Author Topic: Cobby's Methods Questions (Read 40785 times) Tweet Share

GerrySly · « **Reply #135 on:** June 17, 2009, 06:13:30 pm »

$\begin{align*} \frac{dy}{dx}&=\frac{dy}{dt}\frac{dt}{dx}\\ \therefore \frac{dy}{dx}&=2e^{2t}\frac{1}{e^t}\\ &=\frac{2e^{2t}}{e^t}\\ &=2e^{t}\\ \end{align*} $

So therefore when t=0

$\frac{dy}{dx}&=2$

d0minicz · « **Reply #136 on:** June 17, 2009, 06:14:53 pm »

Change $x= e^t + 1$ into $t= log_e (x-1)$
then sub it into $y= e^{2t}$ => $y= e^ {2log_e (x-1)}$
thennn find dy/dx of that
which is : $\frac {2}{x-1} e^{2log_e (x-1)}$
and now u find what x equals to when t=0
sub t=0 into the $x= e^t +1$
and you get x= 2
now sub x=2 into the dy/dx equation
and you'll get 2.

damn beaten
wtf my method is fekd

cobby · « **Reply #137 on:** June 17, 2009, 06:18:07 pm »

Quote from: Mao on June 17, 2009, 06:12:16 pm

$\frac{dy}{dt} = 2e^{2t}$ , $\frac{dx}{dt} = e^t$

$\implies \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = 2e^t$

when t=0, $\frac{dy}{dx} = 2$

Hey mao, why'd you divide the two derivatives???

d0minicz · « **Reply #138 on:** June 17, 2009, 06:19:31 pm »

its the same as saying
$\frac {dy}{dx} = \frac {dy}{dt} . \frac {dt}{dx}$

ilovevce · « **Reply #139 on:** June 17, 2009, 06:21:38 pm »

Alternatively,

$t = \log_{e} {x-1}$

$y = e^{2\log_{e} {x-1}} = (x-1)^2$

$\frac{dy}{dx} = 2(x-1) = 2x-2 = 2e^t$

When $t = 0$ , $\frac{dy}{dx} = 2$

cobby · « **Reply #140 on:** June 17, 2009, 06:27:06 pm »

thanks guys!!!

i have another one now

$y = e^{x}(px^2 +qx +r)$ is such that the tangents at $x = 1$
and $x = 3$ are parallel to the x-axis. The point with co-ordinates $(0,9)$ is on the curve.
Find the values of $p, q$ and $r$

Been trying to do this one for a while...im getting no where

Thanks

ilovevce · « **Reply #141 on:** June 17, 2009, 06:30:27 pm »

It's telling you that there are stationary points at x=1 and x=3 (gradient = 0 therefore tangent is parallel to x-axis).

Differentiate the equation and make equal to 0 to find the stationary points. You should be able to solve for p,q and r from there.

ilovevce · « **Reply #142 on:** June 17, 2009, 06:49:42 pm »

Firstly, substitute in the point on the curve so you get $r = 9$

Now, $\frac{dy}{dx} = e^x(px^2+qx+r+2px+q)$

For stationary points $\frac{dy}{dx} = 0 = px^2+qx+r+2px+q$

Sub in r and x = 1:

$0 = 3p+2q+9$
$\therefore p = -3-\frac{2q}{3}$

Sub in r and x = 3:

$0 = 15(-3-\frac{2q}{3})+4q+9$
$\therefore q = -6$

Finally, $p = -3-\frac{2q}{3} = 1$

cobby · « **Reply #143 on:** July 02, 2009, 04:24:04 pm »

Hey guys

I was trying to find the $\frac {dy}{dx}$ of $tan {2x} + \frac {1}{cos(x)}$ :

When i punched it into the calculator, the result was a super long fraction...is there something wrong with my calc lol?

Did anyone else get this result?

TrueTears · « **Reply #144 on:** July 02, 2009, 04:25:17 pm »

If you use TI-89 go to F2 -> 9 -> 1: texpand or 2: tCollect

try that

zzdfa · « **Reply #145 on:** July 02, 2009, 04:27:01 pm »

try evaluating it at particular point, is it the same as if you did it by hand?

cobby · « **Reply #146 on:** July 02, 2009, 04:27:15 pm »

Quote from: TrueTears on July 02, 2009, 04:25:17 pm

If you use TI-89 go to F2 -> 9 -> 1: texpand or 2: tCollect

try that

hmm it shortens the fraction, but its still long...does the ti-89 have $sec^2$ ?

TrueTears · « **Reply #147 on:** July 02, 2009, 04:27:59 pm »

go to catalog everything (most) is there

or just $\frac{1}{cos^2x}$ it

cobby · « **Reply #148 on:** July 02, 2009, 04:31:26 pm »

Quote from: zzdfa on July 02, 2009, 04:27:01 pm

try evaluating it at particular point, is it the same as if you did it by hand?

I evaluated a point and got the same answer

Quote from: TrueTears on July 02, 2009, 04:27:59 pm

go to catalog everything (most) is there

or just $\frac{1}{cos^2x}$ it

Thanks man

kurrymuncher · « **Reply #149 on:** July 02, 2009, 04:32:06 pm »

Or press 2nd and then 5, which goes to the math menu, then scroll down to Trig.

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