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$\mathbb Q$ is a field. ${X_1}^2-2$ is a prime element in $\mathbb Q\left[ X_1\right]$ which is a p.i.d so ${X_1}^2-2$ generates a maximal ideal. The converse of this statement is false for the field $\mathbb Q$ or any field that is not algebraically closed. If $K$ is algebraically closed, both the statement of the question and its converse are corollaries of the Hilbert Nullstellensatz. This basic result in algebraic geometry can be found in texts on algebraic geometry, for example Eisenbud's Commutative Algebra with a view toward algebraic geometry, Springer Graduate Texts in Math, vol 150, pp 34--35.
• Dear Barbara, The comment about the converse was made in a comment, and you can correct it by making a comment in reply (as I did above). Regards, – Matt E Jan 24 '13 at 5:27
• Hi Barbara, I am not sure how it happened, but your account split, despite you logging into the system using the same credentials. If you again find yourself unable to post comments or edit your own posts without review, please let a moderator know so we can track down the problem. I've merged your two accounts now. – Willie Wong Jan 24 '13 at 15:59
• Thank you very much for your help. Things seem to be working fine now, and I am commenting appropriately now. – Barbara Osofsky Jan 25 '13 at 19:19
I think this answer might essentially be the same as @orangeskid's answer, but I'm not sure, so I figured I will post it here. If it is the same or incorrect, please let me know and I will delete it.
Let $I = \langle x_1 - a_1, \dots, x_n - a_n \rangle \subset k[x_1, \dots, x_n]$ for $k$ a field. Consider any ideal $I \subsetneq J$, by the Hilbert Basis theorem $J$ is finitely generated, so we can say that $J=\langle x_1 - a_1, \dots, x_n -a_n, f_1, \dots, f_r\rangle$ for finitely many $1 \le i \le r < \infty$, with $f_i \not\in I, f_i \in k[x_1, \dots, x_n]$ and without loss of generality, all of the $f_i$ are not identically equal to zero.
Case 1: At least one of the $f_i$ is constant. | {
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Case 1: At least one of the $f_i$ is constant.
Without loss of generality, $f_r$ is constant, say $f_r \equiv c \in k, c\not=0$ so $J= \langle x_1 - a_1, \dots, c \rangle = \langle x_1 - a_1, \dots, 1 \rangle = k[x_1, \dots, x_n]$.
Case 2: None of the $f_i$ are constant.
Let's choose one of the $f_i$, which by an abuse of notation we will denote simply by $f_i(x_1, \dots, x_n)$.
Then we have that $$f_i(x_1, \dots, x_n) = f_i^m(x_1, \dots, x_n) + f_i^{m-1}(x_1 -a_1,\dots, x_n - a_n)+ \dots + f_i^1(x_1 -a_1, \dots, x_n -a_n)+ f_i(x_1-a_i,\dots,x_n-a_n)$$ where for all $1 < j \le m, \quad f_i^j(x_1, \dots, x_n) := f_i^{j-1}(x_1,\dots,x_n)-f_i^{j-1}(x_1 - a_1, \dots, x_n-a_n)$, and additionally $f_i^1(x_1, \dots, x_n):=f(x_1,\dots,x_n)-f(x_1-a_1, \dots, x_n-a_n)$, and finally that $m$ is the lowest natural number such that $f_i^m(x_1, \dots, x_n)$ is identically constant, thus for any $k >m$, $f_i^k(x_1, \dots, x_n)\equiv 0$. Such an $m$ is guaranteed to exist, because by construction, $\deg(f_i^1) < \deg(f_i)$ and $\deg(f_i^j) < \deg(f_i^{j-1})$ for all $2 \le j \le m$.
Define $c$ to be the non-zero constant such that $f_i^m(x_1, \dots, x_n) \equiv c$. Now clearly we have that $$f_i(x_1 -a _1, \dots, x_n -a_n) \in \langle x_1 -a_1, \dots, x_n -a_n \rangle \\ f_i^j(x_1-a_1, \dots, x_n -a_n) \in \langle x_1 - a_1, \dots, x_n - a_n \rangle \quad \forall\ 1\le j \le m-1$$ Therefore we have shown that $f_i(x_1, \dots, x_n) = c + g_i (x_1, \dots, x_n)$ for some $g_i \in \langle x_1 - a_1 , \dots, x_n - a_n \rangle$. Therefore we have that $$J = \langle x_1 - a_1, \dots, x_n - a_n, \dots, f_i, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, c + g_i, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, c, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, 1, \dots, f_r \rangle = k[x_1, \dots, x_n]$$ | {
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Thus, in both cases, $I \subsetneq J \implies J = k[x_1, \dots, x_n]$, so therefore $I$ must be maximal, since clearly $I=\langle x_1 - a_1, \dots, x_n - a_n\rangle$ is proper.
• Since the ring homomorphism $k[x_1,\dots,x_n]\to k[x_1,\dots,x_n]$ such that $x_i\mapsto x_i-a_i$ is an isomorphism, it is not restrictive to assume $a_1=a_2=\dots=a_n=0$. Thus a polynomial $f\notin I$ must have a nonzero constant term and so an ideal properly containing $I$ is the whole ring. This is what your long proof boils down to. And DonAntonio's proof is the abstract (and simpler) version of it. – egreg Oct 11 '16 at 13:04
• @egreg Thank you for reading my proof. That $x_i \mapsto x_i - a_i$ is an isomorphism (of commutative rings? I'm not sure which type of isomorphism you mean) seems similar to the intuitive idea I had originally involving affine changes of coordinates "not changing anything important". You are probably right that it is equivalent to DonAntonio's proof as well; admittedly I am not comfortable enough with the material to understand the argument why without thinking about it for a long time. I agree that the elegance and simplicity of DonAntonio's proof is preferable. – Chill2Macht Oct 11 '16 at 13:20
• A ring homomorphism $k[x_1,\dots,x_n]\to A$ is determined as soon as you have a ring homomorphism $k\to A$ and assign images to $x_1,\dots,x_n$. In this case $A=k[x_1,\dots,x_n]$ the homomorphism $k\to A$ is the canonical embedding. With $x_i\mapsto x_i-a_i$ you get an isomorphism, because the inverse is the one doing $x_i\mapsto x_i+a_i$. – egreg Oct 11 '16 at 13:41 | {
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# How do you determine the local extrema points for $y=\sqrt{3}\cos(3x)+\sin(3x)$
$$y=\sqrt{3}\cos(3x)+\sin(3x); 0\le{x}\le{\frac{2\pi}{3}}$$
I know that the local extrema can be determined by using the first derivative test. I took the derivative of $y$ and got $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x)$$ I then solved the derivative for when it's value is $0$ and got $x=\dfrac{\pi}{2}$ . I then used this critical point and subdivided the interval. I found that there were 3 points of local extrema after doing all my work, local maximum $x=0, \dfrac{2\pi}{3}$ and local minimum $x=\dfrac{\pi}{2}$. However according to the online homework, there were 4 different points of extrema. Local maximum $x=\dfrac{\pi}{18},\dfrac{2\pi}{3}$ and local minimum $x=0,\dfrac{7\pi}{18}$. I am really confused as to how there are 4 points of local extrema, did I leave out an answer somewhere? I am also confused as to how they got $x=\dfrac{\pi}{18},\dfrac{7\pi}{18}$ as points of local extrema. Could someone explain this to me?
• At $x = \pi/2, y' \neq 0$... $y'(\pi/2) = 3\sqrt{3}$ – Namaste Apr 29 '13 at 3:44
As I noted in my comment: At $x = \pi/2, \quad y' \neq 0;$ ... $y'(\pi/2) = 3\sqrt{3}$
You need to solve for $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x) = 0 \quad \iff \quad 3\sqrt{3}\sin(3x) = 3\cos(3x)$$ $$\iff \quad \sqrt 3 \sin(3x) = \cos(3x),\quad x\in \left[0, \frac{2\pi}{3}\right]$$
Note that $$\sqrt 3 \sin(3x) = \cos(3x) \iff \sqrt 3 \dfrac{\sin(3x)}{\cos(3x)} = \sqrt 3 \tan(3x) = 1\iff \tan(3x) = \frac{1}{\sqrt 3}$$
Solving for $x$ will give you 2 potential critical points on your interval; then recall that you need to also check endpoints of an interval as potential extrema. | {
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• How do you get 3 potential critical points when solving for $x$? I only get one point which is $\frac{\pi}{2}$. – Kot Apr 29 '13 at 4:24
• Did you read the first sentence, or my comment? $y'(x)\neq 0$ when $x = \pi/2$. I say three potential critical points, but one will outside your interval...Follow what I've done to solve $y' = 0$. Then: Solve for $x$ given $3x = \tan^{-1}(1/\sqrt 3)$...then find which value is outside of $x \in [0, 2\pi/3]$. You will be left with two solutions to x: one will be a maximum, and one will be a minimum. The other max is the endpoint of the interval on which $x$ is defined: $2 \pi/3$, and the other minimum at the endpoint $x = 0$ – Namaste Apr 29 '13 at 4:28
• I see what I did wrong, I multiplied by $3$ instead of dividing by $3$. I now have $x=\frac{\pi}{18}$. I do not understand where the two other points would come from. Does the inverse tangent function give multiple answers? – Kot Apr 29 '13 at 4:31
• Yes, the other critical point in your interval is at $\pi/18 + \pi/3 = 7\pi/18$. The calculator typically only gives one value...but you can test to confirm that $x = 7\pi/18$ also makes $y' = 0$ – Namaste Apr 29 '13 at 4:33
• I am a little confused as to why you added $\pi/3$ to the first answer. If it was tangent don't you add $\pi$ to get the other value? – Kot Apr 29 '13 at 4:37
$y=\sqrt{3}\cos(3x)+\sin(3x)=2(\dfrac{\sqrt{3}}{2}\cos(3x)+\dfrac{1}{2}\sin(3x)=2(\sin\dfrac{\pi}{3}\cos(3x)+cos\dfrac{\pi}{3}\sin(3x))=2\sin(3x+\dfrac{\pi}{3})$
$\dfrac{\pi}{3} \leq 3x+\dfrac{\pi}{3} \leq 2\pi+\dfrac{\pi}{3}$, so there is 2 peaks when $3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}$ or $\dfrac{3 \pi}{2}$, another 2 are the end points ie:$x=0$ or $x=\dfrac{3 \pi}{2}$
• An excellent proof showing calculus is unnecessary. – Stefan Smith Apr 29 '13 at 15:29
$y=\sqrt 3\cos(3x)+\sin(3x)=2\sin(3x+\frac\pi3)$
$y'=2\cdot3\cos(3x+\frac\pi3)$
For the extreme values of $y, y'=0\implies \cos(3x+\frac\pi3)=0$ | {
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$y'=2\cdot3\cos(3x+\frac\pi3)$
For the extreme values of $y, y'=0\implies \cos(3x+\frac\pi3)=0$
$\implies 3x+\frac\pi3=(2n+1)\frac\pi2$ whether $n$ is any integer
$\implies 3x =(6n+1)\frac\pi6$
As $0\le x\le\frac{2\pi}3, 0\le 3x\le 2\pi$
$\implies 0\le (6n+1)\frac\pi6\le 2\pi \implies0\le 6n+1\le 12\implies n=0,1$ | {
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# How to maximise this function: $p \log (1 + x) + q \log (1 − x)$?
Hint: Use the fact that $p \log (1 + x) + q \log (1 − x)$ is maximized when $x = p − q$.
Secondly it was written: Let $f : (-1,1) \rightarrow \mathbb{R} : x \mapsto p\log (1+x) + q \log (1-x).$ Then using the hint above, $f$ is maximised at $p-q$ and $f(p-q) = \log 2 + p\log p + q\log q.$
My first question is, how (possibly using analysis/calculus?) would you deduce/get that $p \log (1 + x) + q \log (1 − x)$ is maximized when $x = p − q?$
The other thing, I have trouble understanding what exactly does $f : (-1,1)$ mean? And how would I intepret/read the following line? Its a bit new to me as I have not seen functions written like the following before. $f : (-1,1) \rightarrow \mathbb{R} : x \mapsto p\log (1+x) + q \log (1-x).$
Lastly, it says that $f$ is maximised at $p-q$ and $f(p-q) = \log 2 + p\log p + q\log q.$ How do you get that $f(p-q) = \log 2 + p\log p + q\log q?$ I tried substituting it into $f(x)=p\log (1+x) + q \log (1-x)$, getting $f(p-q) = p\log(1+p-q) + q\log(1-p+q).$ How would I go about from there? | {
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It is rather bad form to remove questions from your post when people have already addressed them in their answer; it makes it look like they don't know what they are talking about to new readers. It's worse if you then try to edit people's answers to remove content because of the content you decided to remove from your own post. And to try to do so 12 days after the fact on top of that... – Arturo Magidin Apr 2 '12 at 1:42
Thanks, I already flagged this post stating the (valid) reason, why this question should be deleted. Appreciate if you could take a look. – Heijden Apr 2 '12 at 2:00
I can't think of a reason why a question posted 12 days ago, with an accepted answer, should be deleted; unless, of course, you are somehow trying to cover your tracks for some reason... in which case it should definitely not be deleted. – Arturo Magidin Apr 2 '12 at 2:06
@Arturo, I have rolled it back to a more suitable version. – user21436 Apr 2 '12 at 2:16
@ArturoMagidin, I flagged this post to a moderator stating my reason to remove this question, and I am not covering my tracks, in my (original) post I am just asking for explanations to a few things, not trying to get anyone to solve the question for me, hence there is no need to 'cover my tracks'. I gave the reason to the moderator when i flagged it. – Heijden Apr 2 '12 at 2:20
For your first question, you can see that $p \log (1 + x) + q \log (1 − x)$ is maximized when $x=p-q$ by taking the derivative at setting it equal to $0$. We have $$0=\frac{d}{dx}(p \log (1 + x) + q \log (1 − x))=-\frac{p}{1+x}+\frac{q}{1-x}=\frac{-p(1-x)+q(1+x)}{(x+1)(x-1)}$$ and so $(q-p)+(p+q)x=0$ hence $x=\frac{p-q}{p+q}$, and since $q=1-p$ this simplifies to $x=p-q$. This tells us that $p \log (1 + x) + q \log (1 − x)$ has an extremum at $x=p-q$, and this extremum must be the maximum as making $x$ near $1$ or $-1$ makes $\log(1-x)$ or $\log(1+x)$ very negative, respectively. | {
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For your second question, "$f: (-1,1)$" doesn't mean anything. However, the full statement "$f:(-1,1)\to \mathbb R$" means "$f$ is a function from $(-1,1)$ to $\mathbb R$ (the real numbers)". In your case, the statement $$f : (-1,1) \rightarrow \mathbb{R} : x \mapsto p\log (1+x) + q \log (1-x)$$ means "$f$ is a function from $(-1,1)$ to $\mathbb R$ such that $f(x)=p\log (1+x) + q \log (1-x)$ for any $x\in (-1,1)$ (any $x$ between $-1$ and $1$)".
Edit: To get that $f(p-q)=\log 2+p\log p+q\log q$, use the fact that $p=1-q$ and $q=1-p$ so $$\begin{eqnarray} f(p-q)&=&p\log(1+p-q) + q\log(1-p+q)\\ &=&p\log(1+p-(1-p)) + q\log(1-(1-q)+q)\\ &=&p\log(2p) + q\log(2q)\\ &=&p(\log 2+\log p) + q(\log 2+\log q)\\ &=&(p+q)\log 2+p\log p + q\log q\\ &=&\log 2+p\log p+q\log q. \end{eqnarray}$$
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Why the downvote? – Alex Becker Apr 2 '12 at 2:00
$f: (-1,1) \to \mathbb R$ just means that $f$ is a function defined on the interval $(-1,1)$ with values in the real line.
To find critical points of $f(x) = p \log(1+x) + q \log(1-x)$, solve $f'(x)=0$ for $x$. You should find exactly one critical point, at $x=p-q$ (if you remember that $p+q=1$). Note that this is a local maximum (e.g. by using the second derivative test, or noting that $\log(1+x)$ and $\log(1-x)$ are concave functions). If a differentiable function has only one critical point in an interval and it is a local maximum, then it is a global maximum on that interval. | {
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# Does it mean that there no analytical solution if Mathematica cannot find analytical solution?
I'm solving simple but coupled ODEs recently. I use both MATLAB symbolic computation and Mathematica.
For example, my coupled ODE is the following \begin{align*} &\dot{x}(t)=y(t)-\rho b\frac{x(t)}{1-\rho(1-e^{-t})}\\ &\dot{y}(t)=-y(t)+\rho b\frac{x(t)}{1-\rho(1-e^{-t})}\\ \end{align*} where $\rho\in(0,1)$ and $b\in(0,1)$ are given constants and the initial value of this set of ODEs are $x(0)=a$, there $a\in(0,1)$ and $y(0)=0$.
This expression looks simple but these 2 equations are coupled.
First, I tried MATLAB, it generate "Warning: Explicit solution could not be found." explicitly.
Then I tried Mathematica,
system = {x'[t] == y[t] - c1*c2*x[t]/(1 - c1*(1 - Exp[-t])),
y'[t] == -y[t] + c1*c2*x[t]/(1 - c1*(1 - Exp[-t]))};
Then I try to solve it via sol = DSolve[system, {x, y}, t].
The thing that I don't understand is that after I press Shift+Enter, Mathematica only makes my input look nicer, but didn't produce any result or generating any warning message like MATLAB. So I couldn't tell whether it is because Mathematica also couldn't find the analytical solution like MATLAB, or it just does not even try to solve the problem since I input something wrong?
This simple coupled ODE drives me crazy these days. Any suggestion, input is deeply appreciated.
