MATH2111 Higher Several Variable Calculus (1 Viewer)

leehuan · May 19, 2016

Is there a higher dimensional implicit differentiation hax (sorry I'm doing maths whilst being a bit hyper) to get me a shortcut for this question

$$Find the tangent planes and normal lines to the following surfaces at the points indicated: \\ d) $x^2+y^2+z^2=3$ at the point $(1,1,1)$

Or should I just chain rule it because z = +'ve sqrt(...)

$z=\sqrt{3-x^2-y^2}\Rightarrow \frac{\partial z}{\partial y}=-\frac{y}{\sqrt{3-x^2-y^2}}$

InteGrand · May 19, 2016

Re: Multivariable Calculus

leehuan said:
Is there a higher dimensional implicit differentiation hax (sorry I'm doing maths whilst being a bit hyper) to get me a shortcut for this question

$$Find the tangent planes and normal lines to the following surfaces at the points indicated: \\ d) $x^2+y^2+z^2=3$ at the point $(1,1,1)$

Or should I just chain rule it because z = +'ve sqrt(...)

$z=\sqrt{3-x^2-y^2}\Rightarrow \frac{\partial z}{\partial y}=-\frac{y}{\sqrt{3-x^2-y^2}}$

$\noindent In this case it can be done by inspection. The surface is a sphere centred at the origin (its radius is $\sqrt{3}$ incidentally, though that's not important here), so a normal at a point $\vec{a}$ on the sphere is just $\vec{a}$ itself (this is essentially because of the symmetry of a sphere and the property ``radius is perpendicular to tangent'', or equivalently, ``radius is parallel to normal'' from circle geometry). So at $\vec{a} = (1,1,1)$, a normal is $\vec{n} = \vec{a} = (1,1,1)$. Then an equation for the tangent plane there is $\vec{n}\cdot \vec{x}=\vec{n}\cdot \vec{a}$, or in a Cartesian form, $x+y+z=3$ (as $\vec{n}\cdot \vec{a} = 1+1+1 =3$).$

InteGrand · May 19, 2016

Re: Multivariable Calculus

$\noindent In general, if we're given a smooth surface of the form $F\left(x,y,z\right) = C$ (where $C$ is a constant), a normal to the surface at a point $\vec{a}$ on the surface is given by $\nabla F \left(\vec{a}\right)$. Once we have the normal from this, since we know a point (namely $\vec{a}$), we can find the plane's equation.$

$\noindent This procedure also generalises to functions of higher dimensions, but then it's no longer a tangent \textsl{plane}, but rather \textsl{hyperplane} to the \textsl{hypersurface}. In the case of just two variables, i.e. $F(x,y) = C$, the equation defines a curve in 2-D space (rather than a surface in 3-D space). In this case, the vector $\nabla F \left(\vec{a}\right)$ gives a normal to the tangent \textsl{line} to the curve at the point $\vec{a}$.$

$\noindent Using this method for this particular question, we have $F\left(x,y,z\right) = 3$ as our smooth surface, where $F\left(x,y,z\right) = x^2 + y^2 + z^2$. So $\nabla F \left(\vec{x}\right) = \left(2x,2y,2z\right)$. So at the point $\vec{a}=\left(1,1,1\right)$, we have a normal of $\nabla F \left(\vec{a}\right) = \left(2,2,2\right)$.$

Nailgun · May 19, 2016

Re: Multivariable Calculus

I was kind of hoping InteGrand would use the word 'hax' ahahah

InteGrand · May 19, 2016

Re: Multivariable Calculus

$\noindent In terms of finding partials of $z$, yes, we could use implicit differentiation. Note $x^2 + y^2 + z^2 = 3$ and that $z$ is defined as a differentiable function of $x,y$ near $\vec{a}=\left(1,1,1\right)$. Using implicit differentiation, we have for instance (partially differentiating both sides wrt $y$) $0+2y +2z \frac{\partial{z}}{\partial y}= 0 \Rightarrow \frac{\partial{z}}{\partial y} = -\frac{y}{z} $. Hence at $\vec{a}=\left(1,1,1\right)$ on the surface, we have $ \frac{\partial z}{\partial y} \Big{|} _{\vec{x}=\vec{a}} =-\frac{1}{1}=-1$.$

