Author Topic: TT's Maths Thread (Read 146078 times) Tweet Share

TrueTears · « **Reply #810 on:** January 30, 2010, 12:05:36 am »

Solve for $t$ $\implies$ $2\pi = t - \sin(t)$

Okay, clearly from inspection $t = 2\pi$ is a solution, but how do we do this algebraically? How do we rule out other solutions?

[This question stems from trying to use parametric equations to work out the area under an arch of a cycloid, more precisely... trying to work out the new terminals of the integral...]

kamil9876 · « **Reply #811 on:** January 30, 2010, 12:08:52 am »

cycloids are historically cool

anyway:

$f(t)=t-sin(t)$

$f'(t)=1-cos(t)> 0 almost everywhere$

therefore $f(t)$ is an increasing function and so the equation $f(t)=2\pi$ can only have at most one solution.

edit: terrible calculus mistake (still didn't change the ans tho

)

TrueTears · « **Reply #812 on:** January 30, 2010, 12:11:46 am »

Ahhh, so can the answer $2\pi$ be only done from inspection?

TrueTears · « **Reply #813 on:** January 30, 2010, 05:01:16 pm »

Sketch the following curve with parametric equations $x = 6\sin(t)$ and $y = t^2+t$ without converting to Cartesian form.

So here is what I did...

$\frac{dy}{dx} = \frac{2t+1}{6\cos(t)}$

Now I tried to work out the places where $\frac{dy}{dx} = 0$

So $2t+1 = 0 \implies t = -\frac{1}{2}$ .

So this occurs at $\left(-6\sin\left(\frac{1}{2}\right), - \frac{1}{4}\right)$

Next I tried to find out the vertical tangents of the curve.

So $6\cos(t) = 0 \implies t = \frac{(2n+1)\pi}{2}$

But this doesn't give me any idea of what the curve looks like...

TrueTears · « **Reply #814 on:** January 30, 2010, 08:06:29 pm »

Sketch the curve defined by $x = 2\cos(t)-\cos(2t)$ and $y = 2\sin(t) - \sin(2t)$

$\frac{dy}{dx} = \frac{\cos(t)-\cos(2t)}{\sin(2t)-\sin(t)}$

So I worked out when the graph has horizontal tangents.

This occurs when $\cos(t)-\cos(2t) = 0$

I solved this for $t \in [0, 2\pi]$ since that constitutes 1 period of the graph.

These points are: $\left(1,0\right), \left(-\frac{1}{2}, \frac{3\sqrt{3}}{2}\right), \left(-\frac{1}{2},-\frac{3\sqrt{3}}{2}\right)$

Then I worked out when the graph has vertical tangents.

This occurs when $\sin(2t)-\sin(t) = 0$

Again I solved this for $t \in [0, 2\pi]$ since that constitutes 1 period of the graph.

These points occur at: $\left(1,0\right), \left(\frac{3}{2},\frac{\sqrt{3}}{2}\right), \left(-3,0\right), \left(\frac{3}{2},-\frac{\sqrt{3}}{2}\right)$

So the resultant graph should look like:

My question is: at the point $(1,0)$ , I worked out there is a horizontal tangent but there is ALSO a vertical one? How is that possible? It doesn't look like it has a vertical tangent on the graph?

Also would the integral to work out the surface area of the solid when rotated around the $x$ axis be:

$\int_0^{\pi}2\pi\left(2\sin(t)-\sin(2t)\right)\sqrt{[x'(t)]^2+[y'(t)]^2}dt$

Damo17 · « **Reply #815 on:** January 30, 2010, 09:04:21 pm »

At (1,0), t=0. So we need to find the slope at this point. Use l' Hospital's rule as t approaches 0 and we get 0. So there is a horizontal tangent.

TrueTears · « **Reply #816 on:** January 30, 2010, 09:33:46 pm »

Yeah at t = 0 the slope doesn't exist so there is no horizontal tangent.

Damo17 · « **Reply #817 on:** January 30, 2010, 09:43:43 pm »

Quote from: TrueTears on January 30, 2010, 09:33:46 pm

Yeah at t = 0 the slope doesn't exist so there is no horizontal tangent.

There is a horizontal tangent, using dy/dx gives 0/0 for l' Hospital's, so use the 2nd derivative and you get the limit equaling 0.

