text{And following a similar process, the curve is concave down}\\ \text{when }-1 -\sqrt{3} < x < -1+\sqrt{3}$
This question does not assess anything conceptually hard, It is just ridiculously computationally demanding. If you have any problems with the computations you should post any working out you have achieved thus far.

Remark: An observation that can be made.

If you're thinking that there has to be a better way of solving this problem, there is. First note that the quadratic inequality is inevitable. The quadratic inequality can be solved by proving that the solutions to $x^2+2x-2 = 0$ are $ x = -1\pm \sqrt{3}$, via either the quadratic formula or completing the square. Then, sketch $ y = x^2+2x-2$, and simply read off the graph when $x^2+2x-2 > 0$.

However, the second derivative bit can all be skipped if you know what's going on. In reality, for this curve, it just so happens that
$\text{if }y > 0\text{, i.e. the curve is above the }x\text{-axis, then it is concave up.}$
(And similarly it's concave down when below the $x$-axis. This is somewhat a coincidence. This function was just really "well-behaved" in that regard.

Essentially, it's actually easier to do a sketch of $ y = \frac{1}{x^2+2x-2} $ by computing only the intercepts, asymptotes and stationary points. (ANYTHING is better than the second derivative.) We can infer from the sketch that the curve must be concave up when it's above the $x$-axis. This is because if it were concave down, it would not tend to the vertical and horizontal asymptotes. We need to ensure that our curve obeys the criteria we've found with the stationary points and asymptotes, but it just so turns out that that's all the information we need to deduce its concavity.
$\text{Thus we can jump straight to solving }\frac{1}{x^2+2x-2} > 0\\ \text{without being overly cautious with }\frac{6(x^2+2x+2)}{(x^2+2x-2)^3}>0$
Note that we can't always do this. A function like $ \frac{x}{x^2-1}$ wouldn't be so well behaved.

RuiAce · « **Reply #3421 on:** May 10, 2018, 10:12:51 pm »

Quote from: ravioli on May 10, 2018, 10:03:42 pm

how would I go about finding when a function has a horizontal asymptote?

so my questions asking me to find the values of a that cause f(x) to have a horizontal asymptote when f(x)=1/(x^2+2x+a), and I'm confused because I thought that it would just always have an asymptote at y=0?

Yeah. That definitely holds for any $a$.

K98100 · « **Reply #3422 on:** May 10, 2018, 10:19:26 pm »

Quote from: RuiAce on May 10, 2018, 10:11:26 pm

$\text{The relevant second derivative is}\\ \frac{d^2y}{dx^2} = \frac{6(x^2+2x+2)}{(x^2+2x-2)^3}$
$\text{The curve is concave up when }\frac{d^2y}{dx^2} > 0\\ \text{so we are considering }\frac{6(x^2+2x+2)}{(x^2+2x-2)^3} > 0.$
$\text{It can be shown that }6(x^2+2x+2)\text{ is a positive definite quadratic.}\\ \text{Therefore, we may divide it out without affecting the inequality:}\\ \frac{1}{(x^2+2x-2)^3} > 0.$
$\text{We can then use the square-of-the-denominator trick}\\ \text{and multiply by }(x^2+2x-2)^2\text{, twice, to obtain}\\ \boxed{x^2 + 2x - 2 > 0}.$
$\text{This is now just a quadratic inequality}\\ \text{whose solutions are }x< -1-\sqrt{3}, x >-1+\sqrt{3}.\\ \text{And following a similar process, the curve is concave down}\\ \text{when }-1 -\sqrt{3} < x < -1+\sqrt{3}$
This question does not assess anything conceptually hard, It is just ridiculously computationally demanding. If you have any problems with the computations you should post any working out you have achieved thus far.

Remark: An observation that can be made.
If you're thinking that there has to be a better way of solving this problem, there is. First note that the quadratic inequality is inevitable. The quadratic inequality can be solved by proving that the solutions to $x^2+2x-2 = 0$ are $ x = -1\pm \sqrt{3}$, via either the quadratic formula or completing the square. Then, sketch $ y = x^2+2x-2$, and simply read off the graph when $x^2+2x-2 > 0$.

However, the second derivative bit can all be skipped if you know what's going on. In reality, for this curve, it just so happens that
$\text{if }y > 0\text{, i.e. the curve is above the }x\text{-axis, then it is concave up.}$
(And similarly it's concave down when below the $x$-axis. This is somewhat a coincidence. This function was just really "well-behaved" in that regard.

Essentially, it's actually easier to do a sketch of $ y = \frac{1}{x^2+2x-2} $ by computing only the intercepts, asymptotes and stationary points. (ANYTHING is better than the second derivative.) We can infer from the sketch that the curve must be concave up when it's above the $x$-axis. This is because if it were concave down, it would not tend to the vertical and horizontal asymptotes. We need to ensure that our curve obeys the criteria we've found with the stationary points and asymptotes, but it just so turns out that that's all the information we need to deduce its concavity.
$\text{Thus we can jump straight to solving }\frac{1}{x^2+2x-2} > 0\\ \text{without being overly cautious with }\frac{6(x^2+2x+2)}{(x^2+2x-2)^3}>0$
Note that we can't always do this. A function like $ \frac{x}{x^2-1}$ wouldn't be so well behaved.

