MATH2111 Higher Several Variable Calculus (4 Viewers)

Paradoxica · Mar 19, 2017

Re: Several Variable Calculus

leehuan said:
Just a brief sketch-out please

$f(x)=\begin{cases}0 & (x,y)=(0,0)\\ \frac{2xy}{x^2+y^2} & \text{otherwise}\end{cases}$

$\text{Why is it that }\frac{\partial^2 f}{\partial x \partial y}(0,0)\text{ DNE?}$

I know that f is not continuous at 0 but I'm not sure if that helps since we're talking about partial derivatives here.

If we switch to polar co-ordinates, then the result is:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{-2\cos{4\theta}}{r^6}$

It is then immediately obvious that the limit does not exist as r tends towards zero.

InteGrand · Mar 19, 2017

Re: Several Variable Calculus

leehuan said:
Just a brief sketch-out please

$f(x)=\begin{cases}0 & (x,y)=(0,0)\\ \frac{2xy}{x^2+y^2} & \text{otherwise}\end{cases}$

$\text{Why is it that }\frac{\partial^2 f}{\partial x \partial y}(0,0)\text{ DNE?}$

I know that f is not continuous at 0 but I'm not sure if that helps since we're talking about partial derivatives here.

You can try first computing an expression for ∂f/∂y (using first principles to find the value of this at the origin). Then try computing ∂²f/∂x∂y (i.e. ∂/∂x (∂f/∂y)) at the origin from first principles and show that the limit does not exist.

leehuan · Mar 25, 2017

Re: Several Variable Calculus

Oh damn, looks like the brute way is the only way out of that one. Was hoping for a shortcut
_________________________

I got asked a question but I haven't seen the notation before

$\text{Let }\phi =-4xy^3+z^2x$

$\text{Compute }\nabla \phi\text{ and verify that }\nabla \times (\nabla \phi) = \textbf{0}$

What is that isolated nabla supposed to mean?

seanieg89 · Mar 25, 2017

Re: Several Variable Calculus

Nabla phi I am sure you understand, that is the gradient of phi and is a vector-valued function. Which can be regarded as a triple of functions (f,g,h) from R^3 to R.

Nabla cross (f,g,h) denotes the curl of the vector-valued function (f,g,h).

If you write Nabla as (d/dx,d/dy,d/dz), then Nabla x (f,g,h) is defined the same way as the usual cross product between two 3-d vectors. (Just the first "vector" consists of a triple of differential operators, whilst the second consists of a triple of real-valued functions on R^3. So instead of multiplying real numbers (vector components) together as in the defn of the usual cross product, we are applying differential operators to functions.)

eg first component is dh/dy-dg/dz.

seanieg89 · Mar 25, 2017

Re: Several Variable Calculus

(The notation Nabla dot (f,g,h) is the divergence of a vector field and is understood in the same way.)

MATH2111 · Mar 25, 2017

Re: Several Variable Calculus

If you keep cheating you won't learn. Why are marks so important?

leehuan · Mar 27, 2017

Re: Several Variable Calculus

Seriously? If I am stuck on a question why can I not ask for help? Accusing me of cheating on an assignment if I genuinely can't do my homework is just low.

leehuan · Apr 3, 2017

Re: Several Variable Calculus

So for a continuous mapping f: are the following always true

(I didn't really define the domain and codomain of f so just assume whatever's convenient please)

1) U is closed => f^-1(U) is closed
2) U is open => f^-1(U) is open
3) U is closed => f(U) is closed
4) U is open = f(U) is open

5) U is compact => f(U) is compact
6) U is path-connected => f(U) is path connected

Don't really need proof, just yes/no is sufficient

InteGrand · Apr 3, 2017

Re: Several Variable Calculus

leehuan said:
So for a continuous mapping f: are the following always true

(I didn't really define the domain and codomain of f so just assume whatever's convenient please)

1) U is closed => f^-1(U) is closed
2) U is open => f^-1(U) is open
3) U is closed => f(U) is closed
4) U is open = f(U) is open