leehuan · May 19, 2016

Re: Multivariable Calculus

InteGrand said:
$\noindent In terms of finding partials of $z$, yes, we could use implicit differentiation. Note $x^2 + y^2 + z^2 = 3$ and that $z$ is defined as a differentiable function of $x,y$ near $\vec{a}=\left(1,1,1\right)$. Using implicit differentiation, we have for instance (partially differentiating both sides wrt $y$) $0+2y +2z \frac{\partial{z}}{\partial y}= 0 \Rightarrow \frac{\partial{z}}{\partial y} = -\frac{y}{z} $. Hence at $\vec{a}=\left(1,1,1\right)$ on the surface, we have $ \frac{\partial z}{\partial y} =-\frac{1}{1}=-1$.$

Oh wow that is nifty.
______________________________________________
Feeling somewhat ashamed that I don't know how to do this

$A cylindrical metallic solid is expanding under heat in such a way that its height is increasing at the rate of $0.1\text{cm}\,\text{s}^{-1}$ and its radius is increasing at the rate of $0.05$cm$\,$s$^{-1}$. Find the rate of increase of its volume at the instant when the height is 10 cm and the radius is 5 cm$

InteGrand · May 19, 2016

Re: Multivariable Calculus

leehuan said:
Oh wow that is nifty.
______________________________________________
Feeling somewhat ashamed that I don't know how to do this

$A cylindrical metallic solid is expanding under heat in such a way that its height is increasing at the rate of $0.1\text{cm}\,\text{s}^{-1}$ and its radius is increasing at the rate of $0.05$cm$\,$s$^{-1}$. Find the rate of increase of its volume at the instant when the height is 10 cm and the radius is 5 cm$

$\noindent This requires use of a \textsl{total derivative}. The volume $V$ is given as a function of $h$ (height) and $r$ (radius) as $V=f\left(h,r\right) = \pi r^2 h$. Now, $r=r(t)$ and $h=h(t)$ themselves are functions of time $t$, so we can talk about the total derivative of the volume wrt time. Using the multivariable chain rule, we find that$

$\begin{align*}\frac{\mathrm{d}V}{\mathrm{d}t} &= \frac{\partial f}{\partial h}\frac{\mathrm{d}h}{\mathrm{d}t}+ \frac{\partial f}{\partial r}\frac{\mathrm{d}r}{\mathrm{d}t} \\ &= \pi r^2 h\frac{\mathrm{d}h}{\mathrm{d}t}+2\pi rh\frac{\mathrm{d}r}{\mathrm{d}t}.\end{align*}$

$\noindent At the instant when $h=10$ and $r=5$, $\Big{(}$noting $\frac{\mathrm{d}h}{\mathrm{d}t}=0.1$ and $\frac{\mathrm{d}r}{\mathrm{d}t}=0.05$, which in fact we have at any time since we're given constant rates of change for these$\Big{)}$, we have$

$\begin{align*}\frac{\mathrm{d}V}{\mathrm{d}t} &= \pi \times 5^2 \times 0.05 + 2\pi \times 5 \times 10 \times 0.1 \\ &= 11.25 \pi.\end{align*}$

$\noindent The units of the above quantity are cubic centimetres per second.$

InteGrand · May 19, 2016

Re: Multivariable Calculus

leehuan said:
Ahh, ok problem - my course doesn't actually teach the total derivative itself... How would I derive it?

$\noindent What I used is basically the multivariate chain rule -- have you learnt about this?$

leehuan · May 19, 2016

Re: Multivariable Calculus

InteGrand said:
$\noindent What I used is basically the multivariate chain rule -- have you learnt about this?$

We were probably on our way to it now that I think about it... Thus far he's only mentioned that there's multiple chain rules but not exactly what they are yet...

