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4

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


4

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 ...


4

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


3

$$\lim\limits_{(x,y)\to(0,0)}\frac{2x^2y}{x^4 + y^2}$$, 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.


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 $f(x,y+2\,k\,\pi)$, $k\in\mathbb{Z}$.


2

Consider $$ \lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}}= \lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{x^2+y^2}\sqrt{x^2+y^2} $$ You surely know that $$ \lim_{h\to0}\frac{1-\cos h}{h}=0 $$


2

Here's yet another way $$\lim\limits_{(x,y)→(0,0)}\frac{1-\cos\left(x^2+y^2\right)}{\sqrt{x^2+y^2}} $$ Using polar coordinates, we have $$\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\cos^2\phi+r^2\sin^2\phi\right)}{\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}} $$ $$=\lim\limits_{r\to 0^+} ...


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. ...


2

In general, $v$ does not have to be a unit vector. But if $\|v\| = 1$, then $\|v\|$ doesn't artificially affects our geometric interpretations. In more general contexts (e.g., differential geometry), we define the gradient of $f$ in $p$ as the vector ${\rm grad}\, f(p)$ verifying $${\rm d}f_p(v) = \langle {\rm grad}\,f(p), v\rangle, \quad \forall v$$That is, ...


2

The smoothness and rank of the differential is a red herring. $B$ is compact, so a minimum is attained somewhere on $B$. $\lVert f(0)\rVert < \frac{1}{2}$, since $\lVert f(x) - x\rVert < \frac{1}{2}$ for all $x$. $\lVert x\rVert \geqslant 1 \implies \lVert f(x)\rVert = \bigl\lVert x - \bigl(x-f(x)\bigr)\bigr\rVert \geqslant \lVert x\rVert - \lVert x ...


2

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 ...


2

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 $f_y(x_0,y_0)=72$


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 ...


2

How about $r:=\sqrt{x^2+y^2}$ and $s:=\sqrt{z^2+w^2}$? Let $\theta$ and $\phi$ be the usual angular coordinates of the $xy$-plane and the $zw$-plane, respectively. Then, $\text{d}x\,\text{d}y=r\,\text{d}r\,\text{d}\theta$ and $\text{d}z\,\text{d}w=s\,\text{d}s\,\text{d}\phi$. The ball $B$ is then given by $r^2+s^2\leq 1$, $0 \leq \theta \leq 2\pi$, and ...


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 ...


2

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- ...


2

Hint: Use the approximation of $\sin w $ given by $\lim_{w \to 0} w / \sin w = 1$, obtainable for example with L'Hopital's rule.


2

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 ...


2

Hint: substitute $${ { e } }^{ -t }=\frac { 1 }{ \sqrt { 5 } } \tan { x }$$


1

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 ...


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 $$r'(t)\cdot \hat{n}=0$$ $$\langle0,1,2t\rangle\cdot\langle1,2,3\rangle=0$$


1

$$ f(x,y) = \begin{cases} 0 & \text{if }x=y=0 \\ \bigl(\frac{2xy}{x^2+y^2}\bigr)^2 & \text{otherwise} \end{cases} $$ 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 $t\ne 0$, ...


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 ...


1

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. ...


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 ...


1

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 ...


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) ...


1

It's certainly 2). To see this, note that really we are calculating the derivative of a function $h:\mathbb{R}^2 \rightarrow \mathbb{R}$, where $h(x,y)=f(x,y,g(x,y))$. So we expect the derivative $Dh$ to also be a function from $\mathbb{R}^2$ to $\mathbb{R}$. Your first approach would give a map from $\mathbb{R}^3$, not from $\mathbb{R}^2$.


1

From the first one you have (integrating wrt $x$): $$f(x,y) = x^2 + 2xy + c(y)$$ and taking partial derivatives, $f_y = 2x + c'(y)$, but $f_y = a+3y$, so it must be $c'(y) = 3y$. That it says $a$ is a constant likely means constant wrt $y$, in which case $a=2x$ and it is not a problem. Then, integrating wrt $y$, $$f(x,y) = x^2 + 2xy + 3y^2/2 + K,$$ for some ...



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