5) U is compact => f(U) is compact
6) U is path-connected => f(U) is path connected

Don't really need proof, just yes/no is sufficient

1) True
2) True
3) False
4) False
5) True
6) True

leehuan · Apr 7, 2017

Re: Several Variable Calculus

There was a question in my test that I could not do and I'm seeking a solution please

$\text{Determine with explanations whether or not the point }\textbf{a}=(0,1)\\ \text{is a boundary point for the set}\\ S=\{(x,y)\in \mathbb{R}^2 : x\neq 0, y=\sin \frac1x \}$

InteGrand · Apr 7, 2017

Re: Several Variable Calculus

leehuan said:
There was a question in my test that I could not do and I'm seeking a solution please

$\text{Determine with explanations whether or not the point }\textbf{a}=(0,1)\\ \text{is a boundary point for the set}\\ S=\{(x,y)\in \mathbb{R}^2 : x\neq 0, y=\sin \frac1x \}$

Yeah it is a boundary point. Here's some hints of a possible method. Note that S is just the graph of y = f(x) := sin(1/x) (x =/= 0). Use (or show) the fact that f attains the value 1 for arbitrarily small values of x > 0 to deduce that (0, 1) is a boundary point of S (noting that any ball around (0, 1) contains points not in S also, as there will be points in it with y-value greater than 1).

leehuan · Apr 25, 2017

Re: Several Variable Calculus

I get lost in this Einstein summation convention so can I see how this problem would be approached using it?

$\text{For a general }\textbf{a}=(a_1,a_2,a_3)^T \in \mathbb{R}^3\text{ find the matrix }A\in M_{3,3}(\mathbb{R})\\ \text{such that }\textbf{a}\times \textbf{v}=A\textbf{v}\\\forall \textbf{v}\in \mathbb{R}^3$

InteGrand · Apr 25, 2017

Re: Several Variable Calculus

leehuan said:
I get lost in this Einstein summation convention so can I see how this problem would be approached using it?

$\text{For a general }\textbf{a}=(a_1,a_2,a_3)^T \in \mathbb{R}^3\text{ find the matrix }A\in M_{3,3}(\mathbb{R})\\ \text{such that }\textbf{a}\times \textbf{v}=A\textbf{v}\\\forall \textbf{v}\in \mathbb{R}^3$

$\noindent For $r\in\left\{1,2,3\right\}$, the $r$-th column of $A$ (which will be a $3\times 3$ matrix) will be equal to $A\mathbf{e}_{r} = \mathbf{a}\times \mathbf{e}_{r}$. So try and find an expression for $\mathbf{a}\times \mathbf{e}_{r}$ using the Einstein summation convention, using the formula for a cross product and noting that the $j$-th element of $\mathbf{e}_{r}$ is $\delta_{rj}$.$

leehuan · Apr 30, 2017

Re: Several Variable Calculus

$\text{Express the volume above the cone }z=\sqrt{x^2+y^2}\\\text{and inside the sphere }x^2+y^2+z^2=2az\\ \text{using Cartesian, cylindrical and spherical coordinates}$

Even with the diagram in front of me I still struggle to figure out my boundaries of integration.

$\text{'Progress' with Cartesian coordinates (unsure if right path)}\\ V=\int_{-a}^a \int_{?}^{?} \int_{x}^{a+\sqrt{a^2-x^2-y^2}} dz\,dy\,dx$

InteGrand · Apr 30, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Express the volume above the cone }z=\sqrt{x^2+y^2}\\\text{and inside the sphere }x^2+y^2+z^2=2az\\ \text{using Cartesian, cylindrical and spherical coordinates}$

Even with the diagram in front of me I still struggle to figure out my boundaries of integration.

