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I would say that one of the best texts for advanced calculus is "Vector Calculus, Linear Algebra, and Differential Forms" by J. Hubbard and B. Hubbard. click here You may find this as an additional and helpful resource to "Advanced Calculus" by Buck.


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Off to a good start. Draw the picture! The hard part is determining the integration limits. Really, you just need to understand what the lines of constant $r$ and $s$ mean. Basically, they are lines at 45-degree angles to the axes. So, really, imagine that if $r \in [0,1]$, then what are the limits of $s$? Well, in fact, $s$ ranges from $0$ at the ...


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The standard way is that $f(1,1.9) \approx. f(1,2)+ f_x(1, 2)(x-x_0)+ f_y (1,2)(y-y_0)=5+ 6(1-1.1)+ 2(2-1.9) = 5-0.6+ 0.2= 4.6$. i.e., you consider the change of the function along the tangent plane. The actual value at $(1.1, 1.9)$ is $5.39$, so the approximation is not that good.


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The tangent plane in $x_0, y_0$ is defined as $$z = \nabla_1 f(x_0,y_0)(x-x_0)+\nabla_2 f(x_0,y_0)(y-y_0)+f(x_0,y_0).$$ Now, consider $x_0 = 1$ and $y_0 = 2$. You know that $f(x_0, y_0) = f(1,2) = 5.$ The gradient of your function is: $$\nabla f(x,y)=\left[\begin{array}{c}2x+2y\\2x\end{array} \right]$$ In the point $(1,2)$: $$\nabla f(x_0,y_0)=\nabla ...


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Your sketch in (3) seems exactly right. For example, you could let $A$ be an annulus with the northwest quadrant removed, and $B$ be the same annulus with the southeast quadrant removed. You can extend the same example to 3 dimensions too, simply by ignoring the $z$ coordinate.


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from the 5th property we have: $(a\times b).(c\times d)=((a\times b)\times c).d$ then from the 1th property: $((a\times b)\times c).d=-(c\times (a\times b)).d$ then from the 6th property:$-(c\times (a\times b)).d=-((c.b)a-(c.a)b).d$ so: $$\begin{align}(a\times b).(c\times d)&=((a\times b)\times c).d\\ &=-(c\times ...


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$$\begin{align} (\vec a\times \vec b)\cdot(\vec c\times \vec d)&=((\vec a\times \vec b)\times \vec c)\cdot \vec d \tag 1 \\\\ &=(-\vec c\times (\vec a\times \vec b))\cdot \vec d \tag 2 \\\\ &=((\vec c\cdot \vec a)\vec b-(\vec c\cdot \vec b)\vec a)\cdot \vec d\tag 3\\\\ &=(\vec c\cdot \vec a)(\vec b\cdot \vec d)-(\vec c\cdot \vec b)(\vec ...


1

Property 5 states that: $$\color{blue}{u}\cdot (\color{red}{v}\times\color{green}{w})=(\color{blue}{u}\times\color{red}{v})\cdot\color{green}{w}$$ Applying property 5 to the L.H.S. we get: $\color{blue}{(a\times b)}\cdot (\color{red}{c}\times \color{green}{d}) = (\color{blue}{(a\times b)}\times \color{red}{c})\cdot \color{green}{d}$ Property 1 states ...


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Using property 5, we get $(a \times b) \cdot (c \times d) = ((a \times b) \times c) \cdot d$. Then use property 1 to get $((a \times b) \times c) \cdot d = (-c \times (a \times b)) \cdot d$. Now we can apply property 6 to get $(-c \times (a \times b)) \cdot d = -((c\cdot b)a-(c\cdot a)b)\cdot d$. If you simplify this expression using linearity of the ...


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The parallelogram defined by [2,3,3] and [x,y,z] does not make sense because you don't know what [x,y,z] are permissible (that is what you are trying to find out!) . It is the first one: all solutions to the equation are on the plane (by definition).


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There are many different types of/ definitions of dimensions in Mathematics. As a vector space, the plane (since it goes through the origin) is 2-dimensional, in the sense that every vector in the plane can be written as the linear combination of two vectors. In another sense, the dimension is 2, in that, if you know the value of any two of the coordinates ...


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If you are asking if the plane is unbounded, indeed it is. Equivalently in this case the area of the plane thus described is infinite.


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$$ f(x,y) = \begin{cases} 0 & \text{if }x=y=0 \\ \bigl(\frac{2xy}{x^2+y^2}\bigr)^2 & \text{otherwise} \end{cases} $$ This is obviously smooth in $\mathbb R^2\setminus\{(0,0)\}$, and both first-order partial derivatives are 0 everywhere on the axes. So the Hessian at the origin is zero. Since $f(t,t)=1$ but $f(0,t)=0$ for all $t\ne 0$, ...


