MATH2111 Higher Several Variable Calculus (2 Viewers)

leehuan · May 29, 2016

Re: Multivariable Calculus

My notes were really ambiguous. They just recited the chain rule except replaced all the u's and v's with r's and theta's and I have no idea how to manipulate it.

I attempted the matrix inverse as an exercise for myself in the meantime and I got this:

$\begin{pmatrix}\frac{\partial x}{\partial r} & \frac{\partial y}{\partial r}\\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} \end{pmatrix}^{-1} = \begin{pmatrix}\cos{\theta} & \sin{\theta} \\ -r\sin{\theta} & r\cos{\theta} \end{pmatrix}^{-1} = \begin{pmatrix}\cos{\theta} & -\frac{\sin{\theta}}{r} \\ \sin{\theta} & \frac{\cos{\theta}}{r}\end{pmatrix} = \begin{pmatrix}\frac{\partial r}{\partial x} & \frac{\partial \theta}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial \theta}{\partial y}\end{pmatrix}$

The last step was using what the answers were trying to tell me to prove. I don't actually know that the last step is true.

But I don't get the logic behind it.

These are my notes. Explanations would be greatly appreciated but because idk if the notes are copyrighted I probably can't keep them up for too long.

(Apparently, according to my lecturer, this has never been examined in first year before either...)

InteGrand · May 29, 2016

Re: Multivariable Calculus

leehuan said:
Hm ok yep.
__________________
A lot of the question is omitted for me to have a go at myself
$$Suppose $z=f(x,y)$. We write $x,y$ in polar coordinates so that $x=r\cos{\theta}, y=r\sin{\theta}$. \\ a) Show that $\frac{\partial r}{\partial x}=\cos{\theta},\quad \frac{\partial \theta}{\partial x}=-\frac{-\sin{\theta}}{r}$

No idea at all how to apply what here.

Basically, partials of r and theta wrt x mean we are thinking of r and theta as functions of x and y, and we are differentiating wrt x holding y constant.

If you don't know about Jacobians, note r as a function of x and y is r = sqrt(x^2 + y^2), so (partial)r/(partial)x = x/(sqrt(x^2 + y^2)) = x/r = cos(theta) (sorry for lack of LaTeX, on phone).

For the theta one, tan(theta) = y/x => sec^2 (theta) * (partial)theta/(partial x) = -y/x^2 via chain rule and differentiation wrt x.

Hence since cos^2 (theta) = (x/r)^2, we have by rearrangement

(partial)theta/(partial)x = (-y/x^2)*cos^2 (theta) = - y/r^2 = -(sin(theta))/r, as y = r.sin(theta).

Edit: I see you do know Jacobians it seems. You basically use the 'Inverse Function Theorem' to get derivatives of the inverse map by inverting the Jacobian of the 'forward' map, as seanieg89 was saying.

seanieg89 · May 29, 2016

Re: Multivariable Calculus

The logic behind it is that if f and g are inverse to each other, then the chain rule says that

$I=D(\textrm{Id})(p)=D(f\circ g)(p)=Df(g(p))\cdot Dg(p)$

so the differentials of functions inverse to each other are matrices inverse to each other.

seanieg89 · May 29, 2016

Re: Multivariable Calculus

This is exactly how you would prove the "easy" part of the inverse function theorem. Once you have established the differentiability of the inverse map, the differential of your inverse map is forced to be the inverse of the differential of your original map.

leehuan · May 29, 2016

Re: Multivariable Calculus

Ohh right. Yep thanks.

But I'll admit to another thing. I embarrassingly forgot entirely about what r and theta actually equalled to. So IG's method went right over my head............and was also why I didn't comprehend what Para said

leehuan · May 29, 2016

Re: Multivariable Calculus

So like, this question felt never ending and I aborted.

Part b) is just the chain rule:

$$b) Show that $\frac{\partial f}{\partial x}=\cos{\theta}\frac{\partial f}{\partial r}-\frac{\sin\theta}{r} \frac{\partial f}{\partial \theta}$ and similar for $y$

Is there any shortcut to save me from computing several product and quotient rules in this one

$$c) Show that $\\ \frac{\partial^2 f}{\partial x^2}=\frac{\sin^2{\theta}}{r}\frac{\partial f}{\partial r}+\frac{2\sin{\theta}\cos{\theta}}{r^2} \frac{\partial f}{\partial \theta}+\cos^2{\theta} \frac{\partial^2 f}{\partial r^2} - \frac{2\cos{\theta} \sin\theta}{r}\frac{\partial^2 f}{\partial r \partial \theta} + \frac{\sin^2{\theta}}{r^2} \frac{\partial^2 f}{\partial \theta^2}$

leehuan · May 30, 2016

Re: Multivariable Calculus

This is the hardest question of the semester's homework.. lol

$$Show that the function$\\ f(x,y)=\frac{xy}{x^2+y^2}, \quad (x,y)\neq 0\\ f(0,0)=a$\\ is not continuous at (0,0) for any choice of $a$