$\text{'Progress' with Cartesian coordinates (unsure if right path)}\\ V=\int_{-a}^a \int_?^? \int_{x}^{a+\sqrt{a^2-x^2-y^2}}dz\,dy\,dx$

$\noindent Note that this is an ``ice-cream cone'' shape where the sphere intersects the cone above the $x$-$y$ plane in the horizontal plane $z = a$ (assuming $a > 0$) and makes a circle with $x^{2} + y^{2} = a^{2}$ in this plane in the intersection. So if we cast the shadow of the solid down onto the $x$-$y$ plane, it is the disk $x^{2} + y^{2} \leq a^{2}$. So the iterated integral can have $x$ and $y$ vary over the disk $\mathcal{D} := \left\{(x,y) \in \mathbb{R}^{2}: x^{2} + y^{2} \leq a^{2}\right\}$ and then for each fixed $(x,y)$ in this disk $\mathcal{D}$, we have $z$ go from the cone to the spherical cap (which is just the upper hemisphere, where $z = a + \sqrt{a^{2} - x^{2} - y^{2}}$, which we get from solving the sphere's equation for $z$ and taking the root with $z \geq a$), i.e. from $\sqrt{x^{2} + y^{2}}$ to $a + \sqrt{a^{2} - x^{2} - y^{2}}$. Hence the integral in Cartesian coordinates becomes$

$\begin{align*}\iint_{\mathcal{D}} \int_{z = \sqrt{x^{2} + y^{2}}}^{z = a + \sqrt{a^{2} - x^{2} - y^{2}}} \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x = \int_{x = -a}^{x=a} \int_{y = -\sqrt{a^{2} - x^{2}}}^{y= \sqrt{a^{2} - x^{2}}} \int_{z = \sqrt{x^{2} + y^{2}}}^{z = a + \sqrt{a^{2} - x^{2} - y^{2}}} \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x.\end{align*}$

leehuan · May 4, 2017

Re: Several Variable Calculus

I think spherical is preferred here (could be mistaken). But using the best approach, how would you determine the limits for this

$\text{Find the volume of the solid bounded between the spheres }\\x^2+y^2+z^2=a^2\, \text{and }x^2+y^2+(z-a)^2 = a^2$

InteGrand · May 4, 2017

Re: Several Variable Calculus

leehuan said:
I think spherical is preferred here (could be mistaken). But using the best approach, how would you determine the limits for this

$\text{Find the volume of the solid bounded between the spheres }\\x^2+y^2+z^2=a^2\, \text{and }x^2+y^2+(z-a)^2 = a^2$

Inspection would be the best way to determine the limits probably (you can draw a diagram to help if you decide to use inspection).

leehuan · May 4, 2017

Re: Several Variable Calculus

InteGrand said:
Inspection would be the best way to determine the limits probably (you can draw a diagram to help if you decide to use inspection).

I was able to deduce that $0\le \theta\le 2\pi\text{ and }0\le \phi \le \frac{\pi}{2}$

It's always rho that gets me.

Unless I picked the order of integration wrong.

InteGrand · May 4, 2017

Re: Several Variable Calculus

leehuan said:
I was able to deduce that $0\le \theta\le 2\pi\text{ and }0\le \phi \le \frac{\pi}{2}$

It's always rho that gets me.

Unless I picked the order of integration wrong.

It might be easier to use cylindrical coordinates. The way rho varies is different depending on what range phi is in. When phi is such that your point in the region lies within the "ice cream cone" part of the region, then rho will vary from 0 to a. For phi larger than the angle make by the cone to the vertical, rho will vary from 0 to the value of rho at the point on the sphere x^2 + y^2 + (z-a)^2 = a^2 with this phi value. You can find these using the relations between Cartesian and Spherical Coordinates.

leehuan · May 6, 2017

Re: Several Variable Calculus

$\text{Find the surface area of the part of the plane }2x+y +2z = 16 \text{ bounded by the surfaces }\\x = 0, y = 0\text{ and }x^2+y^2=64$

I might be misinterpreting the question but I thought that they're basically asking for 1/4 * area of curve of intersection, which happens to be an ellipse through (0,0,8) passing through (0,8,4) and (8,0,4).

If I'm right, how do I use surface integrals to get to the answer of $24\pi$ ? And if I'm wrong, how do I get back on the right path?

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MATH2111 Higher Several Variable Calculus (4 Viewers)

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