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You don't even have to argue by contradiction. Start from Taylor's formula at order $2$: \begin{align*} f(x_0+h,y_0+k)&=f(x_0,y_0)+Df(x_0,y_0)\cdot (h,k)+ \frac 12D^2f(x_0,y_0)\cdot (h,k)+o\bigl(\bigl\lVert(h,k)\bigr\rVert^2\bigr)\\ &=f(x_0,y_0)+ \frac 12D^2f(x_0,y_0)\cdot (h,k)+o\bigl(\bigl\lVert(h,k)\bigr\rVert^2\bigr) \end{align*} since at an ...


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It might be that there is no contradiction in the third case. Namely, it can happen that $D^2f(x_0)$ is zero and therefore positive semidefinite and positive semidefinite. You don't really have to work by cases. A matrix $A$ being negative semidefinite means that for all $v$ you have $v^TAv\leq0$. If this is not the case, then there is some $v$ for which ...


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For $x : 0 < x < \sqrt{8-2\sqrt{2}}$, the upper boundary in $y$ is $2\sqrt{2}$ not $8-x^2$ (which is bigger). The $z = 8 - y^2$ cylinder has "chopped off" part of the $y = 8 - x^2$ cylinder. So the integral should be: $I = \displaystyle\int\limits_{0}^{\sqrt{8-2\sqrt{2}}}\int\limits_{0}^{2\sqrt{2}}\int\limits_{0}^{8-y^2}dzdydx\ \ +\ ...


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Consider the function $h:\mathbb R\to\mathbb R$ given by $h(t)=f(tu)$. Check that these hold: This is a $C^1$ function. $h'(t)=u\cdot\nabla f(tu)$. $h(0)=f(0)$ and $h(1)=f(u)$. Can you make progress working with this function $h$? The problem is now one dimensional. If you need help with this problem with $h$, let me know.


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Hint. Apply the chain rule with $f(x)=Ax$ and $g(x,y)=x^Ty$. You have $$h(x)=x^TAx=g(x,f(x))$$ $g^\prime(x,y).(h,k)=h^Ty+xk$ and $f^\prime(x).h=Ah$. Hence $$h^\prime(x).h=h^TAx+x^TAh$$


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If you are really interested in proofs and exploring calculus from first principles, I will highly recommend Michael Spivak's Calculus textbook (available here). It explains things very well in a very thorough and methodical approach. It also has easy as well as very challenging questions that would stretch the minds of even the brightest mathematicians. ...


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$I=\int \int_Vxy\ dx\ dy = \int_{x=-1}^1\int_{y=-\sqrt{1-x^2}}^\sqrt{1-x^2} xy\ dy\ dx$ Change of coordinates - $x=r\cos \theta$ $y=r\sin \theta$ Jacobian - $I=\int \int r^2\cos\theta \sin\theta\ \left\|d\frac{x,y}{r,\theta}\right\|\ d\theta\ dr$ $\left\|d\frac{x,y}{r,\theta}\right\|=\left|\begin{array}{c}\cos \theta&\sin\theta\\ -r\sin ...


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The first integral equality is not correct---you seem to be integrating with respect to $dv$ but over the interval for $u$. Since the integrand does not depend on $v$, we can write $$\int_0^{2 \pi} \int_0^1 u \sqrt{u^2 + 1} du\,dv = \int_0^{2 \pi} dv \int_0^1 u \sqrt{u^2 + 1} du = 2 \pi \int_0^1 u \sqrt{u^2 + 1} du$$


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I think Stewart's or Anton's Calculus are not the right choice for you...unless you have not yet completed a course in university calculus. So, the answer here necessarily depends on your background. If I assume you have take differential and integral calculus of one-variable then I recommend: Susan Colley's Vector Calculus (even the first or second edition ...


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What about all in one ? :-) this book covers almost all what you want. (Sorry I don't have enough reputation for comment)


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I think I have it! The dx,dy & dz terms in the integral represent the derivatives of the x, y and & components of the parametric position vector! With $\frac{d}{dt}(t,1,t) = (1,0,1)$ the integral simplifies to $$\int_1^2 2t\,ln(t)+ 1$$ which has solution $4\,ln(2) - \frac{1}{2}$! I thought I would post it here, in case someone else has a similar ...


2

$$yy'+\frac{y}{x}+k=0 \quad\quad (1)$$ Change of function : $y(x)=\frac{1}{f(x)}$ $\frac{1}{f}\left(-\frac{f'}{f^2}\right)+\frac{1}{xf}+k=0$ $$f'=kf^3+\frac{1}{x}f^2$$ This is an Abel's differential equation of first kind which is knonw as ''non-sovable'' form, meaning that the solutions are not known on the form of a finite number of standard functions. ...