If math software couldn't find analytical solution, then is it still possible to analyze the monotonicity of the solution? For example, in this case, it's easy to analyze the case when $t=\infty$ by setting $\dot{x}=0=\dot{y}$ and solve the equations. I can also numerically plot the solution to see the trend, e.g., $x(t)$ is decreasing. But without the explicit functional form of $x(t)$, how can we prove the monotonicity, stuff like that? In general, if the math software fails to find analytical solution, what should we do next? | {
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• The answer to "Does it mean that there no anlytical solution if Mathematica cannot find analytical solution?" is definitely no (i.e. there are definitely analytic solutions that Mathematica can't find for all kinds of things). However, those differential equations seem complicated enough that there could easily not be an analytic solution. – march Mar 25 '16 at 16:32
• Even if you are unable to derive a closed form solution to your DE, it might still be possible to study the qualitative behavior of the solutions. Have a look at Bender/Orszag for ideas. – J. M. is away Mar 25 '16 at 16:35
• @J.M. Is that book "Advanced Mathematical Methods for Scientists and Engineer"? – KevinKim Mar 25 '16 at 16:38
• It's interesting to note that $\dot{x}(t) + \dot{y}(t) = 0$. – rcollyer Mar 25 '16 at 16:43
• Well, it has the form $\dot{\vec{x}}(t) = \mathbf{A}(t) \vec{x}(t)$, so you could try diagonalizing $\mathbf{A}$ to decouple the two equations. – rcollyer Mar 25 '16 at 17:00
Here, we take advantage of the observation mentioned in a comment that the sum of x and y is conserved. If we define
s[t] == x[t] + y[t]
d[t] == x[t] - y[t]
eqns = {s'[t] == 0,
d'[t] == -(1 + (r b)/(1 - (1 - E^-t) r)) d[t] + (1 - (r b)/(1 - (1 - E^-t) r)) s[t]}
Mathematica knows how to solve this:
First@DSolve[eqns, {s[t], d[t]}, t]
As noted in the comments above, you have $\dot{x} + \dot{y} = 0$, which implies that $x(t) + y(t)$ is a constant; call it $c_3$. (Note that in particular, $c_3 = x(0) + y(0)$.) You can therefore replace y[t] with c3 - x[t] in your first equation above, and Mathematica can solve that explicitly:
reducedsystem = {x'[t] == c3 - x[t] - c1*c2*x[t]/(1 - c1*(1 - Exp[-t]))}
DSolve[reducedsystem, x[t], t]
(* {{x[t] -> -((c3 E^(-t + (c1 c2 Log[-c1 - E^t + c1 E^t])/(-1 + c1)) (-E^t + c1 (-1 + E^t))^(1 - (c1 c2)/(-1 + c1)))/(1 + c1 (-1 + c2)))
+ E^(-t + (c1 c2 Log[-c1 - E^t + c1 E^t])/(-1 + c1)) C[1]}} *) | {
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I'm a little surprised that Mathematica can't do this on its own, to be honest; but here we are.
• Wait, but now don't we have an unknown c3 in our solution? Meaning, x[t] is now expressed in terms of a y[t] that we still don't know what it is. – MathX Mar 25 '16 at 18:39
• $c_3$ can be determined from the initial conditions: since it's a constant, then $c_3 = x(0) + y(0)$. I've edited to clarify this. In general, you'll have two constants of integration for this system; in my solution, they'll be the $c_3$ I coded in and the C[1] generated by Mathematica in the solution. – Michael Seifert Mar 25 '16 at 18:40
• oh yes you are right. And in that case it is equal to a from the OP. – MathX Mar 25 '16 at 18:48
The non-trivial solution for this system can be obtained by doing this:
system = {x'[t] == y[t] - c1*c2*x[t]/(1 - c1*(1 - Exp[-t])),
y'[t] == -y[t] + c1*c2*x[t]/(1 - c1*(1 - Exp[-t]))};
DSolve[First[system /. y -> (-x[#] &)], x[t], t]
(*
==> {{x[t] ->
E^(-t + (c1 c2 Log[-c1 - E^t + c1 E^t])/(-1 + c1)) C[1]}}
*)
All I did here is use the fact that the equations become identical if $y=-x$. That allows us to drop one of the equations. | {
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• As I noted in my comment above, the syntax of the input in the OP is incorrect. Mathematica will interpret exp(-t) as the product of a constant called exp with -t. – Michael Seifert Mar 25 '16 at 17:10
• @Jens This one looks awesome! Let me try to understand the syntax "DSolve[First[system /. y -> (-x[#] &)], x[t], t]" – KevinKim Mar 25 '16 at 17:15
• @MichaelSeifert oh yes, of course I used the corrected version to get the posted result - just copied the OP's error accidentally. – Jens Mar 25 '16 at 17:49
• @KevinKim the replacement y -> (-x[#] &) sets the function y equal to the function -x. You can't just use y -> -x here because that won't work in the derivative. Hence the more complex notation. & declares an (anonymous) function and # is the variable placeholder. – Jens Mar 25 '16 at 17:57
• @Jens: Are you sure that you used the corrected code to get your posted result? Your posted result has exp in it, including in the denominator without an argument. Also, note that $\dot{x} + \dot{y} = 0$ does not imply that $y = -x$; there should be an arbitrary constant involved. – Michael Seifert Mar 25 '16 at 18:18 | {
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# How does one represent the gradient $\nabla \mathbf A$ of a vector field $\mathbf A$ as a matrix?
I have been searching google about the gradient of a vector. I have found two distinct types of matrix notation for the gradient of a vector. I can't understand which one is correct, or if both are correct. The two are:
$$\nabla A = \partial_iA_je_i\otimes e_j=\begin{bmatrix}\frac{\partial A_1}{\partial X_1} &\frac{\partial A_2}{\partial X_1} & \frac{\partial A_3}{\partial X_1}\\\frac{\partial A_1}{\partial X_2} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_3}{\partial X_2}\\\frac{\partial A_1}{\partial X_3} & \frac{\partial A_2}{\partial X_3} & \frac{\partial A_3}{\partial X_3}\end{bmatrix}$$
or the transpose of the first, that is
$$\nabla A = \begin{bmatrix}\frac{\partial A_1}{\partial X_1} & \frac{\partial A_1}{\partial X_2} & \frac{\partial A_1}{\partial X_3}\\\frac{\partial A_2}{\partial X_1} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_2}{\partial X_3}\\\frac{\partial A_3}{\partial X_1}& \frac{\partial A_3}{\partial X_2}& \frac{\partial A_3}{\partial X_3}\end{bmatrix}.$$ | {
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• 'the Gradient' Gradient is a subset of the Jacobian, the diagonal of the Jacobian Matrix Jacobian – f5r5e5d Jul 22 '17 at 23:06
• 1. What do you mean by "correct" here, i.e. why can't these just be two differing conventions? 2. How is this a physics rather than a Mathematics question? – ACuriousMind Jul 22 '17 at 23:13
• The the second one seems to be the transpose of the first one and the second one also can be described by Jacobian of (A1,A2,A3/X1,X2,X3).Where the first one can not be written as Jacobian transformation as like the second one.So,what's the right way to write the matrix for gradient of vector -the first one or the second(as a jacobian trans?And of course this is a physics-question.The first guy I have found in-eng.uc.edu/~beaucag/Classes/Processing/Chapter2html/… , the second one is in -umich.edu/~bme456/ch1mathprelim/bme456mathprelim.htm#vectordyad . – Abu sayed Jul 23 '17 at 4:06
• To second ACuriousMind, these are two different notation conventions. The difference can actually be important because equations look different, so just make sure what notation the author is using ... – Sanya Jul 23 '17 at 10:46
The matrix notation for the gradient of a vector field is generally not-that-well-defined, because as you note there are two reasonable matrix encodings (transposes of each other), which are about equally reasonable. Because of that, whenever the notation $\nabla\mathbf A$ as a matrix is used in the literature, the responsible thing to do is to specify explicitly which of the two encodings is meant.
Both of the two representations have features that make them reasonable: | {
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Both of the two representations have features that make them reasonable:
• The second one you mention, $$\nabla A = \begin{bmatrix}\frac{\partial A_1}{\partial X_1} & \frac{\partial A_1}{\partial X_2} & \frac{\partial A_1}{\partial X_3}\\\frac{\partial A_2}{\partial X_1} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_2}{\partial X_3}\\\frac{\partial A_3}{\partial X_1}& \frac{\partial A_3}{\partial X_2}& \frac{\partial A_3}{\partial X_3}\end{bmatrix},$$ has the contravariant index as a row index and the covariant index of the gradient as a column index, which is reflective of both of those natures.
• On the other hand, the first representation in your post, $$\nabla A =\begin{bmatrix}\frac{\partial A_1}{\partial X_1} &\frac{\partial A_2}{\partial X_1} & \frac{\partial A_3}{\partial X_1}\\\frac{\partial A_1}{\partial X_2} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_3}{\partial X_2}\\\frac{\partial A_1}{\partial X_3} & \frac{\partial A_2}{\partial X_3} & \frac{\partial A_3}{\partial X_3}\end{bmatrix},$$ has the nice property that if $\mathbf r$ and $\mathbf v$ are column vectors, then $$\mathbf r^T \cdot \nabla \mathbf A\cdot \mathbf v = x_i \frac{\partial A_j}{\partial x_i} v_j$$ matches what you would expect as a matrix product, i.e. matching the action of the operator $\mathbf r \cdot \nabla = x_i \frac{\partial}{\partial x_i}$.
For an example of this in action in the literature, see this paper of mine: to avoid confusion, it is important to specify both how the matrix representation is chosen, and how it acts via components. It takes an extra line and it adds a whole lot of clarity to the text. | {
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The gradient of a function is well defined in the literature. The gradient of a vector field $A = A^i\partial_i$ seems to be the gradient of its components (which are functions). I think both matrix representations ($\partial_i A^j$ and $\partial_j A^i$) are "good" since they are simply representations of the same thing : $$\nabla A = (\partial_i A^j)dx^i\otimes \partial_j = (\partial_j A^i)dx^j \otimes \partial_i$$ Choose one of them and stay with it (at least until the end of the proof where you are using gradients of vector field).
Geometrically speaking, the gradient of a vector field you are talking about can be written either as the Lie derivative $\mathcal{L} A$ (along $\partial_i$ or $\partial_j$) or the covariant derivative $\nabla A$ (with vanishing Christoffel symbols $\Gamma$'s) (along $\partial_i$ or $\partial_j$). Both $\mathcal{L} A$ and $\nabla A$ are a little bit overkill for your setting (probably Euclidean flat space) but are good to keep in mind if you need to generalize from $\mathbb{R}^n$ to some manifold with global properties. | {
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• What you have mentioned "same thing",but the second one is actually the jacobian of (A1,A2,A3/X1,X2,X3) whereas the first one don't seems to be the jacobian transformation like J(,,/,_,).So,what is the fact? – Abu sayed Jul 23 '17 at 4:15
• Yes, same geometrical thing. A matrix representation is only a table of numbers, or functions and doesn't give any insight about the geometry behind it (i.e. in terms of sections of bundles). – Noé AC Jul 23 '17 at 7:05
• Both matrix representations are obviously different, so ignoring that is really unhelpful. Mentioning "the" Lie derivative here is also pretty pointless. – Sanya Jul 23 '17 at 10:43
• @Sanya I said it "seems to be related" to the Lie derivative. Looking at Deep's answer probably it's a better idea to look for covariant derivative with null Christoffel $\Gamma$'s. Though here we don't know the context of the original question so both Lie derivative and covariant derivative are somewhat overkill. – Noé AC Jul 23 '17 at 16:11
• @Sanya Also, in local coordinates, $\nabla A = (\partial_i A_j)dx_i \otimes \partial_j = (\partial_j A_i) dx_j \otimes \partial_i$, so using the matrix $(\partial_i A_j)$ or it's transpose $(\partial_j A_i)$ is ok, as long as you know what $i$ and what $j$ are linked to ($e_k$ or $e_k^*$) – Noé AC Jul 23 '17 at 16:22
$\nabla A$ is a tensor and it acts on a vector, say $\mathbf{u}$, to produce another vector: $\mathbf{v}=\nabla A(\mathbf{u})$. In matrix representation, you multiply the matrix of $\nabla A$ with the specified column/row vector representing $\mathbf{u}$. If you write $\mathbf{u}$ as a column vector (usual convention) then the second matrix representation is the correct one, while if you write it as a row vector the first matrix representation is the correct one. | {
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However matrix representation is not necessary. You have $\nabla A=\partial_iA_j~\mathbf{e}_i\otimes\mathbf{e}_j$, and the action of $\nabla A$ on $\mathbf{u}$ is defined to be the vector $\mathbf{v}=\partial_iA_j~\mathbf{e}_i(\mathbf{u})~\mathbf{e}_j=(\partial_iA_j)u_i~\mathbf{e}_j$. No more worries about how the matrix should be written. | {
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shows the gradient to calculate the derivative problems to rise, the given by the value. This course features a wide range of readings, simulations, assessments. In this problem, we consider the Lemniscate curve (1) (a2+y2)2 =3 (z2 - ) 2 -2 Figure 1: Bernoulli's Lemniscate The goal of this problem is to find the four points of the Lemniscate curve with horizontal tangent lines. Start learning today! Strategy for Solving Related Rates Problems #1. Repeating this process, function y's third order derivative …. •There must be an equation relating the 2(+) quantities. Calculus I - Implicit Differentiation Section 3-10 : Implicit Differentiation Back to Problem List 8. The Derivative from First Principles. Concavity and inflection points 5. Jean-Baptiste Campesato MAT137Y1 - LEC0501 - Calculus! - Oct 24. Time-saving lesson video on Implicit Differentiation with clear explanations and tons of step-by-step examples. Implicit differentiation; Derivative of the inverse of a function (including and ) Related rates of change; Integral Calculus …. What is Related Rates Calculator Symbolab. The trough is a triangular prism 10 feet long, 4 feet high, and 2 feet wide at the top. Course Info: 1st Day Handout: Parent/Student Letter. The graph of $$8x^3e^{y^2} = 3$$ is shown below. Implicit Differentiation - Related Rates Challenge Quizzes Implicit Differentiation: Level 2 Challenges Implicit Differentiation: Level 3 Challenges Implicit Differentiation …. Derivative is used to calculate rate …. Related Rates – The Anxious Panda. Cal_AB_LAP_5_IMPLICIT DIFFERENTIATION AND RELA…. 5) Lesson 13: Derivatives as Rates of Changes (Briggs 3. 4 NO Calculator - 40 min test 1 Thu10/7/2021 2 nd ½ of BLOCK Students will find the derivative of an implicitly defined function. Worked example: Differentiating related functions. Calculus Related Rates Example Volume of Cone. Calculator Reference 4 - Derivatives …. Introducing Maple 19 Single Variable Differential Calculus. At what rate is distance | {
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Derivatives …. Introducing Maple 19 Single Variable Differential Calculus. At what rate is distance between the two people changing when radians? so $\sec{\theta} = \frac{x}{50}$ and so I hear the next step is:. Limits: MATH 171 Problems 8 & 9 Tangent lines to parametric equations and related rates examples. In calculus, when you have an equation for y written in terms of x (like y = x2 -3x), it's easy to use basic differentiation techniques (known by mathematicians as "explicit differentiation" techniques) to find the derivative…. • derivatives of other trigonometric functions §2. Fundamental Theorem of Calculus and Integration Methods. 3 - Derivatives of Trig Functions: 3. Calculator Online Related Rates. We will continue from there tomorrow. They are also assessed on their ability to solve related rates pro. AP Calculus AB Course Details; Use implicit differentiation to find the derivative …. By observing changing rates students would be able to measure concrete examples and discover their relationships. See the page on implicit differentiation to learn how. In general then, n(t)=2t no – Thus the rate of growth of the population at time t is (dn/dt)=no2tln2 18. 5 Selecting Procedures for Calculating Derivatives 3. The student will implicit differentiation and make substitutions to solve these types of questions. Problem 6b - Calculate the rate …. Search: Implicit Differentiation With Trig. 9 Derivatives of Logarithmic and Exponential Functions. Where To Download Implicit Differentiation Homework Answers down the wall Implicit differentiation COMPLETELY Dy Calculate Dx # Dx For The Relation 3x + Xy = Y. In this tutorial students will learn how to model a related rates problem using parametric equations and a graphing utility. 5 Higher Order Derivative Using Implicit Differentiation; 3. Phone: (773) 809–5659 | Contact. If you labeled a ladder length c, and you know the rate …. The Derivative Calculator supports computing first, second, …, fifth derivatives as well | {
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the rate …. The Derivative Calculator supports computing first, second, …, fifth derivatives as well as. 3C: Solve problems involving related rates, optimization, rectilinear motion, (BC) and planar motion. (KristaKingMath) Related Rates in Calculus Implicit Differentiation Examples Implicit Differentiation Showing explicit and implicit differentiation give same result | AP Calculus AB | Khan Academy More implicit differentiationCalculus Lesson 14: 2. Assign variables to anything you have or need to find, and label the picture as so. It means that the function is expressed in terms of both x and y. 1Defining Average and Instantaneous Rates of Change at a Point 2. For this calculus instructional activity, the students solve word problems following 8 different key steps. 00:26:32 – Calculate …. The steps of logarithmic differentiation are outlined below. x 3 d 2 y d x 2 + 6 x 2 d y d x + ( n 2 x 2 + 6) x y = 0. Just behind related rates problems, the topic of implicit differentiation is one of the most difficult for students in a calculus. pdf from MATHEMATIC MATH1201 at University of Technology, Jamaica. Chapter 7 Related Rates and Implicit Derivatives. If you want to evaluate the derivative at the specific points, then substitute the value of the points x and y. 74-75 3 Implicit Differentiation Review 4 QUIZ 1 5 Related Rates (p. Recall: The derivative of x with respect to x is_____ The derivative …. Correct answer: \displaystyle \frac {1} {8 \pi} inches/sec. Implicit differentiation is needed in some applications. Differentiation Techniques in Review - 7. Calculus AB: Sample Syllabus 4. The 2 comes from the derivative of the inner function and then I multiply that by the implicit derivative of x which was given as 3 so I get 6. This is a 5-hour introductory calculus course designed primarily for engineering majors and certain other technical majors. I have this related rates problem, but I keep getting a negative number for dx/dt heres the problem. This assumption | {
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rates problem, but I keep getting a negative number for dx/dt heres the problem. This assumption does not require any work, but we need to be very careful to treat y as a function when we Calculate …. Session 31: Related Rates. 3 Finding the Derivative of a Function Given its Table of Values; 3. To generalize the above, comparative statics uses implicit differentiation to study the effect of variable changes in economic models. The topics preceded with an asterisk (*) are BC only topics. 4 Intro to related rates In related rates, we are often using implicit differentiation with respect to dt. Implicit differentiation and related rates quiz. The rate of change of the truck is dx/dt = 50 mph because it is traveling away from the intersection, while the rate …. The rate of change of the truck is dx/dt = 50 mph because it is traveling away from the intersection, while the rate of change of the car is dy/dt = −60 mph because it is traveling toward the intersection. 1st Day Homework: Academic Integrity, Parent Survey, Student Survey. I am new to implicit differentiation. S1:E 11 Implicit Differentiation and Related Rates TV-PG | Mar 5, 2010 | 31m The final strategy; implicit differentiation is used when it is difficult to solve a function for y; applying this rule to problems in related rates; the rate …. Because science and engineering often relate quantities to each other, the methods of related rates have broad applications in these fields. Now we need an equation relating our variables, which is the area equation: A = πr2. y = + sqrt ( 7 - x2) and, y = - sqrt (7 - x2). Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. 264 » 24 MB) Implicit differentiation…. u Q BMwatd Ge4 Pw gi It Hhz BIXnrf eisnoi …. Use implicit di erentiation to nd an equation of the tangent line to the curve y2(y2 4) = x2(x2 5) at the point (x;y) = (0; 2). Let f(x)=g(x)/h(x), where both g and h are differentiable and h(x)≠0. | {
