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TrueTears · « **Reply #135 on:** January 08, 2009, 03:57:02 pm »

When u have something like

Find the cartesian equation which corresponds to the following vector equations and state the domain and range:

$r$ (t) = $\frac{1}{t}$ $i$ + $\frac{1}{t+1}$ $j$ , $t \ne 0, -1$

so Let (x,y) be any point on the cartesian equation

$x = \frac{1}{t}$

$y = \frac{1}{t+1}$

and after eliminating t we get $y = \frac{x}{x+1}$ as cartesian equation.

so range of $x = \frac{1}{t}$ is the domain of the cartesian equation and the range of $y = \frac{1}{t+1}$ is the range of the cartesian equation.

range of $x = \frac{1}{t} \implies R \diagdown {0}$ so domain of cartesian equation is $R \diagdown {0}$

and range of $y = \frac{1}{t+1} \implies R \diagdown {0}$ so range of cartesian equation is $R \diagdown {0}$

But the answer has $R \diagdown {0,-1}$ as domain and $R \diagdown {0,1}$ as range for the cartesian equation. Is this because $t \ne 0, -1$ , so you sub in -1 into the $x = \frac{1}{t}$ and 0 into the $y = \frac{1}{t+1}$ equation and u get the values which are not allowed? If so then there's another question but this doesn't work

Find the cartesian equation and state the domain and range of the following vector equation:

$r(t) = \frac{1}{t+4}i + (t^2+1)j$ $t \ne -4$

$x = \frac{1}{t+4}$

$y = (t^2+1)$

and the cartesian equation is $y = (\frac{1}{x}-4)^2 + 1$

So the domain is $R \diagdown {0}$ for the cartesian equation, but the range is $R^+$ , But shouldn't the range be $R^+ \diagdown {17}$ ? Since $t \ne -4$ and the range for the cartesian equation is deduced from the $y = (t^2+1)$ equation, and if u sub in t = -4 u get y = 17, but u cant have t = -4 so 17 should not be allowed as one of the y values. But why answer just has $R^+$ for the range, why is that?

(all those things after $\diagdown$ should have a { } around them, i dono why it doesn't appear ><)

/0 · « **Reply #136 on:** January 08, 2009, 04:17:08 pm »

Quote from: TrueTears on January 08, 2009, 03:57:02 pm

But the answer has $R \diagdown {0,-1}$ as domain and $R \diagdown {0,1}$ as range for the cartesian equation. Is this because $t \ne 0, -1$ , so you sub in -1 into the $x = \frac{1}{t}$ and 0 into the $y = \frac{1}{t+1}$ equation and u get the values which are not allowed?

Yeah, true

The range for the second problem is $R^+$ because even though $t\neq -4$ , you can have $t=4$ .

Also, for braces, $\{ \}$ , type \{ and \}

TrueTears · « **Reply #137 on:** January 08, 2009, 04:18:16 pm »

ahhh i see, thanks again

TrueTears · « **Reply #138 on:** January 08, 2009, 04:55:00 pm »

Find a vector equation which corresponds to the following:

1 $y = 3-2x$

2 $x^2+y^2 =4$

My book doesn't have any info or steps on how to approach this, is there a systematic way of doing it?

like for 1. i know its just $r(t) = ti + (3-2t)j$ , but that is more or less a bit of trial and error.

/0 · « **Reply #139 on:** January 08, 2009, 05:49:45 pm »

Check out Chapter 1.8, Parametric equations of circles, ellipses and hyperbolas.

$\frac{x^2}{4}+\frac{y^2}{4} = 1$

A substitution you could make is $\frac{x^2}{4}=\sin^2{t}$ , $\frac{y^2}{4}=\cos^2{t}$ , or $\frac{x^2}{4}=\cos^2{t}$ , $\frac{y^2}{4}=\sin^2{t}$

And possible solutions are:
$r(t) = \pm2\sin{t}\mathbf{i}\pm2\cos{t}\mathbf{j}$
$r(t)=\pm2\sin{t}\mathbf{i}\mp2\cos{t}\mathbf{j}$
$r(t) = \pm 2\cos{t}\mathbf{i} \pm 2\sin{t}\mathbf{j}$
$r(t)=\pm 2\cos{t} \mathbf{i} \mp 2\sin{t}\mathbf{j}$

For a linear equation such as $y=3-2x$ , let one of the variables equal the parameter, just like you use parameters simultaneous equations in methods. e.g. $y=t \Rightarrow x = \frac{3-t}{2}$ could be another solution.
Or, to get your solution, you let $x=t$ (and it's probably a better way of putting it)

TrueTears · « **Reply #140 on:** January 08, 2009, 06:41:02 pm »

o yeah nice thx

And just this question

The path of a particle defined as a function of time, t, is given by the vector equation $r(t) = (1+t)i + (3t+2)j$ where $t \ge 0$ . Find:

a) the distance of the particle from the origin when t = 3
b) the times at which the distance of the particle from the origin is one unit.

For part a), i don't know why my answer is wrong, so all u do is sub in t = 3, $r(3) = 4i + 11j$
so $|r(3)| = \sqrt{4^2+11^2} = \sqrt{137}$ but my book answer says its $\sqrt{53}$ , how do u get that?

