MATH2111 Higher Several Variable Calculus (6 Viewers)

leehuan · Aug 21, 2016

Re: Multivariable Calculus

I don't want my friends to just see the word "hyperplane" and run off. (Nor do I have enough of a grasp to translate it myself, even if it seems to make perfect sense to me)

$$By formula I meant using this, but it didn't work when I tried$\\ \Delta A = \frac{\partial A}{\partial a}\Delta a+\frac{\partial A}{\partial b}\Delta b+\frac{\partial A}{\partial C}\Delta C$

(think I meant basic. No idea where I got 'appropriate' from.)

InteGrand · Aug 21, 2016

Re: Multivariable Calculus

$\noindent Note $A = \frac{1}{2}ab \sin C$. Percentage change is given by $\frac{\Delta A}{A}$. Note $\Delta A \approx \frac{1}{2}b \sin C \Delta a + \frac{1}{2}a \sin C \Delta b + \frac{1}{2}ab \cos C \Delta C$ (total differential approximation). So dividing through by $A = \frac{1}{2} a b \sin C$, we get an answer of $\frac{\Delta A}{A} \approx \frac{\Delta a}{a} + \frac{\Delta b}{b} + \cot C \Delta C = 0.05 -0.06 + \cot \frac{\pi}{3}\cdot \left(0.02 \times \frac{\pi}{3}\right) = 0.0020920 \ldots \approx 0.21\%$.$

$\noindent The hyperplane thing is essentially what the total differential approximation is -- we're approximating the change in the function $A$ by moving along its tangent hyperplane by the amounts the variables changed by. It's a higher dimensional analog of the tangent-line approximation from single-variable calculus $f\left(x+\Delta x \Right) \approx f(x) + f^\prime (x) \Delta x$ (or written another way, $\Delta f \approx f^\prime (x)\Delta x$). The reason it's a tangent `hyperplane' and not just a tangent line or tangent plane is that there are three independent variables in this question, so the graph of the function lives in four dimensions, with the tangent hyperplane being a three-dimensional object.$

leehuan · Feb 28, 2017

Several Variable Calculus Marathon & Questions

For some reason I lost the point of intersection (2,0):

$\text{Find the }\textit{two}\text{ points of intersection of the curves}\\ \textbf{r}(t)=(t^2-t,t^2+t)\\ \textbf{r}(t)=(t+t^2,t-t^2)$

I equated the relevant r₁ and r₂ components but only got t=0

InteGrand · Feb 28, 2017

Re: Several Variable Calculus

leehuan said:
For some reason I lost the point of intersection (2,0):

$\text{Find the }\textit{two}\text{ points of intersection of the curves}\\ \textbf{r}(t)=(t^2-t,t^2+t)\\ \textbf{r}(t)=(t+t^2,t-t^2)$

I equated the relevant r₁ and r₂ components but only got t=0

Solve the system of equations

t^2 - t = s + s^2 (1)
t^2 + t = s - s^2 (2).

leehuan · Feb 28, 2017

Re: Several Variable Calculus

InteGrand said:
Solve the system of equations

t^2 - t = s + s^2 (1)
t^2 + t = s - s^2 (2).

Oh. I'll work on that right now but why was it necessary to introduce a new variable?

seanieg89 · Feb 28, 2017

Re: Several Variable Calculus

leehuan said:
Oh. I'll work on that right now but why was it necessary to introduce a new variable?

You have two parametric curves. Their points of intersection don't necessarily have the same parameter as a point on the first curve as they do as a point on the second curve.

seanieg89 · Feb 28, 2017

Re: Several Variable Calculus

Eg, consider the lines (x,y)=(t,0), (x,y)=(2t,0).

These lines coincide exactly, so every point on the x-axis is a point of intersection.

Yet the only place where (t,0)=(2t,0) is at the origin.

leehuan · Feb 28, 2017

Re: Several Variable Calculus

Excellent. Makes sense.
________________

Find the angle between the two curves at the points of intersection.

I'm having a dumb moment now. Which vectors are we taking the dot product of?

InteGrand · Feb 28, 2017

Re: Several Variable Calculus

leehuan said:
Excellent. Makes sense.
________________

Find the angle between the two curves at the points of intersection.

I'm having a dumb moment now. Which vectors are we taking the dot product of?

Find the s and t values at the points of intersection and plug them into the derivatives of the parametric curves. This will give us the "direction vectors" of the curves at the points of intersection. Find the angle between these vectors.

leehuan · Feb 28, 2017

Re: Several Variable Calculus

Strange... That's what I did so maybe there's an error in my computation.