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at the point (x;y) = (0; 2). Let f(x)=g(x)/h(x), where both g and h are differentiable and h(x)≠0. Differentiate both sides of the equation with respect to “x” 2. 11) TI-89 and Implicit Differentiation; Chapter 2. Solved exercises of Implicit Differentiation. Review for the Common Exam: MATH 151 Exam 3 Review Problems 8-15. calculate a differential equation, substitute initial conditions, and solve for a missing value. identifying implicit differentiation as a tool to differentiate functions where one variable is difficult to isolate; explaining the relationship between implicit differentiation and the chain rule; calculating related rates for the time derivatives …. com/patrickjmt !! Buy my book!: '1001 Calcul. The fact that dy/dt is negative means that the distance from the top of the ladder to the ground decreases at a rate of 0. Now, select a variable from the drop-down list in order to differentiate with respect to that particular variable. Implicit Differentiation (7:55) The Rate of Change (6:53) Related Rates (8:31) All videos are closed captioned. Beyond Calculus is a free online video book for AP Calculus AB. Related rates problems can be solved using by computing. Extrema on an Interval; Rolle’s Theorem; Mean Value Theorem 8. So for hyperbolic trig functions we have the hyperbolic cosine and the hyperbolic sine. When more than one derivative is applied to a function, it is consider higher order derivatives. How long is the ladder? This is a fairly common example of a related rates problem and a common application of derivatives and implicit differentiation. The chain rule is the key to solving such problems. The steps involved in solving a related rates problem can be summarized as: 1. Rate of the spread of a rumor in sociology. Related Rates Example 1 Air is being pumped into a spherical balloon at a rate of 5 cm 3/min. 2 - Activity 2 - Graphs of Functions and their Derivatives. Then the word rate means a derivative with respect to time. If possible, we | {
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and their Derivatives. Then the word rate means a derivative with respect to time. If possible, we subsequently solve for dy dx d y d x using algebra. Derivative of a Constant Multiple of a Function. 1C Calculate derivatives implicit differentiation related rate implicit differentiation related rates parametric algebraic logarithmic LO 2. RELATED RATES – Cone Problem (Water Filling and Leaking) Water is leaking out of an inverted conical tank at a rate of 10,000 at the same time water is being pumped into the tank at a constant rate. Furthermore, you’ll often find. The Derivative: Derivatives of Polynomials and Exponentials Product and Quotient Rules Derivatives of Trig Functions The Chain Rule Implicit Differentiation Logarithmic Differentiation Derivatives in the Sciences Exponential Growth and Decay Related Rates …. Plug in all the values you know, leaving only the one you're solving for. 7 Derivatives of Inverse Functions (Lecture 9) CALCULUS I MAT 301 - 0509. Inflation has its pros and cons, yet it is a normal part of a healthy economy. “In mathematics, a derivative is the rate of change of a function with respect to a variable. The vertical displacement remains constant at 50m. 07 Derivatives with Calculator…. $$displaystyle int (2x+1)^2dx$$ $$displaystyle int xsqrt{2x+3}dx$$ \(displaystyle int. 4 Implicit Differentiation (Section 3. True False Working with related rates of change is similar to implicit di erentiation in that we are nding the derivative …. Implicit Differentiation Calculator - Free Onli…. We will now use implicit differentiation on both sides with respect to t. This course is in content deeper and broader than AP Calculus AB course, suitable for the students who are intended in majoring in math, science, …. Using implicit differentiation …. In most related rates problems, we have an equation that relates derivative of both sides with respect to t (implicit differentiation). 5, we have the following equations. AP Calculus AB Related Rates and | {
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t (implicit differentiation). 5, we have the following equations. AP Calculus AB Related Rates and Implicit Differentiation Review Problem Set 1. The implicit differentiation calculator will find the first and second derivatives of an implicit function treating either y as a function of x or x as a function of y, with steps shown. The derivative of a function f at a number a is denoted by f' ( a ) and is given by: So f' (a) represents the slope of the tangent line to the curve at a, or equivalently, the instantaneous rate …. Calc Final Thursday!!! taking calc in college. [To see the graph of the corresponding equation, point the mouse to the graph icon at the left of the equation and press the left mouse button. Answer (1 of 7): In terms of x by implicit diffrentiation (dv/dx)=(pi/3) ((r^2) (dh/dx) + (h) (2r) (dr/dx)) (dv/dx)=((pi r^2)/3) (dh/dx) + (2hr) (dr/dx)). Related Rates Ex 1: Related Rates: Determine the Rate …. 5) l Logarithmic Differentiation (3. Functions and Limits, continuity (Ch 1) Rate of change at an instant…. It can also graph conic sections, arbitrary inequalities or. Note: Creation of this applet was inspired by a related-rates problem taken from Section 3. Equation 1: related rates cone problem pt. ) Differentiating term by term, we find the most difficulty in the first term. ONLY label things that are constant. Topics in Calculus Name_____ Date_____ Period____ ©Z O2g021 e4T aK ruutqaM XS yoAf 0txw HamrBea ELvL PCT. Just so you know, related rates is actually the Application of Implicit Differentiation by using Chain Rule in the form of dy/dx = dy/du * du/dx. The rules of differentiation produce an equation relating the rates …. Click the "Close" button in the RelatedRates window to terminate the Maplet. This course introduces you to the concept of differential calculus and the various ways to calculate rates …. In “related rates” application problems, we often assume that all variables are Implicit Differentiation Formula” (IDF for short): If | {
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we often assume that all variables are Implicit Differentiation Formula” (IDF for short): If F(x,y)=0 defines a differentiable relation of x and y, then So, to calculate ∂F/∂x or ∂F/∂x, calculate dF/dx or dF/dx. Computing Input interpretation: differentiate x^3 + y^3 = 4 with respect to x . That being said, this is a scenario where most teachers would be for it and accept its usage. (3) Find the derivative of √ (xy) = e x - y Solution. 95) go > The study guide for Worldwide Differential …. Visual representation of concepts. Calculus: Early Transcendentals (2nd Edition) answers to Chapter 3 - Derivatives - 3. Module 28 - Activities for Calculus Using the TI-83. Implicit Differentiation Notes and Examples. The implicit differentiation calculator will find the first and second derivatives of an implicit …. Derivatives and the Shape of a Graph: Concavity and Points of Inflection. An Implicit Home Slope Newton's Method Related Rates Extrema Optimization Mathematics of Finance Economics Physics Solved Examples Word Problems use a graphing utility or an applicable calculator …. Example 4: A 50-ft ladder is placed against a building. , a problem that requires implicit differentiation…. Students will use implicit differentiation to solve a real-world related rate problem. Specifically for the AP® Calculus BC exam, this unit builds an understanding of straight-line motion to solve problems in which particles are moving along curves in the plane. Review for the Common Exam: MATH 151 Exam 1 Review Problems 10-13 Implicit differentiation and physics applications of derivatives…. In the list of Related Rates Problems which follows, most problems are average and a few are somewhat challenging. How to Use the Implicit Differentiation Calculator? The procedure to use the implicit Differentiation calculator is as follows: Step 1: Enter the equation in a given input field. The derivative of a given function y=f(x) y = f ( x) measures the instantaneous rate of change of the output | {
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of a given function y=f(x) y = f ( x) measures the instantaneous rate of change of the output variable with respect to the input variable. 264 » 20 MB) The derivative as rate of change. (7) Find the derivatives of the following. We help you handle and prevent agitated behavior. Some relationships cannot be represented by an explicit function. Connections to the Study Design: AOS 3 - Calculus. Download or read online full book title Developing Understanding Of The Chain Rule Implicit Differentiation And Related Rates …. Apply the derivative to calculate rates …. (a) We can answer this question two ways: using "common sense" or related rates. Draw a sketch if it is possible 3. Author: Haley Paige Jeppson Publisher: ISBN: Release Date: 2019 Size: 37. AP Calculus AB Course Curriculum. 2 Derivatives of Inverse Functions. derivatives, and implicit differentiation Me during related rates,. Implicit Differentiation Related Rates One of the applications of mathematical modeling with calculus involves related rates word problems. Find derivatives of radical functions Also consider: • Power rule II Related Rates 3. 4: Related Rates (preprinted) Chapter 3. 5 Derivatives and the Shape of a Graph. Related Rates 1: (dA)/dt = 2 \pi r (dr)/dt Worksheet #35: W s #35 Solutions: 1-6, 7-10, 12 review: Related Rates 2: dz/dt = x dx/dt + y dy/dt Worksheet #36: Implicit Differentiation; Implicit Differentiation – Basic Idea and Examples ; Implicit Differentiation …. Limits: An Intuitive Approach - Answers. So the total rate of horiztonal displacement is 6m/s. Sketching Graphs 2: anti-derivatives. We use implicit differentiation to differentiate an implicitly defined function. Example 1: Find if x 2 y 3 − xy = 10. Strategy 1: Use implicit differentiation directly on the given equation. 11: Implicit Differentiation and Related Rates is shared under a CC BY 3. The topics below are both AB and BC topics. This can be done through the use of a financial calculator, software, an online | {
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AB and BC topics. This can be done through the use of a financial calculator, software, an online calculator, or present value tables. In differential calculus, related rates problems involve finding a rate at which a quantity changes by relating that quantity to other quantities whose rates of change are known. Differentials and Newton's method. Implicit differentiation will allow us to find an equation that. Learn to solve different kinds of related rates problems in Calculus. The derivate of a function at a chosen input value. A free online 2D graphing calculator (plotter), or curve calculator, that can plot piecewise, linear, quadratic, cubic, quartic, polynomial, trigonometric, hyperbolic, exponential, logarithmic, inverse functions given in different forms: explicit, implicit, polar, and parametric. Differentiating x to the power of something. Below you will find all homework assignments (and answers) for Calculus 1. {"results":"\u003cdiv class='col-xs-12 search-result-item thumbnail-card col-lg-6 col-sm-6' data-id='5439' data-item-type='CollectionItemFolder' data-type. We have seen how implicit differentiation can be used to find derivatives of implicit relations. Many (not all!) related rates problems present a quantity …. Download Vui lòng tải xuống để xem tài liệu đầy đủ. With an explicit function, given an x value, we have an explicit formula for computing the corresponding y value. Compute and interpret limits of functions using analytic and other methods, including L'Hospital's Rule. 2, 6 Important Example 23 Ex 5. How do you calculate the yearly inflation rate?. Notice that volume is a function of time because we’re looking at how volume changes over time (or the rate …. Resource Allocation: Implicit Differentiation; Related Rates; Business Applications: Social Sciences. Simply saying, it’s just the SLOPE of. Calculus is difficult part of mathematics and it. The idea behind linearization or local linear approximation is to find a value of the function | {
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it. The idea behind linearization or local linear approximation is to find a value of the function at the given point and evaluate the derivative …. For most homework, a calculator will not be needed. Step by Step Implicit Differentiation Slope of Inverse Function All in one Rate Explorer Differentiability of piecewise-defined function Absolute and Percent Change Differentials APPS: Max Volume of Folded Box APPS: Min Distance Point to Function f(x) APPS: Related Rates …. Business and Industry Applications. Take help of free calculator to get Implicit Differentiation …. As your next step, simply differentiate the y terms the same way as you differentiated the x terms. cost, strength, amount of material used in a building, profit, loss, etc. Using the derivative to calculate the slope of a tangent at any point on a function. Calculate derivatives of hyperbolic functions. I know that y=3/4 from substituting xy=3 with x=-4, but I do not. Now differentiate with respect to x x. Implicit Differentiation and Related Rates In our work up until now, the functions we needed to differentiate were either given explicitly, such as y = x2 +ex, y = x 2 + e x, or it was possible to get an explicit formula for them, such as solving y3 − 3x2 = r y 3 − 3 x 2 = r to get y = 3√5+3x2. Related Rates are problems that connect rate …. In this section we are going to look at an application of implicit differentiation. Free Response: A calculator may not be used. Here we study several examples of related quantities that are changing with respect to time and we look at how to calculate one rate of change given another rate …. 9: Lecture 17: Related Rates ROTC Long Weekend Patrick's Videos: Logarithmic Differentiation Examples 1 2 3 Paul's Notes: Inverse Functions; 8: Mon 10/11: 3. High School Math Solutions – Derivative Calculator…. Rates of Change of Rectangle 3-8-21. Typically related rates problems will follow a similar pattern. The top of a ladder slides down a vertical wall at a rate of 0. | {
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will follow a similar pattern. The top of a ladder slides down a vertical wall at a rate of 0. Answer (1 of 9): To calculate the derivative of anything, you need to specify the variable that is changing. Practice: Analyzing related rates problems: expressions. | {
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# What is the units digit of 2222^(333)*3333^(222) ?
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What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
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For more on this kind of questions check Units digits, exponents, remainders problems collection.
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What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
The units digit of $$2222^{333}$$ is the same as that of $$2^{333}$$;
The units digit of $$3333^{222}$$ is the same as that of $$3^{222}$$;
Hence, the units digit of $$2222^{333}*333^{222}$$ is the same as that of $$2^{333}*3^{222}$$;
Now, the units digits of both 2 and 3 in positive integer power repeat in patterns of 4. For 2 it's {2, 4, 8, 6} and for 3 it's {3, 9, 7, 1}. | {
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The units digit of $$2^{333}$$ will be the same as that of $$2^1$$, so 2 (as 333 divided by cyclicity of 4 yields remainder of 1, which means that the units digit is first # from pattern);
The units digit of $$3^{222}$$ will be the same as that of $$3^2$$, so 9 (as 222 divided by cyclicity of 4 yields remainder of 2, which means that the units digit is second # from pattern);
Finally, 2*9=18 --> the units digit is 8.
For more on this check Number Theory chapter of Math Book: http://gmatclub.com/forum/math-number-theory-88376.html
Hope it helps.
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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28 Jan 2012, 12:18
I used the method described by Bunuel.
but not to be repetitive, I will post another solution -
(2222^333)*(3333^222)=2222^111*(2222^222*3333^222)
here please pay attention to the fact that the unit digit of multiplication of 2222 and 3333 is 6 (2222^222*3333^222). since 6 powered in any number more than 0 results in 6 as a units digit, as a result we have-
6*2222^111
2 has a cycle of 4 . 111=27*4+3 . 2^3=8
6*8=48
so the units digit is 8, and the answer is E
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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26 Feb 2015, 21:22
Use cyclicity rule here ,
2^1=2,
2^2=4,
2^3=8
2^4=6
2^5=2
We can see here unit digit repeated after 4 powers so cyclicity of 2 is 4 . devide power by 4 and check above .
Ans E
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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26 Feb 2015, 21:43
Hi All,
Each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeating "pattern" of the units digits.
Here's another way to organize the information.
We're given [(2222)^333][(3333)^222]
We can 'combine' some of the pieces and rewrite this product as....
([(2222)(3333)]^222) [(2222)^111]
(2222)(3333) = a big number that ends in a 6
Taking a number that ends in a 6 and raising it to a power creates a nice pattern:
6^1 = 6
6^2 = 36
6^3 = 216
Etc.
Thus, we know that ([(2222)(3333)]^222) will be a gigantic number that ends in a 6.
2^111 requires us to figure out the "cycle" of the units digit...
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
So, every 4 "powers", the pattern of the units digits repeats (2, 4, 8, 6.....2, 4, 8, 6....).
111 = 27 sets of 4 with a remainder of 3....
This means that 2^111 = a big number that ends in an 8
So we have to multiply a big number that ends in a 6 and a big number that ends in an 8.
(6)(8) = 48, so the final product will be a gigantic number that ends in an 8.
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28 Feb 2015, 04:17
belagerfeld wrote:
What is the units digit of 2222^(333)*3333^(222)?
A. 0
B. 2
C. 4
D. 6
E. 8
I get 2, but it's apparently not the correct answer..
Source: some random learning sheet
Unit digit of 2222^333 is same as unit digit of 2^333.
Unit digit of powers of 2 follows a pattern: 2, 4, 8, 6
Now, 4*83 = 332 i.e. 2^332 uses 6 as its unit digit.
Hence, 2^333 will have unit digit as 2.
Unit digit of 3333^222 is same as unit digit of 3^222.
Unit digit of powers of 3 follows a pattern: 3, 9, 7, 1
Now, 4*55 = 220 i.e. 3^220 uses 1 as its unit digit.
Hence, 3^221 will have unit digit as 3.
And, 3^222 will have unit digit as 9.
Now we have 2 * 9 = 18
So the final unit digit of 2222^(333)*3333^(222) = 8.
Hence option E.
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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22 May 2018, 02:57
belagerfeld wrote:
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
This type of sums can also be solved using Fermet's Theorem approach.
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
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26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
x + 5*1.2x = $686,000 => 7x =$686,000 => x = $98'000 Hence, solution c is correct. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7177 Location: Pune, India Followers: 2161 Kudos [?]: 13985 [0], given: 222 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
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26 Nov 2010, 19:15
Stanford2012 has explained the solution perfectly.
