MATH2111 Higher Several Variable Calculus (1 Viewer)

InteGrand · May 22, 2016

Re: Multivariable Calculus

leehuan said:
Ah ok so you basically do leave it as is I see. So is this right...?

$$b) $\frac{\partial}{\partial t}f(st^2,s^2+2t)$

$Let $z=f(st^2,s^2+2t),u=st^2, v=s^2+2t, $ then$

$\begin{align*} \frac{\partial z}{\partial t}&=\frac{\partial z}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial t} \\ &= 2st f_1 \left( st^2.s^2+2t \right) +2f_2 \left( st^2,s^2+2t \right) \end{align*}$

Yeah that's correct.

leehuan · May 22, 2016

Re: Multivariable Calculus

Cool. Ok this one is twisting with me but here's my attempt

$$c) $\frac{\partial}{\partial x} f\left(f(x,y),f(x,y)\right)$

$$For some $z= f\left(f(x,y),f(x,y)\right)$ and $u=f(x,y)$

$\begin{align*}\frac{\partial z}{\partial x}&=2\frac{\partial z}{\partial u}\frac{\partial u}{\partial x} \\ &= 2 f_1(x,y) f_1(f(x,y),f(x,y))\end{align*}$

InteGrand · May 22, 2016

Re: Multivariable Calculus

leehuan said:
Cool. Ok this one is twisting with me but here's my attempt

$$c) $\frac{\partial}{\partial x} f\left(f(x,y),f(x,y)\right)$

$$For some $z= f\left(f(x,y),f(x,y)\right)$ and $u=f(x,y)$

$\begin{align*}\frac{\partial z}{\partial x}&=2\frac{\partial z}{\partial u}\frac{\partial u}{\partial x} \\ &= 2 f_1(x,y) f_1(f(x,y),f(x,y))\end{align*}$

$\noindent Need to do it with two variables $u$ and $v$, and make them both equal $f(x,y)$. This is because $f$ may depend differently on its first and second variables, so its partial derivatives won't be the same necessarily.$

InteGrand · May 22, 2016

Re: Multivariable Calculus

leehuan said:
Cool. Ok this one is twisting with me but here's my attempt

$$c) $\frac{\partial}{\partial x} f\left(f(x,y),f(x,y)\right)$

$$For some $z= f\left(f(x,y),f(x,y)\right)$ and $u=f(x,y)$

$\begin{align*}\frac{\partial z}{\partial x}&=2\frac{\partial z}{\partial u}\frac{\partial u}{\partial x} \\ &= 2 f_1(x,y) f_1(f(x,y),f(x,y))\end{align*}$

$\noindent So basically, let $z=f(u,v)$, where $u=f(x,y)$, and $v=f(x,y)$. Then we want to find $\frac{\partial z}{\partial x}$. Note that $\frac{\partial u}{\partial x}=f_{1}(x,y)=\frac{\partial v}{\partial x}$. So we have$

$\begin{align*}\frac{\partial z}{\partial x} &= f_{1}(u,v)\frac{\partial u}{\partial x} + f_{2}(u,v)\frac{\partial v}{\partial x} \\ &= f_{1}(u,v)f_{1} (x,y) + f_{2}(u,v)f_{1} (x,y) \\ &= f_{1}(f(x,y),f(x,y))f_{1} (x,y) + f_{2}(f(x,y),f(x,y))f_{1} (x,y). \end{align*}$

leehuan · May 22, 2016

Re: Multivariable Calculus

InteGrand said:
$\noindent So basically, let $z=f(u,v)$, where $u=f(x,y)$, and $v=f(x,y)$. Then we want to find $\frac{\partial z}{\partial x}$. Note that $\frac{\partial u}{\partial x}=f_{1}(x,y)=\frac{\partial v}{\partial x}$. So we have$

$\begin{align*}\frac{\partial z}{\partial x} &= f_{1}(u,v)\frac{\partial u}{\partial x} + f_{2}(u,v)\frac{\partial v}{\partial x} \\ &= f_{1}(u,v)f_{1} (x,y) + f_{2}(u,v)f_{1} (x,y) \\ &= f_{1}(f(x,y),f(x,y))f_{1} (x,y) + f_{2}(f(x,y),f(x,y))f_{1} (x,y). \end{align*}$

Yep sweet, was about to post that.
__________________________

Last question of this bundle:

$$d) $\frac{\partial}{\partial y}f(yf(x,t),f(y,t))$

$$Let $z=f(u,v)$ where $u=yf(x,t)$ and $v=f(y,t)$

$\begin{align*} \frac{\partial z}{\partial x}&=f_1(u,v) \frac{\partial u}{\partial y} + f_2(u,v) \frac{\partial v}{\partial y} \\ &= f_1(yf(x,t),f(y,t)) f(x,t) + f_2(y f(x,t),f(y,t)) f_1 (y,t)\end{align*}$

