Thread: MATH2111 Higher Several Variable Calculus

1. Re: Multivariable Calculus

Originally Posted by leehuan
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)?.

2. Re: Multivariable Calculus

Originally Posted by dan964
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.

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

3. Re: Multivariable Calculus

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

4. 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*}

5. Re: Multivariable Calculus

Originally Posted by leehuan
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.)

6. Re: Multivariable Calculus

Originally Posted by InteGrand
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 $C^{\infty}$)

7. Re: Multivariable Calculus

Originally Posted by dan964
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.

8. 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*}

9. 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.

10. Re: Multivariable Calculus

Explicitly:

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

11. Re: Multivariable Calculus

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

Originally Posted by seanieg89
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?

12. 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.

13. 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.

14. Re: Multivariable Calculus

Originally Posted by leehuan
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.

15. Re: Multivariable Calculus

Originally Posted by leehuan
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.

16. 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...)

17. Re: Multivariable Calculus

Originally Posted by leehuan
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.

18. 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.

19. 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.

20. 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

21. 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

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

22. Re: Multivariable Calculus

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

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

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

23. 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.

24. 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?

25. Re: Multivariable Calculus

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

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

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

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

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

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

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