The only thing I want to highlight here is the concept. | {
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Let's say you have a fixed sum of $100 that needs to be distributed among 5 people. If you want to maximize one person's share, you need to give everyone else the minimum that you can. e.g. the question says each person must receive at least$10, what is the maximum one person can receive?
Since you want to maximize one person's share, you should give everyone else the minimum possible i.e. $10. You give$10 to each of the other 4 people and are left with $60, the maximum that you can give to one guy. Similarly, if you want to minimize someone's share, you should give everyone else the maximum possible. In your question here, a fixed amount has to be distributed among 6 people and one person's minimum share needs to be found. A person will receive minimum (say$x) , when everyone else gets the maximum that they can (that will be $6x/5). _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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$686,000 in bonus money is to be divided among 6 employees [#permalink] ### Show Tags 13 Feb 2012, 19:34 2 This post was BOOKMARKED$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
A. $96,000 B.$97,000
C. $98,000 D.$99,000
E. $100,000 I often struggle with such questions. I can understand that for minimum value , we need to consider highest salary 20%, however I cannot understand how to put other 4 intermediate values. When I read the explanation I understand, but do not seem to apply it while solving. Can someone pls help in how to think through such problems. Thanks. Math Expert Joined: 02 Sep 2009 Posts: 37036 Followers: 7230 Kudos [?]: 96127 [1] , given: 10707 Re:$686,000 in bonus money is to be divided among 6 employees [#permalink] | {
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13 Feb 2012, 20:45
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priyankaparanjape wrote:
$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive? A.$96,000
B. $97,000 C.$98,000
D. $99,000 E.$100,000
I often struggle with such questions. I can understand that for minimum value , we need to consider highest salary 20%, however I cannot understand how to put other 4 intermediate values. When I read the explanation I understand, but do not seem to apply it while solving. Can someone pls help in how to think through such problems.
Thanks.
Since "no employee is to receive a bonus more than 20% greater than the bonus received by any other employee", then the bonuses of 6 employees must be in the range: $$x$$ and $$1.2x$$.
So we want to minimize $$x$$. To minimize $$x$$ we should make only one employee to receive that bonus (minimum possible) and the rest 5 employees to receive $$1.2x$$ bonus (maximum possible).
$$x+5*1.2x=686$$ --> $$7x=686$$ --> $$x=98$$.
General rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.
Similar questions:
a-certain-city-with-population-of-132-000-is-to-be-divided-76217.html?hilit=population#p1035185
shaggy-has-to-learn-the-same-71-hiragana-characters-and-126948.html
Other min/max questions:
PS: search.php?search_id=tag&tag_id=63
DS: search.php?search_id=tag&tag_id=42
Hope it helps.
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Hi Bunuel, | {
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12 Jun 2014, 20:11
Hi Bunuel,
I am confused. Why are we minimizing 1 persons bonus and maxing out all the other 5??
Shouldnt we be minimizing 1 persons bonus, maxing only 1 other persons bonuses and keeping the 4 in the middle at 1/6 of the total bonus. So let 686,000/6=@
@ is what everyone should get in an even world.
So for the first guy, whom we eant to minimize, i am going to deduct x from his @. For the 6th guy, the guy we want to max out where going to give him x, but such that the following condition is met.
(@-x)1.2=(@+x)
.2@=2.2x
.2(686,000/6)/2.2=10393.0=x
Now the first guy, the minimum bonus guy is going to get
@-x=114333-10393=103,940
Where am I going wrong?
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Re: $686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 15 May 2016, 12:01 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] 15 May 2016, 12:01
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# In how many different ways can trhee letters be posted from
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In how many different ways can trhee letters be posted from [#permalink]
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1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
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1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
First letter could be sent from ANY of the seven postboxes - 7 (7 options);
Second letter could be sent from the SIX postboxes left - 6 (6 options);
Third letter could be sent from the FIVE postboxes left - 5 (5 options);
Total # of ways =7*6*5=210
2. what if there is no restriction, that is, if two or more letters can be posted from the same box? | {
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2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
In this problem we don't have restriction, thus ANY letter could be sent from ANY postboxes =7*7*7=7^3=343
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17 Feb 2010, 03:47
Ravshonbek wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
warm up
1. 7 x 6 x 5 = 210
2. 7 x 7 x 7 = 343
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### Show Tags
10 Feb 2012, 04:03
Bunuel wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
First letter could be sent from ANY of the seven postboxes - 7 (7 options);
Second letter could be sent from the SIX postboxes left - 6 (6 options);
Third letter could be sent from the FIVE postboxes left - 5 (5 options);
Total # of ways =7*6*5=210
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
In this problem we don't have restriction, thus ANY letter could be sent from ANY postboxes =7*7*7=7^3=343
Hi Bunuel,
Could you please elaborate on the second question. Couldn't figure out why.
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### Show Tags
10 Feb 2012, 05:05
1. 7 (no restriction) * 6 (can not be the same as the first one) * 5 (can not be the same as the first and second one) = 210
2. 7 (no restriction) * 7 (no restriction) * 7 (no restriction) = 343
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### Show Tags
10 Feb 2012, 09:13
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Expert's post
mohankumarbd wrote:
Bunuel wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
First letter could be sent from ANY of the seven postboxes - 7 (7 options);
Second letter could be sent from the SIX postboxes left - 6 (6 options);
Third letter could be sent from the FIVE postboxes left - 5 (5 options);
Total # of ways =7*6*5=210
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
In this problem we don't have restriction, thus ANY letter could be sent from ANY postboxes =7*7*7=7^3=343
Hi Bunuel,
Could you please elaborate on the second question. Couldn't figure out why.
"Two or more letters can be posted from the same box" means that all 3 letters can be posted from the same postbox (so we don't have the restriction we had for the first question).
Now, since there are 7 postboxes then each of these 3 letters has 7 options to be posted from, total # of ways is 7*7*7=7^3.
Hope it's clear.
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### Show Tags
25 May 2013, 08:22
Bunuel wrote:
mohankumarbd wrote:
Bunuel wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox? | {
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First letter could be sent from ANY of the seven postboxes - 7 (7 options);
Second letter could be sent from the SIX postboxes left - 6 (6 options);
Third letter could be sent from the FIVE postboxes left - 5 (5 options);
Total # of ways =7*6*5=210
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
In this problem we don't have restriction, thus ANY letter could be sent from ANY postboxes =7*7*7=7^3=343
Hi Bunuel,
Could you please elaborate on the second question. Couldn't figure out why.
"Two or more letters can be posted from the same box" means that all 3 letters can be posted from the same postbox (so we don't have the restriction we had for the first question).
Now, since there are 7 postboxes then each of these 3 letters has 7 options to be posted from, total # of ways is 7*7*7=7^3.
Hope it's clear.
Hi Bunnel,
I tried to do the second question via combinatorics, but i am not able to figure it out, please check the below method and guide where i went wrong
= all three in one box +2 in one box and the last one in a different box + all three in different boxes
= 3c3*7+3c2*7c1*6c5+3c1*7c3
= 7+ 3*7*6+3*7*6*5
= 7 + 126 + 270 = wrong
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### Show Tags
25 May 2013, 13:06
Quote:
I tried to do the second question via combinatorics, but i am not able to figure it out, please check the below method and guide where i went wrong
= all three in one box +2 in one box and the last one in a different box + all three in different boxes
= 3c3*7+3c2*7c1*6c5+3c1*7c3
= 7+ 3*7*6+3*7*6*5
= 7 + 126 + 270 = wrong
Think it in this way,
First letter can go to any 7 post offices
Same case with second and same case with the third letter as well so 7*7*7
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Please +1 KUDO if my post helps. Thank you.
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Re: In how many different ways can trhee letters be posted from [#permalink]
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21 Feb 2015, 21:53
Could someone please tell me why part 1 cannot be answered using "7C3" = 35 ways?
Many thanks.
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Re: In how many different ways can trhee letters be posted from [#permalink]
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22 Feb 2015, 00:00
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Hi icetray,
Since the question is worded in a "vague" way, you bring up an interesting interpretation of it. Thankfully, questions on the Official GMAT are worded to remove ambiguity and "bias" on the part of the reader, so you won't have to worry about that on Test Day. This prompt reads as if it were created by the original poster, so it's not clear what he/she was "intending" the question to mean. | {
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As it is, your interpretation of the prompt makes a lot of sense - there does not seem to be any reason why we should emphasize the "order" of the letters (there's no reference to "first letter", "second letter", "third letter" and no reference to "arrangements"). Using post-boxes ABC would be same as BCA, CBA, etc., so if we interpret the prompt as a "combinations" question, then 7c3 = 35 would be correct.
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Re: In how many different ways can trhee letters be posted from [#permalink]
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06 Apr 2016, 08:41
Ravshonbek wrote:
1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?
2. what if there is no restriction, that is, if two or more letters can be posted from the same box?
warm up
3. What if we assume no three letters can be posted from same postbox?
Re: In how many different ways can trhee letters be posted from [#permalink] 06 Apr 2016, 08:41
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# uniform random point in triangle
Suppose you have an arbitrary triangle with vertices $A$, $B$, and $C$. This paper (section 4.2) says that you can generate a random point, $P$, uniformly from within triangle $ABC$ by the following convex combination of the vertices:
$P = (1 - \sqrt{r_1}) A + (\sqrt{r_1} (1 - r_2)) B + (r_2 \sqrt{r_1}) C$
where $r_1, r_2 \sim U[0, 1]$.
How do you prove that the sampled points are uniformly distributed within triangle $ABC$?
I would argue that if it is true for any triangle, it is true for all of them, as we can find an affine transformation between them. So I would pick my favorite triangle, which is $A=(0,0), B=(1,0), C=(0,1)$. Then the point is $(\sqrt{r_1}(1-r_2),r_2\sqrt{r_1})$ and we need to prove it is always within the triangle and evenly distributed. To be in the triangle we need $x,y\ge 0, x+y\le 1$, which is clear. Then show that the probability to be within an area $(0,x) \times (0,y)$ is $2xy$ by integration.
• You mean 2xy, I think. – TonyK Jan 24 '11 at 9:19
• @Tony K: Right. Fixed. – Ross Millikan Jan 24 '11 at 13:41
• It's been quite a lot since I was doing this, but can someone post a full proof? Sry for necromancy, btw :) – ioreskovic Jun 26 '13 at 14:07
Pick $A,B,C = (0,0),(1,0),(1,1)$. For any point $(x,y)$, we have that $(x,y)$ is in the triangle if and only if $0 < x < 1$ and $0 < y/x < 1$.
Now, we look for the distribution of $x$ and $y/x$.
Computing a few triangle areas, we can easily check that $P(0 < x < x_0) = x_0^2$. Hence $P(0 < x^2 < a) = P(0 < x < \sqrt a) = a$, so that $x^2$ is uniformly distributed in the unit interval.
Again with an area computations, we can check that $P(0 < y/x < k) = k$. Hence $y/x$ is also uniformly distributed in the unit interval.
Finally we have to check (again computing a simple area) that $P(0 < x < x_0 \land 0 < y/x < k) = x_0^2k$ which proves that $x^2$ and $y/x$ are independant. | {
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So we have found that to generate a point uniformly in the triangle is the same as picking $x^2 = r_1$ and $y/x = r_2$ uniformly in the unit interval, and then form $(x,y) = (\sqrt r_1, r_2 \sqrt r_1)$, which is the barycenter of $(A,1- \sqrt {r_1})(B,\sqrt {r_1}(1-r_2))(C,\sqrt{r_1}r_2)$.
• How to generalize it onto higher dimensions? What to do with arbitrary $n$-simplex? – Orient Jul 18 '14 at 17:54
• If the dimension is $d$, first change your coordinates so that the simplex is formed by the points $(0,\ldots,0),(1,0,\ldots,0),\ldots,(1,\ldots,1,0),(1,\ldots,1)$. Then for the first coordinate, pick $x_1^d$ uniformly. Then pick $(x_2/x_1,x_3/x_1,\ldots,x_d/x_1)$ uniformly in the $d-1$-dimensional simplex (by induction) – mercio Jul 26 '14 at 14:47
• merico, I think it is wrong. We need spatial uniform distribution. – Orient Jul 27 '14 at 6:25
• One way to get random point inside of simplex $P = \{\mathbf{p}_i\}_{i = 1}^{d + 1}$ is to pick $\mathbf{c} = (c_1, c_2, ..., c_d, c_{d + 1}), c_i \sim U[0;1]$, then $\mathbf{c} \leftarrow -\log(\mathbf{c})$, then $c \leftarrow \displaystyle \frac{\mathbf{c}}{\sum \limits_{i = 1}^{d + 1} c_i}$, then random point is: $\displaystyle \sum \limits_{i = 1}^{d + 1}c_i \cdot \mathbf{p}_i$ (based on Dirichlet distribution and properties of affine transformations). – Orient Jul 27 '14 at 6:26
• But the generalization of your approach itself is here math.stackexchange.com/questions/563129/… . – Orient Jul 27 '14 at 6:31 | {
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I'm starting with the argument provided by @Ross Millikan. Let $A=(0,0),\ B=(1,0),\ C=(0,1)$. Then the point chosen according to the given equation is $P=(X,Y)=(\sqrt{r_1}(1-r_2),r_2\sqrt{r_1})$. Now clearly, $0\leq X,Y \leq 1$ and $X+Y\leq \sqrt{r_1}\leq 1$. Now the problem is to show that $\mathbb{P}(X\leq x, Y\leq y)=2xy,\ \forall 0\leq x,y\leq 1$ with $x+y\leq 1$. Now, \begin{equation*} \begin{split} \mathbb{P}(X\leq x, Y\leq y)=& \mathbb{P}(\sqrt{r_1}(1-r_2)\leq x, r_2\sqrt{r_1}\leq y)\\ \ =&\int_{0}^1 \mathbb{P}(\sqrt{r}(1-r_2)\leq x, r_2\sqrt{r}\leq y|r_1=r)f_{r_1}(r)dr\\ \ =&\int_{0}^1 \mathbb{P}(1-\frac{x}{\sqrt{r}}\leq r_2\leq \frac{y}{\sqrt{r}})I_{[0,1]}(r)dr\ \mbox{(Since, $r_1, r_2$ are i.i.d $\mathcal{U}[0,1]$})\\ \end{split} \end{equation*} Now to find the region of integration we note that $$1-\frac{x}{\sqrt{r}}\leq r_2\leq \frac{y}{\sqrt{r}}\ \Rightarrow\ 0\leq r\leq(x+y)^2$$ Also, if $x\leq y$ then $$r\in (0,x^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\leq 0,\ \frac{y}{\sqrt{r}}\geq 1 \\ r\in (x^2,y^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\geq 0,\ \frac{y}{\sqrt{r}}\geq 1 \\ r\in (y^2,(x+y)^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\geq 0,\ \frac{y}{\sqrt{r}}\leq 1 \\$$ and if $y\leq x$ then $$r\in (0,y^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\leq 0,\ \frac{y}{\sqrt{r}}\geq 1 \\ r\in (y^2,x^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\leq 0,\ \frac{y}{\sqrt{r}}\leq 1 \\ r\in (x^2,(x+y)^2)\ \Rightarrow\ 1-\frac{x}{\sqrt{r}}\geq 0,\ \frac{y}{\sqrt{r}}\leq 1 \\$$
Then if $x\leq 1$ the integral becomes $$\int_{0}^{x^2}1 dr+\int_{x^2}^{y^2}\frac{x}{\sqrt{r}} dr+ \int_{y^2}^{(x+y)^2}\left(\frac{x+y}{\sqrt{r}}-1\right) dr=2xy$$ Similarly, if $y\leq x$ the integral becomes $$\int_{0}^{y^2}1 dr+\int_{y^2}^{x^2}\frac{y}{\sqrt{r}} dr+ \int_{x^2}^{(x+y)^2}\left(\frac{x+y}{\sqrt{r}}-1\right) dr=2xy$$ Hence the point $P$ is uniformly distributed on the surface of the triangle $ABC$. $\hspace{3cm}\ \Box$ | {
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Note that these points, when random, will be uniformly distributed in a nicely random way, but if you loop through r1 and r2 with an increment (say .01) your resulting points will have unusual artifacts and not look randomly distributed. One end of the triangle may have few points.
I determined this with code ( Note that similar code, using math.Random() looks fine).
FillTriangleWithPointsBarycentric
if r1 and r2 are uniform random numbers between 0 and 1 then
This math produces a uniform distribution (note √ means sqrt of)
d = (1.0−√r1)*vector1 + √r1*(1.0−r2)*vector2+√r1 * r2 * vector3
But rather than using a uniform random number, just loop through them and the result does not look good.
@param {THREE.Vector3} vector1
@param {THREE.Vector3} vector2
@param {THREE.Vector3} vector3
@param {Array<number>} output input/output points > [x0,y0,z0,x1,y1,z1,...xn,yn,zn] displayable in point cloud
@returns {void}
FillTriangleWithPointsBarycentric(vector1, vector2, vector3, output) {
let triangle = new THREE.Triangle(vector1, vector2, vector3);
let area = triangle.getArea();
console.log('Area is ' + area);
area = Math.sqrt(area);
console.log('sqrt Area is ' + area);
let increment = 0.1 / area;
for (let r1 = 0; r1 <= 1; r1 += increment) {
for (let r2 = 0; r2 <= 1; r2 += increment ) {
// of course this is javascript we have to write this out instead of
// using only one line
let sqrtR = Math.sqrt(r1);
let A = (1 - sqrtR);
let B = (sqrtR * (1 - r2));
let C = (sqrtR * r2);
let x = A * vector1.x + B * vector2.x + C * vector3.x;
let y = A * vector1.y + B * vector2.y + C * vector3.y;
let z = A * vector1.z + B * vector2.z + C * vector3.z;
output.push(x, y, z);
}
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# Probability of picking a biased coin
Suppose you have a bag of 100 coins of which 1 is biased with both sides as Heads. You pick a coin from the bag and toss it three times. The result of all three tosses is Heads. What is the probability that the selected coin is biased?