There might be an accidental typo

dan964 · May 22, 2016

Re: Multivariable Calculus

leehuan said:
Yep sweet, was about to post that.
__________________________

Last question of this bundle:

$$d) $\frac{\partial}{\partial y}f(yf(x,t),f(y,t))$

$$Let $z=f(u,v)$ where $u=yf(x,t)$ and $v=f(y,t)$

$\begin{align*} \frac{\partial z}{\partial x}&=f_1(u,v) \frac{\partial u}{\partial y} + f_2(u,v) \frac{\partial v}{\partial y} \\ &= f_1(yf(x,t),f(y,t)) f(x,t) + f_2(y f(x,t),f(y,t)) f_1 (y,t)\end{align*}$

There might be an accidental typo

Isn't the question asking for $\frac{\partial z}{\partial y}$

Apart from that it seems fine, although the notation $f_{1}$ and $f_{2}$ seems a bit odd especially it seems a bit inconsistent, the first one is $\frac{\partial z}{\partial u}$ evaluated at the function value f(u,v) I think and the second in $\frac{\partial z}{\partial v}$ evaluated at f(u,v) as well.
and the last one is just $\frac{\partial v}{\partial y}$ not the same as the first one, evaluated at (u,v)?.

leehuan · May 22, 2016

Re: Multivariable Calculus

dan964 said:
Isn't the question asking for $\frac{\partial z}{\partial y}$

Apart from that it seems fine, although the notation $f_{1}$ and $f_{2}$ seems a bit odd.

It is, my bad.

I'm just using notation I was presented with just now

InteGrand · May 22, 2016

Re: Multivariable Calculus

I used subscript numerals for the notation because otherwise it'd be more confusing due to so many variables being present.

leehuan · May 22, 2016

Re: Multivariable Calculus

Is this correct

$$Q: If $z=f(x,y)$ where $x=s+2t, y=2s-t$, find $\frac{\partial^2 z}{\partial s^2}$

$$First derivatives:$\\ \frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} = \frac{\partial z}{\partial x} + 2\frac{\partial z}{\partial y}\\ \frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} = 2\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y}\\$

$$Since $\frac{\partial^2 z}{\partial x \partial y}=\frac{\partial^2 z}{\partial y \partial x}$

$\begin{align*}\frac{\partial^2 z}{\partial s^2} &= \frac{\partial}{\partial s} \left[ \frac{\partial z}{\partial x} + 2\frac{\partial z}{\partial y} \right] \\ &= \frac{\partial}{\partial x} \left( \frac{\partial z}{\partial s} \right) + 2 \frac{\partial}{\partial y} \left( \frac{\partial z}{\partial s} \right) \\ &=\frac{\partial}{\partial x} \left[ \frac{\partial z}{\partial x} + 2\frac{\partial z}{\partial y} \right] + 2 \frac{\partial}{\partial y} \left[ \frac{\partial z}{\partial x} + 2\frac{\partial z}{\partial y} \right] \\ &= \frac{\partial^2 z}{\partial x^2} + 4 \frac{\partial^2 z}{\partial x \partial y} + 4\frac{\partial^2 z}{\partial y^2}\end{align*}$

InteGrand · May 22, 2016

Re: Multivariable Calculus

leehuan said:
Is this correct

$$Q: If $z=f(x,y)$ where $x=s+2t, y=2s-t$, find $\frac{\partial^2 z}{\partial s^2}$

$$First derivatives:$\\ \frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} = \frac{\partial z}{\partial x} + 2\frac{\partial z}{\partial y}\\ \frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} = 2\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y}\\$

$$Since $\frac{\partial^2 z}{\partial x \partial y}=\frac{\partial^2 z}{\partial y \partial x}$

$\begin{align*}\frac{\partial^2 z}{\partial s^2} &= \frac{\partial}{\partial s} \left[ \frac{\partial z}{\partial x} + 2\frac{\partial z}{\partial y} \right] \\ &= \frac{\partial}{\partial x} \left( \frac{\partial z}{\partial s} \right) + \frac{\partial}{\partial y} \left( \frac{\partial z}{\partial s} \right) \\ &=\frac{\partial}{\partial x} \left[ \frac{\partial z}{\partial x} + 2\frac{\partial z}{\partial y} \right] + \frac{\partial}{\partial y} \left[ \frac{\partial z}{\partial x} + 2\frac{\partial z}{\partial y} \right] \\ &= \frac{\partial^2 z}{\partial x^2} + 4 \frac{\partial^2 z}{\partial x \partial y} + 4\frac{\partial^2 z}{\partial y^2}\end{align*}$

I think you lost a factor of 2 in your second line in the part where you're finding a second derivative (the 2 in front of the ∂z/∂y), but the method is right. (Or maybe you typo'ed it and left it out but did the remaining parts right, haven't checked it too closely.)

dan964 · May 24, 2016

Re: Multivariable Calculus

InteGrand said:
I think you lost a factor of 2 in your second line in the part where you're finding a second derivative (the 2 in front of the ∂z/∂y), but the method is right. (Or maybe you typo'ed it and left it out but did the remaining parts right, haven't checked it too closely.)