1

The smoothness and rank of the differential is a red herring. $B$ is compact, so a minimum is attained somewhere on $B$. $\lVert f(0)\rVert < \,?$ $\lVert x\rVert \geqslant 1 \implies \lVert f(x)\rVert > \,?$


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Hint. Your partial derivatives are correct, now the tangent plane at $(x_0,y_0,f(x_0,y_0))$ is: $$ z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) $$ and you have $x_0=-1$ , $y_0=4$, so $f_x(x_0,y_0)= 18$ and $f_y(x_0,y_0)=72$


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First look at the elliptical cross-section in xy-plane: $\{ (x,y): x^2 + xy + y^2 = 1 \}$ It is clear that this is symmetric in $x$ and $y$ so write $x^2 + xy + y^2 = a(x+y)^2 + b(x-y)^2$ so by equating coefficients $a=\frac{3}{4}, b=\frac{1}{4}$ Therefore $\boxed{\dfrac{3}{4}(x+y)^2 + \dfrac{1}{4}(x-y)^2 = 1}$ Changing to new coordinate system: $X = ...


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$$\lim\limits_{(x,y)\to(0,0)}\frac{2x^2y}{x^4 + y^2}$$, taking $y=mx^2$ ,we have, $$\lim\limits_{x\to0}\frac{2mx^4}{x^4 + m^2x^4}$$ $$\lim\limits_{x\to0}\frac{2m}{1 + m^2}=\frac{2m}{1 + m^2}$$, which shows that limit depends upon the $m$ and changes with value of $m$ . Hence limit does not exist.


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Try $y=kx^2$ and show that the limit doesn't exist.


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You cannot decide this locally. A Jordan curve in Euclidean two space is an oriented $1$-submanifold, and chosing the normal amounts to the choice of an orientation. This, on the other hand, means, that if you know an interior normal at one point and have a global regular parametrization, you can figure out at that single point whether it is orientation pre- ...


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Ignore the $t$-dependence and consider this linear equation and its differential $$\eqalign{ y &= W \cdot x \cr dy &= dW \cdot x \cr &= I\cdot dW \cdot x \cr &= (Ix):dW \cr }$$ where $(Ix)$ is a third-order tensor with components $(Ix)_{ijk}=\delta_{ij}x_k$. I've written an explicit dot for the dot product, so as to reserve ...


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After a while I figured it out. Consider the function $$\begin{align}\psi: [0,1] &\to \mathbb R\\ t &\mapsto \langle f(a + tv), v \rangle\end{align}$$ defined at $[0,1]$, differentiable with $$\begin{align}\psi' (t) \cdot v &= \langle f'(a + tv)\cdot v ,v\rangle + \langle f(a+tv),0 \rangle \\ &= \langle f'(a + tv)\cdot v \rangle > ...


1

Claim: $f(x,y) \to 0 \iff m/p + n/q >1.$ Proof: We only need to work in the open first quadrant. Note the following: $$ (x_n,y_n) \to (0,0) \iff ((x_n)^{1/p},(y_n)^{1/q}) \to (0,0).$$ This tells us that as $(x,y) \to (0,0), f(x,y)\to 0 \iff f(x^{1/p},y^{1/q})\to 0.$ Now $$f(x^{1/p},y^{1/q}) = \frac{x^{m/p}y^{n/q}}{x+y}.$$ So let's look at the ...


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Even restricted, $y^*$ need not be a local maximum. Necessary would be $$ w^* \nabla_{yy} \mathcal L(y^*, \lambda^*) w \le 0 $$ for all $w^* = (w_1, w_2, w_3, 0)$ in the tangent space, that is $$ w_1 - y_3^*w_2 - y_2^* w_3 = 0. $$ Strictly less for sufficient condition. Since the (1,1) element of the hessian is $1$, it depends on $y_2^*$ and $y_3^*$.


1

You are correct that you will need to find $r'(t)$, the derivative of your function. This will give you the direction vector of your curve at an arbitrary time $t$. From there, all we have to do is determine when this direction vector is perpendicular to the normal of the plane and solve for $t$. Start by finding $r'(t)$. Simply differentiate each ...


1

A point on a curve $r(t)=\langle 1,t,t^2\rangle$ in which the tangent line is parallel to plane $x+2y+3z=0$ occurs where the derivative of $r(t)$ is orthogonal to the normal of the plane. That is, when $$r'(t)\cdot \hat{n}=0$$ $$\langle0,1,2t\rangle\cdot\langle1,2,3\rangle=0$$


0

The inverse function theorem and implicit function theorem are "cousins" of each other. You can prove one and then deduce the other. Your intuition is guiding you from the inverse function theorem towards the implicit function theorem. Your description is slightly inaccurate (but easily fixable) in the sense that a function $f: \mathbb{R}^2 \to \mathbb{R}$ ...