$Show that, if $a=0$, then $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exists$

seanieg89 · May 30, 2016

Re: Multivariable Calculus

1. Approaching along the line x=0 gives you a limit of zero, and approaching along the line y=x gives you a limit of 1/2, so you cannot extend f continuously to the full plane.

2. Just literally partially differentiate by first principles, you get zero for both of the limits (the limits defining the partial derivatives at the origin, which is the only potentially problematic point). The difference quotients are identically zero.

Moral of the story:

Saying that all partials of a multivariable function exist at a point is much weaker than saying a function is differentiable at that point. In fact as this example shows, it is even weaker than continuity! It makes sense when you think about it, being "nice" in the coordinate directions does not say anything about how potentially badly behaved you might be in the infinitude of other directions.

Here is a followup question for you:

Suppose f:R^2 -> R has directional derivatives in every direction at the origin. Ie f(tx,ty) is a differentiable function in the single variable t, for any fixed point (x,y) in the plane.

1. Is f necessarily continuous?

2. Is f necessarily differentiable?

Where differentiability at (0,0) is the statement that there exists a linear map f'(0,0) from R^2 to R with

f(x,y)=f(0,0)+f'(0,0)(x,y)+E(x,y)

where E(x,y)/sqrt(x^2+y^2) -> 0 as (x,y)-> 0.

leehuan · Jun 7, 2016

Re: Multivariable Calculus

I need guidance (except for part a) please

$$For $r \neq -1$ and $\textbf{a}=\left(a_0,a_1,a_2\right)$, let $\\ f(r,\textbf{a})=\sum_{j=0}^{2}{\frac{a_j}{(1+r)^j}}=a_0+\frac{a_1}{1+r}+\frac{a_2}{(1+r)^2}$

$The equation $f(r,\textbf{a})=0\quad --- (*)\\ $ defines $r$ implicitly as a function of $\textbf{a}$. If we interpret $\textbf{a}$ as the cash flow stream of an investment then $r$ is called the $\textit{internal rate of return}$ for the investment.$

$$a) Find $\frac{\partial f}{\partial r}(r,\textbf{a})\\ $which was easy, because the answer is just $\\ -\frac{a_1}{(1+r)^2}-\frac{2a_2}{(1+r)^3}$

$$b) By implicitly differentiating equation $(*)$ show that $\\ \frac{\partial r}{\partial a_j}(\textbf{a})=-\frac{\frac{\partial f}{\partial a_j}(r, \textbf{a})}{\frac{\partial f}{\partial r}(r,\textbf{a})}$

I need it seriously broken down because the fact that a is a vector scares me

$$c) You are given, at some cash flow stream $\textbf{a}$ and rate $r, \\ \frac{\partial f}{\partial r}(r,\textbf{a})=-523, \quad \frac{\partial f}{\partial a_0}(r, \textbf{a}) =1\\ \frac{\partial f}{\partial a_1}(r, \textbf{a}) =0.966, \quad \frac{\partial f}{\partial a_2}(r, \textbf{a}) =0.932$

$\\ $Suppose that changes to the tax laws cause the cash flow stream to change by $\\ \Delta \textbf{a}=(-10,-10,10). \\ $Estimate the change $\Delta r$ in the internal rate of return.$$

I always forget when to apply the total differential approximation and how to apply it. Any tips with respect to this question?

InteGrand · Jun 7, 2016

Re: Multivariable Calculus

leehuan said:
I need guidance (except for part a) please

$$For $r \neq -1$ and $\textbf{a}=\left(a_0,a_1,a_2\right)$, let $\\ f(r,\textbf{a})=\sum_{j=0}^{2}{\frac{a_j}{(1+r)^j}}=a_0+\frac{a_1}{1+r}+\frac{a_2}{(1+r)^2}$

