Author Topic: Bucket's Questions (Read 56713 times) Tweet Share

bucket · « **Reply #180 on:** June 21, 2008, 11:32:52 pm »

Ah thanks.
This question confused me, i didn't know exactly what rate of change I was looking for XD.
This helped.

lwine · « **Reply #181 on:** June 21, 2008, 11:42:40 pm »

When you encounter these kind of related rates questions (there usually isn't too much variety in our textbooks), begin by writing what I call the KING EQUATION.

That is, set out exactly what rate they're looking for (whether its the change in height or the change in radius or etc; its almost always with respect to time), and write down the rate they give you (also with respect to time) and try to "fill in" the missing relation, based on the chain rule.

Then you should set out to find that missing relation, and throughout the process, keep referring to your KING EQUATION at the top to make sure you're on the right track.

bucket · « **Reply #182 on:** June 21, 2008, 11:58:23 pm »

$f(x)=x^2e^x$ , Find $x:f'(x)<0$

lwine · « **Reply #183 on:** June 22, 2008, 12:06:08 am »

Using the product rule:
$f'(x)=2xe^x+x^2e^x$ (As my methods teacher says: diff the first one, leave the second one alone, then diff the second one, and leave the first one alone)

$2xe^x+x^2e^x<0$

$e^x(2x+x^2)<0$ (we know e^x cannot equal 0)

$2x+x^2<0$

$x(2+x)<0$

$-2<x<0$

bucket · « **Reply #184 on:** June 22, 2008, 12:11:25 am »

ah thanks heaps! your explanations are really easy to follow (Y).

bucket · « **Reply #185 on:** June 22, 2008, 12:28:27 am »

gah, is it possible to find the equation of an asymptote, say for $f(x)=x+e^{-x}$ without graphing it?

/0 · « **Reply #186 on:** June 22, 2008, 12:45:20 am »

Take limits as $x \to \pm \infty$ . It's an intuitive kind of thing, you need to predict how parts of the function will behave with large numbers.

As $x \rightarrow \infty$ , $e^{-x}$ becomes negligibly small, so $f(x) \rightarrow x$ .

As $x \rightarrow -\infty$ , both $x$ and $e^{-x}$ become very pronounced, and you can't really neglect either one, so there is no asymptote as $x \rightarrow -\infty$ .

So the only asymptote is $y=x$

As another example,

$y = \frac{x^3+2x+1}{(x-1)(x+3)}=\frac{8}{x+3}+\frac{1}{x-1}+x-2$

As $x \to \pm \infty$ , the x in the denominator of the fractions will be so big that they practically disappear, leaving the asymptote $y=x-2$ .

bucket · « **Reply #187 on:** June 22, 2008, 12:49:27 am »

O_O I don't get it... lol.
Guess i'll just stick with graphing it then >.<, thanks anyway man.

bucket · « **Reply #188 on:** June 22, 2008, 02:51:36 pm »

Given that $\frac{dy}{dx}=6x-4$ , find the approximate change in y as x increases from -2 to -1.97

and also

"A manufacturing company has a daily output y on day t of a production run given by $y=600(1-e^{-0.5t})$ . Find the instantaneous rate of change of output y with respect to t on the 10th day."

I'm so fucked for this SAC >.<.

lwine · « **Reply #189 on:** June 22, 2008, 03:45:48 pm »

Mmm, I hate approximation questions.

The best formula for this scenario is $\Delta y\approx\dfrac{dy}{dx}\times\Delta x$

Considering the values given, this imples:

$\Delta y\approx(6x-4)\times\Delta x$

$\Delta y\approx\left(6(-2)-4\right)\times 0.03$

$\therefore\Delta y\approx -0.48$

The other question:

Differentiate $y=600-600e^{-0.5t}$

$\dfrac{dy}{dt}=300e^{-0.5t}$

$\text{Therefore, on the tenth day, when } t=10$

$\dfrac{dy}{dt}=300e^{-0.5\cdot 10}$

$\dfrac{dy}{dt}=2.02\text{ y per day}$

bucket · « **Reply #190 on:** June 22, 2008, 04:13:58 pm »

ah thanks a lot man

bucket · « **Reply #191 on:** June 22, 2008, 07:45:18 pm »

$f(x)=e^{ax}$ , find an approximation for f(h) where h is small in terms of h and a.

and I have another question, when the hell do we use the formula $f(x+h) \approx h \cdot f'(x) + f(x)$ ??
:S

Mao · « **Reply #192 on:** June 22, 2008, 08:46:10 pm »

Quote from: bucket on June 22, 2008, 07:45:18 pm

$f(x)=e^{ax}$ , find an approximation for f(h) where h is small in terms of h and a.

and I have another question, when the hell do we use the formula $f(x+h) \approx h \cdot f'(x) + f(x)$ ??
:S

answered your own question

$f(h)=f(0+h)\approx f(0)+h\cdot f'(0)$

$\implies f(h)\approx 1+a\cdot h$

bucket · « **Reply #193 on:** June 22, 2008, 09:09:59 pm »

wow wtf lol. thanks.

bucket · « **Reply #194 on:** June 22, 2008, 10:14:07 pm »

$y=sin2x, x=0$ , Find the tangent at the given point.

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