Need help, URGENT maths question: (4 Viewers)

1008 · Mar 30, 2016

Hey guys, just had another question:

6) Write down the sets of points corresponding to the following:
a) A parallelogram with three vertices A(1,3,4,2), B(-2,1,0,5), C(-4,0,6,8). Assume B & C are adjacent to A
b) The triangle with three vertices in part a of this question.
c) All three parallelograms which have three vertices given in part a) of this question.

betelnut · Apr 2, 2016

1008 said:
Hey guys, just had another question:

6) Write down the sets of points corresponding to the following:
a) A parallelogram with three vertices A(1,3,4,2), B(-2,1,0,5), C(-4,0,6,8). Assume B & C are adjacent to A
b) The triangle with three vertices in part a of this question.
c) All three parallelograms which have three vertices given in part a) of this question.

a) I guess what you can do is consider 4 cases such that AB is parallel to and equal to CD, AB is parallel to and equal to DC, AD is parallel to and equal to BC and AD is parallel to and equal to CB. That'll give you all possibilities for D
b) Just find the equation of the intersection of the lines, AB, BC and CA
c) Same as a.

1008 · Apr 3, 2016

betelnut said:
a) I guess what you can do is consider 4 cases such that AB is parallel to and equal to CD, AB is parallel to and equal to DC, AD is parallel to and equal to BC and AD is parallel to and equal to CB. That'll give you all possibilities for D
b) Just find the equation of the intersection of the lines, AB, BC and CA
c) Same as a.

Thanks betelnut!

1008 · Apr 23, 2016

Hey everyone, I've got another question I'd like help with. It someone could help ASAP that would be great, as I have another Calculus test coming up next week.

$\text{Suppose that f is a continuous function in } \mathbb{R} \text{ and that}\lim_{x \rightarrow\infty}f(x) =\lim_{x \rightarrow-\infty}f(x) = 0 \\\text{a) Give an example of such a function which has both a maximum value and a minimum value.}\\\text{b) Given an example of such a function which has a minmum value but no maximum value.}\\\text{c) Show that if there is a real number } \epsilon\text{ such that }f(\epsilon)>0\text{ then f attains a maximum value on }\mathbb{R}.\\\text{[Note that a maximum-minimum theorem only applies to }finite\text{ close intervals [a,b].]}$

InteGrand · Apr 23, 2016

1008 said:
Hey everyone, I've got another question I'd like help with. It someone could help ASAP that would be great, as I have another Calculus test coming up next week.

$\rightarrow-\infty}f(x) = 0.\\\text{a) Give an example of such a function which has both a maximum value and a minimum value.}\\\text{b) Given an example of such a function which has a minmum value but no maximum value.}\\\text{c) Show that if there is a real number } \epsilon\text{ such that }f(\epsilon)>0\text{ then f attains a maximum value on }\mathbb{R}.\text{[Note that a maximum-minimum theorem only applies to }finite\text{ close intervals [a,b].}$

The top half of the question seems to be missing.

1008 · Apr 23, 2016

Sorry, my bad. It's fixed now!

leehuan · Apr 23, 2016

I swear I did this question before.

$\\$a) $f(x)=\frac{x}{1+{x}^{2}}\\$b) $f(x)=-e^{-{x}^{2}}\\ $c) Consider seperately the intervals $(-\infty, a], [a,b], [b,\infty)$

I don't remember the full proof to c) right now nor do I have it with me.

InteGrand · Apr 23, 2016

1008 said:
Hey everyone, I've got another question I'd like help with. It someone could help ASAP that would be great, as I have another Calculus test coming up next week.

$\text{Suppose that f is a continuous function in } \mathbb{R} \text{ and that}\lim_{x \rightarrow\infty}f(x) =\lim_{x \rightarrow-\infty}f(x) = 0 \\\text{a) Give an example of such a function which has both a maximum value and a minimum value.}\\\text{b) Given an example of such a function which has a minmum value but no maximum value.}\\\text{c) Show that if there is a real number } \epsilon\text{ such that }f(\epsilon)>0\text{ then f attains a maximum value on }\mathbb{R}.\\\text{[Note that a maximum-minimum theorem only applies to }finite\text{ close intervals [a,b].]}$

a) We can use a function whose graph dips below the x-axis and above it and attains max and min. like that, before tending to 0. You can use this idea to find the formula for such a function.

b) We can use a function that has a minimum below the x-axis but doesn't reach any max. E.g. f(x) = -1/(1+x^2).

Edit: done above.

InteGrand · Apr 23, 2016

I think the same Q as part c) was asked here and answered: http://community.boredofstudies.org...271/laters-maths-help-thread.html#post7133125.

1008 · Apr 23, 2016

Right, thanks. I've got another one.