2

Make your life easier writing$$B=\frac{2hf^3}{c^2\left(e^\frac{hf}{kT}-1\right)}=\frac{\alpha}{e^{\beta}-1}$$ using $\alpha=\frac{2hf^3}{c^2}$ and $\beta=\frac{hf}{kT}$. So $$\frac{dB}{dT}=\frac{dB}{d\beta}\times\frac{d\beta}{dT}$$ Now $$\frac{dB}{d\beta}=-\frac{\alpha e^{\beta }}{\left(e^{\beta }-1\right)^2}$$ $$\frac{d\beta}{dT}=-\frac{hf}{kT^2}$$ So ...


1

I think that you have already solved the problem. Pick $x, y\in U$, define your $v=y-x$, and proceed. Note that $||f(y)-f(x)||\cdot||y-x||>0$. As you picked $x\neq y$, then $f$ must be injective. Sorry for my english.


0

What you are referring to are called directional cosines. They can be seen in this picture except $a$ would be $\alpha$, $b$ would be $\beta$, and $c$ would be $\gamma$, Without explaining multivariable calculus, the best way to do this is to create a vector $v$=$< x,y,z >$ where each component ($x$, $y$, or $z$) is the difference between where the ...


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A gradient $$ \mbox{grad } \Psi = \nabla \Psi $$ is either zero or orthogonal to the level curves $$ C: \Psi = c = \mbox{const} $$ in particular to the tangent vectors $$ t = \frac{dr}{ds} $$ along $C$. The function $\Psi$ is constant along a level curve $C$, thus $d\Psi = 0$. Using the relation between total differential, gradient and displacement we ...


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consider the function $$f(a,b,c,\lambda)=a^2b+b^2c+c^2a+abc+\lambda(a+b+c-3)$$


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In basic geometry you take by definition that (for $a,b$ numbers and $C,D$ curves): $$ \int_{aC+bD}\omega\,dx := a\int_{C}\omega\,dx + b\int_{D}\omega\,dx. $$ The motivation for this comes from algebraic topology. So: $$ \oint_{C_1-C_2}F\,dr =\oint_{C_1}F\,dr - \oint_{C_2}F\,dr. $$ You can think of the part $S$ of the cylinder between the two curves as ...


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We define $g$ as the function inside the absolute value. (i.e: $f(x,y) = |g(x,y)|$) $$\frac{\partial f(x,y)}{\partial x} = \frac{g(x,y)\frac{\partial g(x,y)}{\partial x}}{|g(x,y)|}$$ Therefore, applying this to our function (i.e: $g(x,y) = 2x^2-y$) we get: $$\frac{\partial |2x^2-y|}{\partial x} = \frac{(2x^2-y)(4x)}{|2x^2-y|},$$ $$\frac{\partial ...


2

Here is a method continuing your idea. You should use the negative of the gradient since you are looking for the direction in which the temperature decreases the fastest. Then look for the projection of that direction onto the tangent plane of the mountain. The gradient direction of the temperature is $\vec{t}=(-2,-2,-6)$. The normal vector of the ...


1

The task is to walk away from the present location such that the temperature decreases as much as possible. We are at $(1,1,3)$, the temperature gradient points at $(2,2,6)$. If we move by $dx$ and $dy$ the change in position is $$ dr = (dx, dy, z_x dx + z_y dy) = (dx, dy, -2(dx +dy)) $$ The change in temperature moving there is $$ dT = (T_x, T_y, T_z) ...


3

Use the Inverse function theorem. The jacobian of $f$ is $e^x\ne0$. This gives local invertibility. To show that $f$ is not a injective on $\mathbb{R}^2$, consider $f(x,y)$ and $f(x,y+2\,k\,\pi)$, $k\in\mathbb{Z}$.


0

Yes this is true as the derivative of a linear map $F$ at a point $a \in \mathbb R^n$ is a linear map equal to $F$. Therefore the derivative $F^\prime$ is the constant function that maps $x \in \mathbb R^n$ to the constant linear map $F$. Hence the second derivative is equal to $0$ as all further order derivatives.


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Like @Chilango commented, the vector $f_x(x_0, y_0)i+f_y(x_0,y_0)j$ exists even if $f$ is not differentiable at $(x_0, y_0)$. The link you gave conflates differentiability with existence of the gradient vector, but both my calculus textbook and my real analysis textbook give a different definition for differentiability: A function $f$ is differentiable at ...



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