$\text{The values for }(s,t)\text{ were }(1,-1)\text{ and }(0,0)\text{ and the latter easily worked}$

$\\ \textbf{r}_1^\prime (t) = (2t-1,2t+1)\\ \textbf{r}_2^\prime (s) = (1+2s, 1-2s)$

$\text{So }\\\textbf{r}_1^\prime (-1)=(-5,-1)\\ \textbf{r}_2^\prime(1)=(5,-1)$

$\begin{align*}\textbf{r}_1^\prime\cdot \textbf{r}_2^\prime&=|\textbf{r}_1^\prime||\textbf{r}_2^\prime|\cos \theta\\ -25 + 1 &= \sqrt{26}\sqrt{26}\cos \theta\end{align*}$

But the answer was arccos(0.8)

EDIT: Ouch. I know what I did now.

InteGrand · Feb 28, 2017

Re: Several Variable Calculus

$\noindent So e.g. the first curve is $\mathbf{r}_{1}(t) = \left(t^{2} -t, t^{2} + t\right)$, so the derivative is $\mathbf{r}_{1}^{\prime}(t) = \left(2t-1, 2t+1\right)$. This direction of this essentially tells us the direction of the curve at $t$ (it's basically the velocity vector of a particle whose position at time $t$ is $\mathbf{r}_{1}(t)$). So for example at the intersection point $(0,0)$ (which is where $t=0$ for $\mathbf{r}_{1}(t)$), the velocity is $(-1,1)$ (subbing $t=0$ into the ``velocity'' function found above). Find the velocity vector also for the other curve at the origin (done in a similar manner), and the find the angle between that vector and $(-1,1)$.$

InteGrand · Feb 28, 2017

Re: Several Variable Calculus

leehuan said:
Strange... That's what I did so maybe there's an error in my computation.

$\text{The values for }(s,t)\text{ were }(1,-1)\text{ and }(0,0)\text{ and the latter easily worked}$

$\\ \textbf{r}_1^\prime (t) = (2t-1,2t+1)\\ \textbf{r}_2^\prime (s) = (1+2s, 1-2s)$

$\text{So }\\\textbf{r}_1^\prime (-1)=(-5,-1)\\ \textbf{r}_2^\prime(1)=(5,-1)$

$\begin{align*}\textbf{r}_1^\prime\cdot \textbf{r}_2^\prime&=|\textbf{r}_1^\prime||\textbf{r}_2^\prime|\cos \theta\\ -25 + 1 &= \sqrt{26}\sqrt{26}\cos \theta\end{align*}$

But the answer was arccos(0.8)

You appear to have miscalculated the velocity vectors when subbing in the values of t and/or s (check the first components, noting you're subbing in s = 1 (not 2) and t = -1 (not -2)).

leehuan · Feb 28, 2017

Re: Several Variable Calculus

$\text{Proven earlier: For a particle with velocity }\textbf{v}(t)\\ \text{If }\textbf{v}(t)\cdot \textbf{v}^\prime(t)=0\text{ then the speed }v=\sqrt{v_1^2+\dots+v_n^2}\text{ is constant}$

$\\ \text{A particle of mass }m\text{ with position vector }\textbf{r}(t)\text{ at time }t\text{ is acted on by a total force}\\ \textbf{F}(t)=\lambda \textbf{r}(t)\times \textbf{v}(t)\\ \text{where }\lambda\text{ is a constant and }\textbf{v}(t)\text{ is the velocity of the particle.}$

$\text{Show that the speed }v\text{ of the particle is constant.}$

Can be assumed: F=ma

InteGrand · Feb 28, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Proven earlier: For a particle with velocity }\textbf{v}(t)\\ \text{If }\textbf{v}(t)\cdot \textbf{v}^\prime(t)=0\text{ then the speed }v=\sqrt{v_1^2+\dots+v_n^2}\text{ is constant}$

$\\ \text{A particle of mass }m\text{ with position vector }\textbf{r}(t)\text{ at time }t\text{ is acted on by a total force}\\ \textbf{F}(t)=\lambda \textbf{r}(t)\times \textbf{v}(t)\\ \text{where }\lambda\text{ is a constant and }\textbf{v}(t)\text{ is the velocity of the particle.}$

$\text{Show that the speed }v\text{ of the particle is constant.}$

Can be assumed: F=ma

$\noindent Since there's a cross product, we are in $\mathbb{R}^{3}$ here. We have $\mathbf{F} (t) = m\mathbf{v}'(t) = \lambda \mathbf{r}(t)\times \mathbf{v}(t)$. Dotting both sides with $\mathbf{v}(t)$ yields $m\mathbf{v}'(t) \cdot \mathbf{v}(t) = 0$, which implies by the earlier proven result that the speed is constant.$