Edit: Ok yep looking at the lecture slides it seems like they were to come. Ahh thanks though

dan964 · May 20, 2016

Re: Multivariable Calculus

leehuan said:
We were probably on our way to it now that I think about it... Thus far he's only mentioned that there's multiple chain rules but not exactly what they are yet...

Edit: Ok yep looking at the lecture slides it seems like they were to come. Ahh thanks though

Aah, the good old chain rule. I have a whole subject of Multivariable calculus this session (UOW for the win). And basically the chain rule reduces down in some measure to matrix multiplication which is pretty cool.

leehuan · May 20, 2016

Re: Multivariable Calculus

dan964 said:
Aah, the good old chain rule. I have a whole subject of Multivariable calculus this session (UOW for the win). And basically the chain rule reduces down in some measure to matrix multiplication which is pretty cool.

Saw it in my lecture slides as well, but again to come. Being taught it today.

seanieg89 · May 20, 2016

Re: Multivariable Calculus

leehuan said:
Thus far he's only mentioned that there's multiple chain rules but not exactly what they are yet...

Perhaps in some notations it will seem that way, but they can be unified remarkably well once one learns to think of the derivative of a function f at a point p as nothing more than a linear map between the vector space of tangents at p and the vector space of tangents at f(p) (pretty much the most flexible and abstract way to think of the classical derivative).

What it then boils down to is that if we have two smooth functions $f:X\rightarrow Y$ and $g: Y\rightarrow Z$ , then $h=g\circ f:X\rightarrow Z$ is smooth and $(Dh)(x)=(Dg)(f(x))\circ (Df)(x)$ .

seanieg89 · May 20, 2016

Re: Multivariable Calculus

(Where, X,Y,Z are open subsets of arbitrary dimension Euclidean spaces, or more generally of curves/surfaces/manifolds.)

Sidebar: Spivak has a less well known book "Calculus on Manifolds", that is amazing for learning multivariable calculus. From introducing partial derivatives and Riemann integration to differential forms and the modern version of Stokes theorem. It is quite compact and terse though and considerably less accessible than his "Calculus".

leehuan · May 22, 2016

Re: Multivariable Calculus

Getting stuck on how to incorporate the 0.02 correctly

$$Let $z=\frac{x+1}{y^2+1}.$ The measured values of $x$ and $y$ are 3cm and 1cm respectively and each of the measurements is made with an error whose absolute value is at most 0.02cm. Use the total differential approximation of $z$ to estimate the maximum error in the calculated value of $z$

Progress so far:
$$If $z=f(x,y), \\ \nabla f(3,1)=\begin{pmatrix}\frac{1}{2}\\-2\end{pmatrix}$

$f(x,y)\approx f(3,1)+\nabla f(3,1) \cdot \begin{pmatrix}0.02\\0.02\end{pmatrix}$

InteGrand · May 22, 2016

Re: Multivariable Calculus

leehuan said:
Getting stuck on how to incorporate the 0.02 correctly

$$Let $z=\frac{x+1}{y^2+1}.$ The measured values of $x$ and $y$ are 3cm and 1cm respectively and each of the measurements is made with an error whose absolute value is at most 0.02cm. Use the total differential approximation of $z$ to estimate the maximum error in the calculated value of $z$

Progress so far:
$$If $z=f(x,y), \\ \nabla f(3,1)=\begin{pmatrix}\frac{1}{2}\\-2\end{pmatrix}$

$f(x,y)\approx f(3,1)+\nabla f(3,1) \cdot \begin{pmatrix}0.02\\0.02\end{pmatrix}$

$\noindent The total differential approximation tells us that an estimate for the max. error will be $\left |\Delta z \right |_{\text{max.}} \approx \left |\frac{\partial f}{\partial x} \right |\left |\Delta x\right | + \left |\frac{\partial f}{\partial y} \right | \left |\Delta y \right |$, where the partials are evaluated at the point where we know the function's value (so they're the components of the gradient vector you found).$