Thanks for the reply!
So the graph looks like this. Would f(x) on both extremes of the graph be considered concave up since it would eventually get very close to y=0

RuiAce · « **Reply #3423 on:** May 11, 2018, 12:18:20 am »

Quote from: K98100 on May 10, 2018, 10:19:26 pm

Thanks for the reply!
So the graph looks like this. Would f(x) on both extremes of the graph be considered concave up since it would eventually get very close to y=0

Yeah, $f(x)$ is still concave up as $x\to \infty$ and as $x \to -\infty$. That's still implied by the solutions $ x< -1-\sqrt{3}$, $x > -1+\sqrt{3}$.

(By itself, the fact that it's approaching $y=0$ actually doesn't mean much. It just means a lot when you consider that along with the other facts about the graph.)

mirakhiralla · « **Reply #3424 on:** May 11, 2018, 06:26:10 pm »

Hi (:,
I'm doing a rate of change question:

A vessel, containing water, has the shape of an inverted right circular cone with base 2m in radius and height 5m. Water flows from the bottom of the bottom of the cone at a constant rate of 0.35m^3/min.

a) the water forms the shape of a smaller cone. Find a relationship between the radius and height of the cone formed by water at any t minutes.
I got this: r=2h/5

b) At what rate is the water level falling when the water level in the vessel is 2.5m?
I couldn't do this one right.

RuiAce · « **Reply #3425 on:** May 11, 2018, 07:04:05 pm »

Quote from: mirakhiralla on May 11, 2018, 06:26:10 pm

Hi (:,
I'm doing a rate of change question:

A vessel, containing water, has the shape of an inverted right circular cone with base 2m in radius and height 5m. Water flows from the bottom of the bottom of the cone at a constant rate of 0.35m^3/min.

a) the water forms the shape of a smaller cone. Find a relationship between the radius and height of the cone formed by water at any t minutes.
I got this: r=2h/5

b) At what rate is the water level falling when the water level in the vessel is 2.5m?
I couldn't do this one right.

$\text{We know that the volume of the cone is }V = \frac13 \pi r^2 h\\ \text{so subbing in what you've found, }\boxed{V = \frac{4}{75} \pi h^3}.$
$\text{Therefore }\frac{dV}{dh} = \frac{4}{25} \pi h^2\\ \text{and we're told that }\frac{dV}{dt} = -0.35$
$\text{The chain rule states that }\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}\\ \text{so }\frac{dh}{dt} = \frac{25}{4\pi h^2}\times -0.35\\ \text{Then sub in }h=2.5$

tatkh · « **Reply #3426 on:** May 13, 2018, 02:12:57 pm »

Hello,
I have a rates involving two or more variables question attatched.

I can't seem to get the answer for this.

Thank you

RuiAce · « **Reply #3427 on:** May 13, 2018, 02:30:32 pm »

Quote from: tatkh on May 13, 2018, 02:12:57 pm

Hello,
I have a rates involving two or more variables question attatched.

I can't seem to get the answer for this.

Thank you

$\text{We leave for later that }\boxed{\frac{dV}{dt} = 23}$
$\text{Let the side of the cube be }x.\\ \text{Then it has volume }V = x^3\text{ and surface area }S = 6x^2.$
$\text{This implies that }\boxed{\frac{dV}{dx} = 3x^2}\text{ and }\boxed{\frac{dS}{dx} = 12x}.\\ \text{So by using the chain rule, twice, we have}\\ \begin{align*}\frac{dS}{dt} &= \frac{dS}{dx}\, \frac{dx}{dV}\, \frac{dV}{dt}\\ &= 12x \times \frac{1}{3x^2}\times 23.\end{align*}$
Then sub $x=140$ in.

beeangkah · « **Reply #3428 on:** May 14, 2018, 09:44:17 am »

Could I please get some help with part b?

RuiAce · « **Reply #3429 on:** May 14, 2018, 10:21:15 am »

Quote from: beeangkah on May 14, 2018, 09:44:17 am

Could I please get some help with part b?

$\text{Keep in mind we're rotating about the }\textbf{y}\text{-axis}.$
\begin{align*}y&= \frac{1}{\sqrt{1-x^2}}\\ y^2 &= \frac{1}{1-x^2}\\ 1-x^2 &= \frac{1}{y^2}\\ x^2 & = 1 - \frac{1}{y^2}\end{align*}
$\text{Note that when }x=0, y = 1\\ \text{and when }x = \frac12, y = \frac{2}{\sqrt3}$

$\text{Looking at the graph,}\\ \begin{align*}V&= \text{Volume of cylinder} - \text{Volume of the little slice}\\ &= \pi \left(\frac12\right)^2 \left( \frac{2}{\sqrt3}\right) - \int_1^{2/\sqrt3} \left(1 - \frac1{y^2}\right)\end{align*}$
(Note: This is of course, far beyond 3U. The graph of $ y = \frac{1}{\sqrt{1-x^2}}$ is a 4U exercise.