$\text{Using the Mean Value Theorem show that}\\\frac{4}{27}<3-23^{\frac{1}{3}}<\frac{3}{16}$

leehuan · Apr 23, 2016

1008 said:
Right, thanks. I've got another one.

$\text{Using the Mean Value Theorem show that}\\\frac{4}{27}<3-23^{\frac{1}{3}}<\frac{3}{16}$

Try applying the mean value theorem on f(x) = x^1/3 over the interval [23, 27]

1008 · Apr 24, 2016

leehuan said:
Try applying the mean value theorem on f(x) = x^1/3 over the interval [23, 27]

Thanks. I had one of my friends help me out with this one as I was already doing what you suggested. But it got all weird on the Right hand side (3/16). But how would you do this question? Especially sorting the RHS out.

Also, I've got another question

$\text{The function }f:\mathbb{R}\rightarrow\mathbb{R}\text{ is defined by}\\f(x)=\begin{cases}\sqrt{x}\text{ sin}\sqrt{x} & x \geq 0\\ax+b & x < 0\end{cases}\\\text{where }a\text{ and }b\text{ are real numbers.}\\\text{ Find all values of }a \text{ and }b\text{ (if any) such that }f\text{ is differentiable at 0}$

I understand that for a function to be differentiable, it has to be continuous first, i.e.
$\sqrt{0}\text{ sin}\sqrt{0} = a(0)+b\\\therefore b= 0$

But the answer say that for function to be differentiable, a has to equal 1.

InteGrand · Apr 24, 2016

1008 said:
Thanks. I had one of my friends help me out with this one as I was already doing what you suggested. But it got all weird on the Right hand side (3/16). But how would you do this question? Especially sorting the RHS out.

Also, I've got another question

$\text{The function }f:\mathbb{R}\rightarrow\mathbb{R}\text{ is defined by}\\f(x)=\begin{cases}\sqrt{x}\text{ sin}\sqrt{x} & x \geq 0\\ax+b & x < 0\end{cases}\\\text{where }a\text{ and }b\text{ are real numbers.}\\\text{ Find all values of }a \text{ and }b\text{ (if any) such that }f\text{ is differentiable at 0}$

For the MVT one, I think the user laters asked the same Q. It was answered here: http://community.boredofstudies.org...15/mean-value-theorem-proofs.html#post7141393 .

1008 · Apr 24, 2016

InteGrand said:
For the MVT one, I think the user laters asked the same Q. It was answered here: http://community.boredofstudies.org...15/mean-value-theorem-proofs.html#post7141393 .

Thanks InteGrand! Any ideas on the other question?

InteGrand · Apr 24, 2016

1008 said:
Thanks. I had one of my friends help me out with this one as I was already doing what you suggested. But it got all weird on the Right hand side (3/16). But how would you do this question? Especially sorting the RHS out.

Also, I've got another question

$\text{The function }f:\mathbb{R}\rightarrow\mathbb{R}\text{ is defined by}\\f(x)=\begin{cases}\sqrt{x}\text{ sin}\sqrt{x} & x \geq 0\\ax+b & x < 0\end{cases}\\\text{where }a\text{ and }b\text{ are real numbers.}\\\text{ Find all values of }a \text{ and }b\text{ (if any) such that }f\text{ is differentiable at 0}$

1008 said:
Thanks InteGrand! Any ideas on the other question?

$\noindent For your latest Q., recall that a function can only be differentiable at a point if it is continuous there. Since $f(x)\to 0$ as $x \to 0^{+}$, we also need (and it is sufficient for) the left limit to be 0 for continuity. So we must have $b=0$ (so that the left limit is 0)$.$

$\noindent In terms of differentiability, recall that the continuous function $f$ will be differentiable at $0$ iff its left and right derivatives there exist and match. For the right derivative, note $f(x)=\sqrt{x}\sin \sqrt{x}$ for $x\geq 0$. So for positive $x$, $f^{\prime} (x)=\frac{1}{2\sqrt{x}}\sin \sqrt{x} + \frac{1}{2}\cos \sqrt{x}$. As $x\to 0^{+}$, we have $f^\prime (x)\to \frac{1}{2} +\frac{1}{2} = 1$, using standard limits. Since $f$ is continuous at 0, the right derivative is 1. So we need the left derivative to be 1 for differentiability. So we need $a=1$. With $a=1$, (and $b=0$) $f$ is differentiable at $0$.$

Semi-differentiability isn't really covered in the HSC, so here's the Wikipedia link to it for anyone curious: https://en.wikipedia.org/wiki/Semi-differentiability .