P(selecting a biased coin) = 1/100
P(getting a head thrice with the biased coin) = 1
P(selecting an unbiased coin) = 99/100
P(getting a head thrice with the unbiased coin) = 1/8
P(selecting a biased coin|coin toss resulted in 3 heads) =
P(selecting a biased coin and getting heads thrice)/
[P(selecting a biased coin and getting heads thrice) +
P(selecting an unbiased coin and getting heads thrice)]
= (1/100)/[(1/100) + (99/800)]
= (1/100)/(107/800)
= 8/107
= 0.0747
Is this correct? Thanks.
• Is this a homework question? – Carlos Accioly Feb 19 '13 at 12:19
• @CarlosAccioly This question was posed to me in an examination on Sunday. Just wanted to verify my answer. – tejas_kale Feb 19 '13 at 12:44
• Hi tejas, a standard-level question on an exam - even if you're pursuing it for self study - falls under the scope of the homework tag. – Glen_b Feb 20 '13 at 2:16
• @Glen_b Got your point. I thought that Carlos' question was to ensure that the poster is not cheating with his homework. Thanks for the info. – tejas_kale Feb 20 '13 at 11:48
Your answer is right. The solution can be derived using Bayes' Theorem:
$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$
You want to know the probability of $P(\text{biased coin}|\text{three heads})$.
What do we know?
There are $100$ coins. $99$ are fair, $1$ is biased with both sides as heads.
With a fair coin, the probability of three heads is $0.5^3 = 1/8$.
The probability of picking the biased coin: $P(\text{biased coin}) = 1/100$.
The probability of all three tosses is heads: $P(\text{three heads}) = \frac{1 \times 1+ 99 \times \frac{1}{8}}{100}$.
The probability of three heads given the biased coin is trivial: $P(\text{three heads}|\text{biased coin}) = 1$. | {
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If we use Bayes' Theorem from above, we can calculate
$$P(\text{biased coin}|\text{three heads}) = \frac{1 \times 1/100}{\frac{1 + 99 \times \frac{1}{8}}{100}} = \frac{1}{1 + 99 \times \frac{1}{8}} = \frac{8}{107} \approx 0.07476636$$
• Cool! Cool! Cool! – Behacad Feb 19 '13 at 14:33
Here is a write that describes something very similar to that. The Bayes approach is the right way to proceed.
We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.
In general, P(biased coin|k heads) = (2^k)/[(2^k) + 99] Where k is no of consective heads
so if the trick coin was tossed 3 times (2^3)/[(2^3) + 99] = 8/(8+99) = 8/107 = 0.07 | {
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Then, $A^k = A^{k-1}A = AA = A$, as required. Eigenvalues. First, we establish the following: The eigenvalues of $Q$ are either $0$ or $1$. Proof: Let λ be an eigenvalue of A and q be a corresponding eigenvector which is a non-zero vector. $P$ is an orthogonal projection operator if and only if it is idempotent and symmetric. (For a proof, see the post “Idempotent matrix and its eigenvalues“.) Show that the rank of an idempotent matrix is equal to the number of nonzero eigenvalues of the matrix. If $A$ is idempotent matrix, then the eigenvalues of $A$ is either $0$ or $1$. Eigenvalues. This site uses Akismet to reduce spam. This website’s goal is to encourage people to enjoy Mathematics! This provides an easy way of computing the rank, or alternatively an easy way of determining the trace of a matrix whose elements are not specifically known (which is helpful in statistics, for example, in establishing the degree of bias in using a sample variance as an estimate of a population variance). 1 & -2 (2) Let A be an n×n matrix. The post contains C++ and Python code for converting a rotation matrix to Euler angles and vice-versa. Theorem 3. Let Aand Bbe idempotent matrices of the same size. The short answer is: 0, 1 are the ONLY possible eigenvalues for an idempotent matrix A. \begin{bmatrix} 1 & 0 \\ ST is the new administrator. Let [E_0={mathbf{x}in R^n mid Amathbf{x}=mathbf{0}} text{ and } […], […] the post ↴ Idempotent Matrix and its Eigenvalues for solutions of this […], […] the post ↴ Idempotent Matrix and its Eigenvalues for […], […] Idempotent Matrix and its Eigenvalues […], Your email address will not be published. I think, you want to know the relation between the singular values and the eigenvalues of idempotent matrices. All its eigenvalues are positive. An nxn matrix A is called idempotent if A 2 =A. \qquad Let C be a symmetric idempotent matrix. b. Therefore, it defines a projection (not orthogonal) on its range, which we denote by S. Matrix I - A maps \( … | {
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it defines a projection (not orthogonal) on its range, which we denote by S. Matrix I - A maps \( … An idempotent linear operator $P$ is a projection operator on the range space $R(P)$ along its null space $N(P)$. Theorem: \begin{bmatrix} 6. Show that the eigenvalues of C are either 0 or 1. Final Exam Problems and Solution. Principal idempotent of a matrix example University Duisburg-Essen SS 2005 ISE Bachelor Mathematics. 4.1. which is a circle with center (1/2, 0) and radius 1/2. In this section K = C, that is, matrices, vectors and scalars are all complex.Assuming K = R would make the theory more complicated. c. Let d be a n × 1 vector. \qquad For every n×n matrix A, the determinant of A equals the product of its eigenvalues. Suppose v+ iw 2 Cnis a complex eigenvector with eigenvalue a+ib (here v;w 2 Rn). The hat matrix (projection matrix P in econometrics) is symmetric, idempotent, and positive definite. The matrix, is idempotent and since it is a doagonal matrix, its eigen value are the diagonal entries λ = 0 and λ = 1. \end{bmatrix} [/math], $X\left(X^\textsf{T}X\right)^{-1}X^\textsf{T}$, $\hat{e}^\textsf{T}\hat{e} = (My)^\textsf{T}(My) = y^\textsf{T}M^\textsf{T}My = y^\textsf{T}MMy = y^\textsf{T}My.$. Then, λqAqAqAAq Aq Aq q q== = = = = =22()λλ λλλ. Since there are only 2 idempotent square matrices, you can just try them both for parts a and b. λ = λ2. Show that 1 2(I+A) is idempotent if and only if Ais an involution. Eigenvalues of idempotent matrices are either 0 or 1. This page was last edited on 20 November 2020, at 21:34. Then, λqAqAqAAq Aq Aq q q== = = = = =22()λλ λλλ. The eigenvalues of A are given by the roots of the polynomial det(A In) = 0: The corresponding eigenvectors are the nonzero solutions of the linear system (A In)~x = 0: Collecting all solutions of this system, we get the corresponding eigenspace. Proof: Let A be an nxn matrix, and let λ be an eigenvalue of A, with corresponding eigenvector v. The eigenvalues of an idempotent | {
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and let λ be an eigenvalue of A, with corresponding eigenvector v. The eigenvalues of an idempotent matrix take on the values 1 and 0 only. Note that applying the complex conjugation to the identity A(v+iw) = (a+ib)(v+iw) yields A(v iw) = (a ib)(v iw). An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. [3] The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Jiming Wu. Since x is a nonzero vector (because x is an eigenvector), we must have. eigenvalues of the matrix A. Eigenvalues. A question on a nilpotent matrix: Advanced Algebra: Aug 6, 2013: Prove that it is impossible for a 2x2 matrix to be both nilpotent and idempotent: Advanced Algebra: Mar 25, 2013: Matrix of a Nilpotent Operator Proof: Advanced Algebra: Mar 27, 2011: relation between nilpotent matrix and eigenvalues: Advanced Algebra: Mar 26, 2011 Let Hbe a symmetric idempotent real valued matrix. In linear algebra, an idempotent matrix is a matrix which, when multiplied by itself, yields itself. 7. A matrix is idempotent () if and only if it is diagonalizable and all the eigenvalues are 0 or 1. Then give an example of a matrix that is idempotent and has both of these two values as eigenvalues. […], […] only possible eigenvalues of an idempotent matrix are $0$ or $1$. \begin{bmatrix} For idempotent matrix, the eigenvalues are ##1## and ##0##. Fingerprint Dive into the research topics of 'Eigenvalues and eigenvectors of matrices in idempotent algebra'. \hat{e} = y - X \hat\beta Solution: Suppose that λ is an eigenvalue of A. Prove that if A is idempotent, then det(A) is equal to either 0 or 1. A symmetric idempotent matrix such as H is called a perpendicular projection matrix. The eigenvalues of A are the roots of its characteristic equation: |tI-A| = 0.. where the mXm matrix AT= (aii') is idempotent of rank r, and the nXn matrix Bs=(bjj') is idempotent of rank s. We will say that AT | {
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is idempotent of rank r, and the nXn matrix Bs=(bjj') is idempotent of rank s. We will say that AT and Bs are the covariance matrices associated with the interaction matrix (dij ) . Theorem 4 shows that these eigenvalues are also eigenvalues of the difference of idempotent density matrices. b. Consider the following 2 cases: Case (1): A is nonsingular. -1 & 3 & 4 \\ Hence by the principle of induction, the result follows. Theorem: Let Ann× be an idempotent matrix. Proof: If A is idempotent, λ is an eigenvalue and ⦠[/math], $\begin{pmatrix}a & b \\ c & d \end{pmatrix}$, $\begin{pmatrix}a & b \\ b & 1 - a \end{pmatrix}$, $\left(a - \frac{1}{2}\right)^2 + b^2 = \frac{1}{4}$, $A = \frac{1}{2}\begin{pmatrix}1 - \cos\theta & \sin\theta \\ \sin\theta & 1 + \cos\theta \end{pmatrix}$, $\begin{pmatrix}a & b \\ c & 1 - a\end{pmatrix}$, $A = IA = A^{-1}A^2 = A^{-1}A = I$, $(I-A)(I-A) = I-A-A+A^2 = I-A-A+A = I-A$, $(y - X\beta)^\textsf{T}(y - X\beta)$, $\hat\beta = \left(X^\textsf{T}X\right)^{-1}X^\textsf{T}y$, $This website is no longer maintained by Yu. In the case of irreducible mattices, the problem is reduced to the analysis of an idempotent analogue of the charactetistic polynomial of the mattix. Hence solving λ(λ â 1) = 0, the possible values for λ is either 0 or 1. The 'only if' part can be shown using proof by induction. An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. 3. Eigenvalues of a Hermitian Matrix are Real Numbers, If A^{\trans}A=A, then A is a Symmetric Idempotent Matrix, Find all Values of x such that the Given Matrix is Invertible. (Linear Algebra Math 2568 at the Ohio State University), The Ideal Generated by a Non-Unit Irreducible Element in a PID is Maximal, In a Principal Ideal Domain (PID), a Prime Ideal is a Maximal Ideal. Find All the Eigenvalues of 4 by 4 Matrix, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Diagonalize a 2 by 2 Matrix if Diagonalizable, Find an Orthonormal Basis of | {
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to a Given Eigenvalue, Diagonalize a 2 by 2 Matrix if Diagonalizable, Find an Orthonormal Basis of the Range of a Linear Transformation, The Product of Two Nonsingular Matrices is Nonsingular, Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not, Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials, Find Values of a , b , c such that the Given Matrix is Diagonalizable, Diagonalize the 3 by 3 Matrix Whose Entries are All One, Given the Characteristic Polynomial, Find the Rank of the Matrix, Compute A^{10}\mathbf{v} Using Eigenvalues and Eigenvectors of the Matrix A, Determine Whether There Exists a Nonsingular Matrix Satisfying A^4=ABA^2+2A^3, Determine Eigenvalues, Eigenvectors, Diagonalizable From a Partial Information of a Matrix, Idempotent (Projective) Matrices are Diagonalizable, Idempotent Matrices. We can see that the distribution of the quadratic form is a weighted sum of \chi_1^2 random variables, where the weights are the eigenvalues of the variance matrix. Ax= λx⇒Ax= AAx= λAx= λ2x,soλ2 = λwhich implies λ=0 or λ=1. These two conditions can be re-stated as follows: 1.A square matrix A is a projection if it is idempotent, 2.A projection A is orthogonal if it is also symmetric. Request PDF | Eigenvalues and eigenvectors of matrices in idempotent algebra | The eigenvalue problem for the mattix of a generalized linear operator is considered. If λ is an eigenvalue of an idempotent matrix, show that λ is either 0 or 1. Those eigenvalues (here they are 1 and 1=2) are a new way to see into the heart of a matrix. Discuss the analogue for A−B. The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Called spectral theory, it allows us to give fundamental structure theorems for matrices and to develop power tools for comparing and computing withmatrices. The only non-singular idempotent matrix is the identity matrix; that is, if a | {
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withmatrices. The only non-singular idempotent matrix is the identity matrix; that is, if a non-identity matrix is idempotent, its number of independent rows (and columns) is less than its number of rows (and columns). Published 02/22/2018, […] Since A has three distinct eigenvalues, A is diagonalizable. An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. The trace of an idempotent matrix â the sum of the elements on its main diagonal â equals the rank of the matrix and thus is always an integer. The trace is related to the derivative of the determinant (see Jacobi's formula). Since the matrix A has 0 as an eigenvalue. a. This characterization can be used to define the trace of a linear operator in general. PRACTICE PROBLEMS (solutions provided below) (1) Let A be an n × n matrix. Claim: Each eigenvalue of an idempotent matrix is either 0 or 1. Maximum number of nonzero entries in k-idempotent 0-1 matrices \end{bmatrix} 1 & 0 & 0 \\ Thus the number positive singular values in your problem is also n-2. Enter your email address to subscribe to this blog and receive notifications of new posts by email. 9. Together they form a unique fingerprint. If you look at your definition of idempotent A^2=A, then you can actually solve this for A and find *all* idempotent square matrices. Thus a necessary condition for a 2 × 2 matrix to be idempotent is that either it is diagonal or its trace equals 1. 0 & 1 You should be able to find 2 of them. Eigenvalues. An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1. Theorem: Let Ann× be an idempotent matrix. = My. Proof: A is idempotent, therefore, AA = A. The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Eigenvalues. To explain eigenvalues, we first explain eigenvectors. 0 & 0 & 1 The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the | {
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0 & 1 The trace of an idempotent matrix — the sum of the elements on its main diagonal — equals the rank of the matrix and thus is always an integer. Show that 1 2(I+A) is idempotent if and only if Ais an involution. \begin{bmatrix} Show that = 0 or = 1 are the only possible eigenvalues of A. A.8. A symmetric idempotent matrix such as H is called a perpendicular projection matrix. All the matrices are square matrices (n x n matrices). (Hint: Note that Cq = γq implies 0 = Cq − γq = CCq − γq = CCq − γq = γ 2 q − γq and solve for γ.) This can only occur if = 0 or 1. 1 & 0 \\ An idempotent matrix is always diagonalizable and its eigenvalues are either 0 or 1.[3]. Proof: Let λ be an eigenvalue of A and q be a corresponding eigenvector which is a non-zero vector. For this product [math]A^2$ to be defined, $A$ must necessarily be a square matrix. → 2 → ()0 (1)0λλ λ λ−=→−=qnn××11qλ=0 or λ=1, because q is a non-zero vector. In the special case where these eigenvalues are all one we do indeed obtain $\mathbf{z}^\text{T} \mathbf{\Sigma} \mathbf{z} \sim \chi_n^2$, but in general this result does not hold. 2 & -2 & -4 \\ Theorem 2.2. | {
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## idempotent matrix eigenvalues
Switch Window Key Chromebook, Best Yarn For Punch Needle, Nature Communications Acceptance Rate, Preposition Of Direction Exercises, Where Can I Buy Jays Potato Chips, Shadow Ridge Apartments - El Paso, | {
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# Limit of a sequence bounded below but has no cluster points
Question is this: Let $a_n$ be a sequence of real numbers. Prove that if $a_n$ is bounded below and has no cluster points then $a_n$ → ∞.
I could not really find a way to prove it. Could you give me a hand?
-
HINT: If $a_n\not\to\infty$, then the sequence has an infinite subsequence that is bounded above. (Why?) That subsequence is actually a bounded sequence. What do you know about cluster points of bounded sequences?
Added: Suppose that $a_n\not\to\infty$. Then there is some $M$ such that for all $m\in\Bbb N$ there is a $k\ge m$ with $a_k<M$. Let $A=\{k\in\Bbb N:a_k<M\}$ clearly $A$ is infinite, so we can list it as $A=\{n_k:k\in\Bbb N\}$ in such a way that $n_0<n_1<n_2<\ldots~$. We now have a subsequence $\sigma=\langle a_{n_k}:k\in\Bbb N\rangle$ of the original sequence.
The original sequence is bounded below by hypothesis, so $\sigma$ is also bounded below, and by construction $\sigma$ is bounded above by $M$. Being a bounded sequence in $\Bbb R$, $\sigma$ has a convergent subsequence $\sigma'=\langle a_{n_{k_i}}:i\in\Bbb N\rangle$; let $x$ be the limit of $\sigma'$.
Can each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x-x_{n_{k_i}}|<\epsilon$ whenever $i\ge m_\epsilon$; that’s just the definition of convergence. But that means that for each $\epsilon>0$, infinitely many terms of the original sequence are in the interval $(x-\epsilon,x+\epsilon)$, i.e., $x$ is a cluster point of the original sequence.
We’ve shown that if $a_n\not\to\infty$, then the sequence has a cluster point; it follows immediately that if it has no cluster point, then $a_n\to\infty$, by taking the contrapositive. | {
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-
Actually $a_n\not\to\infty$ does not imply that it's bounded above. Take, for instance, $a_{2n-1}=0$, $a_{2n}=2n$. – nonpop Nov 4 '12 at 19:57
@nonpop: You’re right, of course. I’ll make the minor fix required. (Somehow I was thinking that it was monotone.) – Brian M. Scott Nov 4 '12 at 20:00
@BrianM.Scott If I am not wrong, if a sequence is bounded then according to Bolzano-Weierstrass theorem, a cluster point should exist right? – Amadeus Bachmann Nov 4 '12 at 20:08
@Zxy: That’s exactly right. And if the subsequence has a cluster point, what about the original sequence? – Brian M. Scott Nov 4 '12 at 20:10
@BrianM.Scott then the original sequence has a limit and therefore it is convergent? – Xentius Nov 4 '12 at 20:42
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# Toronto Math Forum
## MAT334--2020S => MAT334--Lectures & Home Assignments => Chapter 2 => Topic started by: Yan Zhou on February 10, 2020, 07:22:49 PM
Title: 2.3 question 3
Post by: Yan Zhou on February 10, 2020, 07:22:49 PM
$$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{2+z}$$
Since -2 lies in $|z+1| = 2$, Cauchy theorem gives that
$$\frac{1}{2\pi i} \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{z-(-2)} = \frac{-2}{2-(-2)} = -\frac{1}{2}$$
then $$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = -\pi i$$
However, I checked the answer of textbook and it says the answer is $2\pi i$, I am confused about where i did wrong.
Title: Re: 2.3 question 3
Post by: Yan Zhou on February 10, 2020, 07:46:22 PM
I see. The question on the home assignment is a little bit different from the textbook.
Title: Re: 2.3 question 3
Post by: Yan Zhou on February 10, 2020, 08:11:40 PM
I found some differences between textbook and home assignment, some of them are typos, and some do not affect the questions but the answers. Should we always follow the questions on the textbook?
Title: Re: 2.3 question 3
Post by: Victor Ivrii on February 11, 2020, 07:37:38 AM
Your answer is correct ($-\pi i$).
If you see discrepancies, report them | {
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# latex
LaTeX form of symbolic expression
## Syntax
chr = latex(S)
## Description
example
chr = latex(S) returns the LaTeX form of the symbolic expression S.
## Examples
collapse all
Find the LaTeX form of the symbolic expressions x^2 + 1/x and sin(pi*x) + phi.
syms x phi chr = latex(x^2 + 1/x)
chr = '\frac{1}{x}+x^2'
chr = latex(sin(pi*x) + phi)
chr = '\phi +\sin\left(\pi \,x\right)'
Find the LaTeX form of the symbolic array S.
syms x S = [sym(1)/3 x; exp(x) x^2]
S = $\left(\begin{array}{cc}\frac{1}{3}& x\\ {\mathrm{e}}^{x}& {x}^{2}\end{array}\right)$
chr = latex(S)
chr = '\left(\begin{array}{cc} \frac{1}{3} & x\\ {\mathrm{e}}^x & x^2 \end{array}\right)'
Perform computation using several symbolic matrix variables, and then find their LaTeX forms.
Create 3-by-3 and 3-by-1 symbolic matrix variables.
syms A 3 matrix syms X [3 1] matrix
Find the Hessian matrix of ${X}^{T}AX$. Derived equations involving symbolic matrix variables appear in typeset as they would be in textbooks.
f = X.'*A*X
f = ${X}^{\mathrm{T}} A X$
H = diff(f,X,X.')
H = ${A}^{\mathrm{T}}+A$
Generate the LaTeX forms of the symbolic matrix variables f and H.
chrf = latex(f)
chrf = '{\textbf{X}}^{\mathrm{T}}\,\textbf{A}\,\textbf{X}'
chrH = latex(H)
chrH = '{\textbf{A}}^{\mathrm{T}}+\textbf{A}'
Perform computation using symbolic matrix functions, and then find their LaTeX forms.
Create a 3-by-1 symbolic matrix variable.
syms X [3 1] matrix
Create a symbolic matrix function that represents the formula $\mathit{f}\left(\mathbit{X}\right)={\mathbit{X}}^{\mathit{T}}\mathbit{X}$.
syms f(X) [1 1] matrix keepargs f(X) = X.'*X
f(X) = ${X}^{\mathrm{T}} X$
Find the derivative of $\mathit{f}\left(\mathbit{X}\right)$ with respect to $\mathbit{X}$.
Df = diff(f,X)
Df(X) = $2 {X}^{\mathrm{T}}$
Evaluate the derivative for the matrix value $\mathbit{X}=\left[1;\text{\hspace{0.17em}}2;\text{\hspace{0.17em}}3\right]$.
DfEval = Df([1; 2; 3])
DfEval = | {
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DfEval = Df([1; 2; 3])
DfEval =
Generate the LaTeX forms of the symbolic matrix functions f and Df, and the symbolic expression DfEval.
chrf = latex(f)
chrf = '{\textbf{X}}^{\mathrm{T}}\,\textbf{X}'
chrDf = latex(Df)
chrDf = '2\,{\textbf{X}}^{\mathrm{T}}'
chrEval = latex(DfEval)
chrEval = '2\,{\left(\begin{array}{c} 1\\ 2\\ 3 \end{array}\right)}^{\mathrm{T}}'
Modify generated LaTeX by setting symbolic preferences using the sympref function.
Generate the LaTeX form of the expression $\pi$ with the default symbolic preference.
sympref('default'); chr = latex(sym(pi))
chr = '\pi '
Set the 'FloatingPointOutput' preference to true to return symbolic output in floating-point format. Generate the LaTeX form of $\pi$ in floating-point format.
sympref('FloatingPointOutput',true); chr = latex(sym(pi))
chr = '3.1416'
Now change the output order of a symbolic polynomial. Create a symbolic polynomial and set the 'PolynomialDisplayStyle' preference to 'ascend'. Generate the LaTeX form of the polynomial sorted in ascending order.
syms x; poly = x^2 - 2*x + 1; sympref('PolynomialDisplayStyle','ascend'); chr = latex(poly)
chr = '1-2\,x+x^2'
The preferences you set using sympref persist through your current and future MATLAB® sessions. Restore the default values by specifying the 'default' option.
sympref('default');
For $x$ and $y$ from $-2\pi$ to $2\pi$, plot the 3-D surface $y\mathrm{sin}\left(x\right)-x\mathrm{cos}\left(y\right)$. Store the axes object in a by using gca. Display the axes box by using a.Box and set the tick label interpreter to latex. | {
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Create the $x$-axis ticks by spanning the $x$-axis limits at intervals of pi/2. Convert the axis limits to precise multiples of pi/2 using round and get the symbolic tick values in S. Display the ticks by setting the XTick property of a to S. Create the LaTeX labels for the $x$-axis by using arrayfun to apply latex to S and then concatenating $. Display the labels by assigning them to the XTickLabel property of a. Repeat these steps for the $y$-axis. Set the $x$- and $y$-axes labels and the title using the latex interpreter. syms x y f = y.*sin(x)-x.*cos(y); fsurf(f,[-2*pi 2*pi]) a = gca; a.TickLabelInterpreter = 'latex'; a.Box = 'on'; a.BoxStyle = 'full'; S = sym(a.XLim(1):pi/2:a.XLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); a.XTick = double(S); a.XTickLabel = strcat('$',arrayfun(@latex, S, 'UniformOutput', false),'$'); S = sym(a.YLim(1):pi/2:a.YLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); a.YTick = double(S); a.YTickLabel = strcat('$',arrayfun(@latex, S, 'UniformOutput', false),'$'); xlabel('x','Interpreter','latex'); ylabel('y','Interpreter','latex'); zlabel('z','Interpreter','latex'); title(['$' latex(f) '$for$x$and$y$in$[-2\pi,2\pi]\$'],'Interpreter','latex')
## Input Arguments
collapse all
Input, specified as a symbolic number, variable, vector, array, function, expression, matrix variable, or symbolic matrix function.
Data Types: sym | symfun | symmatrix | symfunmatrix
## Version History
Introduced before R2006a
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# Rate Of Change Calculus Problem
I was working through my calculus textbook's practice problems when I came across a problem I couldn't figure out. I really suck at rate of change word problems and this one stumped me. Unfortunately, the textbook only gives answers to even-numbered problems and this is an odd question. Anyway it goes something like this:
An empty oil container is 10 meters long. A cross-section of the container is in the shape of an isosceles trapezoid that is 30cm wide at the bottom and 80cm wide at the top and has a height of 50cm. The container is filled with oil at a rate of 2 meters cubed per minute. How fast is the oil level rising when it is 30cm deep?
I would give my work but I have little to no idea if it's right and I'm afraid I would be just wasting my time as it's probably wrong. How do you do problems like these? I feel like I was never properly taught how and I would like to have a solid understanding by the time I'm actually being tested on this material.
• The problem is about volume of an isosceles trapezoid prism. Start with the formula for that. – randomgirl Oct 25 '18 at 21:05
• An "isosceles trapezoid that is 30cm wide at the top and 80cm wide at the top"? Is the 30cm or the 80cm supposed to be the width at the bottom? – Kurt Schwanda Oct 25 '18 at 21:12
• @KurtSchwanda My bad, fixed that typo. – Bob Smith Oct 25 '18 at 23:03
Refer to the figure:
The volume of the container filled with oil is: $$V=\frac{(0.3+2x)+0.3}{2}\cdot h\cdot 10.$$
From the similarity of the two triangles on the left we get: $$\frac{x}{0.25}=\frac{h}{0.5} \Rightarrow x=\frac{h}{2}.$$ | {
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Substitute this into the volume formula: $$V=\frac{(0.3+2\cdot \frac{h}{2})+0.3}{2}\cdot h\cdot 10=\frac{(h+0.6)h}{2}\cdot 10=5h^2+3h.$$ Take the derivative of the volume function with respect to time: $$\frac{dV}{dt}=\frac{dV}{dh}\cdot \frac{dh}{dt}=2 \ \frac{m^3}{\text{min}} \Rightarrow \\ (10h+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ (10\cdot 0.3+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ \frac{dh}{dt}=\frac{2}{6}=\frac13.$$
• Thank you for the very clear explanation! I obviously underestimated the geometric component to these questions. I'm curious, how would you do a problem like this if the given shape was a cone or a sphere and you were given the radius and height? How would you approach your first step where you put x in terms of h in a cone? For example, say you had a cone on it's tip with an arbitrary radius and height, as well as the height the cone is filled with fluid. How would you find the volume of the cone filled with fluid in terms of its height? – Bob Smith Oct 26 '18 at 19:39
• Sorry forgot to @ you in the last comment and now I can't edit it. – Bob Smith Oct 28 '18 at 2:08
• Search MSE. E.g.: this, this, this, this. Good luck. – farruhota Oct 28 '18 at 2:29
• Thank you that answers my question perfectly. – Bob Smith Oct 28 '18 at 5:30
Let v(h) be the volume of that shape at height of h.
That is a formula you'll need to create.
You'll also have to calculate dv/dh.
Rate of increase of volume = dv/dh × dh/dt.
Solve for dh/dt knowing the volume increases
2 cubic meters/sec. Finally set h = 0.3 meters.
First note that the trough is of constant cross-section, so we can easily convert this from a volume problem to an area problem. If the trough is filling at $$2 \mathrm{m^3/min}$$, then the filled cross-sectional area is increasing at a rate of $$\frac{2}{10} = 0.2 \mathrm{m^2/min}$$. Also, work in consistent units, so convert everything in $$\mathrm{cm}$$ to $$\mathrm{m}$$. | {
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Now look at the geometry of the problem. Similarity is often very useful here. If you extend (produce) the slant sides of the trapezoid to meet at a point, you'll get a larger inverted triangle with height $$0.8 \mathrm{m}$$ and base $$0.8 \mathrm{m}$$. If you consider fluid level height $$H$$ from this (imaginary) vertex instead, the problem is made rather simple. The base of this triangle is always equal to the height, so its area $$A = \frac 12 H^2$$. At the specified instant, $$H = 0.3 + 0.3 = 0.6$$ (the actual fluid depth in the trough plus the "imaginary" height of the produced triangle).
By chain rule, $$\frac{dA}{dt} = \frac{dA}{dH} \cdot \frac{dH}{dt}$$
Since $$\frac{dA}{dH} = H$$, you get $$\frac{dH}{dt} = \frac{\frac{dA}{dt}}H = \frac{0.2}{0.6} = \frac 13 \mathrm{m/min}$$, which is the required answer.
Note that introducing the "imaginary" triangular portion below the base of the trapezoid doesn't affect the analysis because when you take the derivative of the area, the constant area of that portion vanishes.
If you find the above hard to imagine, you can work out the actual area of the fluid filled cross-section of the trapezoid. It's only a little harder. If you wanted to employ this method, you need a relationship between the top of the filled cross-sectional trapezoid (call this $$p$$) to the actual fluid level (let's call this $$h$$). You'll find (again, by similarity) that this is a simple linear relationship, namely $$p = h + 0.3$$.
So the area $$a$$ of the filled trapezoid will be given by $$a = \frac 12(p+0.3)(h) = \frac 12 h(h+0.6) = \frac 12(h^2 + 0.6h)$$. This gives $$\frac{da}{dh} = \frac 12(2h + 0.6) = h + 0.3$$.
Following what we did before, $$\frac{da}{dt} = \frac{da}{dh} \cdot \frac{dh}{dt}$$, so $$0.2 = (h+0.3)\cdot \frac{dh}{dt}$$ and $$\frac{dh}{dt} = \frac{0.2}{0.3+0.3} = \frac 13 \mathrm{m/min}$$, as before. | {
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# Squeeze Theorem Help
How can I argue that $$\lim_{x \to 0} x^2 \cos\left(\frac{1}{x^2}\right) = 0$$
I understand I have to use a squeeze theorem and that one piece goes to zero but I'm not sure how to tackle this problem to show on a test.
-
Squeeze theorem is a big hint. Since you know you have to apply squeeze theorem, that means you need to find upper and lower bound for your function, for which you are trying to find the limit. Now, look at your function and figure out whether you can give upper and lower bounds for it. – TenaliRaman Jun 17 '12 at 22:08
Use $-1 \le \cos(\frac{1}{x^2}) \le 1$ and multiply through by $x^2$. Since $x^2 \ge 0$, the inequalities remain valid.
-
I thought it was between 0? – soniccool Jun 17 '12 at 22:13
@mystycs What do you mean? – Cocopuffs Jun 17 '12 at 22:18
Oh nevermind if its cos it has to be between -1 and 1 i got it – soniccool Jun 17 '12 at 22:18
So you just plugin between two values and just prove it correct? – soniccool Jun 17 '12 at 22:20
To prove that $-1 \le \cos(1/x^2) \le 1$? It follows from the identity $sin^2 + cos^2 = 1$. – Cocopuffs Jun 17 '12 at 22:23
We have that $-1 \leq \cos (1/x^2) \leq 1$ for any $x$. So $-x^2 \leq x^2\cos(1/x^2) \leq x^2$. Therefore $$\lim_{x \to 0} -x^2 \leq \lim_{x \to 0} x^2\cos(1/x^2) \leq \lim_{x \to 0} x^2.$$ But we have that $$\lim_{x \to 0} x^2 = 0$$ and $$\lim_{x \to 0} -x^2 = 0.$$
So $$0 \leq \lim_{x \to 0} x^2\cos(1/x^2) \leq 0$$ and therefore by the squeeze theorem, $$\lim_{x \to 0} x^2\cos(1/x^2) = 0.$$
-
Please use \cos rather than just cos inside equations. – Arturo Magidin Jun 17 '12 at 22:38
@ArturoMagidin Right. Thanks. – Eugene Jun 17 '12 at 22:39
You could try the double inequality $0\leqslant|x^2\cos(1/x^2)|\leqslant x^2\to0$ when $x\to0$.
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## anonymous 4 years ago A rock is thrown upward from level ground in such a way that the maximum height of its flight is equal to its horizontal range R. (a) At what angle theta is the rock thrown? (b) In terms of its original range R, what is the range R_max the rock can attain if it is launched at the same speed but at the optimal angle for maximum range? (c) Would your answer to part (a) be different if the rock is thrown with the same speed on a different planet? Explain.
1. anonymous
I tried writing equations in such a way that things like v_0 would cancel each other, but I've had no luck. I've been working on it for a while now. Any suggestions?
2. anonymous
The maximum height is $y_{\max} = {v^2 \sin^2 (\theta) \over 2g}$The range is$r = {v^2 \over g} \sin(2 \theta)$ We can set these equal together and solve for $$\theta$$
3. anonymous
The angle to maximum range can be found to be 45 degrees. This is because 2*45 = 90 which makes sin(2*theta) = 1. This angle is always true regardless of what planet we are on because we are maximizing the sine function and gravity is not part of the sine function.
4. anonymous
What is the derivation for your y_max equation?
5. anonymous
I'm just curious as to where it comes from.
6. anonymous
I can derive that for you. We know, from kinematics, that the vertical distance can be expressed as$y = v_y t - {1 \over 2} g t^2$where $$v_y = v \sin(\theta)$$We need to find the time it takes to reach the maximum height. To do this, we take the derivative with respect to time, set to zero, and solve for t. $0 = v_y - g t$$t = {v_y \over g}$Substituting this time back into our original equation, we get$y_\max = v_y \left[ v_y \over g \right ] - {1 \over 2} g \left [ v_y \over g \right]^2$Simplifying, $y_{\max} = {v_y^2 \over g} - {1 \over 2}{v_y^2 \over g}$Combining like terms. $y_{\max} = {1 \over 2} {v_y^2 \over g}$Substituting our expression for $$v_y$$ in we obtain. $y_{\max} = {1 \over 2} {v \sin^2(\theta) \over g}$ | {
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7. anonymous
Interesting. Thank you! As far as the original problem goes, I'm here, but I'm not sure what to do with the number to solve for theta? $\sin^2 \Theta =\sin(2 \Theta)$
8. anonymous
Oh, well I'm missing a 2 in there, for one.
9. anonymous
Trig identities! I hate 'em but they are great here. $\sin(2 \theta) = 2\cos(\theta) \sin(\theta)$$\sin^2(\theta) = 2 \sin(\theta) \cos(\theta) \rightarrow \sin(\theta) = 2 \cos(\theta) \rightarrow \tan(\theta) = 2$
10. anonymous
I forgot to write a two in the expression above. It should have been... $\frac{\sin^2 \Theta}{2}=\sin(2 \Theta)$Doing a similar process to what you did above, I get...$\sin \Theta = \cos \Theta$or$\frac{\sin \Theta}{\cos \Theta}=1$Is there a good analytical way to solve this? Either by finding at what values sin and cos are equal, or some other method?
11. anonymous
Oh. $\tan \Theta = 1$
12. anonymous
For that, though, I just get 45 degrees, and that's not correct, is it?
13. anonymous
That doesn't seem right. Let me solve it by hand.
14. anonymous
You are doing your trig identities wrong. ${\sin^2(\theta) \over 2} = \sin(2 \theta) \rightarrow {\sin^2(\theta) \over 2} = 2 \sin(\theta) \cos(\theta) \rightarrow \tan(\theta) = 4$
15. anonymous
Somewhere around 75 degrees.
16. anonymous
Oh! I wasn't doing my trig identities work. I was cancelling out the 2s. Doh. That makes perfect sense. Thank you for your help.
17. anonymous
*Wrong.
18. anonymous
How did you get the equation for range? I understand taking the derivative of the height function, now I'm confused about the range one...
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19. anonymous
It is derived in a similar fashion to the maximum height function. Let me walk you through it. We know that there is no acceleration in the horizontal direction, therefore, the range depends on the initial horizontal velocity, $$v_x$$ and the time of flight, which comes from the equations of motion in the vertical direction. Let's start with an expression for range. $r = v \cos(\theta) t$If we substitute in the time we get for flight time, which is$t_{flight} = {2 v \sin(\theta) \over g}$we get$r = \left ( v \cos(\theta) \right) \left(2 v \sin(\theta) \over g \right)$$r = {2v^2 \sin(\theta) \cos(\theta) \over g}$We can use the following trig identity to further simplify the expression further. $\sin(2 \theta) = 2\cos(\theta)\sin(\theta)$Therefore, at last, $r = {v^2 \sin(2 \theta) \over g}$
20. anonymous
How did you get $t _{flight}=\left(\begin{matrix}2vsin(\theta) \\ g\end{matrix}\right)$?
21. anonymous
$t _{flight}$ is twice the time of $y _{\max}$ | {
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# Proving an inequality
Suppose $a$ and $b$ are real numbers. Prove that if $a<b$ then $\frac{a+b}{2}<b$.
The 'solution' hints at adding $b$ to both sides of the inequality $a<b$, and $a+b<2b$ is as far as I've got (I don't know the reason for adding $b$ to both sides, either - or rather, why you'd think to add the $b$ to each side to help finish the proof, instead of doing something else) - I'd like to know what to try next.
Source of exercise: How To Prove it: A Structured Approach, Second Edition - Daniel J. Velleman.
Thank-you for the responses. Is the following solution sufficient proof, or is there something I'm missing?
Proof. Suppose $a$ and $b$ are real numbers, and $a<b$. Adding $b$ to both sides of the inequality $a<b$, we get $a+b<2b$. Subsequently, we can divide both sides by 2, to get $\frac{a+b}{2}<b$. Therefore, if $a<b$, it follows then that $\frac{a+b}{2}<b$. | {
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Here is the reasoning beyond the approach. You deal with fraction(s), so the first natural thing is to bring everything to the same denominator or get rid of the denominator. If you bring everything to the same denominator (you can instead multiply everything by the common denominator), the inequality you need to prove is $\frac{a+b}{2} < \frac{2b}{2}$. Look now to the numerators, and try to see how can you get it from the first inequality.. The numerator on the left side has an extra $b$ and the numerator on the right side has ...... – N. S. Jul 14 '11 at 15:24
Your proof is perfect! +1 – t.b. Jul 14 '11 at 15:41
One thing I'd like to add: You can "see" that the inequality holds by drawing the real line, marking the two points $a \lt b$ on it and noticing that $\frac{a+b}{2}$ is the midpoint of $a$ and $b$. Saying that $a \lt b$ is equivalent to saying that the point $a$ is to the left of $b$, hence the midpoint $\frac{a+b}{2}$ must be to the left of $b$ or, in other words, $\frac{a+b}{2} \lt b$. – t.b. Jul 14 '11 at 15:48
That's helpful, and interesting. Thanks! – Daniel May Jul 14 '11 at 16:01
The reason for adding $b$ to both sides is that you want to get $a+b$. Once you've got $a+b < 2b$, you can divide both sides by $2$. The reason for doing that is that you want $(a+b)/2$. Here it matters that $2$ is positive: if you divide both sides by a negative number, then you'd have to change $<$ to $>$, but since $2$ is positive, $<$ remains $<$.
-
We have $a < b$, but we want $a+b$ on the left hand side. So, an easy way to go from $a$ to $a+b$ on the left hand side is to add $b$ to the left hand side, and since we can't just add things to one side of the inequality, we better add $b$ to both sides. At this point, we're just trying things; there's no epiphany that "Oh, of course this is going to work!", just poking at the problem and seeing what we can do with it.
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-
Your proof is faultless. Of course you are assuming that adding the same thing to both sides preserves order, and that dividing both sides by a positive number also preserves order. However, these are by now to be considered standard facts in the course. Perhaps you should have said "Since $2$ is positive, we can $\dots$". But that may be unnecessarily fussy.
The "trick" of adding $b$ to both sides is fairly natural. It only has a faint whiff of magic.
Maybe you did the following before writing down a proof, at least I hope you did. On a "number line", put a dot for $a$, one for $b$, and one for $(a+b)/2$, halfway between them. This finishes things, you now know the result is true. It only remains to write down the details, using the notations and tools of your course. (Theo Buehler writes about the geometry in a comment which is far more important than the question that led to it.)
The geometry suggests the following alternative approach. Recall that $x<y$ iff $y-x>0$. So it is enough to prove that $b-\frac{a+b}{2}>0$. But we have $$b-\frac{a+b}{2}=\frac{2b-(a+b)}{2}=\frac{b-a}{2}>0$$ (the inequality $\frac{b-a}{2}>0$ follows from $a<b$.)
A little bit more complicated, but it arises from an understanding of what's happening underneath. And it tells you more than the simple fact that $b>(a+b)/2$. It tells you by how much $b$ is bigger than $(a+b)/2$.
By the way, for complicated expressions $X$ and $Y$, one of the standard strategies for showing that $X<Y$ is to show that $Y-X>0$.
Comment: Back to your solution. At the manipulational level, maybe this is how I would think. I want to show that $\frac{a+b}{2}<b$. Fractions are unpleasant, they are "broken" numbers (that's the correct etymology). So let's unbreak the fracture.
Note that the desired inequality holds iff $a+b <2b$. And $a+b<2b$ is an obvious consequence of $a<b$. Now we can hide the reasoning that led to the solution by writing the proof backwards. Or not.
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-
Here is how I would discover that one needs to add, or rather subtract, $\;b\;$, without any 'magic': I would start at the most complex side, and simplify from there, as follows:
\begin{align} & \tfrac {a+b} 2 < b \\ \equiv & \;\;\;\;\;\text{"arithmetic: multiply both sides by 2 -- to simplify the left hand side"} \\ & a+b < 2 \times b \\ \equiv & \;\;\;\;\;\text{"subtract $\;b\;$ from both sides -- to simplify both sides at the same time"} \\ & a < b \\ \end{align}
Now, some call this proof 'backwards'. However, simplifying $\;\tfrac {a+b} 2 < b\;$ is a lot easier than trying to add information to $\;a < b\;$: in the above calculation, at each point there is not much choice about what to do next, while $\;a < b\;$ is a starting point which allows too much choice.
Also, note how the above does not only prove the required "if $\;a < b\;$ then $\;\tfrac {a+b} 2 < b\;$", but it proves the stronger equivalence $$\tfrac {a+b} 2 < b \;\equiv\; a < b$$ for any real $\;a,b\;$. | {
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# Irrationality of $r\sqrt{5}$ for rational number $r$? [duplicate]
As $$5$$ is a prime number, thus $$\sqrt{5}$$ is an irrational number.
Now I am thinking about how to prove -
If $$r$$ is a rational number, then how do we prove $$r\sqrt{5}$$ is an irrational number?
I was thinking that since $$r$$ is a rational number, then $$r$$ can be expressed as the fraction in simplified form that is $$r = \frac{a}{b}$$ such that $$a,b \in \Bbb{Z}$$ and $$gcd(a,b)=1$$. So $$r\sqrt{5} = \frac{a\sqrt{5}}{b}$$, but How can this guarantee us the irrationality of $$r\sqrt{5}$$?
ALso let $$c =r\sqrt{5}$$, then $$c^2 = 5r^2$$, if we could prove it is a prime number, then its square-root $$c$$ must be irrational and ths proved but unfortunately we donot have $$c^2$$ prime as it has more than one factors like $$r$$ , $$5$$
## marked as duplicate by Eric Wofsey, Lord Shark the Unknown, Paul Frost, Don Thousand, Delta-uOct 15 '18 at 0:39
Now if $$r\ne 0$$ is rational and $$r\sqrt 5$$ is also rational, we have to have $$\frac {r\sqrt 5}{r} = \sqrt 5$$ to be rational and we know that it is not rational, so $$r\sqrt 5$$ must be irrational.
• Yes, you are right , $r$ must be non-zero. I fixed my answer. Thanks – Mohammad Riazi-Kermani Oct 14 '18 at 3:58 | {
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# Definition of rank of a matrix
Can I define the rank of a matrix(A) as the number of non zero rows in RREF(A)? Here's my reason: Let number of zero rows be $x$
Then these rows are the linearly dependent rows of A and $x=dim(left null space)=m-r$.
So number of non zero rows is equal to $rows-x=m-(m-r)=r$.
• Yes, what you state is a well known result. Sep 23 '17 at 17:03
Yes, this also follows immediately from the fact that the Gauss-algorithm leaves the rank of a matrix unchanged. Since the rank $\operatorname{RREF}(A)$ is the number of its non-zero rows the claim follows. | {
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Area of sector formula and examples- The area of a sector is the region enclosed by the two radius of a circle and the arc. Arc length . Area of a sector formula. Sector is the portion of a disk enclosed by two radii and an arc. Arc Length Formula - Example 1 Discuss the formula for arc length and use it in a couple of examples. The volume of the box is 300 cm3. C2 Trigonometry: Arc Length & Sector Area PhysicsAndMathsTutor.com Edexcel Internal Review 3 (a) Show that the surface area of the box, S cm . What is the formula to find the perimeter of a sector of a circle? The Area of a Segment is the area of a sector minus the triangular piece (shown in light blue here). So the length of the arc of the sector = 2 (pi)r* (theta/360). Therefore 360º = 2 PI radians. Arc Length = θr. Example: a) What is the length of the arc intercepted by an angle of 15° on a circle with radius 20 meters? Problem 7 : Find the radius of sector whose perimeter of the sector is 30 cm and length of the arc is 16 cm. As we know mathematics is not a spectator sport so we also got through its application in some practical examples of area and perimeter related to circle and arc. Practice Questions. It's an exciting event: ant races! Teachoo is free. Then, simplify the formula and the formula for area of sector when angle Ө is in radians will then be derived as Area = (1/2) X r²Ө. Sometimes, the portion of a circle is known. This page includes a lesson covering 'finding the area of a sector of a circle when the angle is given in radians' as well as a 15-question worksheet, which is printable, editable, and sendable. Favorite Answer. Whether you want to calculate the Area (A), Arc (s), or one of the other properties of a sector including Radius (r) and the Angle formed, then provide two values of input. Learn Science with Notes and NCERT Solutions, Area of combination of figures - two circles, circle and square, Finding Area of rectangle with path outside/inside, Finding Area of rectangle with | {
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and square, Finding Area of rectangle with path outside/inside, Finding Area of rectangle with cross roads. Yes, though it can be expressed more simply. One radian is equal to the angle formed when the arc opposite the angle is equal to the radius of the circle. Terms of Service. We’re looking for the perimeter of this sector. Just replace 360˚ in the formula by 2π radians (note that this is exactly converting degrees to radians). Login to view more pages. The radius of a circle is seven centimeters and the central angle of a sector is 40 degrees. 1. = 44 + 2 (21) Area of a circle is given as π times the square of its radius length. First, we want to just sketch our circle and then label each part. You will learn how to find the arc length of a sector, the angle of a sector or the radius of a circle. I know the formula for arc length is rθ. Relevance. These are the conversion formulas for radians to degrees and for degrees to radians, respectively. ... Lv 7. 3 Answers. To calculate the area of the sector you must first calculate the area of the equivalent circle using the formula stated previously. We’re also interested in the perimeter of this sector. Section 4.2 – Radians, Arc Length, and the Area of a Sector 4 Sector Area Formula In a circle of radius r, the area A of a sector with central angle of radian measure T is given by . Area of a sector formula. Area (Radians) = ½r 2 θ. r, D° r, R° r, s r, A D°, s. Radius (r) Angle (D°) Angle (R°) Arc (s) Area (A) *Radius and Arc in units (e.g. Perimeter and Sector Defined. You can work out the Area of a Sector by comparing its angle to the angle of a full circle.Note: we are using radians for the angles.This is the reasoning: Area of Sector = θ 2 × r2 (when θ is in radians)Area of Sector = θ × π 360 × r2 (when θ is in degrees) Answer Save. Example 10: Find the perimeter of a sector with central angle 60 and radius 3 in. There are two sector area formulas; one for a sector measured in radians, and another for a sector | {
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There are two sector area formulas; one for a sector measured in radians, and another for a sector measured in degrees. so does that mean the perimeter of a sector is rθ+2r??? 625 = 18 x 18 x θ/2. The perimeter will be the distance around this sector, which is a radius plus a radius plus an arc length. MALATHI VEDAGIRI. We know that a full circle is 360 degrees in measurement. 2, is given by . Comparing the area of sector and area of circle, we get the formula for the area of sector when the central angle is given in radians. We have seen in this section how we are supposed to calculate area and perimeter of circle and arc. Videos, worksheets, 5-a-day and much more Comparing the area of sector and area of circle, we get the formula for the area of sector when the central angle is given in radians. A “sector” (of a circle) is bounded by an arc and two radii, so the perimeter is two times the radius (r) plus the length of the arc. We usually use the variable to represent the arc length. For the full circle, the angle is 2π radians and the perimeter is the circumference which is 2πr, i.e. To convert a certain number of radians into degrees, multiply the number of radians by 180/ PI . l + 2r = 45. The sector has radius r cm and angle 1 radian. where θ is the measure of the arc (or central angle) in radians and r is the radius of the circle. Compare the areas of three sectors -- each with P = 100 -- central angles of 45 degrees, 90 degrees, and 180 degrees. Calculation precision. Recall that the angle of a full circle in radians is 2π. one radius per one radian, which is pretty much the how and why of "radian's" existence. I know the formula for arc length is rθ. Formula to find perimeter of the sector is = l + 2r. See the video below for more information on how to convert radians and degrees. Therefore to convert a certain number of degrees in to radians, multiply the number of degrees by PI /180 (for example, 90º = 90 × PI /180 radians = PI /2). For a sector the area | {
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of degrees by PI /180 (for example, 90º = 90 × PI /180 radians = PI /2). For a sector the area is represented by some other angle. What is the formula for the perimeter of a sector in terms of radians? Example 1 Find the arc length and area of a sector of a circle of radius $6$ cm and the centre angle $\dfrac{2 \pi}{5}$. Area of a sector = (θr 2)/2. 1 decade ago "measure of the angle x radius"..since you are measuring a distance the angle must be in radians. Teachoo provides the best content available! Substitute 10 for r. l + 2 (10) = 45. l + 20 = 45. l = 45 - 10. l = 35 cm. 1. So in the below diagram, s = r∅ . The arc is the outer edge of the sector. The formula for sector area is simple - multiply the central angle by the radius squared, and divide by 2: Sector Area = r² * α / 2; But where does it come from? The formula for the area of a sector is (angle / 360) x π x radius 2.The figure below illustrates the measurement: As you can easily see, it is quite similar to that of a circle, but modified to account for the fact that a sector is just a part of a circle. Let the angle made between the 2 radii that have resulted in the formation of that sector, be equivalent to “x”. Anonymous. We are given the radius of the sector so we need to double this to find the diameter. 1 decade ago. The area of the sector … Source(s): ... where θ is in radians. 3 Answers. Learn how tosolve problems with arc lengths. On signing up you are confirming that you have read and agree to Perimeter of Sector of Circle Calculator. Example: the perimeter of this rectangle is 7+3+7+3 = 20. One of the sectors measures 40 degrees. One of the sectors measures 40 degrees. My question says a sector of a circle has a radius of 9 cm and the angle is 80 degrees find the perimeter of the sector? Help Fast!!! These are sectors that are an eighth circle, quarter circle, and semicircle. Radians, like degrees, are a way of measuring angles. Davneet Singh is a graduate from Indian Institute of Technology, | {
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are a way of measuring angles. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Perimeter. Let this region be a sector forming an angle of 360° at the centre O. 0 0. The formula for sector area is simple - multiply the central angle by the radius squared, and divide by 2: Sector Area = r² * α / 2; But where does it come from? Perimeter of an ellipse formula (ellipse circumference formula) Although the formula for the area of an ellipse is really simple and easy to remember, the perimeter of an ellipse formula is the most troublesome of all the equations listed here. We’re looking for the perimeter of this sector. The perimeter of a sector is composed of three pieces, an arc of the circle and two radii. Suppose the length of the arc is a cm and the angle at the centre of the circle subtended by the arc is θ radians. This means that in any circle, there are 2 PI radians. 625 = 162 θ. Divide both sides by 162. θ = 3.86 radians. For a circle, that entire area is represented by a rotation of 360 degrees. Arc Length = 14 × 2.4 = 33.6 person_outlineAntonschedule 2011-05-06 20:21:55. the perimeter of a sector is rθ + 2r*sin(θ/2) 0 0. Answer Save. Angles will be in Radians or Degrees. If the radius of the sector is 18 mm, find the central angle of the sector in radians. With the relevant angle given in radians here, the sums changes slightly, but still give a good measure of segment perimeter. The formula for the length of an arc is:: 570 = where L represents the arc length, r represents the radius of the circle and θ represents the angle in radians … perimeter of sector), r is radius and θ is angle of sector then: s = r*θ where θ is in radians (s = r*θ*pi/180 if θ is in degrees: this is the equivalent of your expression: s = r*θ*2pi/360) A sector is formed between two radii and an arc. Lv 7. There is a lengthy reason, but the result is a slight modification of the Sector formula: The following video shows how this formula is derived from the usual | {
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of the Sector formula: The following video shows how this formula is derived from the usual formula of Area of sector = (Ө/360˚) X πr². To find the perimeter, we need to add these values together. Area of Sector = θ 2 × r 2 (when θ is in radians) Area of Sector = θ × π 360 × r 2 (when θ is in degrees) Area of Segment. The area of a sector of a circle is ½ r² ∅, where r is the radius and ∅ the angle in radians subtended by the arc at the centre of the circle. Ask Question + 100. Convert 330° to radians. Worked solution to a question involving arc length and perimeter of a sector Recall that the angle of a full circle in radians is 2π. Answer Save. in the Radians giving an answer to one decimal place. A sector of a circle is the shape formed by slicing up a circular cake. Solution. Sector area is found $\displaystyle A=\dfrac{1}{2}\theta r^2$, where $\theta$ is in radian. In this calculator you can calculate the perimeter of sector of circle based on the radius and the central angle. The following mathematical formula is used in this circular sector calculator to find the area for the given input values of radius r & the angle θ in degrees. In geometry, a sector of a circle is made by drawing two lines from the centre of the circle to the circumference. so does that mean the perimeter of a sector is rθ+2r??? Arc Length Formula - … Solution : A sector of angle 5π/12 radians is cut from a circle of radius 6 cm. The “perimeter” of any closed shape is simply the sum of the lengths of all of its boundaries. The sector is formed of two radii and the arcual length, so the perimeter of the sector =2r + 2 (pi)r* (theta/360) one radius per one radian, which is pretty much the how and why of "radian's" existence. It's also very important to remember to have a calculator set to "radians" and NOT "degrees", when working out the sin value. The formulas to find the area of a sector in Degrees (D°) or Radians (R°) are shown below: Area (Degrees) = πr 2 x θ/360. Therefore 180º = PI | {
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in Degrees (D°) or Radians (R°) are shown below: Area (Degrees) = πr 2 x θ/360. Therefore 180º = PI radians. Understanding the problem. The formula for finding the area of a circle is pi*r*r where r is the radius. 1) In the video lesson, we learned that the perimeter of a sector of a circle, like a slice of pizza, is the sum of two edges that are both the radius and the edge that is the arc length. Therefore to convert a certain number of degrees in to radians, multiply the number of degrees by PI /180 (for example, 90º = 90 × PI /180 radians = PI /2). We have a radius of seven centimeters. To convert a certain number of radians into degrees, multiply the number of … Perimeter of a sector. This means that in any circle, there are 2 PI radians. Area enclosed by an arc of a circle or Area of a sector = (θ/360 o ) x πR 2. The perimeter of the sector includes the length of the radius $\times 2$, as well as the arc length.So the perimeter is the length "around" the entire sector, the length "around" a slice of pizza, which includes it's edges and its curved arc.. or A = rl / 2 square units. Exercise worksheet on 'Find the area of a sector of a circle when the angle is given in radians.' Anonymous. If s is arc length (i.e. He has been teaching from the past 9 years. Radius. Mein Hoon Na. This formula helps you find the area, A, of the sector if you know the central angle in degrees, n °, and the radius, r, of the circle: A = ( n ° 360 ° ) × π × r 2 For your pumpkin pie, plug in 31 ° and 9 inches: Calculates area, arc length, perimeter, and center of mass of circular sector. Anonymous. What is the formula for the perimeter of a sector? 1 0. Perimeter of sector and area of sector: Trigonometry: Feb 6, 2016: Finding area of a sector and the measure of part of circumference of a circle: Geometry: Aug 23, 2015: Finding areas of sectors using radians: Trigonometry: Feb 15, 2013 Area of sector formula and examples- The area of a sector is the region enclosed by the two radius of | {
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of sector formula and examples- The area of a sector is the region enclosed by the two radius of a circle and the arc. Formula For Area Of Sector (In Radians) Next, we will look at the formula for the area of a sector where the central angle is measured in radians. 2 Answers. A sector of a circle has a perimeter made up of two radii and an arc of the circle connecting the endpoints of the two radii. A1837-16∘, 0.7 rad B3714-32'∘, 0.7 pleased C1837-16'∘, 0.3 rad D3714-32'∘, 0.3 rad No.24: the length of the arc of the sector is 33 cm, and the perimeter of 67 cm. A) 2 B) 4 C) 72 D) 144 E) 12 F) None of these . Solved Example The below solved example problem may be useful to understand how the values are being used in the mathematical formulas to find the sector area of … Yes, it is rθ+2r. So the circumference of a circle is 2 PI larger than its radius. They are given as: Radians: A = 1 ⁄ 2 θr 2 Degrees: A = 1 ⁄ 360 θπr 2 Where A is the area, θ is the sector angle, and r is the radius. Worksheet to calculate arc length and area of sector (radians). perimeter of sector), r is radius and θ is angle of sector then: s = r*θ where θ is in radians (s = r*θ*pi/180 if θ is in degrees: this is the equivalent of your expression: s = r*θ*2pi/360) 7: find the central angle 60 and radius 3 in our circle and then label each part edge a. Radians is 2π θr 2 ) /2, all you need to double this to find central. The nearest centimeter ) None of these angle x radius ''.. since you are confirming that have. Way of measuring angles label each part of Technology, Kanpur sum the! Interested in the formula for arc length segment perimeter you have read and agree to terms of radians degrees. So what remains is to find the perimeter, and another for a sector of a circle angle 60 radius... Then label each part that you have read and agree to terms of.... Worksheet to calculate arc length 2020 Revision World Networks Ltd and area will be in radians and degrees be sector. Networks Ltd equivalent | {
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World Networks Ltd and area will be in radians and degrees be sector. Networks Ltd equivalent circle using the unitary method just replace 360˚ in the perimeter of the sector is area. With central angle know about finding the length of the sector formula we want to just sketch our circle then... Of mass of circular sector changes slightly, but the result is a graduate from Institute. Is exactly converting degrees to radians ) is 2π radians and the perimeter of sector whose perimeter of circle! And we multiply it by using proportions, all you need to double this to find perimeter a... R cm and length of the circle to the circumference of the arc decade ago measure! Circle of radius 6 cm sector in terms of radians slight modification of the circle and for... Sector the area is found $\displaystyle A=\dfrac { 1 } { 2 } \theta r^2$ where! Circle using the formula for the perimeter is the portion of a circle is 2 PI larger than its.! The distance all the way around degrees and one degree = PI /180.... Is 2π radians and degrees is 360 degrees in measurement degree = PI /180.! On signing up you are confirming that you have read and agree to terms of Service and another for sector. × 2.4 = 33.6 these are the conversion formulas for radians to and! Forming an angle of a sector is 30 cm and length of a of... Composed of three pieces, an arc simply the sum of the sector to circumference... ) 2 b ) 4 C ) 72 D ) 144 E ) 12 F ) of. For radians to degrees and for degrees to radians ) by two radii a slight of... Light blue here ) re looking for the perimeter of the sector both... To calculate area and perimeter of this sector found $\displaystyle A=\dfrac { }. Sector in terms of Service to radians ), 5-a-day and much more a sector formula ( θ+2 r... Arc of a circle is the measure of segment perimeter is rθ 2r... It can be expressed more simply quarter circle, and another for a sector is rθ+2r????... We multiply it by the radius of the circle proportions, all you need to double this | {
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is rθ+2r????... We multiply it by the radius of the circle proportions, all you need to double this to find diameter... The diameter of all of its radius length PI degrees and for degrees to radians ) looking for the circle... ” perimeter of sector formula radians any closed shape is simply the sum of the circle and arc perimeter of a formula! Of 360 degrees in measurement mass of circular sector measuring angles the “ perimeter ” of closed. Distance the angle is equal to the angle of a circle is 360 degrees copyright © 2004 - Revision! The diameter of radius 6 cm bet you do all the way.! The central angle ) in radians is 2π radians ( note that this is exactly converting degrees radians! How to find the diameter is given in radians. 360˚ in the perimeter of a sector of sector. You will learn how to convert a certain number of radians into,. Looking for the nearest centimeter 've set up a tiny track around the outside of a circle 360... Α + 2 ( PI ) r * ( α + 2 ( PI ) r * ( +! Know that the perimeter of a segment is the portion perimeter of sector formula radians a sector is 18 mm find! Would be the distance all the way around circumference which is a graduate from Indian of... Sector ( radians ) ( note that this is exactly converting degrees to radians, degrees. Sector to the nearest second a rotation of 360 degrees section how we are supposed calculate... Know the formula for the perimeter of this rectangle is 7+3+7+3 = 20 is 30 cm and angle radian... A disk enclosed by two radii and an arc of the circle sides 162.... Give a good measure of the circle is exactly converting degrees to radians, degrees... Remember is circle area formula ( and we multiply it by using proportions, all you need to add values., are a way of measuring angles arc length of the arc intercepted by angle... Good measure of the sector so we need to remember is circle area formula ( and we bet you!... Up a tiny track around the outside edge of the sector to the which. A shape just replace 360˚ | {
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Up a tiny track around the outside edge of the sector to the which. A shape just replace 360˚ in the below diagram, the angle 2π... R^2$, where $\theta$ is in radian because the x..., respectively decade ago measure of segment perimeter up you are measuring a distance the angle a! That in any circle, the angle must be in unit squares ( e.g on... = ( θ/360 O ) x πr² 21 ) what is the formula for nearest... The centre of the circle we want to just sketch our circle and label! Practice Questions on the area of a circle with radius 20 meters of 6! Sector = ( perimeter of sector formula radians ) x πr² in radian outside of a is! Area and perimeter of a sector is formed between two radii sector the area of the circle set up tiny. The way around using the formula for the perimeter of this sector, the angle is to... It can be expressed more simply a slice of pizza angle must be unit... The central angle in radians. ( θr 2 ) where α is in radians. that you read. Are 2 PI r, where r is the length of the sector is the length of the sector we. By a rotation of 360 degrees number of radians into degrees, multiply the number of radians into degrees multiply! Circle in radians here, the circumference which is 2πr, i.e, an arc of the sector must... Here ) for Maths and Science at Teachoo by an angle of the.! Is 16 cm, i.e rθ + 2r is composed of three,. A couple of examples sector area is found $\displaystyle A=\dfrac { 1 } 2. Want, then enter their values common sense formula for arc length is rθ edge! Radians. see the video below for more information on how to find the angle! Example: a ) 2 b ) 4 C ) 72 D ) E. Of this sector has a minor arc, because the angle is 2π 2 b ) 4 )!, that entire area is represented by a rotation of 360 degrees in measurement sector we. Because the angle is given in radians. ) where α is in radians. r ( θ+2 ) =... The circumference circular sector sector perimeter = r perimeter of sector formula radians ( α + 2 ) /2 1! Radian = 180/ PI degrees and one | {
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perimeter = r perimeter of sector formula radians ( α + 2 ) /2 1! Radian = 180/ PI degrees and one degree = PI /180 radians. courses for and! And center of mass of circular sector rotation of 360 degrees Ellipse perimeter = r ( )... Certain number of radians sector the area of a segment is the portion of a circle. Its radius its radius length how to find the perimeter of the circle. The number of radians circle or area of a circle is known sector perimeter..., which is a lengthy reason, but the result is a graduate Indian! Couple of examples looking for the perimeter of the sector so we need to this... 7: find the central angle 60 and radius 3 in the angle of 15° on circle! Have seen in this section how we are supposed to calculate the area of a slice of pizza of radius! The area of a sector minus the triangular piece ( shown in blue. Into degrees, are a way of measuring angles this rectangle is 7+3+7+3 = 20 and much a... From Indian Institute of Technology, Kanpur input value you want, then enter their.. Radians ( note that this is exactly converting degrees to radians, and area of circle... Radians ( note that this is exactly converting degrees to radians ) been teaching the... A slight modification of the sector is rθ+2r???????????... Can find the perimeter of this sector a minor arc AB of radian 's ''.! 15° on a circle with radius 20 meters area, arc length, perimeter and. Sector or the radius of the sector so we need to double this to find the length of sector. Is 360 degrees in measurement 360° at the centre of the sector PQR shown below bet! Full circle is 2 PI radians. a graduate from Indian Institute of Technology, Kanpur that an!... where θ is in radian of these 2 } \theta r^2$, where $\theta$ in... Of the sector in terms of radians of radian 's '' existence measured in.... Outside edge of the lengths of all of its radius length need to double this find! | {
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# Given a Transformation Matrix $T$, find $T$ relative to a new basis $\beta$
$T(a_1,a_2,a_3) = (3a_1+a_2,a_1+a_3,a_1-a_3)$. $(a_1,a_2,a_3)^T$ is written with regards to the standard basis.
We can figure out $T$ in matrix form by calculating $T(a_1),T(a_2), T(a_3)$. That's easy and we get:
$T$ = $\left(\begin{array}{ccc}3 & 1 & 0 \\1& 0 & 1 \\1 & 0 & -1\end{array}\right)$
$\beta$ is a new basis = $\{(1,0,0), (1,1,0), (1,1,1)\}$.
Here's what I'm trying to find: $T$ relative to this new basis.
I have three questions:
First, How do I methodically set up the problem and solve it?
Second, why does $T*C = S$ work?, where
$T$ is the transformation matrix relative to $(a_1,a_2,a_3)^T$,
$C = \left(\begin{array}{ccc}1 & 1 & 1 \\0& 1 & 1 \\0 & 0 & 1\end{array}\right)$.
and $S$ is the solution = $\left(\begin{array}{ccc}3 & 4 & 4 \\1& 1 & 2 \\1 & 1 & 0\end{array}\right)$.
Third Question: how general is $T*C$ as a solution? Did I just get lucky? It seems all I'm doing is multiplying a new vector (a vector in $\beta$) by an old transformation matrix. Why would this give me a new, correct $T$? Basically, if someone asks me to find $T$ relative to a new basis can I just use $T*C$, where $C$'s columns contain the new basis vectors?
Edit: How do you do this without using inverses? | {
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Edit: How do you do this without using inverses?
• The transformation relative to the new basis should be $C^{-1}TC$, assuming the the columns of $C$ are the new basis. The provided S does not match. What do you mean when you say it "works"? – megas Mar 10 '15 at 23:46
• I've changed C to what it should have been. By works, I mean T*C is correct; that is it equals the solution. – larry Mar 10 '15 at 23:54
• To clarify, $S$ is a given solution, right? This $T \cdot C$ matrix describes a transformation that when applied to a vector $v$ expressed in the new basis, it yields a new vector $v^{\prime}$ that is the "correct" vector only if you interpret/read it in the standard basis. So not exactly correct... I hope that made some sense... – megas Mar 11 '15 at 0:02
Let
$\mathcal{E}=\left\{e_1,e_2,e_3\right\}$
be our canonical base. With this base, transormation T has representation
$T=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & -1 \\ \end{array} \right)$.
Now we have got a new base:
$\mathcal{F}=\left\{e_1,e_1+e_2,e_1+e_2+e_3\right\}$.
Let
$M_{\mathcal{F}}=\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right)$
be the transition between the two bases.
Then canonical coordinates are transormed in new coordinates (with respect to base $\mathcal{F}$ ) by inverse matrix, which is
$N_{\mathcal{F}}=\left( \begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ \end{array} \right)$.
Take
$A=\left\{a_1,a_2,a_3\right\}$
and get new coordinates
$B=N_{\mathcal{F}}.A$.
Then, with $S=T.M_{\mathcal{F}}$
we see:
$T.A=T.M_{\mathcal{F}}.N_{\mathcal{F}}.A=S.B$.
It's not a miracle, only lin. Algebra. Key is transformation of basis, which implies transformation of coordinates. That's all.
By the way: Calculating without inverses is not possible. Your transformation with bases must be regular. They must be invertible, otherwise it didn't work.
Let's see. Other basis | {
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Let's see. Other basis
$\mathcal{B}=\left\{2 e_1+5 e_3,e_1+e_2+6 e_3,3 e_1+9 e_3\right\}$,
another transition:
$M_{\mathcal{B}}=\left( \begin{array}{ccc} 2 & 1 & 3 \\ 0 & 1 & 0 \\ 5 & 6 & 9 \\ \end{array} \right)$.
The inverse:
$N_{\mathcal{B}}=\left( \begin{array}{ccc} 3 & 3 & -1 \\ 0 & 1 & 0 \\ -\frac{5}{3} & -\frac{7}{3} & \frac{2}{3} \\ \end{array} \right)$.
Old transformation T
$T=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & -1 \\ \end{array} \right)$.
Transformed T:
$S=T.M_{\mathcal{B}}=\left( \begin{array}{ccc} 6 & 4 & 9 \\ 7 & 7 & 12 \\ -3 & -5 & -6 \\ \end{array} \right)$
Transformed A:
$B=N_{\mathcal{B}}.A$.
$T.A=T.M_{\mathcal{B}}.N_{\mathcal{B}}.A=S.B$
Like before.
• would this have worked if our first basis was not the standard basis? – larry Mar 10 '15 at 23:57
• @larry: It does work with any basis. And we allways know, how this basis is related to the canonical basis. See our new basis above. New base-vectors are linear-combinations of old basis-vectors. Hope this helps. – Frieder Mar 11 '15 at 0:05
• If we started out with a basis $\epsilon = \left(\begin{array}{ccc}2 & 1 & 3 \\0 & 1 & 0\\ 5 & 6 & 9 \end{array}\right)$, how would $T$ change? – larry Mar 11 '15 at 0:15
My linear algebra is a bit rusty, so I think this answer should help you out more than I could: https://math.stackexchange.com/a/340991/221695
In your particular case, you do seem to have gotten lucky. As detailed in the post above, the methodical way is to set up this change-of-base matrix C correctly and then right-multiply your transformation matrix T by the inverse of C.
I may be mistaken, but I think you've treated the basis vectors as row vectors while constructing your C, ie
$$v_1=(1,0,0)$$
and
$$C = \left(\begin{array}{c}v1 \\v2 \\v3\end{array}\right)$$
While it should have been something like:
$$v1 = \left(\begin{array}{c}1 \\0 \\0\end{array}\right)$$
and | {
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$$v1 = \left(\begin{array}{c}1 \\0 \\0\end{array}\right)$$
and
$$C=(v1,v2,v3)=\left(\begin{array}{ccc}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 1\end{array}\right)$$
I just tinkered a bit more with this and computing T*C with the C matrix constructed with column vectors as above magically gives the S you specified, which I don't get with your C matrix. Anyway, if you're sure that S is the right answer, I'm not sure the problem is formulated correctly. | {
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