InteGrand · Feb 28, 2017

Re: Several Variable Calculus

$\noindent (Note that $\left(\mathbf{r}\times \mathbf{v}\right)\cdot \mathbf{v} = 0$, recalling that in general a scalar triple product like this is always zero if two of the vectors are the same.)$

leehuan · Feb 28, 2017

Re: Several Variable Calculus

InteGrand said:
$\noindent (Note that $\left(\mathbf{r}\times \mathbf{v}\right)\cdot \mathbf{v} = 0$, recalling that in general a scalar triple product like this is always zero if two of the vectors are the same.)$

Completely forgot about this haha

leehuan · Mar 4, 2017

Re: Several Variable Calculus

$\text{Two metrics }\rho\text{ and }\delta \text{ are said to be topologically equivalent}\\ \text{iff every }\rho\text{-ball contains a }\delta\text{ ball}\\ \text{and every }\delta\text{-ball contains a }\rho\text{-ball}$

$\text{Two metrics }\rho\text{ and }\delta\text{ are said to be equivalent}\\ \text{iff }\exists c_1,c_2 > 0:\\ c_1\rho (\textbf{x},\textbf{y}) \le \delta (\textbf{x},\textbf{y}) \le c_2 \rho(\textbf{x},\textbf{y})$

$\text{Prove that equivalent metrics are topologically equivalent}$

InteGrand · Mar 4, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Two metrics }\rho\text{ and }\delta \text{ are said to be topologically equivalent}\\ \text{iff every }\rho\text{-ball contains a }\delta\text{ ball}\\ \text{and every }\delta\text{-ball contains a }\rho\text{-ball}$

$\text{Two metrics }\rho\text{ and }\delta\text{ are said to be equivalent}\\ \text{iff }\exists c_1,c_2 > 0:\\ c_1\rho (\textbf{x},\textbf{y}) \le \delta (\textbf{x},\textbf{y}) \le c_2 \rho(\textbf{x},\textbf{y})$

$\text{Prove that equivalent metrics are topologically equivalent}$

Here's some hints.

$\noindent Assume the given equivalence, i.e. $\exists c_1,c_2 > 0:\\ c_1\rho (\textbf{x},\textbf{y}) \le \delta (\textbf{x},\textbf{y}) \le c_2 \rho(\textbf{x},\textbf{y})$ for all $\mathbf{x},\mathbf{y}$.$

$\noindent Consider an arbitrary $\delta$-ball $B_{\delta}\left(\mathbf{a},\varepsilon\right)$ (some radius $\varepsilon > 0$, centred at $\mathbf{a}$). Now show that the $\rho$-ball $B_{\rho}\left(\mathbf{a}, \frac{\varepsilon}{c_{2}}\right)$ is a subset of the aforementioned $\delta$-ball. The other thing you need to show will follow by symmetry.$

leehuan · Mar 4, 2017

Re: Several Variable Calculus

InteGrand said:
Here's some hints.

$\noindent Assume the given equivalence, i.e. $\exists c_1,c_2 > 0:\\ c_1\rho (\textbf{x},\textbf{y}) \le \delta (\textbf{x},\textbf{y}) \le c_2 \rho(\textbf{x},\textbf{y})$ for all $\mathbf{x},\mathbf{y}$.$

$\noindent Consider an arbitrary $\delta$-ball $B_{\delta}\left(\mathbf{a},\varepsilon\right)$ (some radius $\varepsilon > 0$, centred at $\mathbf{a}$). Now show that the $\rho$-ball $B_{\rho}\left(\mathbf{a}, \frac{\varepsilon}{c_{2}}\right)$ is a subset of the aforementioned $\delta$-ball. The other thing you need to show will follow by symmetry.$

Is it basically just this?

$\begin{align*}\text{Let }\textbf{x} &\in B_\delta (\textbf{a},\epsilon)\\ \implies \delta (\textbf{x},\textbf{a}) &< \epsilon \\ \implies c_1 \rho (\textbf{x},\textbf{a}) &< \epsilon\text{ for some }c_1\in \mathbb{R}^+\\ \implies \rho (\textbf{x},\textbf{a}) &< \frac{\epsilon}{c_1}\\ \implies x&\in B_\rho\left(\textbf{a},\frac{\epsilon}{c_1}\right)\end{align*}$

Although that being said I had c_1 instead of c_2 but I feel that won't matter

leehuan · Mar 6, 2017

Re: Several Variable Calculus

$\text{Bit confused. How is it possible for }\lim_{x\to 0}\lim_{y\to 0}f(x,y)\\ \text{and the reverse order of limiting to not exist}$

$\text{but then }\lim_{(x,y)\to (0,0)}f(x,y)\text{ exists}\\ \text{for }f(x,y)=(x+y)\sin \frac1x \sin \frac1y$

MATH2111 Higher Several Variable Calculus (6 Viewers)

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