TrueTears · « **Reply #818 on:** January 30, 2010, 09:44:37 pm »

There is a horizontal tangent approaching x = 1, but there is no horizontal tangent at x=1

TrueTears · « **Reply #819 on:** January 31, 2010, 03:51:44 pm »

Sketch $r = \sqrt{\theta}$ for $0 \le \theta \le 16\pi$

Damo17 · « **Reply #820 on:** January 31, 2010, 05:25:20 pm »

Quote from: TrueTears on January 30, 2010, 09:44:37 pm

There is a horizontal tangent approaching x = 1, but there is no horizontal tangent at x=1

Yes, I should have thought about x '(t) when t=0, as x '(0)=0 a horizontal tangent does not exist.

Quote from: TrueTears on January 31, 2010, 03:51:44 pm

Sketch $r = \sqrt{\theta}$ for $0 \le \theta \le 16\pi$

You should recognise that this is an Euler Spiral, and because of the domain it will have 8 revolutions.
Just 'plug in' main points such as $\theta = 0, \frac{\pi}{2}, \pi ,\frac{3\pi}{2}, etc$ and connect the points to form a spiral. As $\theta$ increases each revolution will become closer to the last. Long process because of domain.

TrueTears · « **Reply #821 on:** January 31, 2010, 09:08:03 pm »

yeah, I got the graph, pretty cool.

TrueTears · « **Reply #822 on:** January 31, 2010, 10:58:53 pm »

Find the points on the curve where the tangent line is horizontal and vertical.

$r^2 = \sin(2\theta)$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

$x = r\cos(\theta)$ and $y = r\sin(\theta)$

$\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin(\theta) + r\cos(\theta)$

$\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)$

But $2r\frac{dr}{d\theta} = 2\cos(2\theta)$

so $\frac{dr}{d\theta} = \frac{\cos(2\theta)}{r}$

$\frac{dy}{d\theta} = \frac{\cos(2\theta)}{r}\sin(\theta) + r\cos(\theta)$

$\frac{dx}{d\theta} = \frac{\cos(2\theta)}{r}\cos(\theta)-r\sin(\theta)$

$\frac{dy}{dx} = \frac{\frac{\cos(2\theta)}{r}\sin(\theta) + r\cos(\theta)}{\frac{\cos(2\theta)}{r}\cos(\theta)-r\sin(\theta)}$

$\implies \frac{dy}{dx} = \frac{\cos(2\theta)\sin(\theta)+r^2\cos(\theta)}{\cos(2\theta)\cos(\theta)-r^2\sin(\theta)}$

$\implies \frac{dy}{dx} = \frac{\cos(2\theta)\sin(\theta)+\sin(2\theta)\cos(\theta)}{\cos(2\theta)\cos(\theta)-\sin(2\theta)\sin(\theta)}$

$\implies \frac{dy}{dx} = \frac{\sin(3\theta)}{\cos(3\theta)}$

So for horizontal tangents: $\sin(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$

For vertical tangents: $\cos(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$

A few questions: First how do you know which domain to solve for $\theta$ ? Because initially I thought if you sketched the cartesian equation of $r^2 = \sin(2\theta)$ the period is $\pi$ . But solving $\theta$ over $[0, \pi]$ doesn't give all the points for which there is a horizontal tangent?

Also when $\theta = 0$ there is a horizontal tangent at the origin, but when $\theta = \frac{\pi}{2}$ there is a vertical tangent at the origin. How is this possible?

Damo17 · « **Reply #823 on:** January 31, 2010, 11:36:13 pm »

Quote from: TrueTears on January 31, 2010, 10:58:53 pm

So for horizontal tangents: $\sin(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$

For vertical tangents: $\cos(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$

Values of $\theta$ do not exist in the 2nd and 4th quadrants for this graph. That should clear up many questions.

Damo17 · « **Reply #824 on:** February 01, 2010, 12:04:39 am »

Quote from: TrueTears on January 31, 2010, 10:58:53 pm

Also when $\theta = 0$ there is a horizontal tangent at the origin, but when $\theta = \frac{\pi}{2}$ there is a vertical tangent at the origin. How is this possible?

It is possible that at a point, both a Horizontal and Vertical tangent exist.
In the case of $r^2=sin(2 \theta)$ , the origin (0,0) is passed twice during the course of 1 period. In one of these cases a Horizontal tangent exists and on the other pass through (0,0) a Vertical tangent exists.

Well.. that is the way I see it.

Search the forums now!

We have moved!

Author Topic: TT's Maths Thread (Read 146078 times) Tweet Share

TrueTears

Re: TT's Maths Thread

kamil9876

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Damo17

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Damo17

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Damo17

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Damo17

Re: TT's Maths Thread

Damo17

Re: TT's Maths Thread

Recent Posts

User:
Password: