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0

The direction towards the origin $O$ from $P$ is represented by the vector $$\mathbf u = O - P = (-1, 1).$$ The unit vector will then be $$\mathbf{\hat u} = \frac{\mathbf u}{\|\mathbf u\|} = \left(-\frac1{\sqrt{2}}, \frac1{\sqrt{2}}\right) = \frac{-i + j}{\sqrt 2}.$$ We then have $$\frac{\mathrm df}{\mathrm d{\mathbf{\hat u}}} = \nabla f \cdot \mathbf{\hat ...


0

Hint. $\overrightarrow{PQ} \times \overrightarrow{b} = \overrightarrow{0}$ iff $P$ lies on the line passing through $Q$ and parallel to $\overrightarrow{b}$. $\overrightarrow{PQ} \cdot \overrightarrow{c} = 5$ is the equation of a plane. At the end $P$ is the intersection point of the line and the plane.


0

If we consider the dot product result (which you provided), and cross product equation (which returns the result $\overrightarrow{PQ}×\overrightarrow{b}=<-p_1,2p_2,-p_3>$), we get a system of equations: $$ 2p_1 + 2p_2 + 5p_3 = 14 \\ -p_1+2p_2-p_3=0 $$ after much algebraic manipulation, we get: $p_1=\frac{14}{3}, p_2=\frac{14}{6}, p_3=0$


1

If you read the question carefully... you will notice it says find "a" vector b such that ... NOT find "the" vector b such that... The obvious choice is the easiest one, namely: Let $b=\frac{5}{|a|}a$ Then $$\frac{a\cdot b}{\vert a\vert} =\frac{a\cdot(\frac{5}{|a|}a)}{|a|}= 5\frac{|a|^2}{|a|^2}=5$$


0

The OP was a bit loose with showing their work, so I've expanded the derivations here, more-or-less trusting their numerically-checked derivations along the way: $$ \begin{align} \theta_{ba} &= \theta_b \cdot \nabla_a^\intercal \\ &= \gamma(b,c)(\cot\theta\, \mathbf{u}_{ab} - \csc\theta\, \mathbf{u}_{ac}) \cdot \nabla_a^\intercal \\ \\ ...


1

Here's an argument that should work in $\mathbb{R}^n$ for $n \geq 3$. I guarantee that it is not the most elegant approach. Assume $\gamma$ contains no segment along a line through the origin. (If $\gamma$ contains one or more such segments, shift it by a translation $\tau$ along a vector not in one of those lines. Then carry out the following procedure, ...


2

Since \begin{eqnarray} \frac{\partial F}{\partial x}(0,0)&=&\frac{\partial f}{\partial x}(0,0)+2\frac{\partial f}{\partial y}(0,0)=4+2(-3)=-2\\ \frac{\partial F}{\partial y}(0,0)&=&3\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)=3(4)-(-3)=15, \end{eqnarray} we have \begin{eqnarray} \frac{\partial ...


1

$f'(c,u)=\nabla f(c) . u=(4,-3). (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})= 4\frac{\sqrt{2}}{2}-3\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$ ( "." is inner product symbole.)


0

Directional derivative of $f$ at $c$ in the direction $u$ is given by $$f'(c,u)=\lim_{h\to 0}\frac{f(c+hu)-f(c)}{h}.$$


0

This is about an unknown function $(x,t)\mapsto U(x,t)$ (called $V'$ in your question) whose domain $\Omega$ is (part of) the $(x,t)$-plane. Consider a point $(x_0,t_0)\in\Omega$. If you look at this function along the horizontal line $t=t_0$ then you obtain a function $$\phi(x):=U(x,t_0)$$ of the single variable $x$, which presumably has a derivative at ...


2

Why do we need the linearity of $L$ wrt $h$? What do we lose if $L$ is not linear? The derivation at a point shall be the best linear approximation of the function at this point. More precisely $L$ is the best linear approximation of the difference function $g(h):=f(x+h)-f(x)$ for a fixed point $x$. Why $x$ is the argument of $L$, isn't it fixed? Shouldn't ...


0

You could also look at the function’s directional derivatives at the origin: If $\mathbf u=(\cos \theta,\sin \theta)$, then $$D_{\mathbf u}(0,0) = \lim_{h \to 0} \frac{f(h\cos \theta,h\sin \theta) - f(0,0)}{h}=(\cos^3\theta+\sin^3\theta)^{1/3}$$ But if $f$ is differentiable, then we must also have $D_\mathbf u=\nabla f\cdot \mathbf u=\cos \theta+\sin ...


0

Your book seems to assume that $y$ is a function of $x$, and not independent on $x$. In this case when you derive you'll also have the terms with $y'(x)$. But if you set $y'(x) =0$ (treating $y$ as a constant wrt to $x$) then you'll recover your result.


0

Let $U \subset \gamma^{-1}(N)$ be a neighborhood of $0$. Choose a smooth function $b : \mathbb{R} \to \mathbb{R}$ which is equal to $1$ outside $U$ and equal to $0$ in a neighborhood of $0$. Define $$\tilde{\gamma}(t) = b(t) \gamma(t).$$ Then $\tilde{\gamma}$ agrees with $\gamma$ outside of $\gamma^{-1}(N)$ ; it is smooth on $\mathbb{R}$ since it is constant ...


2

We are given $\nabla f=C\vec r$, where $\vec r$ is the position vector and $C$ is the constant of proportionality. Then, by integration, we find that $f$ is given by $$f(x,y,z)=\frac12 C(x^2+y^2+z^2)+C'$$ where $C'$ is an integration constant. Evaluating $f$ at $(0,0,a)$ and $(0,0,-a)$ reveals that $$f(0,0,a)=\frac12Ca^2+C'$$ and $$\begin{align} ...


0

Because $\nu$ is very small, rearranging things thus seems to be helpful: $$ 2\pi\int^a_b\nu \cdot e^{x^2}\int_{-\infty}^x\nu \cdot erfcx(-y)dydx. $$ However, the integration still fails for some values of $\nu$, $a$, and $b$.


0

In any neighborhood of $(0,0)$ there is a point of the form $(a,-a)$ with $0<a<1.$ Think of what happens as we approach $(a,-a)$ from the right. We have $$f(a+h,-a) = [\dots ]\cdot \left ( \frac{1}{(a+h)^3 +(-a)^3}\right)^{1/3},$$ where the expression in brackets has a nonzero limit as $h\to 0^+.$ But the other factor looks like $1/0^+$ as $h\to ...


0

Rather than using your trigonometric expressions, I'd opt to stick with the vector and polynomial calculations throughout. I think the definition $$ g(a,b,c) = (b-a ) \cdot (c-a)/{||b-a|| ||c-a||}$$ along with the identities $$ \nabla_b g(a,b,c) = \frac{(c-a) ||b-a||^2 - \left( (b-a) \cdot (c-a) \right) (b-a)}{||b-a||^3 ||c-a||},$$ and similarly for ...


0

We have $$v\times w=\langle v\times w,u\rangle u+\langle v\times w,v\rangle v +\langle v\times w,w\rangle w=\langle v\times w,u\rangle u=-\langle w, v\times u\rangle u=\langle w,u\times v\rangle u=\langle w,w\rangle u=u.$$


0

Recall the vector identity ORA-OAR : $$ A \times (B \times C) = (A \cdot C) B - (A\cdot B) C $$ Therefore, if $\{ u,v,w \}$ is an orthonormal basis, and $w = u \times v$, then $v\cdot v = 1$ and $v\cdot u=0$, which gives $$ v\times w = v \times (u \times v) = (v\cdot v)u-(v\cdot u)v=u. $$ Similarly, $$ w\times u = (u\times v)\times ...


0

Suppose f'(x) denotes the derivative of f at x,which is practically a linear transformation.Now as you vary x you can think of f' is a mapping from R^n to L(R^n,R^n),the set of all linear transformation from R^n to R^n.Now how you give metric structure on L(R^n,R^n)?Define for A in L(R^n,R^n) |A|=sup A(x) where |x|<=1(it's a compact set and linear maps ...


0

We will use the Schwarz theorem for the second derivatives, which states that $$ \frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2f}{\partial y\partial x} $$ for a function $f:\mathbb{R}^2\to\mathbb{R}$. Set the Cauchy-Riemann equations $$ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, ~~~~~(1) \\ \frac{\partial u}{\partial ...


4

Hint: replace $1+xy$ by $1$ in the numerator (why?) and reduce to $$ \lim_{(x,y) \to (0,0)} \left( \frac{x^3 y^3}{x^3+y^3} \right)^{1/3}. $$ Put $\alpha =x^3$, $\beta=y^3$ and consider the constraint $\alpha + \beta = \alpha^3$ with $\alpha \to 0$. Then $$ \frac{\alpha \beta}{\alpha+\beta} = \frac{\alpha^4-\alpha^2}{\alpha^3} $$ which becomes unbounded as ...


0

This is prossible? $ 0\leq|xy\left(\frac{1+xy}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3y^3+x^4y^4}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3y^3}{x^3+y^3}+\frac{x^3y^3(xy)}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3}{x^3+y^3}y^3+\frac{x^3}{x^3+y^3}y^3(xy)\right)^{1/3}|\leq|(y^3+y^3(xy))^{1/3}|\leq (y^2|y|+y^4|x|)^{1/3}\leq(|y|+|x|)^{1/3}$ so like the function is ...


-1

Try Polar coordinates maybe $x=r \cos \phi$ and $y=r \sin \phi$ and then check the limit $r \to 0$.


1

After calculating $P_x,Q_y,R_z$, deduce that $\nabla \cdot F$ (the sum of these) is zero, which means (by the divergence theorem) that the total flux (which is an integral of $\nabla \cdot F$) is zero. There's nothing wrong with your reasoning. You could calculate the flux without the theorem with a parametrized surface integral, but this approach would be ...


1

$\sin(x^2)\sim x^2$, and: $$ |x^3+y^3| \leq |x|^3+|y|^3 \leq (|x|+|y|)(x^2+y^2), $$ hence the limit is trivially $\color{red}{0}$.


0

Many years ago I had similar tasks in computer geometry. The formula is too complicated and it is too easy to miss it. So I decided to develop a C++ template class that holds some quantity --- a number --- together with its first two derivatives with respect to some real parameters. (The number of the real parameters is a template parameters) So the class ...


2

I would use the squeeze theorem. Since in a neighbouhood of $0$ you have $$\sin 4t^2 \ge t^2$$ you can substitute $x=2t$ and get $$0 \le \frac{x^3+y^3}{\sin x^2+y^2} = \frac {8t^3+y^3}{\sin 4t^2+y^2} \le \frac {8t^3+y^3}{t^2+y^2} \to 0$$ so the limit is $0$.


3

Divide top and bottom by $x^2$ to get rid of the sine, as $\sin x^2 / x^2 \to 1$, and take $y$ out of the new numerator to end up with $$\lim_{(x,y)\to(0,0)}y\frac{x/y + y^2 / x^2}{1 + y^2 / x^2}, $$ which "could" not exist or be non-zero only if $\lvert x/y \rvert \to \infty$, but then again the limit would be the same as $$ \lim_{(x,y)\to(0,0)}y ...


0

i think it goes something like this x and y are concentrations $iii)$ gene A produces protein A at rate $\frac{1}{1+y}$ $ii)$ A degrades at rate x $i)$ $f(x,y)$ is difference between production and degradation rate so $f(x,y)=\frac{1}{1+y}-x$ $iv)$ i guess when B is absent its concentration is $0$ ($y=0$) so $f(x,0)=3-x$ ( production is 3 and i ...


0

I recommand that $U:=\cup_1^n \text{interior}\ D_i$. Now for compact set $D_i\subset U$, there exist non negative $C^{\infty} $ function $\psi_i: U\rightarrow [0,1]$ which is positive on $D_i$ and 0 outside of some closed set ( i.e. $\text{supp}(\psi_i)$) contained in $U$.


2

The induced metric is just the Euclidean metric on $\mathbb R^{n+1}$, i.e. $g_{ij} = \delta_{ij}$, but restricted to act on vectors tangent to $S^n$. You need to choose a system of $n$ coordinates parametrizing $S^n$ alone (or an $n$-frame tangent to $S^n$) if you want to get an $n \times n$ matrix of components for $g$.


0

For your third question: the bottom of page 63 says that the $\psi_i$ are nonnegative functions, so the only question is why at least one of them is positive at each point. Well, since the interiors of $D$ cover the space, every point is in some $D_i$, and $\psi_i$ is positive on $D_i$, so you're done. For question 2, since the (finite) sum of the $\psi$s ...


4

Let we set $A=x^2,B=y^2,C=z^2$ and: $$ S_n(A,B,C) = \frac{A^n}{(A-B)(A-C)}+\frac{B^n}{(B-A)(B-C)}+\frac{C^n}{(C-A)(C-B)} .\tag{1}$$ By partial fraction decomposition, it is straightforward to check that $S_0=S_1=0$ and $S_2=1$. For every $n>2$ induction gives: $$ S_n(A,B,C) = h_{n-2}(A,B,C)\tag{2} $$ where $h_k$ is a complete homogeneous symmetric ...


0

The most general approach is to define two sequences that converge to your desired point such that the sequence of function values (evaluated on these sequences) converge to two different values, or at least one of the limits of function values doesn't exist. Often by inspecting your function, you can parametrize your two sequences, as in your ...


1

Your conclusion is wrong. If the Hessian is positive definite, the critical point is a local minimizer.


0

HINT: adding the two given equations we obtain:$$4x+y=1$$ or $$x=\frac{1}{4}-\frac{y}{4}$$ plugging this in the second equation we get $$\frac{1}{4}-\frac{y}{4}+2y-z=-1$$ thus $$y=\frac{3}{7}+4z$$ plugging this in the second equation above and we get $$x=\frac{1}{7}-z$$ and $$y=\frac{3}{7}+4z$$ and $$z=z$$


-1

Hint: Start by trying to write $f$ as a function of $r$ and $\theta$ (as in polar coordinates).


0

After doing further research throughout the readings of 18.02 and 18.02sc, I've found that the above is indeed true in a non-simply connected region. The nuance I was missing was almost notational, in a sense - much like how $ \int_C \vec F \cdot d \vec r = \int_C M dx + N dy$ takes a dot product and carries it into a differential, $ \int_C \nabla f \cdot d ...


0

Given $A, H\in GL(n,\Bbb R)$, let $C = A^{-1}H$ and consider $E(t) := t^2 C^2(I + tC)^{-1}$. Then $E(t) = o(t)$ as $t\to 0$ and $(I + tC)^{-1} = I - tC + E(t)$. So $$(A + tH)^{-1} = (I + tC)^{-1}A^{-1} = A^{-1} - tCA^{-1} + E(t)A^{-1} = A^{-1} - tCA^{-1} + o(t),$$ which implies $$df(A)(H) = -CA^{-1} = -A^{-1}HA^{-1}.$$


1

Hint: $x=r\cos(a)$, $y=r\sin(a)$, $r>0$ and the limit becomes: $$\lim_{r\to0}\frac{r^{m+n}\cos^m(a)\sin^n(a)}{r^{2p}}$$


0

Even though there is an inaccuracy in your computation, your conclusion about the derivative of $A^{-1}$ is correct. Directional derivative in the direction $H$ just means that$$\frac{dA}{dt}=H.$$Substituting in your formula we get$$df(A)(H)=\frac{dA^{-1}}{dt}=-A^{-1}HA^{-1}.$$ Edit: When differentiating $AA^{-1}=I$, the right equation ...


0

Another way to view a circle is as the image of the function $f:[0,1)\to \mathbb R^2, \quad t\mapsto (a+r\cos 2\pi t,b+r\sin 2\pi t)$. This approach has the advantage that it extends the idea to curves in $\mathbb R^n$.


3

A circle, i.e, an equation of the form $(x-a)^2+(y-b)^2=r^2$ is a relation on $\Bbb R$, that is, a subset of $\Bbb R^2$, which relates an element x with an element y, with some criteria, in this case, the circle equation. To write it explicitly, the relation would be $C=\left\{(x,y)\in\Bbb R^2: (x-a)^2+(y-b)^2=r^2\right\}$.


1

$f(x,y)$ isn't discontinuous at $(0,0)$. How do I know? First of all, can you get a picture of this function in your head? I can. Since $f: \Bbb R^{2} \to \Bbb R$, you should think of this function as first taking the $XY$-plane in $3$D space (let it be the horizontal axes), and warp that plane. In this case, we are warping the plane like a taco. If ...


1

$\forall h\neq 0,\quad \left | \frac{f(0,0+h)-f(0,0)}{h} \right |<\left | \frac{h^{2}}{h} \right |=\vert h\vert $ so if we let $h\to 0$, we see that $f_y(0,0)=0$. Likewise, $\forall h\neq 0,\quad \left | \frac{f(0+h,0)-f(0,0)}{h} \right |=\left | \frac{0}{h} \right |=0$ so $f_x(0,0)=0$. Now, $\vert f(x,y)-f(0,0)\vert <y^{2}$ which is $<\epsilon ...


2

HINT: Note that we have $$\frac{\partial g}{\partial \nu}=\frac{2}{\epsilon}$$ on $\Gamma_{\epsilon}$. Therefore, $$\lim_{\epsilon\to 0}\int_{\Gamma_{\epsilon}}\left(g\frac{\partial\phi}{\partial\nu}-\phi\frac{\partial g}{\partial\nu}\right)\,ds=-4\pi\phi(0)$$


1

If $R$ is large enough $\varphi$ and $\nabla \varphi$ are equal to $0$ on $\Gamma_2$, and so is the RHS in the last formula.


0

As alluded to in another answer of mine (which I added only once I understood that these were the same issues), the representation $(u_1,u_2)$ is indeed w.r.t. unit vectors, not the polar parameters. The key relation is that $\dot \theta = u_2/r$ which does indeed implies (in agreement with the chain rule) that $df/dt = u_1f_r + (u_2/r)f_\theta = \boldsymbol ...



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