kaustubh.patel · « **Reply #3430 on:** May 14, 2018, 08:31:39 pm »

Hey guys need a bit of help with 3U trig-induction question

RuiAce · « **Reply #3431 on:** May 14, 2018, 09:10:04 pm »

Quote from: kaustubh.patel on May 14, 2018, 08:31:39 pm

Hey guys need a bit of help with 3U trig-induction question

$\text{Note that the formula you found in part a) rearranges to}\\ -\cos (A + 2B) = 2 \sin (A+B) \sin B - \cos A.\\ \text{Perform the substitutions: }A = 2k\theta, B = \theta\\ \text{Then we have }\boxed{-\cos (2k\theta + 2\theta) = 2 \sin (2k\theta+\theta) \sin \theta - \cos 2k\theta}$
__________________________________________________________________
$\text{When }n=1,\\ \begin{align*}\frac{1-\cos 2\theta}{2\sin \theta} &= \frac{2\sin^2\theta}{2\sin\theta}\\ &= \sin\theta\end{align*}\\ \text{as required.}$
$\text{Suppose for the sake of induction}\\ \boxed{\sin \theta + \sin 3\theta + \sin 5\theta + \dots + \sin \left[ (2k-1)\theta\right] = \frac{1 - \cos 2k\theta}{2\sin \theta}}.$
$\text{Then,}\\ \begin{align*}&\quad\sin \theta + \sin 3\theta + \sin 5\theta + \dots + \sin \left[ (2k-1)\theta\right]+\sin \left[(2k+1)\theta\right]\\ &= \frac{1-\cos 2k\theta}{2\sin\theta} \tag{assumption} + \sin(2k\theta+\theta)\\&=\frac{1-\cos 2k\theta + 2\sin(2k\theta+\theta)\sin\theta}{2\sin\theta}\\ &=\frac{1 - \cos (2k\theta +2\theta)}{2\sin\theta}\tag{from above}\\ &=\frac{1-\cos [2(k+1)\theta]}{2\sin\theta} \end{align*}\\ \text{as required}$

owidjaja · « **Reply #3432 on:** May 14, 2018, 10:56:22 pm »

Hey guys,
This is probably a random question but how does mathematical induction apply to real-life situations? Is there a reason why the working out is so long?

This is probably me running on frustration over induction proofs.

RuiAce · « **Reply #3433 on:** May 14, 2018, 11:34:18 pm »

Quote from: owidjaja on May 14, 2018, 10:56:22 pm

Hey guys,
This is probably a random question but how does mathematical induction apply to real-life situations? Is there a reason why the working out is so long?

This is probably me running on frustration over induction proofs.

The examples given in high school math mostly exist to illustrate a point with induction, and is really just long to write. The first step up is its applications in quite bizarre university-level mathematical proofs. It is favoured for proofs that involve only the positive integers numbers (1, 2, 3, 4, ...) in that it uses the technique of 'recursion'.

The main idea behind an induction is to say that "suppose everything in the past has already happened". Can you then force the same thing to happen again in the "present", and as a consequence make it happen again in the "future"?

(In fact, in 4U many students will realise that induction allows you to prove a formula for the Fibonacci numbers. All we need to do is show that we have a base case, and then we can keep building upon it.)

The standard real-life example is this one. Suppose you have a chain of dominoes. You first need to check that the first domino will fall.
Now suppose that the $k$-th domino falls. What you're really doing in the inductive step is then asking, does the next domino fall as well? (i.e. the $(k+1)$-th domino.)

If it does (i.e. TRUE), then you instantly have a pattern. The fall of the first domino implies the fall of the second domino. The fall of the second domino implies the fall of the third domino. Which implies the fall of the fourth, and so on, and that's how we know that we have a whole sequence of dominoes falling.

For a story-like one, you could consider this. The first person in line at a new food outlet orders a certain type of food. The person behind overhears it, and they go for it as well. And so does the next one. And the next one. AND the NEXT one................

Another story-like one: A series of really good books are being published (say, one every year). A bookworm, who happens to love the genre, decides to read the first book. If they like the first book, then they'll read the second (well, assuming that the bookworm has a library or just enough money). If they like the second, then they'll read the third. And so on.

A fancier practical use involves using it to prove Dijkstra's algorithm, which several computer scientists use a lot these days to find shortest paths between two locations.

TLDR it's useful, you just don't see why in high school. As far as I know, the only 3U induction that has ever interested me was this one, and it no longer interests me anyway: $ 1+2+\dots+n^2 = \frac16n(n+1)(2n+1)$.

stels · « **Reply #3434 on:** May 17, 2018, 08:22:03 pm »

ABC is an acute-angled triangle. Squares BAHK and CAXY are constructed on AB and AC respectively and outside the triangle. The diagonals HB and XC meet when produced at P. Prove that angle BPC + angle BAC= 90

(Yes, this question has been asked on this forum by someone else but I wasn't able to access the solution).

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