And also for part b), without even working my book's answer says $t = \frac{-2}{5}$ and $t = -1$ , but its already stated $t \ge 0$ , how can u get negative answers? and besides t is time, u can have negative time? i don't get this question at all ><

TrueTears · « **Reply #141 on:** January 08, 2009, 07:14:22 pm »

and another Q

Express $r(t) = (e^t + e^{-t})i + (e^{t} - e^{-t})j, t \ge 0$ in cartesian form.

Mao · « **Reply #142 on:** January 08, 2009, 07:38:12 pm »

did you mean, $r(t)=(e^t+e^{-t})i+(e^{t}-e^{-t})j$ ?

$(e^t+e^{-t})^2 - (e^{t}-e^{-t})^2 = 4$

$\implies x^2 - y^2 = 4$

looking at the expression of x and y in terms of t, we find that $x\ge 2$ (minimum value, t=0) and that $y\ge 0$ (minimum value, t=0)

$\implies y=\sqrt{x^2-4}, x\ge 2$ [only the positive stream is taken, as $y\ge 0$ ]

The following is NOT on the course, however it is still very useful

the hyperbolic functions:

$\sinh (x) = \frac{e^{x}-e^{-x}}{2}$ , $\cosh (x) = \frac{e^{x}+e^{-x}}{2}$
graph those two and learn what they look like

and this identity should be useful: $\cosh^2 (x) - \sinh^2 (x)=1$

also, $\frac{d}{dx}\sinh (x) = \cosh (x)$ , $\frac{d}{dx}\cosh (x) = \sinh (x)$

TrueTears · « **Reply #143 on:** January 08, 2009, 08:01:42 pm »

ahh clever mao, XD yeah soz typo btw lol

TrueTears · « **Reply #144 on:** January 08, 2009, 08:26:05 pm »

Also just another small Q

Theres a vector function like $r(t) = -cos(2 \pi t)i + sin(2 \pi t)j$
And my book says the particle moves around the circle with a period of one unit.
How do u work that out? Is it just $\frac{2\pi}{2\pi} = 1$ ? ie $\frac{2\pi}{n}$ where n is whatever is in the $cos(2 \pi t)$ or $sin(2 \pi t)$ ?

But what about these functions $r(t) = cos^2(3 \pi t)i + 2cos^2(3 \pi t)j, t \ge 0$

and $r(t) = cos(2 \pi t)i + cos(4 \pi t)j$

In general, how do u work out the 'time' it takes to do one cycle of whatever vector function it is?

And also how would u get the cartesian equation of $r(t) = cos(2 \pi t)i + cos(4 \pi t)j$ ?

(answered by mao already XD)

TrueTears · « **Reply #145 on:** January 09, 2009, 12:16:54 am »

If $r(t) = ati + \frac{a^2t^2}{4}j$ where $a \ge 0$ and $t \ge 0$
Find when the magnitude of the angle between $\dot r (t)$ and $\ddot r(t)$ is 45 degrees.

EDIT: forgot the angle between LOL

Mao · « **Reply #146 on:** January 09, 2009, 01:35:07 am »

$\dot{r} (t)=a \textbf{i} + \frac{a^2t}{2}\textbf{j}$

$\ddot{r} (t)=0 \textbf{i} + \frac{a^2}{2}\textbf{j}$

$\cos \theta = \frac{1}{\sqrt{2}}=\frac{\dot{r}(t)\cdot \ddot{r}(t)}{\left|\dot{r}(t)\right|\cdot \left|\ddot{r}(t)\right|}=\dfrac{\frac{a^4t}{4}}{\sqrt{a^2+\frac{1}{4}a^4t^2}\times \frac{a^2}{2}}=\frac{at}{\sqrt{4+a^2t^2}}$

$\implies \frac{1}{\sqrt{2}}=\frac{at}{\sqrt{4+a^2t^2}}$

$\implies \sqrt{4+a^2t^2} = \sqrt{2}\cdot at$

$\implies 4+a^2t^2 = 2a^2t^2$

$\implies a^2t^2 = 4 \implies at=2$ (discarding negative as a>0 and t>0)

$\impleis t=\frac{2}{a}$

at this point, $\textbf{r}\left(\frac{2}{a}\right)=2\textbf{i}+\textbf{j}$

TrueTears · « **Reply #147 on:** January 09, 2009, 01:37:25 am »

and...
The position r(t) of a projectile at time $t ( t \ge 0 )$ is given by $r(t) = 400ti + (300t - 4.9t^2)j$ , if the projectile is initially at ground level find the initial angle of projection from the horizontal

Mao · « **Reply #148 on:** January 09, 2009, 02:02:17 am »

$\dot{r} (t)=400\textbf{i}+(300-9.8t)\textbf{j}$

$\dot{r} (0)=400\textbf{i}+300\textbf{j}$

$\cos \theta = \frac{\dot{r}(0)\cdot \textbf{i}}{|\dot{r}(0)|\cdot |\textbf{i}|}=\frac{4}{5}$

$\theta = \cos^{-1}\frac{4}{5}= 36.87^o$ i think...

TrueTears · « **Reply #149 on:** January 09, 2009, 01:46:01 pm »

thanks mao, but how do u know to use the $\dot{r}(t)$ function?

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