1008 · Apr 24, 2016

InteGrand said:
$\noindent For your latest Q., recall that a function can only be differentiable at a point if it is continuous there. Since $f(x)\to 0$ as $x \to 0^{+}$, we also need (and it is sufficient for) the left limit to be 0 for continuity. So we must have $b=0$ (so that the left limit is 0)$.$

$\noindent In terms of differentiability, recall that the continuous function $f$ will be differentiable at $0$ iff its left and right derivatives there exist and match. For the right derivative, note $f(x)=\sqrt{x}\sin \sqrt{x}$ for $x\geq 0$. So for positive $x$, $f^{\prime} (x)=\frac{1}{2\sqrt{x}}\sin \sqrt{x} + \frac{1}{2}\cos \sqrt{x}$. As $x\to 0^{+}$, we have $f^\prime (x)\to \frac{1}{2} +\frac{1}{2} = 1$, using standard limits. Since $f$ is continuous at 0, the right derivative is 1. So we need the left derivative to be 1 for differentiability. So we need $a=1$. With $a=1$, (and $b=0$) $f$ is differentiable at $0$.$

Semi-differentiability isn't really covered in the HSC, so here's the Wikipedia link to it for anyone curious: https://en.wikipedia.org/wiki/Semi-differentiability .

Thanks! Here's another one:

$\\\text{The function } f \text{ is differentiable at }a\text{. Find:}\\\\\lim_{h \rightarrow 0}\frac{f(a+ph)-f(a-ph)}{h}$

InteGrand · Apr 24, 2016

1008 said:
Thanks! Here's another one:

$$\noindent The function $f$ is differentiable at $a$. Find:\\\lim_{h \rightarrow 0}\frac{f(a+ph)-f(a-ph)}{h}$

$\noindent Since $f$ is differentiable at $a$, we have $f^{\prime} (a) = L$ for some real number $L$. We recall (see link below) using symmetric derivatives that $L = \lim _{\delta \to 0} \frac{f(a+\delta) -f(a-\delta)}{2\delta}$. Now, use the substitution $\delta = ph$, $p\neq 0$, $p$ real. Then $\frac{f(a+\delta)-f(a-\delta)}{2\delta}= \frac{f(a+ ph)-f(a-ph)}{2ph}$. This tends to $L$ as $ph\to 0$, i.e. as $h\to 0$. Hence using the algebra of limits, we have that $\frac{f(a+ph)-f(a-ph)}{h}$ tends to $2pL$ as $h\to 0$. So the answer is $2pL = 2p f^{\prime}(a)$.$

Symmetric derivatives: https://en.wikipedia.org/wiki/Symmetric_derivative .

leehuan · Apr 24, 2016

1008 said:
Thanks! Here's another one:

$\\\text{The function } f \text{ is differentiable at }a\text{. Find:}\\\\\lim_{h \rightarrow 0}\frac{f(a+ph)-f(a-ph)}{h}$

If you happen to be doing MATH1151 a simple solution is given at the back of the book for this question without knowledge of the concepts InteGrand provided

1008 · Apr 24, 2016

InteGrand said:
$\noindent Since $f$ is differentiable at $a$, we have $f^{\prime} (a) = L$ for some real number $L$. We recall (see link below) using symmetric derivatives that $L = \lim _{\delta \to 0} \frac{f(a+\delta) -f(a-\delta)}{2\delta}$. Now, use the substitution $\delta = ph$, $p\neq 0$, $p$ real. Then $\frac{f(a+\delta)-f(a-\delta)}{2\delta}= \frac{f(a+ ph)-f(a-ph)}{2ph}$. This tends to $L$ as $ph\to 0$, i.e. as $h\to 0$. Hence using the algebra of limits, we have that $\frac{f(a+ph)-f(a-ph)}{h}$ tends to $2pL$ as $h\to 0$. So the answer is $2pL = 2p f^{\prime}(a)$.$

Symmetric derivatives: https://en.wikipedia.org/wiki/Symmetric_derivative .

Thanks. There is this question that's been bugging me for a while now.
$\\\text{(a) A container in the shape of a right circular cone, of semi-vertical angle }\tan^{-1}{\frac{1}{2}}\text{,}\\\text{ is placed vertex downward with its axis vertical. Water is poured in at the rate of }10mm^{3}\text{ per}\\\text{sec. Find the rate at whice the depth, }h\text{ mm, is increasing when the depth of water in the }\\\text{cone is }50mm. \\\\\text{The cone is filled to a depth of 100mm and pouring is then stopped. A hole is then}\\\text{at the vertex of the cone and water flows out of the hole at the rate of }50\sqrt{h}\text{ }{mm}^{3}\\\text{ per second,where }h\text{ is the depth at time }t.\text{Show that it takes 200s to empty the cone.}$

I get part (a). The answer is $\frac{2}{125\pi}$ . But the answer to part (b) that I get is 166.67 seconds.

1008 · Apr 24, 2016

Hey leehuan. Yes, I am doing MATH1151. Do you mean the red booklet? I can't find the answers in the purple booklet anywhere....

Need help, URGENT maths question: (4 Viewers)

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