$\noindent For the specific problem at hand, we know that $ \left |\frac{\partial f}{\partial x} \right | = \frac{1}{2}$ and $\left | \frac{\partial f}{\partial y} \right | = 2$ (from your calculations) and we are given $\left |\Delta x\right |, \left |\Delta y \right | \leq 0.02$. So the max. error in $z$ will be approximately: $\frac{1}{2} \times 0.02 + 2\times 0.02 = 0.01 + 0.04 = 0.05$.$

leehuan · May 22, 2016

Re: Multivariable Calculus

This thread may be packed with questions...cause my book doesn't give answers to almost anything. Sorry!

$$Find the indicated derivatives, assuming that the function $f(x,y)$ has continuous first order partial derivatives.\\ a) $\frac{\partial}{\partial x}f(3x,2y)$

I did something like this and I'm not sure if I can break it down any further

$$Let $g(x,y)=f(3x,2y), u=3x, v=3y\\ \begin{align*} \frac{\partial}{\partial x}f(3x,2y) &= \frac{\partial g}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial g}{\partial v}\frac{\partial v}{\partial x}\\ &= 3 \frac{\partial g}{\partial u} + 0 \frac{\partial g}{\partial v} \end{align*}$

InteGrand · May 22, 2016

Re: Multivariable Calculus

leehuan said:
This thread may be packed with questions...cause my book doesn't give answers to almost anything. Sorry!

$$Find the indicated derivatives, assuming that the function $f(x,y)$ has continuous first order partial derivatives.\\ a) $\frac{\partial}{\partial x}f(3x,2y)$

I did something like this and I'm not sure if I can break it down any further

$$Let $g(x,y)=f(3x,2y), u=3x, v=3y\\ \begin{align*} \frac{\partial}{\partial x}f(3x,2y) &= \frac{\partial g}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial g}{\partial v}\frac{\partial v}{\partial x}\\ &= 3 \frac{\partial g}{\partial u} + 0 \frac{\partial g}{\partial v} \end{align*}$

$\noindent Consider $z=f(u,v)$, where $u=3x$, $v=2y$. Then from the chain rule, $\frac{\partial z}{\partial x}= f_{1}\left(u,v\right) \frac{\partial u}{\partial x}+ f_{2} \left(u,v\right) \frac{\partial v}{\partial x} = f_{1}\left(u,v\right) \times 3 + 0 \Rightarrow \frac{\partial z}{\partial x} = 3f_{1}\left(3x,2y\right)$.$

leehuan · May 22, 2016

Re: Multivariable Calculus

InteGrand said:
$\noindent Consider $z=f(u,v)$, where $u=3x$, $v=2y$. Then from the chain rule, $\frac{\partial z}{\partial x}= f_{1}\left(u,v\right) \frac{\partial u}{\partial x}+ f_{2} \left(u,v\right) \frac{\partial v}{\partial x} = f_{1}\left(u,v\right) \times 3 + 0 \Rightarrow \frac{\partial z}{\partial x} = 3f_{1}\left(3x,2y\right)$.$

Hmm, makes sense but I'm having a dumb. What is the difference between f₁(u,v) and f₂(u.v)?

InteGrand · May 22, 2016

Re: Multivariable Calculus

leehuan said:
Hmm, makes sense but I'm having a dumb. What is the difference between f₁(u,v) and f₂(u.v)?

$\noindent The difference is that$f_{2}$ is the partial derivative of the function wrt its \textsl{second} designated variable ($v$ here), whereas $f_{1}$ is that wrt the first variable. The reason that term is 0 though is because $\frac{\partial v}{\partial x}=0$.$

leehuan · May 22, 2016

Re: Multivariable Calculus

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$

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

MATH2111 Higher Several Variable Calculus (1 Viewer)

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