$The equation $f(r,\textbf{a})=0\quad --- (*)\\ $ defines $r$ implicitly as a function of $\textbf{a}$. If we interpret $\textbf{a}$ as the cash flow stream of an investment then $r$ is called the $\textit{internal rate of return}$ for the investment.$

$$a) Find $\frac{\partial f}{\partial r}(r,\textbf{a})\\ $which was easy, because the answer is just $\\ -\frac{a_1}{(1+r)^2}-\frac{2a_2}{(1+r)^3}$

$$b) By implicitly differentiating equation $(*)$ show that $\\ \frac{\partial r}{\partial a_j}(\textbf{a})=-\frac{\frac{\partial f}{\partial a_j}(r, \textbf{a})}{\frac{\partial f}{\partial r}(r,\textbf{a})}$

I need it seriously broken down because the fact that a is a vector scares me

$$c) You are given, at some cash flow stream $\textbf{a}$ and rate $r, \\ \frac{\partial f}{\partial r}(r,\textbf{a})=-523, \quad \frac{\partial f}{\partial a_0}(r, \textbf{a}) =1\\ \frac{\partial f}{\partial a_1}(r, \textbf{a}) =0.966, \quad \frac{\partial f}{\partial a_2}(r, \textbf{a}) =0.932$

$\\ $Suppose that changes to the tax laws cause the cash flow stream to change by $\\ \Delta \textbf{a}=(-10,-10,10). \\ $Estimate the change $\Delta r$ in the internal rate of return.$$

I always forget when to apply the total differential approximation and how to apply it. Any tips with respect to this question?

$\noindent For (b), the vector notation $\bold{a}$ is just shorthand notation to say we are evaluating the partial derivative at the point $\left(r,a_0,a_1,a_2\right)$ or $\left(a_0,a_1,a_2\right)$. In other words, the notation $\bold{a} := \left(a_0,a_1,a_2\right)$ is just used to shorten things so we don't need to write $\left(a_0,a_1,a_2\right)$ every time. It's nothing to be scared of (like, we're not differentiating wrt a vector or anything like that).$

$\noindent So the question is treating $r$ as a function of $a_0,a_1,a_2$, and asking to show that the partial derivative of $r$ wrt each $a_j$ is the expression given, using implicit diff.$

$\noindent For (c), recall the total differential approximation will be$

$\Delta r = r_{a_{0}}\Delta a_{0}+r_{a_{1}}\Delta a_{1}+r_{a_{2}}\Delta a_{2},$

$\noindent where the partials of $r$ wrt each $a_j$ here are evaluated at the given point $\bold{a}$ (well it's not given, but we are given the values of the relevant partial derivatives for this point), and the $\Delta a_j$ are the components of the given $\Delta \bold{a}$ vector. We can calculate each $r_{a_{j}}$ by using the formula for $r_{a_{j}}$ derived in part (b) and subbing in the given values of partial derivatives.$

$\noindent $\Big{(}$And $r_{a_{j}}$ is a relatively standard notation that just means $\frac{\partial r}{\partial a_j}$.$\Big{)}$$

leehuan · Jun 14, 2016

Re: Multivariable Calculus

I'm so shit at this topic.

$Let $F(u,v)=\int_1^u{g(v+2t)dt}$ where $g$ has a continuous derivative.$

$The answers found $\frac{\partial F}{\partial u}=g(v+2u)$ using FTC1 which I perfectly get$

$$But then they find $\\ \begin{align*}\frac{\partial F}{\partial v}&=\int_1^u{g^\prime (v+2t) dt} \quad \text{which appears ok at first sight; it's just Leibniz's rule}\\ =\left[\frac{1}{2}g(v+2t)\right]_1^u\end{align*}$
____________

$$So like, this only partially makes sense. The $g^\prime$ was forced out by taking a derivative with respect to $v$. So why is it ok that when integrating with respect to $t$, the prime can get dropped off again?$

Edit: Why is LaTeX broken on this site again?

Ok basically the main question is

They had d/dv g(v+2t) = g'(v+2t)

So why was it safe to integrate with respect to t to get 1/2 g(v+2t) again? I'm missing something elementary

InteGrand · Jun 14, 2016

Re: Multivariable Calculus

leehuan said:
I'm so shit at this topic.

$Let $F(u,v)=\int_1^u{g(v+2t)dt}$ where $g$ has a continuous derivative.$

$The answers found $\frac{\partial F}{\partial u}=g(v+2u)$ using FTC1 which I perfectly get$

$$But then they find $\\ \begin{align*}\frac{\partial F}{\partial v}&=\int_1^u{g^\prime (v+2t) dt} \quad \text{which appears ok at first sight; it's just Leibniz's rule}\\ =\left[\frac{1}{2}g(v+2t)\right]_1^u\end{align*}$
____________

$$So like, this only partially makes sense. The $g^\prime$ was forced out by taking a derivative with respect to $v$. So why is it ok that when integrating with respect to $t$, the prime can get dropped off again?$

Edit: Why is LaTeX broken on this site again?

Ok basically the main question is

They had d/dv g(v+2t) = g'(v+2t)

So why was it safe to integrate with respect to t to get 1/2 g(v+2t) again? I'm missing something elementary

They're integrating the derivative of a function, so they used FTC. (The 1/2 coming out due to reverse chain rule basically.)

Note that (∂/∂v)(g(v+2t)) = g'(v+2t), where g' just means the derivative of the function g.

E.g. if g(y) was cos(y), we'd have ∂/∂v (g(v+2t)) = ∂/∂v (cos(v+2t)) = -sin(v+2t) (i.e. g' evaluated at v+2t, which is what g'(v+2t) means).

Then integrating -sin(v+2t), we'd get back to (1/2)cos(v+2t).

leehuan · Jun 14, 2016

Re: Multivariable Calculus

InteGrand said:
They're integrating the derivative of a function, so they used FTC. (The 1/2 coming out due to reverse chain rule basically.)

Note that (∂/∂v)(g(v+2t)) = g'(v+2t), where g' just means the derivative of the function g.

E.g. if g(y) was cos(y), we'd have ∂/∂v (g(v+2t)) = ∂/∂v (cos(v+2t)) = -sin(v+2t) (i.e. g' evaluated at v+2t, which is what g'(v+2t) means).

Then integrating -sin(v+2t), we'd get back to (1/2)cos(v+2t).

Oops thanks.

It makes sense with an example; I think I just got scared of too many variables here

seanieg89 · Jun 14, 2016

Re: Multivariable Calculus

leehuan said:
I'm so shit at this topic.

$Let $F(u,v)=\int_1^u{g(v+2t)dt}$ where $g$ has a continuous derivative.$

$The answers found $\frac{\partial F}{\partial u}=g(v+2u)$ using FTC1 which I perfectly get$

$$But then they find $\\ \begin{align*}\frac{\partial F}{\partial v}&=\int_1^u{g^\prime (v+2t) dt} \quad \text{which appears ok at first sight; it's just Leibniz's rule}\\ =\left[\frac{1}{2}g(v+2t)\right]_1^u\end{align*}$
____________

$$So like, this only partially makes sense. The $g^\prime$ was forced out by taking a derivative with respect to $v$. So why is it ok that when integrating with respect to $t$, the prime can get dropped off again?$

In the integrand we are applying g (a function of a single variable) to the expression v+2t.

Partially differentiating g(v+2t) with respect to v and partially differentiating with respect to t just differ by a factor of 2, as computed using the chain rule. (Write h(v,t)=v+2t as a function from R^2 to R and carefully apply the chain rule to the composition (g o h) if you are still confused after this post.)

So to summarise, Leibniz lets us differentiate the integral w.r.t. v by differentiating the integrand w.r.t. v. The new integrand can be replaced by the t-partial derivative of g(v+2t) up to the constant factor of 2 which we account for by division. Then we are just integrating the t-derivative of something w.r.t t which the FTC takes care of.

seanieg89 · Jun 25, 2016

Re: Multivariable Calculus

Another exercise without much required knowledge.

Suppose we have a smooth function $X:\mathbb{R}^n\rightarrow \mathbb{R}^n$ .

Suppose further, that given any starting point $x\in\mathbb{R}^n$ , we can find a smooth curve $c(t)$ with $c(0)=x$ and $c'(t)=X(c(t))$ .

(Intuitively, the vector field X is the infinitesimal generator of the curve c.)

Under some simple assumptions, the function $p_t(x): \mathbb{R}^n \rightarrow \mathbb{R}^n$ that maps a pair $x$ to $c(t)$ where c is the curve starting at x, is a smooth function with smooth inverse for any fixed t.

So we can consider how sets in R^n "flow". Of course, in general this will distort volume and other geometric quantities.

One way of quantifying volume distortion locally about a point p is as follows:

$E_X(p):= \lim_{\epsilon\rightarrow 0}\frac{\frac{d}{dt}(|p_t(B(p,\epsilon))|)|_{t=0}}{|B(p, \epsilon)|}$

where |S| denotes the volume of a subset S in R^n, or equivalently, the integral of the constant function 1 over this set, which we assume is a well defined Riemann integral for the sets involved in this question and B(p,r) denotes the ball about p of radius r.

(So E_X measures the limiting rate of change of volumes of small balls about p relative to the size of these balls.)

1. Compute an expression for E_X(p) in terms of the components of X and it's derivatives.

Followups to come if anyone answers this question.

leehuan · Jul 24, 2016

Re: Multivariable Calculus

leehuan said:
So like, this question felt never ending and I aborted.

Part b) is just the chain rule:

$$b) Show that $\frac{\partial f}{\partial x}=\cos{\theta}\frac{\partial f}{\partial r}-\frac{\sin\theta}{r} \frac{\partial f}{\partial \theta}$ and similar for $y$

Is there any shortcut to save me from computing several product and quotient rules in this one

$$c) Show that $\\ \frac{\partial^2 f}{\partial x^2}=\frac{\sin^2{\theta}}{r}\frac{\partial f}{\partial r}+\frac{2\sin{\theta}\cos{\theta}}{r^2} \frac{\partial f}{\partial \theta}+\cos^2{\theta} \frac{\partial^2 f}{\partial r^2} - \frac{2\cos{\theta} \sin\theta}{r}\frac{\partial^2 f}{\partial r \partial \theta} + \frac{\sin^2{\theta}}{r^2} \frac{\partial^2 f}{\partial \theta^2}$

Aha thanks InteGrand, I gave this question much more thought today and finally got the proof out (one and a half months late

)

leehuan · Aug 21, 2016

Re: Multivariable Calculus

Three months later and I still don't know how to use total differential approximation. Correct answer = 0.21% (possibly approximated)

$A triangle has two sides of length $a$ and $b$ with an included angle measuring $\frac{\pi}{3}$ radians. Given that $a$ increases by 5\%, $b$ decreases by 6\%, and the included angle increases by $2\%$, estimate the percentage increase of area of the triangle.$

$$Deductions so far:$\\ \begin{align*}\frac{\partial A}{\partial a}&=\frac{1}{2}b\sin C\\ \frac{\partial A}{\partial b}&=\frac{1}{2}a\sin C\\ \frac{\partial A}{\partial C}&=\frac{1}{2}ab\cos C\end{align*}$

Paradoxica · Aug 21, 2016

Re: Multivariable Calculus

leehuan said:
Three months later and I still don't know how to use total differential approximation. Correct answer = 0.21% (possibly approximated)

$A triangle has two sides of length $a$ and $b$ with an included angle measuring $\frac{\pi}{3}$ radians. Given that $a$ increases by 5\%, $b$ decreases by 6\%, and the included angle increases by $2\%$, estimate the percentage increase of area of the triangle.$

$$Deductions so far:$\\ \begin{align*}\frac{\partial A}{\partial a}&=\frac{1}{2}b\sin C\\ \frac{\partial A}{\partial b}&=\frac{1}{2}a\sin C\\ \frac{\partial A}{\partial C}&=\frac{1}{2}ab\cos C\end{align*}$

Just find the equation of the tangent hyper-plane at the point ½absinC and plug in the given values...

Here, the tangent hyperplane is given by:

$$\noindent$ w(a,b,C) = \frac{b \sqrt{3}}{4}(a_\Delta - a) + \frac{a \sqrt{3}}{4}(b_\Delta - b) + \frac{ab}{4}(C_\Delta - \frac{\pi}{3}) + \frac{ab \sqrt{3}}{4} \\ $The differential approximation will be given by plugging in the specified values of $ a_\Delta, b_\Delta, C_\Delta$

leehuan · Aug 21, 2016

Re: Multivariable Calculus

Paradoxica said:
Just find the equation of the tangent hyper-plane at the point ½absinC and plug in the given values...

Here, the tangent hyperplane is given by:

$$\noindent$ w(a,b,C) = \frac{b \sqrt{3}}{4}(a_\Delta - a) + \frac{a \sqrt{3}}{4}(b_\Delta - b) + \frac{ab}{4}(C_\Delta - \frac{\pi}{3}) + \frac{ab \sqrt{3}}{4} \\ $The differential approximation will be given by plugging in the specified values of $ a_\Delta, b_\Delta, C_\Delta$

I'm asking for a friend. Can we just break it back down to appropriate use of the formula thanks.

Paradoxica · Aug 21, 2016

Re: Multivariable Calculus

leehuan said:
I'm asking for a friend. Can we just break it back down to appropriate use of the formula thanks.

it's the higher dimensional analogue of construction of a tangent at a point. fairly intuitive to understand if you ask me, and also I have no knowledge of what constitutes "appropriate use"

MATH2111 Higher Several Variable Calculus (2 Viewers)

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