1. ## Several Variable Calculus

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

2. ## Re: Several Variable Calculus

Originally Posted by leehuan
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).

3. ## Re: Several Variable Calculus

Originally Posted by InteGrand
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?

4. ## Re: Several Variable Calculus

Originally Posted by leehuan
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.

5. ## 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.

6. ## 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?

7. ## Re: Several Variable Calculus

Originally Posted by leehuan
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.

8. ## 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*}

EDIT: Ouch. I know what I did now.

9. ## Re: Several Variable Calculus

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

10. ## Re: Several Variable Calculus

Originally Posted by leehuan
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*}

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)).

11. ## 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

12. ## Re: Several Variable Calculus

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

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

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

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

13. ## Re: Several Variable Calculus

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

14. ## Re: Several Variable Calculus

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

15. ## Re: Several Variable Calculus

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

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

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

16. ## Re: Several Variable Calculus

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

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

$\text{Prove that equivalent metrics are topologically equivalent}$
Here's some hints.

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

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

17. ## Re: Several Variable Calculus

Originally Posted by InteGrand
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

18. ## Re: Several Variable Calculus

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

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

19. ## Re: Several Variable Calculus

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

$\text{but then }\lim_{(x,y)\to (0,0)}f(x,y)\text{ exists}\\ \text{for }f(x,y)=(x+y)\sin \frac1x \sin \frac1y$
$\noindent That example is showing that a multivariable limit can exist even though the corresponding iterated limits may not exist. If we hold x fixed and non-zero and limit y to 0, the limit won't exist because f(x, y) = x \sin \frac1x \sin \frac1y + y\sin \frac1x \sin \frac1y. The first term here doesn't have a limit as y\to 0, but the second one does (has limit 0, as you can show by the squeeze law), so overall the limit doesn't exist.$

$\noindent Of course, \lim_{(x,y)\to (0,0)}f(x,y) exists and equals 0, basically because \left| f(x,y)\right| \leq |x+y| for all x\neq 0 and y\neq 0.$

20. ## Re: Several Variable Calculus

Originally Posted by InteGrand
$\noindent That example is showing that a multivariable limit can exist even though the corresponding iterated limits may not exist. If we hold x fixed and non-zero and limit y to 0, the limit won't exist because f(x, y) = x \sin \frac1x \sin \frac1y + y\sin \frac1x \sin \frac1y. The first term here doesn't have a limit as y\to 0, but the second one does (has limit 0, as you can show by the squeeze law), so overall the limit doesn't exist.$

$\noindent Of course, \lim_{(x,y)\to (0,0)}f(x,y) exists and equals 0, basically because \left| f(x,y)\right| \leq |x+y| for all x\neq 0 and y\neq 0.$
Appears so counterintuitive though. I can't visualise what's going on here

21. ## Re: Several Variable Calculus

Originally Posted by leehuan
Appears so counterintuitive though. I can't visualise what's going on here
nobody can.....

22. ## Re: Several Variable Calculus

Consider the two metrics $d(\mathbf{x}, \mathbf{y}) = ||\mathbf{x} - \mathbf{y}||$ and
$\delta(\mathbf{x}, \mathbf{y}) = \frac{d(\mathbf{x}, \mathbf{y})}{1 +d(\mathbf{x}, \mathbf{y})}$ (you may assume they are metrics).
i) Show that d and δ are not equivalent.

23. ## Re: Several Variable Calculus

Originally Posted by QuantumRoulette
Consider the two metrics $d(\mathbf{x}, \mathbf{y}) = ||\mathbf{x} − \mathbf{y}||$ and
$\delta(\mathbf{x}, \mathbf{y}) = \frac{d(\mathbf{x}, \mathbf{y})}{1 +d(\mathbf{x}, \mathbf{y})}$ (you may assume they are metrics).
i) Show that d and δ are not equivalent.
I assume d(x,y) is supposed to be ||x-y|| and the set is some normed vector space V with norm ||.|| (e.g. R^d). (Please specify more if this is not the intended setting.)

Then these two metrics are not (strongly) equivalent because V is bounded with the delta metric but unbounded with the d metric.

That V is bounded with the delta metric follows immediately from the definition of delta, which must always lie in [0,1). On the other hand d(tx,0)=|t|d(x,0) can be made arbitrarily large for nonzero x.

Note that these two metrics ARE topologically equivalent though, in the sense that convergence in one metric implies convergence in the other. This follows from from the map x->x/(1+x) being a homeomorphism from [0,inf) to [0,1).

24. ## Re: Several Variable Calculus

Originally Posted by leehuan
Appears so counterintuitive though. I can't visualise what's going on here
Might not be easy to visualise the graph of the function on all of R^2, but you should certainly be able to visualise what it looks like on the slices x=const. or y=const which is all that matters for seeing/proving the nonexistence of the iterated limit. It is an oscillatory expression that oscillates faster as you approach axes. One of the two summands becomes irrelevant as you get close to the axes, so the other one dominates. This thing behaves like (const).sin(1/x), which of course does not converge unless that const is zero.

The boundedness of sine makes it clear that f(x,y) tends to zero as (x,y) tends to zero though.

Long story short: don't be too hasty to form intuitions in analysis, lots of things can behave weirdly...you kind of have to slowly build up a list of things that ARE true (via proof!) rather than assuming innocuous statements are true and ruling these things out as you come across pathological counterexamples.

And when you are looking at functions like this, try to isolate the terms that actually matter for the property you are trying to prove. A large part of analysis is just approximating ugly things by nice things, throwing away small sets on which a function behaves badly, etc etc.

25. ## Re: Several Variable Calculus

$\text{Prove that }\lim_{(x,y)\to (0,a)}\frac{\sin (xy)}{x}=a$

$\text{This result may be assumed: }\left| \frac{\sin a}{a}-1\right|=|a|^2, \, a\in \mathbb{R}$

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