Does he need to say that f is smooth (in $C^{\infty}$ )

InteGrand · May 24, 2016

Re: Multivariable Calculus

dan964 said:
Does he need to say that f is smooth (in $C^{\infty}$ )

Well yeah it does need to be smooth enough so that we can switch the order of the mixed partials (which doesn't require C-infinity). But that was probably an implicit assumption of the question.

leehuan · May 29, 2016

Re: Multivariable Calculus

$$A function $f(x,y)$ is said to be homogeneous of degree $n$ if $f(tx,ty)=t^nf(x,y) $ for all $t>0$. Show that such a function satisfies the equation$\\ x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf$

Progress so far

$$Let $u=tx, v=ty\\ t^n \frac{\partial}{\partial x}f(x,y)\stackrel{\text{chain}}{=}t \frac{\partial f(u,v)}{\partial u}\\ t^{n-1}\frac{\partial}{\partial y}f(x,y)=\frac{\partial f(u,v)}{\partial v}$
___________

$\begin{align*}x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}&=t^{1-n}x \frac{\partial f(u,v)}{\partial u} + t^{1-n} y \frac{\partial f(u,v)}{\partial v} \\ &= t^{-n} u \frac{\partial f(u,v)}{\partial u} + t^{-n} v \frac{\partial f(u,v)}{\partial v} \\ &= \frac{f(x,y)}{f(u,v)} \left[ u \frac{\partial f(u,v)}{\partial u} + v \frac{\partial f(u,v)}{\partial v} \right] \end{align*}$

seanieg89 · May 29, 2016

Re: Multivariable Calculus

Fix x and y, differentiate with respect to t (using the chain rule to deal with the LHS differentiation), and then set t=1 in the resulting identity.

seanieg89 · May 29, 2016

Re: Multivariable Calculus

Explicitly:

$RHS=nf(x,y)=nt^{n-1}f(x,y)|_{t=1}=\frac{\partial}{\partial t}(t^nf(x,y))|_{t=1}=\frac{\partial}{\partial t}(f(tx,ty))|_{t=1}=(x\frac{\partial f}{\partial x}(tx,ty)+y\frac{\partial f}{\partial y}(tx,ty))|_{t=1}= LHS$

leehuan · May 29, 2016

Re: Multivariable Calculus

Hang on, just one final bit that I don't compute (I think it's just a dumb moment)

seanieg89 said:
Explicitly:

$\frac{\partial}{\partial t}(f(tx,ty))|_{t=1}=(x\frac{\partial f}{\partial x}(tx,ty)+y\frac{\partial f}{\partial y}(tx,ty))|_{t=1}$

If I use the chain rule on this without the t=1 bit treating u=tx and v=ty

$\frac{\partial}{\partial t}(f(tx,ty))= \frac{\partial f}{\partial u}(u,v)\frac{\partial u}{\partial t} + \frac{\partial f}{\partial v}(u,v)\frac{\partial v}{\partial t}= x\frac{\partial f}{\partial u}(u,v) + y\frac{\partial f}{\partial v}(u,v)$

Obviously when t=1, f(u,v)=f(x,y), but I can't justify it in my head why the partial f/partial u becomes partial f/partial x. I just need confirmation that this was an allowed step you used?

seanieg89 · May 29, 2016

Re: Multivariable Calculus

Think about what partial f/partial u means, it just means the derivative of the function with respect to the first variable (and you can replace u with x or whatever your favorite greek letter is). You are differentiating f with respect to its first variable and then evaluating it at the coordinates (x,y).

Try to convince yourself that

$\partial_u (f(u,v))$
and
$\partial_x(f(x,y))$

are the same functions, just with different "dummy variables".

Introducing things like u and v can be more hassle than help.

This is also an example of why some people prefer using numerical subscripts for a function to denote differentiation with respect to the j-th variable.

leehuan · May 29, 2016

Re: Multivariable Calculus

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.

Paradoxica · 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.

You can't use the standard definition of the inverse tangent function, because θ can be anything from -π to π.

For that, you need to use a slightly trickier modification, known as atan2(y,x)

Let α be the principle arctangent of y/x

then atan2(y,x)
= α if x is positive
= α+π if x is negative and y is non-negative
= α-π if x and y are negative
= π/2 if x=0 and y is positive
= -π/2 if x=0 and y is negative
= undefined if x=y=0

Since r is positive, there is no trouble there, as it is simply your standard absolute distance using Pythagoras's Theorem.

seanieg89 · 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.

Do you know how to differentiate x and y as functions of r and theta?

Once you have the differential/Jacobian of the map (r,theta)->(x,y) (which will be a 2x2 matrix), just invert this matrix to get the differential of the inverse map.

MATH2111 Higher Several Variable Calculus (1 Viewer)

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